Data Communications and

Computer Networks

CN_Prelims.indd 1 2/20/2012 6:44:12 PM

 This page is intentionally left blank.

CN_Prelims.indd 2 2/20/2012 6:44:12 PM

 Data Communications and
Computer Networks

ITL Education Solutions Limited

Research and Development Wing
New Delhi

CN_Prelims.indd 3 2/20/2012 6:44:12 PM

 Copyright © 2012 by Dorling Kindersley (India) Pvt. Ltd
Licensees of Pearson Education in South Asia

No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s
prior written consent.

This eBook may or may not include all assets that were part of the print version. The publisher reserves
the right to remove any material present in this eBook at any time.

ISBN 9788131761274
eISBN 9788131776209

Head Office: A-8(A), Sector 62, Knowledge Boulevard, 7th Floor, NOIDA 201 309, India
Registered Office: 11 Local Shopping Centre, Panchsheel Park, New Delhi 110 017, India

CN_Prelims.indd 4 2/20/2012 6:44:12 PM

 Contents

Preface vii
Unit I Introduction
1. Overview of Data Communications and Networking 1
2. Reference Models and Network Devices 23

Unit II Physical Layer

3. Analog and Digital Transmission 40
4. Transmission Media 69
5. Multiplexing and Switching 83

Unit III Data Link Layer

6. Error Detection and Correction 110
7. Flow and Error Control 124
8. Media Access Control 144
9. Ethernet, Virtual Circuit Networks and SONET 161

Unit IV Network Layer

10. Routing and Congestion Control 194
1  1 . Quality of Service and Protocols 223

Unit V Transport Layer

1 2. Internet Transport Protocols 237

CN_Prelims.indd 5 2/20/2012 6:44:12 PM

 vi Contents

Unit VI Application Layer

1 3. Application Layer Protocols 250
1 4. Multimedia 267

Unit VII Security

1 5. Network Security 278

Index 293

CN_Prelims.indd 6 2/20/2012 6:44:12 PM

 Preface

Data communications and computer networks are two aspects of a multifarious field that caters to both
telecommunications and computing industries. Over the past 10 years, an enormous growth has been
seen in this field. Nowadays, when we talk about communications and networking, we are not restricted
to just traditional telephone networks or wired LANs such as Ethernet; rather, we have plentiful new
developments including Wi-Fi, 2G and 3G mobile networks, Bluetooth, WAP and wireless local loops.
Because of these technological advancements, the demand for the courses on data communications
and computer networks has been continuously increasing. Keeping pace with this trend, almost all
universities have integrated the study of data communications and computer networks in B.Tech. (CSE
and IT), M.C.A. and M.B.A. courses. This book, Data Communications and Computer Networks, in its
unique easy-to-understand question-and-answer format, directly addresses the need of students enrolled
in these courses.
The questions and corresponding answers in this book have been designed and selected to cover all
the necessary material in data communications and networking as per the syllabi of most universities.
The text has been designed to make it particularly easy for students having little or even no background
in data communications and networking to grasp the concepts. The students can use it as a textbook for
one- or two-semester course while interested professionals can use it as a self-study guide. The orga-
nized and accessible format allows students to quickly find the questions on specific topic
This book is a part of series named Express Learning Series which has a number of books designed
as quick reference guides.

Unique Features
1. This book is designed as a student-friendly and self-learning guide. In addition, it is written in a
clear, concise and lucid manner.
2. It has been prepared in an easy-to-understand question-and-answer format.
3. The chapters of this book include previously asked as well as new questions.
4. The chapters cover various types of questions such as multiple choice, short and long
­questions.
5. Solutions to numerical questions asked in various examinations are provided in this book.
6. All ideas and concepts included in this book are presented with clear examples.
7. All concepts are well structured and supported with suitable illustrations.
8. To help readers, the inter-chapter dependencies are kept to a minimum.
9. A comprehensive index, which is set at the end of the book, will be useful for readers to ­access
the desired topics quickly.

CN_Prelims.indd 7 2/20/2012 6:44:12 PM

 viii Preface

Chapter Organization
All questions and answers are organized into seven units with each unit consisting of one or more
chapters. A brief description of these units is as follows:
� Unit I: Introduction
This unit covers the introductory concepts of data communications and computer networks in
two chapters. Chapter 1 discusses the differences between communication and transmission,
the components of data communications, the categories and applications of computer networks,
network topologies, protocols and standards and the general idea of the Internet. Chapter 2
introduces two standard networking models, which are open systems interconnection (OSI) and
the Internet model (TCP/IP), along with a brief idea about the different layers of these models. It
also describes various network devices such as switch, hub, bridge and gateway. This unit serves
as a prelude to the rest of the book.
� Unit II: Physical Layer
This unit focuses on the physical layer of reference models and consists of three chapters. Chapter 3
deals with analog and digital transmissions. It describes various techniques to convert digital/
analog data to analog or digital signals. Chapter 4 spells out the characteristics of transmission
media and various guided and unguided media such as twisted-pair cables, coaxial cables, fibre
optic cables, radio waves, microwaves, infrared waves and satellite communication. Chapter 5
discusses how the available bandwidth of a channel can be utilized efficiently using multiplexing
and spreading. It also explains the concept of switching which is not only related to this layer but
also to several layers. The use of two common public networks, telephone and cable TV networks,
for data transfer is also covered in the chapter.
� Unit III: Data Link Layer
This unit discusses, in detail, the data link layer of reference models and consists of four chapters.
Chapter 6 outlines various design issues and types of services provided by the data link layer. It also
discusses how to detect and correct the errors that occur due to certain transmission impairments.
Chapter 7 spells out two main responsibilities of data link layer: flow control and error control. It
first discusses the protocols needed to implement flow and error control and then the discussion
moves on to bit-oriented protocols such as high-level data link control (HDLC) protocol and
byte-oriented protocols such as point-to-point protocol (PPP). Chapter 8 familiarizes the reader
with a sub-layer of data link layer named media access control (MAC) layer that is responsible
for resolving access to shared media. It discusses a number of multiple-access protocols such as
ALOHA, CSMA, CSMA/CD, CSMA/CA, reservation, polling, token passing, FDMA, TDMA
and CDMA that have been devised to control access to shared media. Chapter 9 throws light on
various IEEE 802 standards specified for wired and wireless LANs. It discusses, in detail, the
Ethernet—a wired LAN, Bluetooth—a short-range wireless technology, X.25—a virtual circuit
network, frame relay and asynchronous transfer mode (ATM)—switched WANs, and synchronous
optical network (SONET)—a high-speed WAN that uses fibre-optic technolog .
� Unit IV: Network Layer
This unit is all about the network layer of reference models and consists of two chapters.
Chapter 10 explores major design issues in switched data networks and Internet which are routing
and congestion control. It includes discussion on IP addressing and various routing algorithms for

CN_Prelims.indd 8 2/20/2012 6:44:12 PM

 Preface ix

designing optimal routes through the network, so that the network can be used efficientl . It also
describes what the congestion is, when it occurs, what its effects are and how it can be controlled.
Chapter 11 familiarizes the reader with the meaning of quality of service (QoS) of a network and
the ways to help improve the QoS. It includes discussion on the major protocol defined at the
network layer which is Internet protocol (IP). It also discusses other network layer protocols such
as ARP and ICMP that assist IP to perform its function.
� Unit V: Transport Layer
This unit comprises a single chapter detailing the transport layer of reference models. Chapter 12
provides an overview of transport layer and explains the duties and services of this layer. It also
discusses two transport layer protocols, which are user datagram protocol (UDP) and transmission
control protocol (TCP).
� Unit VI: Application Layer
This unit revolves around the application layer of reference models and consists of two chapters.
Chapter 13 describes a client/server application named domain name system (DNS) that is
responsible for providing name services for other applications. It also expounds on three common
applications in the Internet, which are TELNET, e-mail and file transfer protocol (FTP). The chapter
concludes with a brief discussion on the famous World Wide Web (www), hypertext transfer
protocol (HTTP), which is used to access the Web, and the network management protocol named
as simple network management protocol (SNMP). Chapter 14 details multimedia and some new
protocols that have been developed to deal with specific problems related to multimedia in other
layers.
� Unit VII: Security
This unit consists of a single chapter that discusses, in detail, the security in the network. Chapter 15
describes the importance of security in communications and networking. It discusses cryptography,
different cryptographic algorithms, such as data encryption standard (DES), triple-DES and RSA,
hash functions and digital signatures. It also discusses the role of firewall in the network and the
means of user authentication and message authentication.

Acknowledgements
We would like to thank the publisher, Pearson Education, their editorial team and panel of reviewers
for their valuable contributions towards content enrichment. We are indebted to the technical and edito-
rial consultants for devoting their precious time to improve the quality of the book. In addition, we are
grateful to the members of our research and development team who have put in their sincere efforts to
bring out a high-quality book.

Feedback
For any suggestions and comments about this book, please send an e-mail to itlesl@rediffmail.com.
We hope that you will enjoy reading this book as much as we have enjoyed writing it.
Rohit Khurana
Founder and CEO
ITL ESL

CN_Prelims.indd 9 2/20/2012 6:44:13 PM

 This page is intentionally left blank.

CN_Prelims.indd 10 2/20/2012 6:44:13 PM

 1
Overview of Data
Communications and Networking

1. What is the difference between communication and transmission?

Ans: Both communication and transmission deal with the exchange of information. However, a few
differences between them are listed in Table 1.1.

Table 1.1 Differences Between Communication and Transmission

Communication Transmission
• It refers to exchange of meaningful information • It refers to the physical movement of information.
between two communicating devices.
• It is a two-way scheme. • It is a one-way scheme.

2. What is meant by data communication? What are the characteristics of an efficien data
communication system?
Ans: Data communication refers to the exchange of data between two devices through some form
of wired or wireless transmission medium. It includes the transfer of data, the method of transfer and the
preservation of the data during the transfer process. To initiate data communication, the communicating
devices should be a part of a data communication system that is formed by the collection of physical
equipments (hardware) and programs (software). The characteristics of an efficient data communication
system are as follows:
‰‰ Reliable Delivery: Data sent from a source across the communication system must be delivered
only to the intended destination.
‰‰ Accuracy: Data must be delivered at the destination without any alteration. If the data is altered
or changed during its transmission, it may become unusable.
‰‰ Timely Delivery: Data must be delivered in time without much time lags; otherwise, it may
be useless for the receiver. As in case of video and audio transmissions, timely delivery means

CN_Chapter 01.indd 1 2/20/2012 6:00:13 PM

 2 Data Communications and Computer Networks

delivering data at the same time it is produced, in the same order in which it is produced and also
without any delay.
‰‰ Jitter: It refers to the differences in the delays experienced during the arrival of packets. It is
­uneven delay in the arrival time of audio or video data packets. These packets must be delivered at
a constant rate; otherwise, the quality of the video will be poor.
3. What are the components of data communication system?
Ans: There are five basic components of a data communication system (Figure 1.1)
‰‰ Message: It refers to the information that is to be communicated. It can be in the form of text,
numbers, images, audio or video.
‰‰ Sender: It refers to the device, such as a computer, video camera and workstation, which sends
the message.
‰‰ Receiver: It refers to the device, such as a computer, video camera and workstation, for which the
message is intended.
‰‰ Transmission Medium: It refers to the path which communicates the message from sender to
­receiver. It can be wired such as twisted-pair cable and coaxial cable or wireless such as satellite.
‰‰ Protocol: It refers to a set of rules (agreed upon by the sender and the receiver) that coordi-
nates the exchange of information. Both sender and receiver should follow the same protocol to
­communicate with each other. Without the protocol, the sender and the receiver cannot communi-
cate. For example, consider two persons; one of which speaks only English while another speaks
only Hindi. Now, these persons can communicate with each other only if they use a translator
(protocol) that converts the messages in English to Hindi and vice versa.
Protocol Protocol

P ro t o c ol

Message

Transmission medium

Sender Receiver
Figure 1.1 Components of a Data Communication System

4. What are the different forms of data representation? Explain in detail any two coding
schemes used for data representation.
Ans: Data can be available in various forms such as text, numbers, images, audio and video.
In ­networking, data has to be transmitted from source to destination in binary form. Thus, information
such as alphabets (a–z, A–Z), numbers (0, 1, 2, …, 9), special symbols (!, @, #, $ etc.) and images

CN_Chapter 01.indd 2 2/20/2012 6:00:19 PM

 Overview of Data Communications and Networking 3

(in the form of pixels/picture elements) has to be converted into sequences of bits, that is, 0 and 1.
Audio and video data have to be converted from analog to digital signal for transmission with the help
of different encoding schemes (explained in Chapter 03). There are various coding schemes including
Unicode, ASCII and EBCDIC which are used these days to represent the data. Here, we discuss only
ASCII and EBCDIC.

American Standard Code for Information

Interchange (ASCII)
The standard binary code for the alphanumeric characters is ASCII. This code was originally designed
as a seven-bit code. In addition, ASCII is often used with a parity bit which is the eighth bit. This bit
is used at the most significant bit (MSB) position and is used for detecting errors during transmission.
ASCII is commonly used in the transmission of data through data communication and is used almost
exclusively to represent the data internally in the microcomputers.
In ASCII, uppercase letters are assigned codes beginning with hexadecimal value 41 and continuing
sequentially through hexadecimal value 5A and lowercase letters are assigned hexadecimal values of
61 through 7A. The decimal values 1–9 are assigned the zone code 011 in ASCII. Table 1.2 of ASCII-7
coding chart shows uppercase and lowercase alphabetic characters, numeric digits 0–9 and special
­characters. The standard ASCII-7 code defines 128 character codes (0–127), of which, the first 32 are
control codes (non-printable) and other are printable characters.

Table 1.2 ASCII-7 Coding Chart

Alphabetic characters
Uppercase Lowercase
ASCII-7 code ASCII-7 code
In binary In In binary In
Prints as Zone Digit hexadecimal Prints as Zone Digit hexadecimal
A 100 0001 41 a 110 0001 61
B 100 0010 42 b 110 0010 62
C 100 0011 43 c 110 0011 63
D 100 0100 44 d 110 0100 64
E 100 0101 45 e 110 0101 65
F 100 0110 46 f 110 0110 66
G 100 0111 47 g 110 0111 67
H 100 1000 48 h 110 1000 68
I 100 1001 49 i 110 1001 69
J 100 1010 4A j 110 1010 6A
K 100 1011 4B k 110 1011 6B
L 100 1100 4C l 110 1100 6C
M 100 1101 4D m 110 1101 6D
(Continued )

CN_Chapter 01.indd 3 2/20/2012 6:00:19 PM

 4 Data Communications and Computer Networks

Table 1.2 ASCII-7 Coding Chart–(Continued )

Alphabetic characters
Uppercase Lowercase
ASCII-7 code ASCII-7 code
In binary In In binary In
Prints as Zone Digit hexadecimal Prints as Zone Digit hexadecimal
N 100 1110 4E n 110 1110 6E
O 100 1111 4F o 110 1111 6F
P 101 0000 50 p 111 0000 70
Q 101 0001 51 q 111 0001 71
R 101 0010 52 r 111 0010 72
S 101 0011 53 s 111 0011 73
T 101 0100 54 t 111 0100 74
U 101 0101 55 u 111 0101 75
V 101 0110 56 v 111 0110 76
W 101 0111 57 w 111 0111 77
X 101 1000 58 x 111 1000 78
Y 101 1001 59 y 111 1001 79
Z 101 1010 5A z 111 1010 7A
Numeric characters
0 011 0000 30 5 011 0101 35
1 011 0001 31 6 011 0110 36
2 011 0010 32 7 011 0111 37
3 011 0011 33 8 011 1000 38
4 011 0100 34 9 011 1001 39
Special characters
SPACE 010 0000 20 : 011 1010 3A
! 010 0001 21 ; 011 1011 3B
“ 010 0010 22 < 011 1100 3C
# 010 0011 23 = 011 1101 3D
$ 010 0100 24 > 011 1110 3E
% 010 0101 25 ? 011 1111 3F
& 010 0110 26 [ 101 1011 5B
‘ 010 0111 27 \ 101 1100 5C
( 010 1000 28 ] 101 1101 5D
) 010 1001 29 ^ 101 1110 5E
* 010 1010 2A — 101 1111 5F
+ 010 1011 2B ` 110 0000 60
(Continued )

CN_Chapter 01.indd 4 2/20/2012 6:00:20 PM

 Overview of Data Communications and Networking 5

Table 1.2 ASCII-7 Coding Chart–(Continued )

Alphabetic characters
Uppercase Lowercase
ASCII-7 code ASCII-7 code
In binary In In binary In
Prints as Zone Digit hexadecimal Prints as Zone Digit hexadecimal
’ 010 1100 2C { 111 1011 7B
- 010 1101 2D | 111 1100 7C
. 010 1110 2E } 111 1101 7D
/ 010 1111 2F ~ 111 1110 7E
@ 100 0000 40 111 1111 7F

Extended Binary Coded Decimal Interchange Code (EBCDIC)

EBCDIC uses eight bits for each character; therefore, it is possible to represent 256 different characters
or bit combinations (Table 1.3). This provides a unique code for each decimal value from 0 to 9 (for a
total of 10), each uppercase and lowercase letters (for a total of 52), and for a variety of special char-
acters. Since it is an eight-bit code, each group of the eight bits makes up one alphabetic, numeric, or
special character and is called a byte.
In EBCDIC, the bit pattern 1100 is the zone combination (zone and digit) used for the alphabetic
characters A through I, 1101 is used for the characters J through R and 1110 is the zone combination
used for characters S through Z. The bit pattern 1111 is the zone combination used when representing
decimal digits. Other zone combinations are used when forming special characters.
The concepts and advantages of ASCII are identical to those of EBCDIC. The important difference
between EBCDIC and ASCII coding systems lies in the eight-bit combinations assigned to represent the
various alphabetic, numeric and special characters. While using ASCII eight-bit code, we notice that the
selection of bit patterns used in the positions differs from those used in EBCDIC.

Table 1.3 EBCDIC Coding Chart

Alphabetic characters
Uppercase Lowercase
EBCDIC EBCDIC
In binary In In binary In
Prints as Zone Digit hexadecimal Prints as Zone Digit hexadecimal
A 1100 0001 C1 a 1000 0001 81
B 1100 0010 C2 b 1000 0010 82
C 1100 0011 C3 c 1000 0011 83
D 1100 0100 C4 d 1000 0100 84
E 1100 0101 C5 e 1000 0101 85
(Continued )

CN_Chapter 01.indd 5 2/20/2012 6:00:20 PM

 6 Data Communications and Computer Networks

Table 1.3 EBCDIC Coding Chart–(Continued )

Alphabetic characters
Uppercase Lowercase
EBCDIC EBCDIC
In binary In In binary In
Prints as Zone Digit hexadecimal Prints as Zone Digit hexadecimal
F 1100 0110 C6 f 1000 0110 86
G 1100 0111 C7 g 1000 0111 87
H 1100 1000 C8 h 1000 1000 88
I 1100 1001 C9 i 1000 1001 89
J 1101 0001 D1 j 1001 0001 91
K 1101 0010 D2 k 1001 0010 92
L 1101 0011 D3 l 1001 0011 93
M 1101 0100 D4 m 1001 0100 94
N 1101 0101 D5 n 1001 0101 95
O 1101 0110 D6 o 1001 0110 96
P 1101 0111 D7 p 1001 0111 97
Q 1101 1000 D8 q 1001 1000 98
R 1101 1001 D9 r 1001 1001 99
S 1110 0010 E2 s 1010 0010 A2
T 1110 0011 D3 t 1010 0011 A3
U 1110 0100 E4 u 1010 0100 A4
V 1110 0101 E5 v 1010 0101 A5
W 1110 0110 E6 w 1010 0110 A6
X 1110 0111 E7 x 1010 0111 A7
Y 1110 1000 E8 y 1010 1000 A8
Z 1110 1001 E9 z 1010 1001 A9
Numeric characters
0 1111 0000 F0 5 1111 0101 F5
1 1111 0001 F1 6 1111 0110 F6
2 1111 0010 F2 7 1111 0111 F7
3 1111 0011 F3 8 1111 1000 F8
4 1111 0100 F4 9 1111 1001 F9

5. Explain different modes of data transmission between two devices.

Ans: There are three types of transmission modes: simplex, half-duplex and full-duplex (Figure 1.2).
‰‰ Simplex: This mode of transmission is unidirectional. The information flows in one direction
across the circuit, with no capability to support response in the other direction. Only one of the
communicating devices transmits information; however, the other can only receive it. Television

CN_Chapter 01.indd 6 2/20/2012 6:00:20 PM

 Overview of Data Communications and Networking 7

SIMPLEX

One-way communication

DIRECTION OF DATA

HALF-DUPLEX

Two-way communication
DIRECTION OF DATA AT TIME T1

DIRECTION OF DATA AT TIME T2

(Signal goes in both
directions one-at-a time)

FULL-DUPLEX

Two-way communication
DIRECTION OF DATA AT ALL TIME

(Signal goes in both

directions simultaneously)

Figure 1.2 Data Transmission Modes

transmission can be considered as an example of simplex mode of transmission where the satellite
only transmits the data to the television, vice versa is not possible.
‰‰ Half-Duplex: In this transmission mode, each communicating device can receive and transmit
­information, but not at the same time. When one device is sending, the other can only receive at that
point of time. In half-duplex transmission mode, the entire capacity of the transmission medium is
taken over by the device, which is transmitting at that moment. Radio wireless set is an example of
half-duplex transmission mode where one party speaks and the other party listens.
‰‰ Full-Duplex: This transmission mode allows both the communicating devices to transmit and
­receive data simultaneously. A full-duplex mode can be compared to a two-way road with traffic
flowing in both directions. A standard voice telephone call is a full-duplex call because both parties
can talk at the same time and be heard.
6. Defin computer network. What are the different criteria that a network should meet?
Ans: A computer network refers to a collection of two or more computers (nodes) which are con-
nected together to share information and resources. Nodes are connected if they are capable of exchang-
ing information with each other. To be able to provide effective communication, a network must meet a
certain number of criteria, some of which are as follows:
‰‰ Performance: Performance of a network can be determined by considering some factors such as
transit time, response time, throughput and delay. The amount of time taken by a message to travel

CN_Chapter 01.indd 7 2/20/2012 6:00:22 PM

 8 Data Communications and Computer Networks

from one device to another is known as transit time and the time elapsed between the user initiates
a request and the system starts responding to this request is called the response time. The amount
of work done in a unit of time is known as throughput. To achieve greater performance, we need
to improve throughput and reduce the transit time, response time and delay. However, increasing
the throughput by sending more data to the network often leads to traffic congestion in the network
and thus, increases the delay. Some other factors that affect the performance of a network are the
type of transmission medium, the total number of users connected to the network and the efficiency
of connected hardware and software.
‰‰ Reliability: An efficient network must be reliable and robust. Reliability of a network is deter-
mined by the factors such as how frequently the failure is occurring and how much time is being
spent in recovering from a link failure.
‰‰ Security: A network must also provide security by protecting important data from damage and
­unauthorized access. In addition, there must be procedures and policies to handle theft and ­recovery
of data.

7. What are the various advantages of a computer network?

Ans: Today, computer networks are being used in every facet of life, as they provide the following
advantages.
‰‰ Sharing Information: This is one of the most important advantages of a computer network. In the
absence of a network, transferring information from one computer to another requires the use of a
compact disk, floppy disk, printer, etc. However, if the communicating systems are ­geographically
apart, sharing information becomes even harder. Computer networks solve this problem, as
­computers connected to a network can share information as if they are in the same building even
when they are located geographically apart. For example, when we connect to the Internet and open
a website on our ­computer, we can access information that is not stored in our own ­computers.
In such a networked system, all information is stored on a central and powerful computer known
as a server. All other computers in the network can easily access information from the server as if
it was located on their own computer.
‰‰ Sharing Hardware Resources: A network facilitates sharing of hardware resources in an effec-
tive and user-friendly manner. When computers are connected to a network, they can share periph-
erals such as printer and hard disk drives, with any other computer. For example, in an office
having five to ten computers and one printer, in the absence of network, only the computer that
is connected to the printer can be used to print data. If others have to access the printer, then they
would first need to transfer their data over to the computer connected to the printer. Contrastive to
this, in a networked environment, the printer can be shared on the network and every computer on
the network can easily access the printer without having the need to transfer data.
‰‰ Sharing Software Resources: Software resources are the programs or applications that are used
by computers to perform any useful function or to carry out daily basis task. In an environment
where networking is not available, users will have to install and configure any applications that
they need individually. However, if the computers are connected via a network, the required soft-
ware or application can be installed and configured centrally on a server and shared by all. This
saves the valuable time and disk space.
‰‰ Preserve Information: In addition to sharing information, a networked environment helps to pre-
serve information as well. It is difficult to maintain regular backups of data on a number of stand-
alone computers and without backup, important data can be lost in case of some accident or failure

CN_Chapter 01.indd 8 2/20/2012 6:00:23 PM

 Overview of Data Communications and Networking 9

of computer. However, in a networked environment, a copy of the important data can be kept on
the server as well as on other connected computers on the network. In this case, failure of one com-
puter will not result in loss of information, as the data can still be accessed from other computers
whenever required.
‰‰ Communication: Computer networks have revolutionized the way people communicate. Rather
than exchanging memos and directives on paper, which involves a lot of printing costs and delays,
network users can instantly send messages to others and even check whether or not their messages
have been received.
8. What are the various applications of a computer network?
Ans: Nowadays, computer networks have become an essential part of industry, entertainment world,
business as well as our daily lives. Some of the applications of a computer network in different fields are
as follows:
‰‰ Business Applications: There is a need of effective resource sharing in companies for the ex-
change of ideas. This can be achieved by connecting a number of computers with each other.
It allows transferring of business information effectively without using paper. For example, an
employee of one department can access the required information about another department using
network.
‰‰ Marketing and Sales: Marketing firms utilize networks for conducting surveys to gather and
­analyze data from the customers. This helps them to understand the requirements of a customer
and use this information in the development of the product. Sales professionals can use various
applications such as online shopping, teleshopping and online reservation for airlines, hotel rooms
etc. in order to increase the revenue of their organization.
‰‰ Financial Services: Computer networks play a major role in providing financial services to people
across the globe. For example, the financial application such as electronic fund transfer helps
the user to transfer money without going into a bank. Some other financial applications that are
­entirely dependent on the use of networks include ATM, foreign exchange and investment services,
credit history searches and many more.
‰‰ Directory and Information Services: Directory services permit a large number of files to be stored
a central location thereby speeding up the worldwide search operations. Information ­services of
Internet such as bulletin boards and data banks provide a vast amount of information to the users
within seconds.
‰‰ Manufacturing: Computer networks are widely being used in manufacturing. For example, the
applications such as computer-aided design (CAD) and computer-assisted manufacturing (CAM)
use network services to help design and manufacture the products.
‰‰ E-mail Services: This is one of the most widely used applications of network. With the help of
computer networks, one can send e-mails across the world within a few seconds and without using
paper.
‰‰ Mobile Applications: With the help of mobile applications such as cellular phones and wireless
phones, people wishing to communicate are not bound by the limitation of being connected by
fixed physical connections. Cellular networks allow people to communicate with each other even
while travelling across large distances.
‰‰ Conferencing: With the help of networking, conferencing (teleconferencing or videoconferencing)
can be conducted that allows remotely located participants to communicate with each other as if
they are present in the same room.

CN_Chapter 01.indd 9 2/20/2012 6:00:23 PM

 10 Data Communications and Computer Networks

9. Discuss two types of computer network architectures.

Ans: The computer network architecture specifies how the physical and logical components of
a ­computer network are assembled and connected with each other to facilitate information exchange
and ­resource sharing. The two major types of network architectures are client/server and peer-to-peer
­architectures.

Client/Server Architecture
In this architecture, each computer is either a client or a server. To complete a particular task, there exists
a centralized powerful host computer known as server and a user’s individual workstation known as
client (Figure 1.3). The client requests for services (file sharing, resource sharing etc.) from the server
and the server responds by providing that service. The servers provide access to resources, while the
clients have access to the resources available only on the servers. In addition, no clients can communi-
cate directly with each other in this architecture. A typical example of client/server architecture is accessing
a website (server) from home with the help of a browser (client). When a client makes a request for an
object to the server, then the server responds by sending the object to the client. In addition, it must be
noticed that two browsers accessing the same website, never communicate with each other.
Server

Client
Client

Client Client
Figure 1.3 Client/Server Architecture
An advantage of client/server architecture is that the IP address of the server is always fixed and the
server is always available on the network for clients. However, the disadvantage of this architecture is that
with time as the number of clients starts to increase, the number of requests to the server also ­increases
rapidly. In this scenario, we might need more than one server to serve larger number of requests.

Peer-to-Peer (P2P) Architecture

This architecture does not rely on dedicated servers for communication; instead, it uses direct c­onnections
between clients (peers) (Figure 1.4). A pure P2P architecture does not have the notion of clients or
­servers, but only equal peer nodes that simultaneously function as both “clients” and “servers” to the
other nodes on the network. That is, every node is able to initiate or complete any supported transaction
(file transfer) with the other connected node and every node can directly communicate with anothe .

CN_Chapter 01.indd 10 2/20/2012 6:00:24 PM

 Overview of Data Communications and Networking 11

Client Client

Client Client

Figure 1.4 Point-to-Point Architecture

The upper limit of the number of nodes that can function as both clients and servers on a P2P network
is between 10 and 25. If there are more nodes, then a P2P machine can be used as a dedicated server
with additional high-performance hardware. An advantage of P2P architecture is that it is very scalable,
that is, millions of nodes can be connected to the network to contribute to resources irrespective of the
differences in their local configuration, processing speed, network and storage capacity. However, the
highly decentralized approach of P2P architecture can be tough to manage. For example, during file
sharing with other remote clients, the only client having a specific file might log off from the network.
Examples of P2P networks are file-sharing applications such as Morpheus and Kaaza
10. Differentiate between P2P and client/server architecture.
Ans: Both P2P and client/server networks have associated advantages and disadvantages. These advan-
tages and disadvantages form a part of distinction between the two. These differences are listed in Table 1.4.

Table 1.4 Differences Between P2P Architecture and Client/Server Architecture

Basis P2P architecture Client/server architecture

Centralized There is no central repository for files and Resources and data security are controlled
applications in this architecture. through the server.
Maintenance It incurs low maintenance cost. A large network requires extra staff to
ensure ­efficient operation.
Installation It can be easily installed. It requires experts for proper installation of
the network.
Cost It does not require a dedicated server; thus, it It is expensive, as it requires a dedicated
is not much expensive. server.
Security Lack of proper security policies is the biggest It provides high level of security.
drawback.
Dependence All nodes are independent of each other. Failure When server goes down, it affects
­occurring in one node does not affect the functioning of the entire network.
functioning of other nodes in the network.

CN_Chapter 01.indd 11 2/20/2012 6:00:25 PM

 12 Data Communications and Computer Networks

11. Distinguish between point-to-point and multipoint connections? Give suitable diagrams.
Ans: In order to communicate with each other, two or more devices must be connected using a
link. A link is a physical path using which data can be transferred between two or more devices. The
two possible types of connections between two or more devices are point-to-point and multipoint
­connections.
‰‰ Point-to-Point: In a point-to-point connection, there is a dedicated link between the communicating
devices (Figure 1.5). The link is totally reserved for transmission specifically for those two devices.
Most point-to-point connections are established with cable or wire though satellite or microwave
links are also possible. For example, operating any device such as television using a remote control
establishes a point-to-point connection between the device and the remote control.

Link

Workstation Workstation

Link

Workstation Workstation

Figure 1.5 Point-to-Point Connection

‰‰ Multipoint: In a multipoint (multidrop) connection, a single link is shared between two or more de-
vices, that is, there is no dedicated link between the communicating devices (Figure 1.6). If the shared
link can be utilized by many devices simultaneously, it is known as a spatially shared connection
whereas if devices need to take turns to utilize the link, it is known as a timeshared connection.

Workstation Workstation

Link

Mainframe

Workstation

Figure 1.6 Multipoint Connection

12. What do you understand by the term network topology?

Ans: The term topology refers to the way a network is laid out, either physically or logically.
A ­topology can be considered as the network’s shape. It is the geometric representation of the relation-
ship of all the links. There are five basic topologies: bus, ring, sta , tree and mesh.

CN_Chapter 01.indd 12 2/20/2012 6:00:26 PM

 Overview of Data Communications and Networking 13

13. Discuss bus and mesh topology. Compare them.

Ans: Bus and mesh topology are detailed as under.

Bus Topology
Bus topology uses a common bus or backbone (a single cable) to connect all devices with t­erminators
at both ends. The backbone acts as a shared communication medium and each node (file server,
­workstations and peripherals) is attached to it with an interface connector. Whenever a message is to
be transmitted on the network, it is passed back and forth along the cable, past the stations (computers)
and between the two terminators, from one end of the network to the other. As the message passes each
station, the station checks the message’s destination address. If the address in the message matches
the station’s address, the station receives the message. If the addresses do not match, the bus carries
the message to the next station and so on. Figure 1.7 illustrates how devices such as file servers, work-
stations and printers are connected to the linear cable or the backbone.

Terminator Backbone
File server

Nodes

Figure 1.7 Bus Topology

Mesh Topology
In a mesh topology, every node has a dedicated point-to-point
link to every other node (­Figure 1.8). Messages sent on a mesh
network can take any of several possible paths from source to
destination. A fully connected mesh network has n(n − 1)/2
physical links to link n devices. For example, if an organization
has five nodes and wants to implement a mesh topology, then
5(5−1)/2, that is, 10 links are required. In addition, to accom-
modate those links, every device on the network must have n − 1
Figure 1.8 Mesh Topology
communication (input/output) ports.
A comparison of bus topology and mesh t­opology is given in Table 1.5.

Table 1.5 Comparison of Bus and Mesh Topology

Bus topology Mesh topology

• It uses a single cable to connect all devices with • It has a dedicated point-to-point link to every other node.
terminators at both ends.
• It requires least amount of cabling. • It requires more amount of cabling.
• It incurs less cost. • It is expensive.
(Continued )

CN_Chapter 01.indd 13 2/20/2012 6:00:30 PM

 14 Data Communications and Computer Networks

Table 1.5 Continued

Bus topology Mesh topology
• Entire network shuts down if there is failure in • If one link becomes unusable, it does not disable entire
backbone (cable). system.
• It is easy to install and reconfigure. • It is difficult to install and reconfigure.
• An example of this topology is cable TV network. • An example of this topology is mobile ad hoc network (MANet).

14. Explain ring, star and tree topologies along with their advantages and disadvantages.
Ans: The ring, star and tree topology are detailed as under.

Ring Topology
In this topology, computers are placed on a circle of cable without any terminated ends since there are
no unconnected ends (Figure 1.9). Every node has exactly two neighbours for communication purposes.
All messages travel through a ring in the same direction (clockwise or counterclockwise) until it reaches
its destination. Each node in the ring incorporates a repeater. When a node receives a signal intended for
another device, its repeater regenerates the bits and passes them along the wire.

Figure 1.9 Ring Topology

The advantages of ring topology are as follows:

‰‰ It is easy to install and reconfigure
‰‰ Every computer is given equal access to the ring. Hence, no single computer can monopolize the
network.
The disadvantages of ring topology are as follows:
‰‰ Failure in any cable or node breaks the loop and can take down the entire network.
‰‰ Maximum ring length and number of nodes are limited.

Star Topology
In this topology, devices are not directly linked to each other; however, they are connected via a
­centralized network component known as hub or concentrator (Figure 1.10). The hub acts as a ­central
controller and if a node wants to send data to another node, it boosts up the message and sends the
­message to the intended node. This topology commonly uses twisted-pair cable; however, coaxial cable
or fibre-optic cable can also be used

CN_Chapter 01.indd 14 2/20/2012 6:00:32 PM

 Overview of Data Communications and Networking 15

File server

Figure 1.10 Star Topology

The advantages of star topology are as follows:
‰‰ It is easy to install and wire.
‰‰ The network is not disrupted even if a node fails or is removed from the network.
‰‰ Fault detection and removal of faulty parts are easier in star topology.
The disadvantages of star topology are as follows:
‰‰ It requires a longer length of cable.
‰‰ If the hub fails, nodes attached to it are disabled.
‰‰ The cost of the hub makes the network expensive when compared to bus and ring topologies.

Tree Topology
A tree topology combines characteristics of linear bus and star topologies (Figure 1.11). It consists of groups
of star-configured workstations connected to a bus backbone cable. Not every node plugs ­directly to the cen-

Sec
Secondary
hub
ub
u b

Central
hub

Backbone

Figure 1.11 Tree Topology

CN_Chapter 01.indd 15 2/20/2012 6:00:39 PM

 16 Data Communications and Computer Networks

tral hub. The majority of nodes connect to a secondary hub that, in turn, is connected to the central hub. Each
secondary hub in this topology functions as the originating point of a branch to which other nodes connect.
A tree topology is best suited when the network is widely spread and partitioned into many branches.
The advantages of tree topology are as follows:
‰‰ The distance a signal can travel increases, as the signal passes through a chain of hubs.
‰‰ It allows isolating and prioritizing communications from different nodes.
‰‰ It allows for easy expansion of an existing network, which enables organizations to configure a
network to meet their needs.
The disadvantages of tree topology are as follows:
‰‰ If the backbone line breaks, the entire segment goes down.
‰‰ It is more difficult to configure and wire than other topologie

15. What are the different categories of a network?

Ans: There are no generally accepted criteria to classify the computer networks; however, two di-
mensions are considered more important, which are scale and transmission technology. On the basis
of scale, computer networks can be classified into three types: local area network (LAN), metropoli-
tan area network (MAN) and wide area network (WAN). On the basis of transmission ­technology,
computer networks can be categorized as point-to-point networks and broadcast networks.
16. Explain the categories of networks based on scale.
Ans: A network can be as few as several personal computers on a small network or as large as the Inter-
net, a worldwide network of computers. Today, when talking about networks, we are generally referring
to three primary categories: LAN, MAN, and WAN.

Local Area Network

A LAN is a computer network that covers only a small geographical area (usually within a square mile or
less) such as an office, home or building (Figure 1.12). In a LAN, connected computers have a ­network
operating system installed onto them. One computer is designated as the fil server, which stores all
the software that controls the network. It also stores the software that can be shared by the ­computers
attached to the network. Other computers connected to the file server are called ­workstations.
The workstations can be less powerful than the file server and they may have additional software on
their hard drives. On most LANs, cables are used to connect the computers. Generally, LAN offers a
­bandwidth of 10–100 Mbps.

Figure 1.12 Local Area Network

CN_Chapter 01.indd 16 2/20/2012 6:00:43 PM

 Overview of Data Communications and Networking 17

Figure 1.13 Metropolitan Area Network

Metropolitan Area Network

A MAN is a network of computers spread over a “metropolitan” area such as a city and its suburbs
(Figure 1.13). As the name suggests, this sort of network is usually reserved for metropolitan areas
where the city bridges its LANs with a series of backbones, making one large network for the entire
city. It may be a single network such as a cable television network or it may be a means of connecting
a number of LANs. Note that MAN may be operated by one organization (a corporate with several
­offices in one city) or be shared and used by several o ganizations in the same city.

Wide Area Network

A WAN is a system of interconnecting many computers over a large geographical area such as cities,
states, countries or even the whole world (Figure 1.14). These kinds of networks use telephone lines,
satellite links and other long-range communication technologies to connect. Such networks are designed
to serve an area of hundreds or thousands of miles such as public and private packet switching networks
and national telephone networks. For example, a company with offices in New Delhi, Chennai, and
Mumbai may connect the LANs for each of those locations to each other through a WAN. Although a
WAN may be owned or rented by private business, it is usually a public network designed to connect
small and intermediate sized networks together. The largest WAN in existence is the Internet.

Figure 1.14 Wide Area Network

CN_Chapter 01.indd 17 2/20/2012 6:00:49 PM

 18 Data Communications and Computer Networks

WAN offers many advantages to business organizations. Some of them are as follows:
‰‰ It offers flexibility of location because not all the people using the same data have to work at the
same site.
‰‰ Communication between branch offices can be improved using e-mail and file sharin
‰‰ It facilitates a centralized company wide data backup system.
‰‰ Companies located in a number of small and interrelated offices can store files centrally and access
each other’s information.
17. What are the two types of transmission technology available?
Ans: The two types of transmission technology that are available include broadcast networks and
point-to-point networks.
In broadcast networks, a single communication channel is shared by all the machines of that
­network. When a short message, let us say a packet is sent by any machine, it is received by all the
other machines on that network. This packet contains an address field which stores the address of
the intended recipient. Once a machine receives a packet, it checks the address field. If the address
mentioned in the address field of packet is matched with the address of the recipient machine, it is
processed; otherwise, the packet is ignored. In broadcast systems, there is a special code in the ­address
field of the packet which is intended for all the destinations. When a packet with this code is trans-
mitted, it is supposed to be received and processed by all the machines on that network. This mode
of operation is called broadcasting. Some of the networks also support transmission to a subset of
machines what is called multicasting.
In point-to-point networks, there could be various intermediate machines (such as switching de-
vices called nodes) between the pair of end points called stations. Thus, there could be various pos-
sible routes of different lengths for a packet to travel from the source to the destination. Various routing
algorithms are considered and finall , one of them is chosen for the packets to travel from the source to
the destination.
Generally, for small geographically localized network (such as LAN), broadcasting is considered
favourable while larger networks (such as WAN) use point-to-point networks. If there is specifically one
sender and one receiver in any point-to-point network, it is sometimes considered as unicasting.

18. Defin Internet and write a brief history on Internet.

Ans: The word Internet is derived from two words: Interconnection and Networks. Also known
as “the Net”, Internet is a worldwide system of computer networks, that is, a network of networks,
which allows the participants (users) to share information. It consists of thousands of separately ad-
ministered networks of various sizes and types. Each of these networks comprises tens of thousands
of computers. Moreover, the total number of users of the Internet is known to be in millions. This high
level of connectivity encourages an unparalleled degree of communication, resource sharing and infor-
mation access.
The foundation of Internet was laid in 1969 by the Department of Defense (DOD) of United
States of America. They wanted to create a computer network that could continue to function in the
event of a disaster, such as a nuclear war. Even if a part of the network was damaged or destroyed,
the rest of the system would continue to work. That network was known as ARPANET (Advanced
Research Projects Agency Network), which linked US scientific and academic researchers. It was
the forerunner of today’s Internet. Later in 1980, another agency, the National Science Foundation
(NSF) created a new network of computers based on ARPANET, called NSFNET, which turned

CN_Chapter 01.indd 18 2/20/2012 6:00:50 PM

 Overview of Data Communications and Networking 19

out to be more efficient and capable. Initially, NSFNET was designed to link five super computers
situated at the major universities of NSF and allowed only academic research. Over the time, this
network expanded to include sites for business, universities, government etc. and finally becoming a
network consisting of millions of computers, now known as the Internet. Now, it is probably the most
powerful and important technological advancement since the introduction of the desktop computer.
With the advancement of Internet, the quality, quantity and variety of information also grew. Today,
the Internet is a repository of every type of information. Nowadays, an Internet user can get all sorts
of information ranging from how to add to the design of a functional spaceship to how to choose a
product for personal use.
19. Defin protocol. Describe the key elements of protocols.
Ans: Protocol refers to a set of rules that coordinates the exchange of information between the
sender and receiver. It determines what is to be communicated, and how and when it is to be commu-
nicated. Both sender and receiver should follow the same protocol to communicate data. Without the
protocol, the sender and receiver cannot communicate with each other. The main elements of a protocol
are as follows:
‰‰ Syntax: It refers to the format of the data that is to be transmitted.
‰‰ Semantics: It refers to the meaning and interpretation of each bit of data.
‰‰ Timing: It defines when the data should be sent and with what speed

20. What do you understand by the standards? Explain briefl .

Ans: Standards are crucial for establishing an open and worldwide marketplace for the ­manufacturers,
vendors, suppliers, government agencies and other service providers to provide them worldwide
­interconnectivity. Standards define some common set of rules and guidelines which ensure univer-
sal interoperability of data and communication technology and processes. The standards for data
­communication are classified into two categories, which are de facto and de jure standards.
‰‰ De facto: This is a Latin word which means “from the fact” or “by convention”. This category includes
standards that have not been approved formally by some authority but have been accepted as standards
because of their widespread use. For instance, the UNIX operating system has largely been used in
computer science departments of most universities and thus has been adopted as the standard. ­Similarly,
many manufacturers prefer to copy IBM machines while developing PCs for small office and home
computers and thus, IBM PCs are the de facto standard. The de facto ­standards are usually founded by
the manufacturers in an attempt to specify the functionality of some new technology or product.
‰‰ De jure: This is also a Latin word which means “by law” or “by regulation”. Unlike de facto stan-
dards, the de jure standards are formally declared as legal standards by recognized standardization
authorities. The international standardization authorities governing de jure standards may be estab-
lished by treaty among governments of different nations or may comprise volunteers from different
standard organizations without nay treaty.
21. Write short notes on intranet and extranet.
Ans:
‰‰ Intranet: This is a private network that is set up within an organization and also controlled by the
organization; nobody outside the organization is permitted to access the network. Intranet utilizes
the same protocols as used for accessing the Internet through a web browser. The users of intranet
can access the basic services of the Internet such as e-mail. The difference between an intranet and

CN_Chapter 01.indd 19 2/20/2012 6:00:50 PM

 20 Data Communications and Computer Networks

the Internet is that an intranet user can view only those websites that are owned and maintained
by the organization hosting the intranet. On the other hand, an Internet user may visit any website
without any permission.
‰‰ Extranet: This is an extended intranet owned, operated and controlled by an organization. In addi-
tion to allow access to members of an organization, an extranet uses firewalls, access profiles and
privacy protocols to allow access to users from outside the organization. In essence, an extranet is
a private network that uses Internet protocols and public networks to securely share resources with
customers, suppliers, vendors, partners or other businesses.
22. Represent the message 5A.dat in ASCII code. Assume parity bit position (eighth bit) as 0.
Ans: Using the ASCII coding chart shown in Table 1.2, 5A.dat will be coded as shown here.

Bit positions 7 6 5 4 3 2 1 0
5 0 0 1 1 0 1 0 1
A 0 1 0 0 0 0 0 1
. 0 0 1 0 1 1 1 0
d 0 1 1 0 0 1 0 0
a 0 1 1 0 0 0 0 1
t 0 1 1 1 0 1 0 0

23. Represent HelLO34 in EBCDIC code.

Ans: Using the EBCDIC coding chart shown in Table 1.3, HelLO34 will be coded as shown here.

Bit positions 7 6 5 4 3 2 1 0
H 1 1 0 0 1 0 0 0
e 1 0 0 0 0 1 0 1
l 1 0 0 1 0 0 1 1
L 1 1 0 1 0 0 1 1
O 1 1 0 1 0 1 1 0
3 1 1 1 1 0 0 1 1
4 1 1 1 1 0 1 0 0

24. What is hybrid topology?

(a) Draw a hybrid topology with a star backbone and three bus networks.
(b) Draw a hybrid topology with a star backbone and four ring networks.
Ans: Hybrid topology is the combination of two or more topologies such that the resultant ­network
does not retain the characteristics of any of the basic topologies including star, bus, ring and tree.
A hybrid topology is created when two different basic topologies are connected.

CN_Chapter 01.indd 20 2/20/2012 6:00:50 PM

 Overview of Data Communications and Networking 21

(a) A hybrid topology with star backbone and three bus networks is shown in Figure 1.15.

Station Station Station

Station Station Station

Hub

Station Station Station

Figure 1.15 Hybrid Topology with a Star Backbone and Three Bus Networks

(b) A hybrid topology with star backbone and four ring topologies is shown in Figure 1.16.
Station

Station Station
Station Station

Station

Station
Hub

Station Station Station

Station

Station
Figure 1.16 Hybrid Topology with a Star Backbone and Four Ring Networks

CN_Chapter 01.indd 21 2/20/2012 6:00:52 PM

 22 Data Communications and Computer Networks

25. Assume a network with n devices. Calculate how many links are required to set up this
network with mesh, ring, bus and star topologies?
Ans: The number of links required to set up this network with the various topologies are listed in
Table 1.6.

Table 1.6 Topology along with Number of Links

Topology Number of links

Mesh (n(n − 1))/2
Ring n
Bus (n drop lines + one line for bus)
Star n

Multiple Choice Questions

1. Which of these is the part of a data commu- (a) Delay      
nication system? (b) Response time
(a) Sender      (b) Message (c) Transit time    
(c) Protocol      (d) All of these (d) Throughput
2. ASCII stands for 6. In star topology, devices are connected via a
(a) American standard code for information centralized network component known as
identificatio (a) Node     (b) Client
(b) American standard code for information (c) Bus      (d) Hub
interchange 7. The network topology which uses hierarchy
(c) American standard coding for informa- of nodes is
tion interchange (a) Ring (b) Tree
(d) American standard coding for informa- (c) Bus (d) Fully connected
tion identificatio
8. A MAN is ___________ in size as compared
3. EBCDIC uses __________ bits for each to a LAN.
character. (a) Larger    (b) Smaller
(a) Six        (b) Seven (c) Equal      (d) None of these
(c) Eight        (d) Two
9. Internet is a
4. In _______ transmission mode, the flow of (a) LAN      (b) WAN
information is bidirectional at the same time. (c) MAN      (d) Both LAN and MAN
(a) Half-duplex    (b) Simplex
(c) Full-duplex      (d) None of these 10. Which of these is not the key element of a
protocol?
5. The amount of time taken by a message to (a) Syntax    (b) Standard
travel from one device to another is known as (c) Semantics   (d) Timing
Answers
1. (d) 2. (b) 3. (c) 4. (c) 5. (c) 6. (d) 7. (b) 8. (a) 9. (b) 10. (b)

CN_Chapter 01.indd 22 2/20/2012 6:00:52 PM

 2
Reference Models and
Network Devices

1. What is meant by a reference model? Why do we need it?

Ans: A reference model is a conceptual layout that describes how communication between devices
should occur. For efficient communication, the reference model identifies the tasks involved in inter-
computer communication and divides them in logical groups called layers, with each layer perform-
ing a specific function. A communication system designed in such a manner is referred to as layered
architecture. We need a reference model, as it provides various advantages, some of which are as
follows:
‰‰ It permits different types of network software and hardware to communicate with each other.
‰‰ It defines standards for building network components thereby permitting multiple-vendor
­development.
‰‰ It defines which functions should be performed at each layer of the model thereby promoting the
standardization of network.
‰‰ It divides the overall communication function into simpler and smaller components thus, helps in
component development and reducing trouble shooting.
2. What are the advantages of layering in a network?
Ans: The main aim behind building a layered architecture is to reduce the design complexity of a
computer network. Other advantages of layering in a computer network are as follows:
‰‰ Layering can resolve complicated tasks by breaking it into smaller and manageable pieces.
‰‰ Each layer can be analyzed and tested independently.
‰‰ By layering, the functionalities are carried out in logical sequential manner.
‰‰ Implementation of layers can be changed without disturbing other layers, as the details of all the
layers are hidden from each other.
‰‰ Layering allows reuse in a way that once a common functionality is implemented in a lower layer,
the upper layers can share it.

CN_Chapter 02.indd 23 2/20/2012 6:03:11 PM

 24 Data communications and Computer Networks

3. Why standardization of network architecture is important?

Ans: Due to the advantages of layered architecture, many network vendors and suppliers used the
concept of layered architecture for designing most computer systems, but the set of protocols and inter-
faces defined by each vendor were not same. In addition, the partition of layers defined by each vendor
was alike. All this resulted into the integration incompatibility of different architectures defined by
different vendors. Thus, to allow different vendor’s network architectures to interoperate, the need for
standardization of network architecture was felt. The standardization of architecture also tends to reduce
the effort required to develop interfaces for the networking of different architectures.
4. What is Open Systems Interconnection (OSI) reference model? What are the principles
used in defining the OSI layers
Ans: OSI is a standard reference model for communication Application layer
between end users in a network. By the term open system, we mean a Presentation layer
set of protocols using which a system can communicate with any ­other Session layer
system irrespective of the differences in their underlying hardware
Transport layer
and software. In 1983, International Organization for Standardization
Network layer
(ISO) published a document called ‘The Basic Reference Model for
Open Systems Interconnection’, which visualizes network protocols Data link layer
as a seven-layered model. The OSI model consists of seven separate Physical layer
but related layers, namely, physical, data link, network, transport,
Figure 2.1 Layers in the OSI Model
­session, presentation and application layers as shown in Figure 2.1.
With in a single machine, a layer in OSI model communicates with two other OSI layers. It services
to the layer that is located directly above it while uses the services offered by the layer that is located
directly below it. For example, the data link layer provides services to the network layer while calls upon
the services of the physical layer. In contrast, during communication between two machines, each layer
on source machine communicates with the corresponding layer (called the peer layer) on the destination
machine using a set of protocols that are appropriate for the layer.
In 1983, Day and Zimmermann laid down certain principles which were used in defining the OSI
layers. These principles are as follows:
‰‰ Each layer should perform a well-defined function
‰‰ The functionality of each layer should be defined keeping in mind the internationally standardized
protocols.
‰‰ Changes made in any layer should not affect the other layers.
‰‰ The numbers of layers should be more enough, so that each layer is associated with distinct
functions.
‰‰ A different layer should be made when the abstraction level changes.

5. Explain how data flows between the layers in OSI model

Ans: When data is sent from one machine to another, it travels down sequentially from layer to layer
on the source machine and as it reaches the destination machine, it moves up through the layers. While
the data passes through layers on source machine, each layer adds header (and sometimes, trailer which
is usually added at data link layer) to it and passes the whole unit to the layer directly below it. The
header attached at each layer contains control information such as sequence numbers and size of data.
Figure 2.2 shows the data flow in OSI model during communication between two processes on
­machines A and B. Initially, the application layer (layer 7) of machine A adds an application header
(AH) to the data and passes the package to the presentation layer (layer 6), which further adds its own

CN_Chapter 02.indd 24 2/20/2012 6:03:12 PM

 Reference Models and Network Devices 25

Sending process Data Receiving processing

Application protocol
7 Application AH Data Application 7

Presentation protocol
6 Presentation PH AH Data Presentation 6

Session protocol
5 Session SH PH AH Data Session 5

4 Transport Transport protocol Transport 4

TH SH PH AH Data

Network protocol
3 Network NH TH SH PH AH Data Network 3

Data link
2 Data DH NH TH SH PH AH Data DT Data 2
protocol

Physical Bits Physical

1 DH NH TH SH PH AH Data DT 1

Actual data transmission path

Machine A Machine B

Figure 2.2 Data Flow in OSI Model

header PH to the received data and passes it to the session layer. The same process is repeated at the
­session (layer 5), transport (layer 4) and network layers (layer 3). At the data link layer (layer 2), a trailer
DT is also added to the data received from network layer along with the header DH. Finally, the entire
package (data plus headers and trailer) reaches the physical layer where it is transformed into a form
(electromagnetic signals) that can be transmitted to the machine B.
At machine B, the reverse process happens. The physical layer transforms the electromagnetic signals
back into digital form. As the data travels up through the layers, each layer strips off the header (or
trailer) added by its peer layer and passes the rest package to the layer directly above it. For example, the
data link layer removes DH and DT from the package received from the physical layer and passes the
resultant package to the network layer. The network layer then removes NH and passes the rest package
to the transport layer and so on. Ultimately, the data reaching the application layer of machine B is in a
format appropriate for the receiving process on machine B and thus is passed to the process.
6. Explain the duties of each layer in OSI model.
Ans: The seven layers of OSI model are divided into two groups according to their functionalities.
Physical, data link and network layers are put in one group, as all these layers help to move data between
devices. Transport, session, presentation and application layers are kept in other group, because they
allow interoperability among different software. The functions of each layer are discussed as follows:
1. Physical Layer: This layer defines the physical and electrical characteristics of the network.
It acts as a conduit between computers’ networking hardware and their networking software.

CN_Chapter 02.indd 25 2/20/2012 6:03:12 PM

 26 Data communications and Computer Networks

It handles the transfer of bits (0s and 1s) from one computer to another. This is where the bits
are actually converted into electrical signals that travel across the physical circuit. Physical ­layer
communication media include various types of copper or fibre-optic cable, as well as many
­different wireless media.
2. Data Link Layer: This layer is responsible for reliable delivery of data from node to node and
for providing services to the network layer. At sender’s side, the data link layer divides the pack-
ets received from the network layer into manageable form known as frames. These data frames
are then transmitted sequentially to the receiver. At the receiver’s end, data link layer detects and
corrects any errors in the transmitted data, which it gets from the physical layer. Other functions
of data link layer are error control and flow control. Error control ensures that all frames have
finally been delivered to the destination network layer and in the proper order. Flow control manages
the sender to send frames according to the receiving capability of the recipient.
3. Network Layer: This layer provides end-to-end communication and is responsible for transport-
ing traffic between devices that are not locally attached. Data in the network layer is called packet
(group of bits) which in addition to data contains source and destination address. Packets are sent
from node to node with the help of any of two approaches, namely, virtual circuit (connection-
oriented) and datagram (connectionless). In virtual circuit method, route is decided while estab-
lishing connection between two users and the same path is followed for the transmission of all
packets. In datagram method, there is no connection established; therefore, sequenced packets
take different paths to reach destination. Therefore, virtual circuit method resembles telephone
system and datagram method resembles postal system. Other functions of network layer include
routing, deadlock prevention and congestion control. Network layer makes routing decisions
with the help of routing algorithms to ensure the best route for packet from source to destination.
Congestion control tries to reduce the traffic on the network, so that delay can be reduced and
overall performance can be increased.
4. Transport Layer: The basic function of this layer is to handle error recognition and recovery of
the data packets. It provides end-to-end communication between processes which are ­executing
on different machines. It establishes, maintains and terminates communications between the sender
process and the receiver process. It splits the message at the sender’s end and passes each one onto
the network layer. At the receiver’s end, transport layer rebuilds packets into the original message,
and to ensure that the packets arrive correctly, the receiving transport layer sends acknowledgements
to the sender’s end.
5. Session Layer: The session layer comes into play primarily at the beginning and end of transmis-
sion. At the beginning of the transmission, it lets the receiver know its intent to start transmission.
At the end of the transmission, it determines whether or not the transmission was successful.
This layer also manages errors that occur in the upper layers such as a shortage of memory or
disk space necessary to complete an operation or printer errors. Some services provided by the
session layer are dialog control, synchronization and token management. Dialog control service
allows traffic to flow in both directions or in single direction at a time and also keeps track of
whose turn it is to transmit data. Synchronization helps to insert checkpoints in data streams, so
that if connection breaks during a long transmission then only the data which have not passed the
checkpoint yet need to be retransmitted. Token management prevents two nodes to execute the
same operation at the same time.
6. Presentation Layer: The function of this layer is to ensure that information sent from the applica-
tion layer of one system would be readable by the application layer of another system. ­Therefore,

CN_Chapter 02.indd 26 2/20/2012 6:03:13 PM

 Reference Models and Network Devices 27

presentation layer concerns with the syntax and semantics of the information transmitted. This is
the place where application data is packed or unpacked and is made ready to use by the running
application. This layer also manages security issues by providing services such as data encryp-
tion and compression, so that fewer bits need to be transferred on the network.
7. Application Layer: This layer is the entrance point that programs use to access the OSI model and
utilize network resources. This layer represents the services that directly support ­applications.
This OSI layer is closest to the end users. Application layer includes network software that
directly serves the end users of the network by providing them user interface and application
features such as electronic mail.
7. Explain the terms interfaces and services? Discuss protocol data unit (PDU).
Ans: Each layer contains some active elements called entities, such as process. Entities are named
as peer entities when they are in the same layer on different systems. Between two adjacent layers is an
interface which defines the operations and services of the lower layer that are available to its immediate
upper layer. A well-defined interface in a layered network helps to minimize the amount of traffic passed
between layers. The set of operations provided by a layer to the layer above it is called service. The
service is not concerned about the implementations of operations but defines what operations the layer
can perform for its users. Thus, lower layer implements services that can be used by its immediate upper
layer with the help of an interface. The lower layer is called a service provider and the upper layer is
called a service user.
A layer can request for the services of the lower layer, which
is present below it, through a specific location known as service Application layer PDU
access point (SAP). SAP has a unique address associated with Presentation layer
it. For example, in a fax machine system, SAP is the socket in Session layer
which a fax machine can be plugged and SAP addresses are
Transport layer Segments
the fax numbers which are unique for every user. Therefore, to Packets
Network layer
send a fax, the destination SAP ­address (fax number) must be
Data link layer Frames
known (Figure 2.3).
For communication and information sharing, each layer Physical layer Bits
makes use of PDUs. PDU can be attached in front (header) or Figure 2.3 Layers and Their PDU
end (trailer) of the data and contains control information which
is encapsulated with data at each layer. Depending on the information provided in the header, each
PDU is given a specific name. For example, at transport layer, data plus PDU is called a segment; at
network layer, segment and PDU (added by network layer) is given the name packet or datagram and
at data link layer, packet with data link PDU is called a frame. The PDU information attached by a
specific layer at the sender’s end can only be read by the peer layer at the receiver’s end. After reading
the information, the peer layer strips off the PDU and passes the remaining package to its immediate
upper layer.
8. Write the functions of data link layer in OSI model.
Ans: Data link layer is responsible for the transmission of frames between two nodes and provides
error notificatio to ensure that data is delivered to the intended node. To achieve this, it performs the
following functions:
‰‰ Framing: The data link layer takes the raw stream of bits from the physical layer and divides it into
manageable units called frames. To indicate the start and end of each frame to the receiver, several
methods including character count, bit stuffing and byte stuffing are use

CN_Chapter 02.indd 27 2/20/2012 6:03:13 PM

 28 Data communications and Computer Networks

‰‰ Physical Addressing: The data link layer adds physical address of the sender and/or receiver by
attaching header to each frame.
‰‰ Flow Control: The data link layer provides flow control mechanism, which prohibits a slow receiv-
er from being flooded by the fast sender. If the sender’s transmission speed is faster as compared
to the receiving capability of receiver, it is quite probable that some frames are lost. To avoid such
undesirable events, the data link layer must provide a means to control the flow of data between
sender and receiver.
‰‰ Error Control: The data link layer is responsible for ensuring that all frames are finally delivered
to the desired destination. To achieve this, the error control mechanism of data link layer makes
the receiver to send positive or negative acknowledgement to the sender. Positive acknowledge-
ment gives surety to the sender that frame has been received without any errors and negative
acknowledgement indicates that frame has not been received or has been damaged or lost during
transmission. It is the responsibilty of data link layer to ensure the retransmission of damaged and
lost frames.
‰‰ Access Control: When two or more devices are connected to each other via same link then
data link layer protocol detects which device has control over the link at a point of time. The
Institute of Electrical and Electronis Engineers (IEEE) has divided the data link layer into two
sublayers.
• Logical Link Control (LLC) Sublayer: This sublayer establishes and maintains links ­between
the communicating devices. It also provides SAPs, so that hosts can transfer information from
LLC to the network layer.
• Media Access Control (MAC) Sublayer: This sublayer determines which device to access
the channel next in case the channel is being shared by multiple devices. It communicates
directly with the network interface card (NIC) of hosts. NIC has MAC address of 48 bits that
is unique for each card.
9. Discuss some functions of the session layer and presentation layer.
Ans: Session layer is responsible for the establishing and maintaining session between processes
running on different machines as well as synchronizing the communication between them. Some other
functions of this layer are as follows:
‰‰ It allows the processes to communicate in either half duplex or full duplex mode. That is, infor-
mation can be transmitted between the processes either only in one direction at a time or in both
directions at same time.
‰‰ It makes use of checkpoints for synchronization that helps in identifying which data to retransmit.

The presentation layer is concerned with the syntax and semantics of the information being transmitted
between communicating devices. Other functions of presentation layer are as follows:
‰‰ It converts the representation of information used within the computer to network standard repre-
sentation and vice versa.
‰‰ It encrypts and decrypts data at the sender’s and receiver’s end respectively. It also compresses the
data for reducing the traffic on the communication channel
10. Compare the functionalities of network layer with transport layer.
Ans: Both network and transport layers are responsible for end-to-end communication but still there
are certain differences in the set of services they provide. These differences are listed in Table 2.1.

CN_Chapter 02.indd 28 2/20/2012 6:03:13 PM

 Reference Models and Network Devices 29

Table 2.1 Comparison Between Network and Transport Layer

Network layer Transport layer

• It performs routing and makes the routing decisions • It ensures that the entire message is delivered to
based on the priority of packets. ­destination machine’s transport layer without any error.
• It is a connection-oriented layer. That is, it requires • It provides both connection-oriented and
connection to be established between source and ­connectionless services.
destination before delivering the packets.
• It performs packet sequencing, flow control and error • It provides end-to-end data transport service.
control.

11. Explain in brief the transmission control protocol/internet protocol (TCP/IP) reference model.
Ans: TCP/IP model was developed by the U.S. Department
..
of Defense (DoD) to connect ­multiple networks and preserve TP.
S HT
N UD P
data integrity. TCP/IP protocol model came after the OSI model PD
Application layer FT P TCP P ...
and the ­numbers of layers in TCP/IP differ from that of the OSI T M
Transport layer SC RP IC
model. TCP/IP model comprises of four layers, namely, network A
IP anda r d
Internet layer st ols
access (also called host-to-network layer), Internet, transport Network access
l l
A toc
and ­application layers (see Figure 2.4). pro
layer
The network access layer of TCP/IP model corresponds to the
combination of physical and data link layers of OSI model. The Figure 2.4 TCP/IP Model
Internet layer corresponds to the network layer of OSI model and
the application layer performs tasks of session, presentation and application layers of OSI model with
the transport layer of TCP/IP performing a part of responsibilities of session layer of OSI model.
TCP/IP protocol suite contains a group of protocols forming a hierarchy such that the lower layer
protocols support upper layer protocols.
1. Network Access Layer: This layer does not rely on specific protocol hence supports all standard
protocols. It connects two nodes to the network with the help of some protocol and move data
across two nodes which are connected via same link. The nodes after connection can transfer IP
packets to each other. This layer is also referred to as host-to-network layer.
2. Internet Layer: The main function of this layer is to enable the hosts to transmit packets to
­different networks by taking any of the routes available for reaching the destination. This layer
strengthens the whole architecture and defines the format of packet. The rules to be followed
while delivering the packets are transparent to the users. Internet layer supports many protocols
such as IP, address resolution protocol (ARP), reverse ARP (RARP), Internet control message
protocol (ICMP) and Internet group message protocol (IGMP). IP is an unreliable and connec-
tionless protocol that transmits data in the form of packets called datagrams. Each datagram is
transmitted independently and can travel through a different route. Moreover, the datagrams may
reach the destination not necessarily in the order in which they were sent or may be duplicated.
IP neither keeps track of the routes followed by the datagrams nor does it perform any error ­checking.
It tries its best to deliver the datagrams at their intended destinations; however, it does not ensure
the delivery. ARP is used to identify the physical address of a node whose logical address is
known. RARP performs just the reverse of ARP, that is, it enables a host whose physical address
is known to identify its logical address. This protocol is used when a new node is connected to the

CN_Chapter 02.indd 29 2/20/2012 6:03:14 PM

 30 Data communications and Computer Networks

network or when an already connected node is formatted. ICMP is used to send error messages
to the sender in case the datagram does not reach its destination. IGMP is used to deliver the
same message to a number of recipients at the same time.
3. Transport Layer: The main function of this layer is to deliver a message from a process on
the source machine to a process on the destination machine. This layer is designed to allow
end-to-end conversation between peer entities. It uses three protocols, namely, TCP, user datagram
protocol (UDP) and stream control message protocol (SCMP) to accomplish its responsibilities.
TCP is a connection-oriented protocol which means a connection must be established between
the source and the destination before any transmission begins. It is also a reliable protocol, as it
ensures error-free delivery of data to the destination. UDP is an unreliable and a connectionless
protocol that performs very limited error checking. SCTP is the combination of UDP and TCP
and it supports advanced features such as voice over the Internet.
4. Application Layer: This layer contains all the high-level ­protocols such as file transfer protocol
(FTP) and virtual ­terminal (TELNET). Some more protocols which were added later include
domain name service (DNS), hyper text transfer protocol (HTTP) and many more. With the
help of various protocols, this layer integrates many activities and responsibilities for effective
­communication.
12. Describe various types of addresses associated with the layers of TCP/IP model.
Ans: Each layer in the TCP/IP model uses an address for the efficient delivery of data between
­communicating nodes. The host-to-network layer (physical plus data link layer) relates to physical address,
network layer relates to logical address, transport layer concerns with port address and application layer
defines specifi address. The description of these addresses is as follows:
‰‰ Physical Address: It is the address assigned to a node by the network (LAN or WAN) in which it is
connected. It is the lowest-level address that is included in the frames at the data link layer to identify
the destination node. The size and format of physical address is highly dependent on the underlying
physical network. That is, different networks can have different address formats. Physical address is
also known by other names including link address, MAC address and hardware address.
‰‰ Logical Address: In an inter-networked environment connecting different networks having differ-
ent address formats, the physical addresses are inadequate for communication. Thus, a universal
addressing system is used that assigns each host in the network a unique address called logical
address (also referred to as IP address or software address) which is independent of the underlying
physical network. For example, in Internet, each host connected to the Internet is assigned a 32-bit
IP address and no two hosts connected to the Internet can have the same IP address. It is the respon-
sibility of the network layer to translate the logical addresses into physical addresses.
‰‰ Port Address: The data link layer (using physical address) and network layer (using IP address)
ensure end-to-end delivery of data, that is, data is reached the destination host. Now, since multiple
processes may be running simultaneously on the host machine, there should be some means to iden-
tify the ­process to which data is to be communicated. To enable this, each running process on the host
machine is assigned with a label what is known as port address. Using the port address, the transport
layer ensures process-to-process delivery. In TCP/IP architecture, port address is of 16 bits.
‰‰ Specific Address: Some applications such as e-mail and World Wide Web (WWW) provide user-
friendly addresses designed for that specific address. Some examples of specific address include
an e-mail address that helps to identify the recipient of that e-mail and URL of a website that helps
to search a document on the web.

CN_Chapter 02.indd 30 2/20/2012 6:03:14 PM

 Reference Models and Network Devices 31

13. Compare OSI model with TCP/IP model.

Ans: OSI and TCP/IP are layered models that allow the computer systems to communicate with
each other. The OSI reference model was developed by ISO in order to standardize the protocols being
used in various layers and the TCP/IP model was developed by DoD to connect multiple networks. Both
models have some similarities which are as follows:
‰‰ Both OSI and TCP/IP models use set of independent protocols for enabling communication
between users.
‰‰ In both the models, upper layers focus on application such as web browser and lower layers focus
on end-to-end delivery of data.
The differences between OSI and TCP/IP models are listed in Table 2.2.

Table 2.2 Differences Between OSI and TCP/IP Models

OSI model TCP/IP model

• It is a seven-layer model. • It is a four-layer model.
• It was unable to connect to radio and satellite network. • It had the ability to connect to radio and satellite network.
• It supports only connection-oriented communication • It supports both connection-oriented and connection-
in ­transport layer while both connection-oriented and less communication in transport layer while only
c­onnectionless ­communication in network layer. connectionless communication in network layer.
• It clearly distinguishes services, interfaces and • It does not clearly distinguish services, interfaces
protocols. and protocols.
• It was defined before the invention of the Internet. • It was defined after the invention of Internet.
• T he model was developed before the corresponding • The model was developed after the protocols came
protocols came into existence. into existence.

14. Explain where the following fit in the OSI eference model.
(a) A 4-kHz analog connection across the telephone network.
(b) A 33.6-kbps modem connection across the telephone network.
(c) A 64-kbps digital connection across the telephone network.
Ans: (a) The actual 4-kHz analog signal exists only in the physical layer of the OSI reference model.
(b) A 33.6-kbps modem is used for connecting a user to the switch across the telephone network
and a modem also performs error checking, framing and flow control. Therefore, data link
layer will be used for performing such functionality.
(c) A 64-kbps digital signal carries user information and is similar to the 4-kHz analog ­connection
which makes use of twisted pair cable. Therefore, physical layer will be used for performing
such functionality.
15. Discuss in brief the Novell Netware network model.
Ans: Novell Netware is the popular network system which was designed for replacing mainframes
with a network of PCs thereby reducing the cost of companies. In this network, each user is assigned
a desktop PC operating as client that uses ­services (database, file etc) provided by some powerful PCs
operating as servers. Novell network is the modification of old Xerox network system (XNS) and it
uses a protocol stack (see Figure 2.5). It was developed prior to OSI and looks like TCP/IP model.

CN_Chapter 02.indd 31 2/20/2012 6:03:14 PM

 32 Data communications and Computer Networks

Layer

Application SAP File server ...

Transport NCP SPX
Network IPX
Data link Ethernet Token ring ARCnet
Physical Ethernet Token ring ARCnet

Figure 2.5 Novell Netware Model

The physical and data link layers of Novell Netware network can use any standard protocols includ-
ing Ethernet, IBM token ring and ARCnet. The network layer uses Internet packet exchange (IPX)
protocol, which is an unreliable and connectionless inter-network protocol. IPX transmits packets from
the source to destination transparently regardless of whether the source and destination are on the same
or on the different networks. It provides the same functionality as that of IP. However, the only differ-
ence between the two is that IPX uses 12-byte address while IP uses 4-byte address. The transport layer
uses network core protocol (NCP) which is a connection-oriented protocol. It provides many services
in addition to transporting data. Another protocol which works on transport layer is sequenced packet
exchange (SPX) that ­provides only data transport service. The application can choose from a variety of
protocols such as SAP and file serve .
16. Explain various network devices.
Ans: Networks are becoming more complicated and more pervasive everyday. Therefore, to reduce
complication, some network devices were developed. Network devices help nodes to get connected in a
network for efficient communication. Network devices include NIC, switch, router, bridge and gateway.

Network Interface Card

It is a hardware device that connects clients, servers
and peripherals to the network through a port. Most
network interfaces come as small circuit board that can
be inserted onto one of the computer motherboard’s
slots. Alternatively, modern computers sometimes in-
clude the network interface as part of their main cir-
cuit boards (motherboards). Each network interface
is associated with a unique address called its MAC
address. The MAC address helps in sending informa-
tion to the intended destination. NICs are the major
factor in determining the speed and performance of
a network. It is a good idea to use the fastest network Figure 2.6 Network Interface Card
card available for the type of workstation one is using
(Figure 2.6).

Repeater
It is the most basic device on a network. Signals that carry information within a network can travel a
fixed distance before attenuation endangers the integrity of the data. A repeater installed on the link

CN_Chapter 02.indd 32 2/20/2012 6:03:15 PM

 Reference Models and Network Devices 33

receives signal, regenerates it and sends the refreshed copy back to the link. Doing this means that the
new signal is clean and free from any background noise introduced while travelling down the wire.
In Figure 2.7, two sections in a network are connected by the repeater.

Repeater

11001100 11001100

Figure 2.7 Repeater

Repeaters are most commonly used to extend a network. All network cable standards have maximum
cable length specification. If the distance between two network devices is longer than this specification,
a repeater is needed to regenerate the signal. Without the repeater, the signal will be too weak for the
computers on each end to reliably understand. A good example of the use of repeaters would be in a
LAN using a star topology with unshielded twisted pair cabling. The length limit for unshielded twisted
pair cable is 100 m. The repeater amplifies all the signals that pass through it allowing for the total
length of cable on the network to exceed the 100 m limit. Nonetheless, the repeaters have no in-built
intelligence and they do not look at the contents of the packet while regenerating the signal. Thus, there
is no processing overhead in sending a packet through a repeater. However, a repeater will repeat any
errors in the original signal.

Hub
It is a small box that connects individual devices on a network, so that they can communicate with one
another. The hub operates by gathering the signals from individual network devices, optionally ampli-
fying the signals, and then sending them onto all other connected devices. Amplification of the signal
ensures that devices on the network receive reliable information. A hub can be thought of as the centre
of a bicycle wheel, where the spokes (individual computers) meet.
Nowadays, the terms repeater and hub are used synonymously, but actually they are not same.
Although at its very basic level, a hub can be thought of as a multi-port repeater. Typically, hubs have
anywhere from 4 to over 400 ports. When a signal is received on one port of the hub, it is regenerated
out to all the other ports. It is most commonly used to connect multiple machines to the same LAN.
Administrators connect a computer to each port on the hub, leaving one port free to connect to another
hub or to a higher-level device such as a bridge or router.

Bridge
This device allows the division of a large network into two or more smaller and efficient networks.
It monitors the information traffic on both sides of the network, so that it can pass packets of infor-
mation to the correct location. Most bridges can ‘listen’ to the network and automatically figure out
the address of each computer on both sides of the bridge. A bridge examines each packet as it enters
though one of the ports. It first looks at the MAC address of the sender and creates a mapping between

CN_Chapter 02.indd 33 2/20/2012 6:03:15 PM

 34 Data communications and Computer Networks

the port and the sender’s MAC address. It then looks at the address of the recipient, comparing the
MAC address to the list of all learned MAC addresses. If the address is in the list, the bridge looks
up the port number and forwards the packet to the port where it thinks the recipient is connected. If
the recipient’s MAC address is not in the list, the bridge then does a flood; it sends the signal to all the
ports except the one from where it was received. As a result, a bridge reduces the amount of traffic on
a LAN by dividing it into two segments. It inspects incoming traffic and decides whether to forward
or discard it (Figure 2.8).

LAN 1

LAN 2
Bridge

Figure 2.8 Bridge

Bridges can be used to connect networks with different types of cabling or physical topologies. They
must, however, be used between networks employing the same protocol. Since a bridge examines the
packet to record the sender and looks up the recipient, there is overhead in sending a packet through a
bridge. On a modern bridge, this overhead is miniscule and does not affect network performance.

Switch
It is a multi-port bridge. It connects all the devices on a network, so that they can communicate with one
another. The behaviour of a switch is same as that of a bridge. It is capable of inspecting the data packets
as they are received, determining the source and destination device of that packet, and forwarding that
packet appropriately. The difference is that most switches implement these functions in hardware using
a dedicated processor. This makes them much faster than traditional software-based bridges.

Router
It is an essential network device for interconnecting two or more networks. The router’s sole aim
is to trace the best route for information to travel. As network traffic changes during the day, rout-
ers can redirect information to take less congested routes. A router creates and/or maintains a table,
called a routing table that stores the best routes to certain network destinations. While bridges know
the addresses of all computers on each side of the network, routers know the addresses of computers,
bridges and other routers on the network. Routers can even ‘listen’ to the entire network to determine
which sections are the busiest. They can then redirect data around those sections until they clear up
(Figure 2.9).

CN_Chapter 02.indd 34 2/20/2012 6:03:16 PM

 Reference Models and Network Devices 35

Routers are generally expensive and difficult

INTERNET to configure and maintain. They are critical com-
ponents of a network and if they fail, the network
WAN services will be significantly impaired. Most
Modem
routers operate by examining incoming or outgo-
ing signals for information at the network layer.
Router In addition, they can permit or deny network
communications with a particular network.

Hub
Gateway
LAN
It is an internetworking device, which joins net-
works operating on different protocols together.
It is also known as protocol converter. A gateway
accepts the packet formatted for one protocol
and converts the formatted packet into another
protocol. For example, a gateway can receive
­e-mail message in one format and convert it into
another format. A gateway can be implemented
completely in software, hardware, or as a com-
Figure 2.9 Router bination of both. One can connect systems with
different protocols, languages and architecture
using a gateway (Figure 2.10).

Windows

Netware (Novell) server

Gateway

Figure 2.10 Gateway

17. What are the different types of bridges?
Ans: A bridge is a network device that allows division of a large network into two or more similar
networks. It connects two segments of LANs and monitors the traffic on both sides of the network, so
that it can pass packets of information to the correct location. It maintains a forwarding table that links

CN_Chapter 02.indd 35 2/20/2012 6:03:24 PM

 36 Data communications and Computer Networks

addresses of all stations connected through it and, thus, helps to forward frames from one station to
another. Different types of bridges are as follows.

Transparent Bridge
As the name implies, the existence of bridge is transparent to the stations connected through it. The
transparent bridge is also called learning bridge, as its forwarding table is made automatically by
learning the movement of frames in the network. Initially, the forwarding table contains no entries;
however, as the frame move across the networks, the bridge uses the source addresses to make or
update entries to the forwarding table and the destination address to make forwarding decisions.
Figure 2.11 shows a transparent bridge connecting two networks LAN1 and LAN2 via ports 1 and 2,
respectively.

LAN 1 LAN 2
1 2
Transparent
Station A Station B bridge Station C Station D
Figure 2.11 Transparent Bridge

To understand how transparent bridge works, suppose station A wishes to send frame to station D.
Since there is no entry corresponding to A or D in the forwarding table, the bridge broadcasts the
frames to both the ports (that is, ports 1 and 2). However, at the same time, it learns from the source
address A that the frame has come through port 1 and, thus, adds an entry (the source address A and
port 1) to the forwarding table. Next time, whenever a frame from a station (say, C) destined for A
comes to the bridge, it is forwarded only to port 1 and not elsewhere. In addition, an entry (source
address C and port 2) is added to the forwarding table. Thus, as the frames are forwarded, the learning
process of bridge continues.
An important advantage of transparent bridge is that since stations are not aware of the presence of
bridge, the stations need not be reconfigured in case the bridge is added or deleted

Source Routing Bridge

This bridge is used to connect two or more token ring LANs. In addition to the source and destination
address, each frame includes the address of the all the bridges to be visited as specified by the source
station. Thus, the path of a frame is already defined by the source station and intermediate nodes ­cannot
take any decision on the routes. The source station acquires the addresses of bridges by exchanging
some special frames with the destination station before the transmission of data frame begins. Figure 2.12
shows a source routing bridge.

Remote Bridge
It is used to connect two bridges at remote locations using dedicated links. Remote bridge configuration
is shown in Figure 2.13 in which two bridges are interconnected using WAN.

CN_Chapter 02.indd 36 2/20/2012 6:03:26 PM

 Reference Models and Network Devices 37

Token ring
3
Source-routing
bridge B

Token ring Token ring

1 2 Source-routing
Source-routing bridge D
bridge A

Token ring
4
Source-routing
bridge C

Figure 2.12 Source Routing Bridge

Remote bridge

WAN

Figure 2.13 Remote Bridge

18. Differentiate bridge, router and repeater.

Ans: Some differences among bridge, router and repeater are listed in Table 2.3.

Table 2.3 Comparison Among Bridges, Routers and Repeaters

Bridge Router Repeater

• It operates at the data link layer • It operates at the network layer of • It operates at the physical
of OSI model. OSI model. layer of OSI model.
• It is mostly used in LANs. • It is mostly used in internetworking. • It is used to extend the
physical length of a network.
• It transmits the frames in similar • It transmits the packets using simi- • It is an analog device con-
network. lar protocols in different networks. nected to two cable ­segments
which regenerates the signals.

CN_Chapter 02.indd 37 2/20/2012 6:03:30 PM

 38 Data communications and Computer Networks

19. Differentiate switch and hub.

Ans: The differences between switch and hub are listed in Table 2.4.

Table 2.4 Differences Between Switch and Hub

Switch Hub
A switch operates at the data link layer. A hub operates at the physical layer.
It is a complex device and more expensive than a hub. It is a simple device and cheaper than switch.
It is a full-duplex device and more secured. It is a half-duplex device and less secured.
Each port of switch has its own collision domain, that is, each The entire hub forms a single-collision domain, that
port has buffer space for storing frame; therefore, when two is, when two frames arrive at the same time, they will
frames arrive at the same time, frames will not be lost. always collide and hence frame will be lost.
It is an intelligent device, as it maintains a table to transmit It is a non-intelligent device, as each time frame is
the frame to the intended recipient. broadcast to all the connected nodes.
It utilizes the bandwidth effectively. Wastage of bandwidth is more in the case of hub.

20. Differentiate router and switch.

Ans: Some of differences between the router and switch are listed in Table 2.5.
Table 2.5 Differences Between Router and Switch

Router Switch
• It connects nodes on different networks. • It connects nodes within a same network.
• It operates in network layer. • It operates in data link layer.
• It uses IP address for transmission of packets. • It uses MAC address for transmission of frames.
• It is more intelligent and complex device than switch. • It is less intelligent and simpler device than router.
• Various algorithms are used to forward packets along • No such algorithms are used by switch.
their best path.
• It needs to be configured before using. • M
 ost switches are ready to use and need not be
configured.

Multiple Choice Questions

1. The correct order of corresponding OSI lay- (c) network, presentation, data link and
ers for having the functionalities of packet transport
­priortization, shared access resolution, end-to- (d) network, data link, application,
end flow and error control and socket-based inter presentation
process communication are ______________. 2. The main function of transport layer is
(a) network, physical, transport and _______________.
application
(a) synchronization
(b) network, data link, presentation and
application (b) node-to-node delivery

CN_Chapter 02.indd 38 2/20/2012 6:03:30 PM

 Reference Models and Network Devices 39

(c) process-to-process delivery (c) network layer

(d) updating routing tables (d) application layer
3. IP is responsible for __________ communica- 7. Which of the following internetworking de-
tion while TCP is resposible for __________ vice uses the greatest number of layers in the
­communication. OSI model?
(a) process-to-process, host-to-host (a) Bridge
(b) host-to-host, process-to-process (b) Gateway
(c) process-to-process, node-to-node (c) Router
(d) node-to-node, host-to-host (d) None of these
4. Which of the following is/are an application 8. In Novell Netware, IPX is a/an _________
layer service? byte address.
(a) File transfer and access (a) 8 (b) 12
(b) Domain name service (c) 4 (d) 48
(c) Remote login 9. Which type of bridge builds and updates its
(d) All of these tables from address information of packets?
5. To interconnect two homogeneous WANs, (a) Tranparent
we need a ___________. (b) Source routing
(a) bridge (b) gateway (c) Remote
(c) router (d) hub (d) None of these
6. A bridge recognizes the addresses of 10. Which of the following address(es) is needed
__________. for the effective communication?
(a) physical layer (a) MAC (b) IP
(b) data link layer (c) Port address (d) All of these

Answers
1. (b) 2. (c) 3. (b) 4. (d) 5. (c) 6. (b) 7. (b) 8. (b) 9. (a) 10. (d)

CN_Chapter 02.indd 39 2/20/2012 6:03:30 PM

 3
Analog and Digital
Transmission

1. What do you mean by data and signal?

Ans: Data refers to an entity that expresses some meaning based on rules which are agreed upon by
the sender and the receiver. It can be available in various forms such as text, graphics, audio and video.
Signal refers to the representation of the data in a form (such as electrical, electronic or optical) suitable
for transmission over a transmission medium.
2. Defin analog and digital data.
Ans: Analog data is defined as the data having continuous states. For example, sounds produced
while speaking take continuous values. Digital data is defined as the data having discrete states. For
example, data to be stored in the memory of computer is in the form of 0s and 1s.
3. Distinguish analog and digital signals?
Ans: Data is transmitted across a transmission medium in the form of electromagnetic signals. An
electromagnetic signal can be either analog or digital. A signal that passes through and includes a wide
range of varying values of intensity over a period of time is referred to as analog signal [Figure 3.1(a)].

Amplitude Amplitude

1 1 1
0 0
Time Time

(a) Analog signal (b) Digital signal

Figure 3.1 Analog and Digital Signals

CN_Chapter 03.indd 40 2/20/2012 6:08:01 PM

 Analog and Digital Transmission 41

In contrast, a signal that has only a finite range of values (generally 0 and 1) is referred to as digital signal
[Figure 3.1(b)]. Either of the analog or digital signals can be used to transmit either analog or digital data.
4. What do you understand by periodic and non-periodic signals?
Ans: Both analog and digital signals can be either periodic or non-periodic. A periodic signal
­exhibits a specific signal pattern that repeats over time [Figures 3.2(a) and 3.2(b)]. Sine waves and
square waves are the most common examples of periodic analog and digital signals, respectively. On
the other hand, a non-periodic (or aperiodic) signal does not repeat any specific signal pattern over
time [Figures 3.2(c) and 3.2(d)]. Usually, in data communications, periodic analog and non-periodic
digital signals are used.

Amplitude Amplitude

1 1 1
0 0
Time Time

(a) Periodic analog signal (b) Periodic digital signal

Amplitude Amplitude

Time Time

(c) Non-periodic analog signal (d) Non-periodic digital signal

Figure 3.2 Periodic and Non-periodic Signal

5. Describe the parameters that characterize the sine waves.

Ans: The sine wave is the most fundamental form of periodic analog signal. It is characterized by
the following three parameters.
‰‰ Peak Amplitude: It is the highest value or strength of the signal at any point of time [Figure 3.3(a)].
The unit for amplitude depends on the type of the signal. For example, in case of electrical signals,
the unit of amplitude is normally volts and amperes.
‰‰ Frequency: It refers to the number of cycles a signal completes in 1 s. It is measured in cycles per
second or Hertz (Hz). Another parameter equivalent to frequency is the period which is defined as
the time taken by a signal to complete one cycle [Figure 3.3(a)]. Thus, we can say that frequency is
equal to the number of periods per second. Both frequency (f ) and period (T ) are related to each
other by the following formula.
f = 1/T

CN_Chapter 03.indd 41 2/20/2012 6:08:02 PM

 42 Data Communications and Computer Networks

Amplitude

Peak amplitude
Frequency = 3Hz

Time

1
Period =
3

(a) Peak amplitude and period

Amplitude

Time

(b) Phase shift of 90°

Figure 3.3 Sine Wave

For example, if a signal wave completes one period in 1 s, its frequency is 1 Hz.
‰‰ Phase: It refers to the measure of shift in the position of a signal with respect to time 0. Figure 3.3(b)
shows a shift of 90° in the sine wave shown in Figure 3.3(a). Phase is measured in degree (°) or
­radians (rad) where 1° = 2π/360 rad.

6. Defin wavelength. What is the relation between wavelength and frequency?

Ans: Wavelength refers to the distance covered by the signal in one cycle. In other words, it is
the ­distance between two similar points in corresponding phases of two adjacent cycles of a signal
(Figure 3.4). It is usually denoted by λ.

Amplitude

Maximum
amplitude

Time

Wavelength
(one cycle)
Figure 3.4 Wavelength of a Signal

CN_Chapter 03.indd 42 2/20/2012 6:08:04 PM

 Analog and Digital Transmission 43

Wavelength relates the frequency (f ) of a signal with the speed of propagation of the medium through
which the signal is being transmitted. The relation between frequency and wavelength is expressed
­using the following formula.
λ = propagation speed/f
⇒ λ = propagation speed × period
For example, in vacuum, the signal propagates with speed of light (c), which is equal to 3 × 108 m/s. Thus,
λ = c/f
7. Distinguish time-domain and frequency-domain representations.
Ans: A signal can be represented in two ways: time-domain and frequency-domain. In time-domain
representation, the signal is represented as a function of time. The time-domain plot of a signal depicts
the changes in the amplitude of a signal with time. Figure 3.5 represents the sine wave using time-
domain plot. The horizontal axis represents the time and the vertical axis represents the amplitude.
In frequency-domain representation, a signal is represented as a function of frequency. Unlike time-
domain plot, the frequency-domain plot does not depict the changes in amplitude; rather it depicts only
the peak amplitude of the signal and frequency. In addition, the complete sine wave is represented just
by a spike. Figure 3.6 shows the frequency domain plot for the sine wave shown in Figure 3.5.

Amplitude
Amplitude

Frequency = 3Hz
6 Peak value = 6
6
Time
Frequency (Hz)
1 2 3 4 5 6

Figure 3.5 T ime-domain Representation of a Figure 3.6 F requency-domain Representation of a

Sine Wave Sine Wave
8. What do you understand by composite signal?
Ans: In data communications, single frequency sine waves are not used to communicate data, rather
composite signals are used. A composite signal is a collection of one or more signals having different
frequencies, amplitude and phases (Figure 3.7). It can be periodic or non-periodic. The ­decomposition
of a periodic composite signal produces a series of signals with discrete frequencies, whereas the
­decomposition of a non-periodic composite signal produces a combination of signals with continuous
frequencies in the range of zero to infinit .

Time

Figure 3.7 Composite Signal

CN_Chapter 03.indd 43 2/20/2012 6:08:05 PM

 44 Data Communications and Computer Networks

9. Why digital transmission is preferred over analog transmission?

Ans: Analog transmission is a means of transmitting analog signals (representing either analog or
digital data) while the digital transmission is a means of transmitting digital signals (representing either
analog or digital data). Digital transmission offers certain advantages due to which it is preferred over analog
transmission. These advantages are as follows:
‰‰ To achieve long-distance analog transmission, amplifiers are to be used for boosting up the signal
energy. However, the amplifiers boost up the noise too, thereby resulting in more distorted signal.
In contrast, repeaters used in digital transmission do not contribute to noise and other impairments.
As a result, the data can be transmitted to longer distances while maintaining the data integrity.
‰‰ The digital data as well as the digitized analog data can be easily encrypted using encryption
­techniques.
‰‰ With the advent of large-scale integration (LSI) and very large-scale integration (VLSI) ­technologies,
the size and cost of digital equipments have reduced to a greater extent as compared to analog
equipments.
‰‰ To utilize the large channel capacity effectively, multiplexing is required and it is quite easier and
economical to accomplish this using digital techniques than analog techniques.
‰‰ All data including voice, video and digital data can be integrated and passed through digital circuits
conveniently.
10. How Fourier analysis is important in data communication?
Ans: Fourier analysis is a tool that is used to convert a time domain signal to a frequency domain
signal and vice versa. According to this analysis, a composite periodic signal with a period T can be
decomposed into a series called Fourier series of sine and cosine functions with each function being an
integral harmonic of the fundamental frequency f of the composite signal as shown here.
∞ ∞

S(t) = A­+ ∑ A
0 n
sin(2πnft) + ∑ Bn cos(2πnft),
n=1 n=1

where
T
1
A­
0
=
T ∫ s(t)dt, the average value of a signal over a period
0

T
2
A­
n
=
T ∫
0
s(t)cos(2πnft)dt, the coefficient of nth cosine component

T
= 2 s(t)sin(2πnft)dt, the coefficient of nth sine component.
B­
n
T ∫
0

Fourier series converts time domain of a periodic signal to discrete frequency domain. To convert
time domain of a non-periodic signal to continuous frequency domain, Fourier transform is used as
­expressed below:

S(f) = ∫
-∞
s(t)e­­
j2Πft
dt

CN_Chapter 03.indd 44 2/20/2012 6:08:09 PM

 Analog and Digital Transmission 45

Inverse Fourier transform is expressed by the following formula:

∞

s(t) = ∫
-∞
S(f)e­­
–j2Πft
dt

11. Defin bit interval, bit rate and bit length of a digital signal.
Ans:
‰‰ Bit Interval: It is the time required to transmit one bit.
‰‰ Bit Rate: The number of bits transmitted per second is known as bit rate. It can also be defined as
the number of bit intervals per second. Its unit is bits per second (bps). The bit rate for a bit having
bit interval t will be 1/t.
‰‰ Bit Length: The distance occupied by a single bit while transmission on a transmission medium is
known as bit length. It is related to bit interval as per the formula given below:
Bit length = bit interval × speed of propagation
12. Differentiate baseband and broadband transmissions?
Ans: Though both baseband and broadband transmissions are the approaches used to transmit digital
signals, there are certain differences between the two. These differences are listed in Table 3.1.

Table 3.1 Differences Between Baseband and Broadband Transmissions

Baseband transmission Broadband transmission

• The digital signal is sent over a channel without con- • The digital signal is first converted into analog
verting it into an analog signal. signal and then sent over the channel.
• The entire bandwidth of the cable is consumed by a • Signals are sent on multiple frequencies, allowing
single signal. multiple signals to be sent simultaneously.
• It requires a low-pass channel—a channel whose • It requires a bandpass channel—a channel
bandwidth starts from zero. whose ­bandwidth does not start from zero.

13. Discuss different types of transmission impairments.

Ans: Sometimes, when a signal passes through a transmission medium, it gets deteriorated due to
imperfections in the transmission medium. As a result, the signal received at the receiver’s end differs
from the one being sent. For example, while transmitting digital signals, bit errors may occur such that
in the received signal, some binary 1 is transformed to binary 0 or vice versa. The types of impairments
that may occur during transmission are as follows:
‰‰ Attenuation: While a signal passes through a transmission medium, it loses some of its strength (or
weakens) to get over the resistance of transmission medium. This loss of strength or energy is termed
as attenuation. In case the signal strength becomes very low, it cannot be detected and ­interpreted
properly at the receiver’s end. To compensate the loss of strength, amplifiers or ­repeaters can be used.
The extent of attenuation of a signal depends greatly on its frequency; the effect of attenuation is
different for different frequency components. For example, during the transmission of a composite
signal, some frequency components bypass without any attenuation, some lose little bit strength
while others get blocked. This dependence of attenuation upon frequency leads to a distortion what
is called attenuation distortion, which is more prevalent in analog signals than digital signals. How-
ever, its effect can be reduced by using a suitable equalizer between the channel and the receiver.

CN_Chapter 03.indd 45 2/20/2012 6:08:10 PM

 46 Data Communications and Computer Networks

‰‰ Delay Distortion: This is another type of distortion that results because of difference in the delays
experienced by different frequency components while passing through the transmission medium.
Since the speed of propagation of a signal varies with frequency, different frequency components
of a composite signal arrive at the receiver at different times leading to delay distortion. As a
result, the shape of received signal changes or gets distorted. Like attenuation distortion, the ­effect
of ­delay distortion can be neutralized with the use of equalizers. However, unlike attenuation
­distortion, delay distortion is predominant in digital signals.
‰‰ Noise: When a signal transits through a transmission medium, some types of undesired signals
may mix with it such as intermodulation noise and crosstalk. Intermodulation noise occurs in the
cases where signals with different frequencies pass through the same transmission medium. In such
cases, the frequencies of some signals may combine (add or subtract) to generate new frequency
components which may interfere with the signals of same frequency sent by the transmitter. This
leads to distortion in the signal what is known as intermodulation noise. Crosstalk results when
the electromagnetic signals passing through one wire are induced on other wires due to the close
proximity of wires.

14. What does the signal-to-noise ratio (SNR) determine?

Ans: The SNR is used to determine the effect of noise on the signal. Formally, it is defined as the
ratio of the average power of a signal to the average noise power, as shown here.

SNR = avg(PS)/avg(PN)
More the value of SNR is, the less is the signal distorted due to noise and vice versa. SNR is usually
expressed in decibel (dB) units. SNRdB is defined as

SNRdB = 10 log10 SNR

15. Write down the theoretical formula to calculate data rate for a noiseless channel and a
noisy channel.
Ans: Data rate for a noiseless channel can be determined by using the Nyquist bit rate formula as
given below.
Bit rate = 2 × B × log2L,
where B = bandwidth of channel and L = the number of signal levels used to represent data.
Data rate for a noisy channel can be determined by Shannon capacity, as given below.
C = B × log2(1 + SNR),
where C = data rate and B = bandwidth of channel.
16. Describe the factors used to measure the performance of a network?
Ans: The performance of a network is one of the most important issues in networking. Some major
factors that measure the performance of a network are as follows.
‰‰ Bandwidth: It is one of the main characteristics that measure the performance of a network. It
can be used in two different contexts. First is the bandwidth in hertz which refers to the range of
frequencies in a composite signal and second is the bandwidth in bps which refers to the number
of bits that can be transmitted by a channel or network in 1 s. An increase in bandwidth in hertz
implies the increase in bandwidth in bps.

CN_Chapter 03.indd 46 2/20/2012 6:08:10 PM

 Analog and Digital Transmission 47

‰‰ Throughput: It is the measure of how much data can be transmitted through a network in 1 s.
Though from the definitions throughput sounds similar to bandwidth, it can be less than bandwidth.
For example, a network may have bandwidth of 10 Mbps but only 2 Mbps can be transmitted
through it.
‰‰ Latency (Delay): It refers to the time elapsed between the first bit transmitted from the source and
the complete message arrived at the destination. It is the sum of propagation time, transmission
time, queuing time and processing delay of a frame.
The time taken by a bit to travel from the source to destination is referred to as the propagation
time. It can be calculated using the following formula.
Propagation time = distance/propagation speed
As the propagation speed increases, propagation time decreases.
The transmission time measures the time between the transmission of first bit from the send-
er’s end and the arrival of last bit of the message at the destination.
Transmission time = message size/bandwidth
Greater is the message size is, the more will be the transmission time.
Whenever a message being transmitted arrives at some intermediate device, such as router, it
is kept in a queue (if device is not free) maintained by the device. The device processes the queued
messages one by one. The time a message spends in the queue of some intermediate or end device
before being processed is referred to as the queuing time. It depends on the traffic load in the
­network. More the load of traffic is, the more is the queuing time
‰‰ Bandwidth-Delay Product: It refers to the product of bandwidth and delay of a network which
specifies the maximum number of bits that can be at any time on the link. For example, if a link
has a bandwidth of 8 bps and delay is 5 s, then the maximum number of bits that can fill the link is
40 bits.
‰‰ Jitter: It refers to a problem that occurs when the inter-arrival time between packets is not constant
and the application using the data is time sensitive.
17. Defin line coding. List some common characteristics of line coding schemes.
Ans: Line Coding is defined as the process of converting digital data or sequence of bits to digital
signals. Encoder is used at the sender’s end to create a digital signal from digital data and decoder is
used at the receiver’s end to recreate digital data from digital signal (Figure 3.8).

Sender Receiver

Digital data Digital data

Digital signal
Encoder Decoder

Figure 3.8 Line Coding

CN_Chapter 03.indd 47 2/20/2012 6:08:11 PM

 48 Data Communications and Computer Networks

Some common characteristics of line coding schemes are as follows:

‰‰ Data Element: It refers to the smallest entity (that is, bit) that expresses some information.
It ­describes what is to be sent.
‰‰ Signal Element: It refers to the shortest unit of digital signal. Each signal element carries one
or more data elements. The ratio (r) of number of data elements to number of signal element
­indicates the number of data elements carried by a signal element.
‰‰ Data Rate: Data rate, also called bit rate, refers to the number of data elements that can be trans-
mitted per second. It is usually expressed in bps. The speed of transmission increases with the
increase in data rate.
‰‰ Signal Rate: Signal rate, also called baud rate or modulation rate or pulse rate, refers to the number
of signal elements transmitted per second. It is expressed in baud. An increase in signal rate increases
the demand of bandwidth. The relationship between data rate and signal rate can be expressed as:
S = CF × d × (1/r),
where S = the number of signal elements; CF= case factor; D = data rate and r = the number of
data elements carried by a signal element.
‰‰ Bandwidth: Bandwidth refers to the maximum volume of data that can be transferred over any
communication medium at a given point of time. More the data needed to be transmitted in a given
period is, the more is the bandwidth required. On digital circuits, bandwidth is measured in bps.
The bandwidth is related to signal rate (S) as per the formula shown below.
Bmin= S
⇒ Bmin= CF × d × (1/r)
Similarly, if we are given the bandwidth (B) of channel, we can find out the maximum data rate
(dmax) using the following formula.
dmax= B × r × (1/CF)
‰‰ DC Components: A signal may include very low frequency component in its spectrum. These
components with frequencies around zero (called DC components) are undesirable, as some
­systems do not allow such components to pass through them. Thus, for these systems, the line cod-
ing with no DC component is required.
‰‰ Baseline Wandering: Baseline refers to the average of the received signal power calculated by
the receiver while decoding a digital signal. Baseline wandering refers to the movement or drift
in the baseline. Due to baseline wandering, the process of decoding at the receiver’s end becomes
difficult. Thus, a good line coding scheme must prevent baseline wandering.
‰‰ Error Detection: Line coding scheme should have built-in capability to detect some or all the
­errors that occur during transmission.
‰‰ Self-Synchronization: Bit intervals of the receiver should match with bit intervals of the sender to
interpret correctly the signal at the receiver’s end. For this, the digital signal must include timing
information (that is, self-synchronized) which means the transitions in the digital signal alert the
receiver at different points of the pulse. This information is used to readjust or reset a clock if the
receiver’s clock is not synchronized.
‰‰ Resistance to Noise: Line coding scheme should have ability to create a code that is unsusceptible
to noise or other inferences in the transmission medium.
‰‰ Complexity: Line coding scheme should be simple, as it is quite expensive to implement a ­complex
coding scheme as compared to a simple scheme.

CN_Chapter 03.indd 48 2/20/2012 6:08:11 PM

 Analog and Digital Transmission 49

18. Can the bit rate be less than the pulse rate? Why or why not?
Ans: Bit rate is always greater than or equal to the pulse rate because the relationship between pulse
rate and bit rate is defined by the following formula
Bit rate = pulse rate × log2L,
where L denotes the number of data levels of the signal and log2L denotes the number of bits per level.
If a pulse carries only 1 bit (that is, log2L=1), the pulse rate and the bit rate are the same. How-
ever, if the pulse carries more than 1 bit, then the bit rate is greater than the pulse rate.
19. Discuss various line coding schemes.
Ans: There are various line coding schemes that can be classified into three basic categories namely,
unipolar, polar and bipolar.

Unipolar Scheme
This scheme uses two voltage levels of a signal and both of these voltage levels are on one side of the
time axis (above or below). In this scheme, bit rate and baud rate are equal. However, the encoded signal
includes DC components and there is lack of synchronization in case of long series of 0s and 1s. The
only coding scheme that falls under this category is non-return-
Amplitude
to-zero (NRZ).
‰‰ NRZ: In this scheme, bit 1 is used to define positive volt-
0 1 0 1
age while bit 0 is used to define zero voltage of a signal.
The name NRZ comes from the fact that the signal does not
return to zero during the middle of a bit rather only between
two bits (Figure 3.9). The unipolar NRZ scheme is not gen- Time
erally used for data communication.
Figure 3.9 Unipolar NRZ Scheme
Polar scheme
This scheme uses two voltage levels of a signal, positive and negative, that can be on both sides of the
time axis. The positive voltage may represent 0 while negative voltage may represent 1 or vice versa.
Four different schemes fall under this category, which are discussed as follows.
‰‰ NRZ: It is the most common type of polar coding scheme. In this scheme, positive voltage
is used to represent one binary value and negative voltage is used to represent another. There
are two types of polar NRZ schemes namely, NRZ-level (NRZ-L) and NRZ-invert (NRZ-I).
In NRZ-L, the value of bit is determined by the signal level which remains constant during
bit duration (Figure 3.10). In NRZ-I, the value of bit is determined by inversion or lack of
inversion in the signal level. If there is change in the signal level, the value of bit will be 0,
whereas if there is lack of change in the signal level, the value of bit will be 1 (Figure 3.11).
Both NRZ-L and NRZ-I suffer from synchronization problem in case of long sequence of 0s.
However, NRZ-I suffers in case of long sequence of 1s also. In addition, both NRZ-L and NRZ-I
have DC component problem.
‰‰ Return-to-Zero (RZ): This scheme solves the synchronization problem of NRZ scheme. It uses
three values of voltage namely, zero, positive and negative. Unlike NRZ, the signal changes during
the middle of a bit but not between the bits. Once it changes during the bit, it remains there until

CN_Chapter 03.indd 49 2/20/2012 6:08:12 PM

 50 Data Communications and Computer Networks

Amplitude Amplitude

0 1 0 1 1 0 1 0 1 1

Time Time

Figure 3.10 Polar NRZ-L Scheme Figure 3.11 Polar NRZ-I Scheme

the next bit starts. Thus, to encode a single bit, two signal changes are required (Figure 3.12). There
is no DC component problem in RZ coding scheme. However, it is complex as it requires three
levels of voltage.

Amplitude Amplitude

0 1 0 1 1 0 1 0 1 0

Time Time

Figure 3.12 RZ Scheme Figure 3.13 Manchester Scheme

‰‰ Manchester Scheme: This is a biphase scheme that combines the idea of RZ and NRZ-L
schemes. In this scheme, the level of voltage changes at the middle of bit to provide synchro-
nization (Figure 3.13). The low-to-high transition in the middle of a bit indicates 1 while the
high-to-low indicates 0. This scheme overcomes the disad-
vantages of NRZ-L scheme. Amplitude
‰‰ Differential Manchester: This is also a biphase scheme. It
combines the idea of RZ and NRZ-I schemes. It is ­similar
to the manchester scheme in the sense that it changes ­during
the middle of a bit but the only difference is that the bit 0 1 0 1 1
­values are determined at the starting of bit. Transition
­occurs when the next bit is 0; otherwise, no transition occurs Time
­(Figure 3.14). This scheme overcomes all the disadvantages
of NRZ-I scheme.
Figure 3.14 D ifferential Manchester
Scheme
Bipolar Scheme
This scheme is similar to RZ encoding scheme and has three levels of voltage. The only difference
is that bit 0 is represented by zero level and bit 1 is represented by positive and negative levels of
voltage. In bipolar encoding, a stream of bits containing long sequence of binary 0s creates a constant

CN_Chapter 03.indd 50 2/20/2012 6:08:14 PM

 Analog and Digital Transmission 51

zero voltage. Therefore, bipolar encoding schemes does not have Amplitude
DC component. This scheme is available in two forms, which are
as follows:
‰‰ Alternate Mark Inversion (AMI): This is the most commonly
used bipolar coding scheme. It is so called because it involves 0 1 0 1 1
representation of binary 1s by alternate positive and negative
Time
levels of voltages. Here, bit 0 is represented using zero level
of voltage (Figure 3.15). This scheme is used for communica-
tion between devices placed at a large distance from each other.
Figure 3.15 AMI Scheme
However, it is difficult to achieve synchronization in this scheme
when a continuous stream of bit 0 is present in the data.
‰‰ Pseudoternary: This scheme is a modification of AMI encod- Amplitude
ing scheme. In this scheme, binary 1 is represented using zero
level of voltage while binary 0s are represented by alternate
positive and negative voltages (Figure 3.16).
0 1 0 1 1
20. Explain the concept of block coding.
Time
Ans: Block Coding is an alternative to line coding scheme that is
also used to convert digital data to digital signals. However, it is much
better than line coding, as it ensures synchronization and has built-in
error detecting capability, which results in better performance than Figure 3.16 Psedoternary Scheme
line coding. In this scheme, a three-step process (Figure 3.17) is used
to code the digital data.
1. Division: The original sequence of bits is divided into blocks or groups of n bits each.
2. Substitution: Each n-bit group is replaced with m bits where m >n.
3. Line Coding: An appropriate line coding scheme is used to combine the m-bit groups to form a
stream.

Sequence of bits Division

0110...111 0011..011
0110..11...0011..
Group of n bits each
Substitution

Line coding
Digital signal 011..0001 1011..000
Group of m bits each
Figure 3.17 Block Coding

The block coding is usually represented as nB/mB (n binary/m binary) such as 4B/5B and 5B/6B. For
example, in 4B/5B coding, original bit sequence is divided into 4-bit codes and each 4-bit code is ­replaced
with a 5-bit block and then NRZ-I line coding is used to convert 5-bit groups into digital signal.
21. Explain pulse code modulation (PCM) and delta modulation (DM).
Ans: Both PCM and DM are the techniques used to convert analog signal to digital data.

CN_Chapter 03.indd 51 2/20/2012 6:08:15 PM

 52 Data Communications and Computer Networks

Pulse Code Modulation

This technique involves PCM encoder which encodes analog signal using three steps, namely,
­sampling, quantization and encoding (Figure 3.18).

PCM encoder

PAM Quantized
Analog signal Sampling Quantization Encoding
Signal Signal

Digital data

Figure 3.18 PCM Encoding

1. Sampling: In this process, the numbers of samples of the original analog signal are taken at
regular intervals of time (called sampling period). The inverse of sampling period is referred to
as sampling rate. According to Nyquist theorem, the sampling rate should be at least two times
of the highest frequency contained in the signal to regenerate the original analog signal at the
receiver’s end. For example, during the sampling of voice data, with frequency in the range of
400–5,000 Hz, 10,000 samples per second are sufficient for the coding. The sampling process is
also termed as pulse amplitude modulation (PAM) because it produces a PAM signal—a series
of pulses having amplitude between the highest and the lowest amplitudes of the original analog
signal. Sample and hold is the most common sampling method used today for creating flat-top
samples.
2. Quantization: The sampled PAM pulses may have non-integral values of amplitude which
­cannot be encoded directly. Thus, the sampled pulses need to be quantized and approximated
to integral values using analog-to-digital converter. Considering the original analog signal has
amplitudes between Vmaximum and Vminimum, the following steps are used for quantizing the signal.
i) The range of amplitudes of analog signal is partitioned into L levels or zones each of height h,
where
h = (Vmaximum - Vminimum)/L.
The value of L is chosen depending on the range of amplitudes of the original analog signal
as well as the extent of accuracy required in the recovered signal. The signal having more
amplitude values require more number of quantization levels.
ii) The quantized values of 0 to L-1 are assigned at the midpoint of each zone.
iii) The value of the sample amplitude is approximated to the quantized value.
Since the output values of the quantization process are only the approximated values, an
error known as quantization error may occur due to which it may not be possible to re-
cover the original signal exactly. The quantization error also affects the signal-to-noise ratio
(SNRdB) of the signal and the amount of effect depends on the value of L or the number of
bits per sample (nb) as shown in the following formula.
SNRdB = 6.02 nb + 1.76 dB.

CN_Chapter 03.indd 52 2/20/2012 6:08:16 PM

 Analog and Digital Transmission 53

The effect of quantization error can be minimized by using a process called companding and
expanding. This process uses a compressor before encoding and uses an expander after decoding.
Companding refers to decreasing the instantaneous voltage amplitude for larger values while
expanding refers to increasing the instantaneous voltage amplitude for smaller values. This helps
to improve the SNRdB of the signal.
3. Encoding: After quantization, encoding is done in which each sample is converted to m-bit codeword
where m is equal to number of bits per sample (nb). The value of nb depends on the value of L as
shown in the following formula.
m = nb = log2 L
The relationship between bit rate and the number of bits per sample (nb) can be expressed as:
Bit rate = nb × sampling rate
At the receiver’s side, original signal is recovered using PCM decoder which converts the codeword
into a staircase signal. The staircase signal is formed by changing the codeword into a pulse that main-
tains the amplitude till the next pulse. Then, a low-pass filter is used to smoothen the staircase signal
into an analog signal. Figure 3.19 depicts this process.

PCM Decoder

Make and Low-pass

Digital data connect samples Analog signal
filter

Figure 3.19 PCM Decoding

Delta Modulation
This is an alternative to PCM technique with much reduced quantization error. In this technique, a
modulator is used at the sender’s side that produces the bits from an analog signal and these bits are sent
one after another; only one bit is sent per sample. The modulator generates a staircase signal against
which analog signal is compared. At each sampling interval, the amplitude value of analog signal is
compared with last amplitude of staircase signal to determine the bit in the digital data. If amplitude of
analog signal is smaller, the next bit will be 0. However, if the amplitude of analog signal is greater,
the next bit will be 1. Figure 3.20 shows the components of DM.

Modulator

Analog signal
Comparator Digital data

Delay unit Staircase maker

Figure 3.20 Delta Modulation Components

CN_Chapter 03.indd 53 2/20/2012 6:08:17 PM

 54 Data Communications and Computer Networks

To reproduce analog signal from digital data, demodulator is used at the receiver’s end. The
­ emodulator uses staircase maker and delay unit to generate analog signal which is then passed through
d
a low-pass filter for smoothing. Figure 3.21 depicts the delta demodulation process

Demodulator

Delay unit

Staircase Low-pass
Digital data Analog signal
maker filter

Figure 3.21 Delta Demodulation

22. What are the differences between serial and parallel transmissions?
Ans: The transmission of binary data across a link connecting two or more digital devices can be
performed in either of two modes: serial and parallel (Figure 3.22). There are certain differences
between these modes which are listed in Table 3.2.

Parallel to Serial to
serial converter parallel converter

1 1

1 0 1
Sender 0 0 Receiver
device device

1 1

(a) Serial transmission

0
Sender Receiver
device device

(b) Parallel transmission

Figure 3.22 Serial and Parallel Transmissions

CN_Chapter 03.indd 54 2/20/2012 6:08:18 PM

 Analog and Digital Transmission 55

Table 3.2 Serial and Parallel Transmissions

Serial transmission Parallel transmission

• The data is transmitted by sending one bit per each • The stream of bits is divided into groups and one
clock pulse [see Figure 3.21(a)]. group is sent per each clock pulse [see Figure
3.21(b)].
• It is a slower mode of transmission, as only one • It is a faster mode of transmission, as several
bit can be transmitted at a time. bits can be transmitted at a time.
• It requires only one communication channel between • To send n bits at a time, it requires
communicating devices; thus, it is a cheaper mode of n ­communication channel between
transmission. ­communicating devices. As a result, cost is
increased by a factor of n as compared to serial
transmission.
• A s communication within devices is parallel, both sender • No such converters are required.
and receiver require converter at the interface between
the device and the communication channel. The con-
verter at the interface between sender device and com-
munication channel converts parallel transmission to
serial transmission while the converter at the interface
between communication channel and receiver device
converts serial transmission to parallel transmission.

23. Discuss the three types of serial mode transmissions.

Ans: Serial mode transmission can be classified into three types, namely, synchronous, asynchronous
and isochronous.
‰‰ Asynchronous Transmission: In this transmission, the entire bit stream is divided into groups of
8   bits (that is, one byte) each (Figure 3.23). Each byte is treated independently and transmitted when-
ever ready regardless of the timer. To let the receiver know about the arrival of a byte, a start bit

Direction of flow

Stop bit Start bit

Data
1 11001011 0

01011 0 1 11001011 0 1 00110111 0 1 10

Gaps between
data units
Figure 3.23 Asynchronous Transmission

CN_Chapter 03.indd 55 2/20/2012 6:08:22 PM

 56 Data Communications and Computer Networks

(usually ­represented by 0) is added to the starting of each byte. Similarly, one or more stop bits
(usually represented by 1) are added to indicate the end of each byte. Though the transmission is
asynchronous at the byte level, still some synchronization is needed during the transmission of bits
within a byte. To achieve this synchronization, the receiver starts the timer after finding the start bit
and counts the number of bits until it finds the stop bit. There may be gap between the transmission
of two bytes which is filled with a series of stop bits or idle channel. Asynchronous transmission is
slow as it adds control information such as start bits, stop bits, and gap between bytes. However,
it is a cheaper and effective mode of transmission and thus is suitable for communication between
devices which do not demand fast speed.
‰‰ Synchronous Transmission: In this transmission, timing source is used for synchronization, so
that the receiver can receive the information in the order in which it is sent. Multiple bytes are
combined together to form frames. Data is transmitted as a continuous sequence of 1s and 0s
with no gap in between. However, if there is any gap, that gap is filled with a special sequence
of 0s and 1s called idle (Figure 3.24). At the receiver’s end, the receiver keeps counting the bit
and separates them into byte groups for decoding. Synchronous transmission is fast as compared
to asynchronous transmission, as it does not use any extra bits such as start bits and stop bits.
Thus, it is best suitable for applications requiring high speed. However, this transmission can-
not be used for real-time applications such as television broadcasting, as there can be uneven
delays between the arrival of adjacent frames at the receiver’s end which in turn results in poor
quality of video.

Direction of flow

10101011 11001011 11100110 11100111 01010100 101

Figure 3.24 Synchronous Transmission

‰‰ Isochronous Transmission: This transmission provides synchronization in the entire stream of

bits and not only at the byte level thereby ensuring the fixed arrival rate of data. This characteristic
makes this transmission suitable for real-time applications for which synchronous transmission is
not suitable.
24. What do you mean by carrier signal?
Ans: Carrier Signal, also called carrier frequency, refers to the high-frequency signal produced
by the sender device to which receiver device is tuned and is used as a base for the information signal.

CN_Chapter 03.indd 56 2/20/2012 6:08:26 PM

 Analog and Digital Transmission 57

25. Defin digital-to-analog conversion. Discuss various techniques for digital-to-analog

conversion.
Ans: The process of changing one or more of the attributes of analog signal based on information in
digital data is referred to as digital-to-­analog conversion. It is also called the modulation of a digital
signal. Depending on whether the amplitude, frequency or phase of the carrier signal is modified, there
are three basic techniques for digital-to-analog conversion, which include amplitude shift keying (ASK),
frequency shift keying (FSK) and phase shift keying (PSK). In addition, one more efficient technique
for digital-to-analog conversion is quadrature amplitude modulation (QAM) (explained in Q27) that
involves changing of both amplitude and phase of the carrier signal.

Amplitude Shift Keying

It involves changing the amplitude of the carrier 0 1 0 1
signal without changing its frequency and phase.
The amplitude of a carrier signal is multiplied by
binary 0 or 1. A special case of ASK is binary ASK
(BASK), also known as on-off keying (OOK), Digital data
where peak amplitude for one binary digit is taken as
0 while the other binary digit has amplitude equal to
the peak amplitude of carrier frequency (Figure 3.25). Carrier frequency
Another variation of ASK is the multilevel ASK
(MFSK) which involves more than two levels of
amplitude.
The bandwidth (B) for ASK is directly propor- Modulated signal
tional to signal rate (S) as shown in the formula
given below: Figure 3.25 Amplitude Shift Keying
B = (1 + fc) × S
Here, fc is the factor whose value lies between
0 and 1 and it depends on the modulation and
filtering process
It is clear from the above formula that Bmin = S 0 1 0 1
(when fc = 0) and Bmax = 2 × S (when fc = 1).
Though ASK is the simplest technique, it is
Digital data
highly susceptible to noise and thus is an inef-
ficient modulation technique
Carrier frequency 1
Frequency Shift Keying
In this technique, the frequency of the carrier sig- Carrier frequency 2
nal is changed without changing its amplitude
and phase. The simplest form of FSK is binary
FSK (BFSK) in which two different frequencies Modulated signal
(say f1 and f2) close to the carrier frequency are
taken to represent two binary values in digital data Figure 3.26 Binary Frequency Shift Keying
­(Figure 3.26).

CN_Chapter 03.indd 57 2/20/2012 6:08:27 PM

 58 Data Communications and Computer Networks

The bandwidth for BFSK is calculated by the following given formula:

B = (l + fc) × S + (f1 – f2),
where, f1 – f2 = 2∆f
Another variation of FSK is multilevel FSK (MFSK) which involves using more than two carrier frequen-
cies. For example, four frequencies can be used to transmit two bits, eight frequencies to transmit three bits
and so on. The only requirement is that the distance between any two carrier frequencies should be 2∆f.
FSK is less susceptible to noise than ASK and thus is suitable for high-frequency radio transmission.

Phase Shift Keying

In this technique, the phase of the carrier signal
is changed without changing its amplitude and 0 1 0 1
­frequency. There are two forms of PSK tech-
niques; namely, binary PSK (BPSK) and quadra-
ture PSK (QPSK ). BPSK, also called as two- Digital data
level PSK, uses two phase values. To represent
one binary value, the signal is sent with the same
phase as that of its predecessor and to represent
another binary value; the signal is sent with op- Carrier frequency
posite phase from its predecessor (Figure 3.27).
The bandwidth of BPSK is equal to BASK, but
less than that for BFSK.
QPSK, also called as four-level PSK, offers Modulated signal
an efficient use of bandwidth by using two bits at a
time in each signal element. It is so called, because Figure 3.27 Binary Phase Shift Keying
it involves two BPSK modulations of which one is
in-phase and another is out-of-phase (quadrature). In this technique, the stream of bits is first passed through
a converter which converts serial transmission into parallel transmission and then, one bit is transmitted to
in-phase modulator and other bit is transmitted to out-of-phase modulator. Both modulators produce com-
posite signals which have the same frequency but different phases. When these signals are combined, the
resultant signal may involve one of four different phases: 45°, –45°, 135° and –135°.
PSK is more commonly used than ASK and FSK. It is less susceptible to noise than ASK as phase is not
likely to be affected easily by the noise. Moreover, it does not require two carriers as required in FSK.
26. What is constellation diagram? Explain with a suitable example.
Ans: Constellation Diagram is a 2D diagram used to define the amplitude and phase of a signal element
especially in the cases where two carrier signals (one in-phase and other quadrature) are being used. The
horizontal axis of the diagram represents the in-phase carrier while the vertical axis represents the quadrature
­carrier signal (Figure 3.28). Each signal element is represented by a point (black dot) with bit or combina-
tions of bits written next to it. The projection of the point on the horizontal axis defines the peak a­ mplitude
of in-phase component while the projection of point on the vertical axis defines the peak amplitude of the
quadrature component. The line joining the point to origin indicates the peak amplitude of signal element
and the angle that this line makes with the horizontal axis indicates the phase of the signal element.
For example, the constellation diagram for QPSK as discussed in previous question may involve four
phases as shown in Figure 3.29.

CN_Chapter 03.indd 58 2/20/2012 6:08:28 PM

 Analog and Digital Transmission 59

Quadrature
carrier

em de
t
en
el itu

Peak amplitude
al pl

Bits Phase 01 11
gn m

of quadrature
si k a

00 45°
of ea

component
P

01 135°
Phase
10 −135°
In-phase carrier
Peak amplitude 11 −45° 00 10
of in-phase
component

Figure 3.28 Constellation Diagram Figure 3.29 Constellation Diagram for QPSK Signal

27. Write a short note on quadrature amplitude modulation (QAM).

Ans: QAM technique combines the idea of ASK
and PSK. That is, instead of changing only one at- 011
tribute of a carrier, two attributes including ampli- 010
tude and phase of the carrier signal are changed. Like
QPSK, it uses two carriers (one is in-phase and the 100
101 000 001
other is quadrature) with different amplitude lev-
els. Figure 3.30 shows the constellation diagram of
110
a QAM signal with two amplitude levels and four
phases. 111
The minimum bandwidth required by QAM is Figure 3.30 Constellation Diagram for QAM Signal
equal to that for ASK and PSK. It is less susceptible
to noise than ASK as well as does not require two carriers as required in FSK.
28. Define analog-to-analog conversion. Discuss various techniques for analog-to-analog
conversion.
Ans: The process of representing analog data by analog signal is known as analog-to-analog
conversion (also called analog modulation). The techniques used for analog-to-analog conversion
include amplitude modulation (AM), frequency modulation (FM) and phase modulation (PM).

Amplitude Modulation
In this modulation, the amplitude of a carrier wave is varied in accordance with the characteristic of the
modulating signal. The frequency of the carrier remains the same, only the amplitude changes to follow
variations in the signal. In simpler words, the two discrete binary digits (0 and 1) are represented by
two different amplitudes of the carrier signal. Figure 3.31 depicts how the modulating signal is super-
imposed over the carrier signal that results in an amplitude-modulated signal.

CN_Chapter 03.indd 59 2/20/2012 6:08:31 PM

 60 Data Communications and Computer Networks

The bandwidth of the amplitude-modulated signal is twice of that of modulating signal. That is,
BAM = 2B,

Amplitude
Modulating signal

Time

Amplitude
Carrier frequency

Time

Amplitude
AM signal
Time

Figure 3.31 Amplitude Modulation

where B = bandwidth of modulating signal.

Frequency Modulation
In this modulation, the instantaneous frequency of carrier wave is caused to depart from the centre
­frequency by an amount proportional to the instantaneous value of the modulating signal. In simple
words, FM is the method of impressing modulating signal onto a carrier signal wave by varying its
instantaneous frequency rather than its amplitude (Figure 3.32).
The total bandwidth that is needed for the frequency-modulated signal can be calculated from the
below-given formula.
BFM = 2 (1 + β) B,

Amplitude
Modulating signal

Time

Amplitude
Carrier frequency

Time

Amplitude
AM signal

Time

Figure 3.32 Frequency Modulation

CN_Chapter 03.indd 60 2/20/2012 6:08:32 PM

 Analog and Digital Transmission 61

where β is a factor whose value depends on the modulation technique with a common value of 4 and β
is the bandwidth of modulating signal.

Phase Modulation
This is the encoding of information into a carrier wave by variation of its phase in accordance with an
input signal. In this modulation technique, the phase of sine wave carrier is modified according to the
amplitude of the message to be transmitted (Figure 3.33).

Amplitude

Modulating signal
Time

Amplitude
Carrier frequency

Time

Amplitude
AM signal

Time

Figure 3.33 Phase Modulation

The total bandwidth that is needed by a PM signal can be calculated from the bandwidth and maximum
amplitude of the modulating signal as shown here.
BPM = 2 (1 + β) B,
where β is a factor whose value is lower in PM than FM. For narrowband, its value is 1 and for wideband,
its value is 3. B is the bandwidth of modulating signal.
29. What does a decibel (dB) measure? Give example.
Ans: Decibel (dB) is a measure of comparative strength of two signals or one signal at two differ-
ent points. It is used by engineers to determine whether the signal has lost or gained strength.
The positive value of a signal indicates gain in strength while negative value indicates that the signal
is attenuated.
dB = 10 log10(P2/P1),
where P1 and P2 are the powers of signal at two different points or the powers of two different signals.
30. For the following frequencies, calculate the corresponding periods. Write the result in
seconds (s), milliseconds (ms) and microseconds (ms): 24 Hz; 8 MHz.
Ans: As frequency = 1/period. Thus, for frequency 24 Hz, period = 1/24 s = 0.041 s
As 1 s = 103­ ms ⇒ 0.041 s = 0.041 × 103­ = 41 ms
As 1 s = 106 µs ⇒ 0.041 s = 0.041 × 106­ = 41000 ms.

CN_Chapter 03.indd 61 2/20/2012 6:08:32 PM

 62 Data Communications and Computer Networks

Similarly, for frequency 8 MHz, period = 1/8000000 s = 0.000000125 s

0.000000125 × 103­ ms = 0.000125 ms
0.000000125 × 106­ ms = 0.125 ms.
31. A sine wave completes one cycle in 25 microseconds. What is its frequency? Express the
frequency in hertz.
Ans: Frequency = 1/period. Here, period = 25 ms = 25 × 10−6 s.
Therefore, frequency = 106/25 = 40000 Hz.
32. A digital signal has a bit interval of 40 microseconds. What is the bit rate? Express the bit
rate in kbps.
Ans: Bit rate = number of bit intervals per second. Therefore, bit rate = 1/(40 × 10–6) = 25000 bps.
⇒ Bit rate = 25 kbps.
33. A signal has a bandwidth of 30 Hz. The highest frequency is 80 Hz. What is its lowest
frequency? Draw the spectrum if the signal contains all the frequency of same amplitude.
Ans: Bandwidth = the highest frequency – the lowest frequency. Bandwidth = 30 Hz (given).
The highest frequency = 80 Hz. Thus, the lowest frequency = 80–30 = 50 Hz. The spectrum of the signal
containing all the frequencies of same amplitude is shown in Figure 3.34.

Amplitude

Bandwidth = 30Hz

50 51 52 53 80
Frequency
(Hz)

Figure 3.34 Spectrum of a Signal

34. An image has the size of 1024 × 786 pixel with 256 colours. Assume the image is uncompressed.
How does it take over a 56-kbps modem channel?
Ans: To represent 256 colours, we need log2256 = 8 bits. Therefore, the total number of bits that
are to be transmitted will be equal to 1024 × 786 × 8 = 6438912 bits. Thus, bit rate = 6438912/
(56 × 1000) = 115 bps.
35. An analog signal carries four bits in each signal element. If 1000 signal elements are sent
per second, fin the baud rate and bit rate.
Ans: Baud rate = the number of signal elements per second = 1000. As we know that baud rate = bit
rate × (1/number of data elements carried in one signal element). Bit rate = 1000 × 4 = 4000 bps.

CN_Chapter 03.indd 62 2/20/2012 6:08:33 PM

 Analog and Digital Transmission 63

36. The power of a signal is 10 mW and the power of the noise is 1 mW. What is the value of
SNR in dB?
Ans: SNR = signal power/noise power. Thus, SNR = (10 × 10–3)/(1 × 10–6) = 10,000.
SNRdB = 10 log10 SNR = 10 log10 10000 =10 × 4 = 40.
37. Calculate the highest bit rate for a telephone channel, given the bandwidth of a line to be
3000 Hz and the SNR being 35 dB.
Ans: As SNRdB = 10 log10 SNR. SNR = 10 (SNRdB/10) = 103.5 = 3162 (rounded off). According to
Shannon’s capacity, The highest bit rate = bandwidth × log2 (1 + SNR). Therefore, the highest bit rate
= 3000 log2 (1 + 3162) = 3000 × log2 3163 = 3000 × 11.6 = 34800 bps.
38. What is the Nyquist minimum sampling rate for the following?
(a) A complex low-pass signal with a bandwidth of 300 kHz.
Ans: According to Nyquist theorem, the sampling rate is two times of the highest frequency
in the signal. As the frequency of a low-pass signal starts from zero, the highest frequency in the
signal = bandwidth; that is, 300 kHz. Thus, sampling rate = 2 × 300 kHz = 600000 samples per
second.
(b) A complex bandpass signal with a bandwidth of 300 kHz.
Ans: The sampling rate cannot be found in this case because we do not know the maximum frequency
of the signal.
(c) A complex bandpass signal with bandwidth of 300 kHz and the lowest frequency of
100 Hz.
Ans: The highest frequency = 100 +300 = 400 kHz. Thus, sampling rate = 2 × 400 kHz = 800000
samples per second.
39. What is the maximum bit rate for a noiseless channel with a bandwidth of 6000 Hz
­transmitting a signal with four signal levels?
Ans: Bit rate = 2 × channel bandwidth × log2 (number of signal levels). Therefore, bit rate = 2 × 6000
× log2 4 = 24000 bps.
40. What is the total delay (latency) for a frame size of 10 million bits that is being set up
on a link with 15 routers, each having a queuing time of 2 ms and a processing time of 1 ms? The
length of link is 3000 km. The speed of light inside the link is 2 × 108 m/s. The link has bandwidth
of 6 Mbps.
Ans: Here, propagation time = distance / propagation speed
⇒ (3000 × 1000)/(2 × 108) = 1.5 × 10–2 s.
Transmission time = message size / bandwidth
⇒ (10 × 106)/(6 × 106) = 1.7 s (approx).
As there are 15 routers, total queuing time = 15 × 2 × 10–6 = 30 × 10–6 s.
Processing time = 15 × 1 × 10–6 = 15 × 10–6 s. Now, latency = propagation time + transmission time +
queuing time + processing time
⇒ 1.5 × 10–2 + 1.7 + 30 × 10–6 s + 15 × 10–6 s = 1.715045 s.

CN_Chapter 03.indd 63 2/20/2012 6:08:34 PM

 64 Data Communications and Computer Networks

41. A data stream is made of 10 alternating 0s and 1s. Encode this stream using the following
encoding schemes.
(a) Unipolar
Ans:
Amplitude

0 1 0 1 0 1 0 1 0 1

Time

(b) Polar NRZ-L

Ans:
Amplitude

0 1 0 1 0 1 0 1 0 1

Time

(c) NRZ-I
Ans:
Amplitude

0 1 0 1 0 1 0 1 0 1

Time

(d) RZ
Ans:
Amplitude

0 1 0 1 0 1 0 1 0 1

Time

CN_Chapter 03.indd 64 2/20/2012 6:08:36 PM

 Analog and Digital Transmission 65

(e) Manchester
Ans:
Amplitude

0 1 0 1 0 1 0 1 0 1

Time

(f) Differential Manchester

Ans:
Amplitude

0 1 0 1 0 1 0 1 0 1

Time

(g) AMI
Amplitude

0 1 0 1 0 1 0 1 0 1

Time

(h) Pseudoternary
Amplitude

0 1 0 1 0 1 0 1 0 1

Time

CN_Chapter 03.indd 65 2/20/2012 6:08:38 PM

 66 Data Communications and Computer Networks

42. A signal is quantized using 10-bit PCM. Find the SNR in dB.
Ans: SNRdB = 6.02 nb + 1.76. Here, nb is the number of bits used for quantization = 10.
Thus, SNRdB = 6.02 × 10 + 1.76 = 61.96 dB.
43. A system is designed to sample analog signals, convert them to digital form with a 4-bit
converter, and transmit them. What bit rate is required if the analog signal consists of frequencies
between 400 Hz and 3400 Hz.
Ans: Bandwidth = 3400 – 400 = 3000 Hz. Bit rate = 2 × bandwidth × log2 L = 2 × 3000 × log2 4
     ⇒ 2 × 2 × 3000 = 12000 bps = 12 kbps.
44. Given the bit pattern 01100, encode this data using ASK, BFSK, and BPSK.
Ans: Bit pattern 01100 can be encoded using ASK, BFSK and BPSK as shown in Figure 3.35(a),
3.35 (b) and 3.35 (c), respectively.

Amplitude Amplitude

0 1 1 0 0 0 1 1 0 0

Time Time

(a) Amplitude shift keying (ASK) (b) Binary frequency shift keying (BFSK)

Amplitude

0 1 1 0 0

Time

(c) Binary phase shift keying (BPSK)

Figure 3.35 Encoding of Bit Pattern Using ASK, BFSK and BPSK

45. Find the maximum bit rate for an FSK signal if the bandwidth of the medium is 12,000 Hz
and the difference between the two carriers must be at 2000 Hz. Transmission is in full-duplex mode.
Ans: As we know that in FSK, bandwidth = (1 + fc) × signal rate + difference between carrier
frequencies.
⇒ 12000 = (1 + 1) × signal rate + 2000
⇒ Signal rate = 5000 baud.
Now, as we know that signal rate = bit rate × (1/r). For FSK, one data element is carried by one
signal element, that is, r = 1
⇒ Bit rate = 4000 kbps.

CN_Chapter 03.indd 66 2/20/2012 6:08:39 PM

 Analog and Digital Transmission 67

46. Find the bandwidth for the following situations if we need to modulate a 5-kHz voice.
(a) AM
Ans: BAM = 2 × B = 2 × 5 = 10 kHz.
(b) PM (set b = 5)
Ans: BFM = 2 × (1 + b) × B = 2 × (1 + 5) × 5 = 60 kHz.
(c) PM (set b =1)
Ans: BPM = 2 × (1 + b) × B = 2 × (1 + 1) × 5 = 20 kHz.

Multiple Choice Questions

1. Which of the following is a characteristic of 6. Shannon’s noisy channel capacity theorem
an analog quantity? states that C = B log2 (1+ SNR) bps. In this
(a) Changes in quantity are represented by equation, B stands for
discrete values. (a) Bandwidth of the signal
(b) Its values follow a logarithmic response (b) Bandwidth of the channel
curve. (c) Strength of the signal
(c) It can be described with a finite number (d) None of these
of steps. 7. In which of the following encoding schemes
(d) It has a continuous set of values over a three levels of voltage are used?
given range. (a) Unipolar     (b)   Polar
2. In which of the following form data has to be (c) Bipolar      (d)   None of these
transformed so that it can be transmitted. 8. In differential manchester encoding, transi-
(a) Electromagnetic signals tion at the middle of the bit is used for
(b) Aperiodic signals (a) Bit transfer    (b)   Synchronization
(c) Low-frequency sine waves (c) Baud transfer   (d)   None of these
(d) None of these
9. In which of the following schemes change
3. A sine wave is or lack of change in the value of voltage
(a) Periodic and continuous determines the value of bit
(b) Aperiodic and continuous (a) NRZ-I        (b)   NRZ-L
(c) Periodic and discrete (c) Both (a) and (b)  (d)   None of these
(d) Aperiodic and discrete
10. A 6-MHz analog video signal can be sam-
4. What is the bandwidth of a signal that ranges pled and later reconstructed if it was sampled
from 200 MHz to 500 MHz? at a rate of at least (in M samples/s)?
(a) 400 MHz     (b) 100 MHz (a) 3 (b) 6
(c) 300 MHz     (d)   None of these (c) 12 (d) 9
5. A sine wave has a frequency of 8 kHz. What 11. One factor in the accuracy of a reconstructed
is its period? PCM signal is the
(a) 124 microseconds (a) Signal bandwidth
(b) 150 microseconds (b) Carrier frequency
(c) 125 microseconds (c) Number of bits used for quantization
(d) 156 microseconds (d) Baud rate

CN_Chapter 03.indd 67 2/20/2012 6:08:40 PM

 68 Data Communications and Computer Networks

12. Amplitude modulation is a technique used 14. How many carrier frequencies are used in
for BFSK?
(a) Analog-to-digital conversion (a) 2 (b) 1
(b) Digital-to-analog conversion (c) 0 (d) None of these
(c) Analog-to-analog conversion 15. How many dots are present in the constellation
(d) Digital-to-digital conversion diagram of 8-QAM?
13. Which of the following factors of a carrier (a) 4
frequency is varied in QAM?
(b) 8
(a) Frequency (b) Amplitude
(c) 16
(c) Phase (d) Both (b) and (c)
(d) None of these

Answers
1. (d)   2. (a)   3. (a)   4. (c)   5. (c)   6. (b)   7. (c)   8. (b)   9. (b)   10. (c)
11. (c) 12. (c) 13. (d) 14. (a) 15. (b)

CN_Chapter 03.indd 68 2/20/2012 6:08:40 PM

 4
Transmission Media

1. What are transmission media? What are the different categories of transmission media?
Ans: Transmission media refer to the media through which data can be carried from a source
to a destination. Data is transmitted from one device to another through electromagnetic signals.
­Transmission media are located under and controlled by the physical layer as shown in Figure 4.1.
Physical layer (sender) Physical layer (receiver)

Transmission media
(cable or air)

Figure 4.1 Transmission Media and Physical Layer

The different categories of transmission media include guided (or wired ) and unguided (or wireless) ­media
as shown in Figure 4.2. Guided transmission media use a cabling system that guides the data signals along a
specific path. It consists of a cable composed of metals such as copper, tin or silver. The data signal in guided
medium is bound by the cabling system; hence, the guided medium is also known as bound medium. There
are three basic types of guided transmission media: twisted-pair cable, c­ oaxial cable and fibre-optic cable

Transmission media

Guided (wired) Unguided (wireless)

Twisted pair Coaxial Fibre-optic Infrared

Radiowaves Microwaves Satellite
cable cable cable waves

Figure 4.2 Categories of Transmission Media

CN_Chapter 04.indd 69 2/20/2012 6:11:03 PM

 70 Data Communications and Computer Networks

Unguided transmission media facilitate data transmission without the use of a physical conduit.
The electromagnetic signals are transmitted through earth’s atmosphere (air, water or vacuum) at a much
faster rate covering a wide area. The electromagnetic waves are not guided or bound to a fixed channel
to follow. There are basically four types of unguided transmission media including radio waves, micro-
waves, satellite transmission and infrared waves.
2. Differentiate guided and unguided transmission media.
Ans. In order to transmit a message or data from a source to a destination, we need a ­transmission
medium. Transmission medium can be broadly classified into two categories, which are guided and
­unguided media. The differences between these two transmission media are listed in Table 4.1.

Table 4.1 Differences Between Guided and Unguided Media

Guided media Unguided media
• Signal is transmitted by establishing a physical path • No physical path is established between the source
between the source and destination. and destination; signals are propagated through air.
• Signals propagate in the form of current or voltage. • Signals propagate in the form of electromagnetic
waves.
• G uided media are well suited for point-to-point com- • Unguided media are well suited for broadcast com-
munication. munication.
• Examples of guided media are twisted-pair cables, • Examples of unguided media are microwave satel-
coaxial cables and fibre-optic cables. lites, infrared waves and radio waves.

3. Write in short the design factors for a data transmission system.

Ans: The major concerns during the design of a data transmission system are achieving a higher data
rate and covering maximum transmission distance. The key design factors that affect the data rate and
transmission distance are as follows:
‰‰ Bandwidth: The greater the bandwidth of the signal is, the higher will be the data rate.
‰‰ Transmission Impairments: Impairments such as attenuation, limit the transmission distance and
hence, they are undesirable.
‰‰ Interference: Different signals in overlapping frequency bands result in interference which can
distort or cancel out a signal. Interference is a major issue with unguided media but it also affects
the guided media.
‰‰ Number of Receivers: In a guided medium, as the number of receivers on a shared link increases,
more attenuation and distortion are introduced which limit the distance and/or data rate.
4. Explain in detail the various types of guided transmission media.
Ans: There are three basic types of guided media—twisted-pair cable, coaxial cable and fib e-optic cable.

Twisted-pair Cable
It is one of the most common and least expensive transmission media. A twisted-pair cable consists of
two insulated copper conductors that are twisted together forming a spiral pattern. A number of such
pairs are bundled together into a cable by wrapping them in a protective shield. One of the wires in each
twisted pair is used for receiving data signal and another for transmitting data signal. Twisted pairs are

CN_Chapter 04.indd 70 2/20/2012 6:11:04 PM

 Transmission Media 71

used in short-distance communication (less than 100 metres). The biggest network in the world, the
telephone network, originally used only twisted-pair cabling and still does for most local connections.
A twisted-pair cable has the capability of passing a wide range of frequencies. However, with the in-
crease in frequency, attenuation also increases sharply. As a result, the performance of a twisted-pair
cable decreases with the increase in frequency. A twisted-pair cable comes in two forms: unshielded and
shielded with a metal sheath or braid around it. Accordingly, they are known as unshielded twisted-pair
(UTP) and shielded twisted-pair (STP) cables.
‰‰ UTP Cable: This cable has four pairs of wires inside the jacket (Figure 4.3). Each pair is twisted
with a different number of twists per inch to help eliminate interference from adjacent pairs and
other electrical devices. The tighter the twisting is, the higher will be the supported transmission
rate and greater will be the cost per foot. Each twisted pair consists of two metal conductors (usu-
ally copper) that are insulated separately with their own coloured plastic insulation. UTP cables
are well suited for both data and voice transmissions; hence, they are commonly used in telephone
systems. They are also widely used in DSL lines, 10Base-T and 100Base-T local area networks.

Copper
wire
Jacket

Insulator

Figure 4.3 UTP Cable

‰‰ STP Cable: This cable has a metal foil or braided-mesh covering that covers each pair of insulated
conductors (Figure 4.4). The metal foil is used to prevent infiltration of electromagnetic noise.
This shield also helps to eliminate crosstalk. An advantage of STP cables over UTP cables is that
they are suitable for the environments with electrical interference. In addition, they provide better
performance at higher data rates. However, the extra shielding makes the STP cables quite bulky
and more expensive than UTP cables.

Copper
wire
Jacket
Shield

Insulator

Figure 4.4 STP Cable

Coaxial Cable
Coaxial cables (or coax) have a single central conductor, which is made up of solid wire (usually, cop-
per) (Figure 4.5). This conductor is surrounded by an insulator over which a sleeve of metal mesh is

CN_Chapter 04.indd 71 2/20/2012 6:11:13 PM

 72 Data Communications and Computer Networks

woven to block any outside interference. This metal Jacket Center

Shield Insulator
mesh is again shielded by an outer covering of a thick conductor
material (usually PVC) known as jacket.
Although coaxial cabling is difficult to install, it is
highly resistant to signal interference. It can support Figure 4.5 Coaxial Cable
greater cable lengths between network devices and
can offer greater bandwidth than twisted-pair cable. However, attenuation in coaxial cables is much
higher as compared to twisted-pair cables due to which the signal weakens rapidly. As a result, repeaters
are to be used frequently to boost up the signals. Coaxial cables are capable of transmitting data at a fast
rate of 10Mbps. Some of the applications that use coaxial cables include analog and digital telephone
networks, cable TV networks, Ethernet LANs, and short range connections.

Fibre-optic Cable
Fibre-optic cable or optical fib e consists of thin glass fibres that
Cladding
can carry information in the form of visible light. The typical ­optical Jacket
fibre consists of a very narrow strand of glass or plastic called the Core
core. Around the core is a concentric layer of less dense glass or
plastic called the cladding. The core diameter is in the range of
8–50 microns (1 micron = 10−6 metres) while cladding generally has Side view End view
a diameter of 125 microns. The cladding is covered by a protective
Figure 4.6 Optical Fibre
coating of plastic, known as jacket (see Figure 4.6).
Optical fibres transmit a beam of light by means of total inter-
nal reflectio . When a light beam from a source enters the core, the core refracts the light and guides
the light along its path. The cladding reflects the light back into the core and prevents it from escaping
through the medium (see Figure 4.7). These light pulses, which can be carried over long distances via
optical fibre cable at a very high speed, carry all the information

Cladding

Light rays

Core

Figure 4.7 Signals Carried over an Optical Fibre

Optical fibre has the capability to carry information at greater speeds, higher bandwidth and data rate.
A single optical fibre can pack hundreds of fibres, where each fibre has the capacity equivalent to that of
thousands of twisted-pair wires. This capacity broadens communication possibilities to include services
such as video conferencing and interactive services. In addition, fibre optic cables offer lower attenuation
and superior performance and require fewer repeaters as compared to coaxial and twisted-pair cables.
The major applications of the fibre-optic cables are cable TV, military applications, long-haul trunks,
subscriber loops, local area networks, metropolitan trunks and rural trunks.
5. Why twisting of wires is necessary in twisted-pair cables?
Ans: Twisting is done in twisted-pair cables because it tends to minimize the interference (noise)
between the adjacent pair of wires in cable thereby reducing the crosstalk. In case the two wires are

CN_Chapter 04.indd 72 2/20/2012 6:11:18 PM

 Transmission Media 73

parallel, they may not get affected equally by the electromagnetic interferences (noise and crosstalk)
from nearby sources due to their different locations relative to the source. As a result, the receiver would
receive some unwanted signals. On the other hand, if the wires are twisted, both wires are probable to be
get affected equally by the external interferences thereby maintaining a balance at the receiver. To under­
stand, suppose in one twist, one of the twisted wires is closer to noise source and the other is farther, then
in the next twist, the opposite will be true. As a result, the unwanted signals of both the wires cancel out
each other and the receiver does not receive any unwanted signal. Thus, crosstalk is reduced.
6. What are the different categories and connectors of UTP cables?
Ans: As per the standards developed by the Electronic Industries Association (EIA), UTP cables
have been classified into seven categories, from 1 to 7. Each category is based on the quality of the cable
and the higher number denotes higher quality. Table 4.2 lists the categories of UTP cables along with
their specification and data rate
To connect UTP cables to network devices, UTP connectors are required. The most common
c­ onnector for the UTP cables is RJ45 (RJ stands for registered jack) as shown in Figure 4.8. Being a
keyed ­connector, the RJ45 can be inserted in only one way.
Table 4.2 Categories of UTP Cables
Category Specification Data rate
(CAT) (in Mbps)
1 UTP cables used in telephones <0.1
2 UTP cables used in T-lines 2
3 Improved CAT2 used in LANs 10
4 Improved CAT3 used in token ring network 20
5 Cable wire is normally 24AWG with a jacket and outside sheath; used in LANs 100
5E Extension to category 5 that reduces crosstalk and interference; used in LANs. 125
6 A new category with matched components coming from the same ­manufacturer. 200
This cable must be tested at a data rate of 200 Mbps. This is used in LANs.
7 This is the shielded screen twisted-pair cable (SSTP). Shielding increases data rate 600
and reduces crosstalk effect. This cable is used in LANs.

7. What are the different categories and

connectors of coaxial cables?
Ans: According to the ratings provided by ­radio
government (RG), the coaxial cables have been divided
into three categories. Each category has a specific RG
number that indicates a unique set of physical specifi-
cations. These specifications include the wire gauge of
inner conductor, shield, type and size of outer casing
and type and thickness of inner insulator. Different cat-
egories of the coaxial cables are listed in Table 4.3.
To connect coaxial cables to other devices, ­coaxial
connectors are required. The three most popular
­coaxial connectors include the BNC connector, BNC Figure 4.8 RJ45 UTP Connector

CN_Chapter 04.indd 73 2/20/2012 6:11:19 PM

 74 Data Communications and Computer Networks

Table 4.3 Categories of Coaxial Cables

Category Use Impedance
RG-59 Cable TV 75 Ω
RG-58 Thin Ethernet 50 Ω
RG-11 Thick Ethernet 50 Ω

­terminator and BNC T connector as shown in Figure 4.9. The Bayone-Neill-Concelman (BNC) ­connector
is the most commonly used connector that connects the coaxial cable to a device such as an amplifier or
television set. The BNC terminator is used at the end of the cable to prevent the reflection of the signal and
the BNC T connector is often used in Ethernet networks for branching out connections to other devices.

BNC connector BNC terminator BNC T

Figure 4.9 Coaxial Cable Connectors

8. Explain the different fib e-optic propagation modes.

Ans: Fibre-optic cables support two modes for propagating light, which are multimode and single
mode. Each mode requires fibre with di ferent physical characteristics.

Multimode Propagation
In this mode, many beams from a light source traverse the fibre along multiple paths and at multiple angles
as shown in Figure 4.10(a). Depending upon the structure of core inside the cable, multimode can be
implemented in two forms: step-index and graded-index. In multimode step-index fib e, the core’s den-
sity is constant from the centre to the edges. A light beam moves through the core in a straight path until it
meets the interface of the core and cladding. As the interface has a lower density than the core, there comes
a sudden change in the angle of the beam’s motion further adding to distortion of the signal as it moves on.
The multimode graded-index fib e reduces such distortion of signal through the cable. As the density is
high at the centre of the core, the refractive index at the centre is high which causes the light beams at the
centre to move slower than the rays that are near the cladding. The light beams curve in a helical manner
[see Figure 4.10(b)], thus, reducing the distance travelled as compared to zigzag movement. The reduction
in path and higher speed allows light to arrive at the destination in almost the same time as straight lines.

Single-mode Propagation
This mode employs step-index fibre of relatively small diameter and less density than that of multimode
fibre and a much focused light source. Because of the focused light source, the beams spread out to a
small range of angles and propagate almost horizontally. Since all beams propagate through the fibre
along a single path, distortion does not occur. Moreover, all beams reaching at the destination together
can be recombined to form the signal. The single-mode propagation is well suited for long-distance
­applications such as cable television and telephones.

CN_Chapter 04.indd 74 2/20/2012 6:11:24 PM

 Transmission Media 75

Source Destination

(a) Multimode, step-index

Source Destination

(b) Multimode, graded-index

Source Destination
(c) Single-mode

Figure 4.10 Propagation Modes for Fibre-optic Cable

9. What are the different fib e sizes and connectors available?

Ans: The type of optical fibre is specified by the ratio of the diameter of its core to the diameter of
its cladding. The commonly available fibre sizes are listed below in Table 4.4.
Fibre-optic cables use three types of connectors, which are SC connector, ST connector and
MT-RJ connector as shown in Figure 4.11. SC (subscriber channel) connector uses a push–pull
locking mechanism and is primarily used to connect fibre-optic cables for cable television. The ST
(straight tip) connector is used to connect fibre-optic cables to network devices. This connector is
more reliable than SC connector. MT-RJ (mechanical transfer-registered jack) connector is a small
size connector which looks similar to RJ-45 connector. It is widely used for networking applications.

Table 4.4 Fibre Types

Type (core/cladding) Mode
50/125 Multimode, graded
index
62.5/125 Multimode, graded
index
100/125 Multimode, graded
index
7/125 Single mode

CN_Chapter 04.indd 75 2/20/2012 6:11:27 PM

 76 Data Communications and Computer Networks

SC connector ST connector MT-RJ connector

Figure 4.11 Fibre-optic Cable Connectors

10. What are the advantages and disadvantages of fib e optic cables?
Ans: Fibre-optic cables are widely used in many domains such as telephone network and cable
­television network. Some advantages of fibre-optic cables are as follows
‰‰ Since transmission is light-based rather than electricity, it is immune to noise interference.
‰‰ Transmission distance is greater than other guided media because of less signal attenuation
(degradation in quality over distance).
‰‰ It is extremely hard to tap into, making it desirable from the security viewpoint.
‰‰ They are smaller and lighter than copper wire and are free from corrosion as well.
‰‰ Fibre optic offers, by far, the greatest bandwidth of any transmission system.
‰‰ Transmission through fibre optic cable requires lesser number of repeaters for covering larger
transmission distances.
The disadvantages of fibre-optic cables are as follows
‰‰ The installation and maintenance of fibre-optic cables are quite expensive
‰‰ The propagation of light is unidirectional and often requires precise alignment.
‰‰ Extending the fibre-optic cables by joining them together is a tough task
‰‰ It is more fragile when compared to copper wires.

11. What are the two kinds of light sources used in fib e-optic cables?
Ans: In fibre-optic cables, the light sources generate a pulse of light that is carried through the fibre
medium. The two kinds of light sources used with the fibre-optic cables include light-emitting diodes
(LEDs) and semiconductor laser diodes. These light sources are used to perform signalling and can be
tuned in wavelengths by inserting Mach–Zehnder or Fabry–Pérot interferometers between the source
and the fibre media. Table 4.5 lists the comparison between these two light sources.

Table 4.5 Comparison Between LEDs and Semiconductor Diodes

Characteristics LED Semiconductor laser diode
Data rate Low High
Cost Low cost Expensive
Fibre type Multimode Multimode/single mode
Lifetime Long life Short life
Temperature sensitivity Minor Substantial
Distance Short Long

CN_Chapter 04.indd 76 2/20/2012 6:11:29 PM

 Transmission Media 77

12. Explain the use of electromagnetic spectrum for communication.

Ans: Electromagnetic signals include power, voice, radio waves, infrared light, visible light, ultra-
violet light, X-rays and gamma rays. Each of these forms a portion of the electromagnetic spectrum (see
Figure 4.12). The portions of the electromagnetic spectrum that can be used for transmitting information
include radio wave, microwave, infrared light and visible light. Information can be transmitted using
these portions by modulating the frequency, amplitude or phase of the electromagnetic waves. Though us-
ing UV, X-rays and gamma rays for transmitting information is a better choice due to their high frequency
(HF), they are not generally used. This is because it is difficult to produce and modulate these rays as well
as they cannot penetrate obstacles such as buildings. Moreover, they are dangerous for the living beings.

f(Hz) 100 102 104 106 108 1010 1012 1014 1016 1018 1020 1022 1024

Radio Microwave Infrared UV X-ray Gamma ray

Visible
light

f(Hz) 104 105 106 107 108 109 1010 1011 1012 1013 1014 1015 1016
Twistedpair Satellite Fibre
Coax Terrestrial optics
AM FM microwave
Maritime radio radio

TV

Figure 4.12 Electromagnetic Spectrum and Its Uses for Communication

The portion of the electromagnetic spectrum that can be used for transmitting information is parted
into eight different ranges. These ranges are regulated by the government authorities and are known as
bands. Some of the properties of bands are listed in Table 4.6.

Table 4.6 Properties of Bands

Band Use Range Propagation
VLF (very low frequency) Long-range radio navigation 3–30 kHz Ground wave
LF (low frequency) Locators, radio beacons 30–300 kHz Ground wave
MF (middle frequency) AM radio 300 kHz–3 MHz Sky
HF (high frequency) Citizens band, ship/aircraft com- 3–30 MHz Sky
munication
VHF (very high frequency) FM radio, VHF TV 30–300 MHz Sky and line-of-sight
UHF (ultra high frequency) Cell phones, satellite, UHF TV, paging 300 MHz–3 GHz Line-of-sight
SHF (super high frequency) Satellites 3–30 GHz Line-of-sight
EHF (extremely high frequency) Satellite, radars 30–300 GHz Line-of-sight

CN_Chapter 04.indd 77 2/20/2012 6:11:30 PM

 78 Data Communications and Computer Networks

13. Explain the different propagation methods for unguided signals.

Ans: Unguided signals can propagate in three ways, which are ground wave, ionosphere and line-of-sight.
‰‰ Ground Wave Propagation: In this propagation method, the radio waves pass through the lowest
portion of the atmosphere, that is, the curvature of the earth (Figure 4.13). These low frequency
radio waves when transmitted by an antenna disperse in all directions following the curvature of
earth. The distance travelled is directly proportional to the power of the signal. That is, greater the
amount of power in the signal is, the more will be the distance covered.
Atmosphere

House
Radio Earth
tower

Figure 4.13 Ground Wave Propagation

‰‰ Ionospheric Propagation: In this propagation method, the higher frequency radio waves
transmitted by antenna travel upwards into ionosphere layer in the upper portion of atmosphere
and bounced off by the layer towards earth; a low power signal can travel a greater distance
(Figure 4.14). It operates in the frequency range of 2–30 MHz. As this type of propagation depends
on the earth’s ionosphere, it changes with the day timings and weather. This method of propagation
is also known as sky wave propagation.
Ionosphere

Home
Radio
tower Earth

Figure 4.14 Ionosphere Propagation

‰‰ Line-of-Sight Propagation: In this propagation method, very high frequency signals are transmit-
ted which travel exactly in straight line (Figure 4.15). This method demands both transmitting and
receiving antennas to be in line of sight of each other, that is, the receiving antenna must be in view
of the transmitting antenna. It is sometimes called space waves or tropospheric propagation. It is
limited by the ­curvature of the earth for ground-based stations (50 km).

lonosphere

Radio Home
tower
Earth

50 km
Figure 4.15 Line-of-Sight Propagation

CN_Chapter 04.indd 78 2/20/2012 6:11:35 PM

 Transmission Media 79

14. Explain in detail the different types of unguided media.

Ans: Unguided media are used in those cases when transmission of data through guided media is
difficult. The three main types of unguided media are discussed in the following sections:

Radio Waves
The electromagnetic waves with frequency in the range of 3 kHz to 1 GHz are generally known as radio
waves. These waves are omnidirectional, that is they are propagated in all directions when transmitted by
an antenna. Thus, the antennas that send and receive the signals need not be aligned. However, the radio
waves transmitted by two antennas using the same band or frequency may interfere with each other.
Radio waves present different characteristics at different frequencies. At low (VLF, LF) and medium
(MF) frequencies, they follow the curvature of earth and can penetrate walls easily. Thus, a device such
as a portable radio inside a building can receive the signal. At high
frequencies (HF and VHF bands), as the earth absorbs the radio
waves, they are propagated in sky mode. The higher frequency
radio waves can be transmitted up to greater distances and thus
are best suited for long-distance broadcasting. However, at all fre-
quencies, radio waves are susceptible to interference from electri-
cal equipments.
An omnidirectional antenna is used to transmit radio waves
in all ­directions (see Figure 4.16). Due to their omnidirectional
character­istics, radio waves are useful for multicast (one sender,
many receivers) communication. Examples of multicasting are Figure 4.16 Omnidirectional Antenna
cordless phones, AM and FM radios, paging and maritime radio.

Microwaves
The electromagnetic waves with frequency in the range of 1–300 GHz are known as microwaves. Un-
like radio waves, microwaves are unidirectional. The advantage of the unidirectional property is that
multiple transmitters can transmit waves to multiple receivers without any interference. Since micro-
waves are transmitted using line-of-sight propagation method, the towers with mounted antennas used
for sending and receiving the signal must be in direct sight of each other (Figure 4.17). In case, the
antenna towers are located far away from each other, the towers should be quite tall so that the signals
do not get block off due to curvature of earth as well as other obstacles. Moreover, repeaters should be
often used to ­amplify the signal strength. Microwaves at lower frequencies cannot penetrate buildings
and also, during propagation, refraction or delays can occur due to divergence of beams in space. These

Repeater
Transmitter
Receiver

Figure 4.17 Microwave Transmission

CN_Chapter 04.indd 79 2/20/2012 6:11:37 PM

 80 Data Communications and Computer Networks

delayed waves can come out of phase with the direct wave leading to cancellation of the signal. This
phenomenon is known as the multipath fading effect.
Microwaves require unidirectional antennas that transmit signal only in one direction. Two such
antennas are dish antenna and horn antenna (see Figure 4.18). A dish antenna works based on the
geometry of a parabola. All the lines parallel to the line of sight when hit the parabola, they are reflected
by the parabola curve at angles such that they converge at a common point called the focus. The dish
parabola catches many waves and directs them on the focus. As a result, the receiver receives more of
the signal. In a horn antenna, outgoing transmission is sent through a stem and as it hits the curved
head, the transmission is deflect d outward as a series of parallel beams. The received transmissions are
collected by a horn similar to the parabolic dish, which deflects them back into the stem
Since microwaves are unidirectional, they are best suited for unicast communication such as in
cellular networks, wireless local area networks and satellite networks.

Focus

Waveguide

(a) Dish antenna (b) Horn antenna

Figure 4.18 Unidirectional Antennas

Infrared Waves
The electromagnetic waves with frequency in the range of 300 GHz to 400 THz are known as infrared
waves. These waves are widely used for indoor wireless LANs and for short-range communication; for
example, for connecting a PC with a wireless peripheral device, in remote controls used with stereos,
VCRs, TVs, etc. (Figure 4.19). Infrared waves at high frequencies are propagated using line-of-sight
method and cannot penetrate solid objects. Therefore, a short range infrared system in a room will not
be interfered by such a system present in an adjacent room. Furthermore, infrared waves cannot be
used outside a building because the infrared rays coming from the sun may interfere with it and distort
the signal.
The use of infrared waves has been sponsored by an association, known as Infrared Data
­Association (IrDA). This association has also established standards for the usage of infrared signals
for ­communication between devices such as keyboards, printers, PCs and mouses. For example, some
wireless keyboards are attached with an infrared port that enables the keyboard to communicate with

CN_Chapter 04.indd 80 2/20/2012 6:11:38 PM

 Transmission Media 81

Figure 4.19 Infrared Waves

the PC. Since infrared signals transmit through line-of-sight mode, the infrared port must be pointed
towards the PC for communication.

Communication Satellites
A communication satellite can be referred to as a microwave relay station. A satellite links two or more
ground-based (earth) stations that transmit/receive microwaves. Once a satellite receives any signal on a
frequency (uplink), it repeats or amplifies that signal and sends it back to earth on a separate frequency
(downlink). The area shadowed by the satellite (see Figure 4.20) in which the information or data can
be transmitted and received is called the footprint.
Satellites are generally set in geostationary orbits directly over the equator, which rotates in
synchronization with the earth and hence looks stationary from any point on the earth. These geostationary
orbits are placed approximately 36,000 km above the earth’s surface and satellites placed in this orbit
are known as geostationary satellites.
Satellite transmission is also a kind of line-of-sight transmission and the best frequency for it is in
the range of 1–10 GHz. The major applications for satellites are long-distance telephone transmission,
weather forecasting, global positioning, television, and many more.

CN_Chapter 04.indd 81 2/20/2012 6:11:42 PM

 82 Data Communications and Computer Networks

Satellite
36,000 km
orbit

Antenna

Downlink
Uplink
Downlink
Earth stations

Footprint
Figure 4.20 Satellite Transmission

Multiple Choice Questions

1. A transmission medium is located under and (a) total internal reflection
controlled by the (b) total internal refraction
(a) transport layer (b) application layer (c) Both (a) and (b)
(c) physical layer (d) session layer (d) None of these
2. Guided transmission media include 7. In multimode step-index fibre, the density
(a) coaxial cable (b) fibre-optic cabl of the core ______ from the centre to the
(c) twisted-pair cable (d) All of these edges.
(a) increases (b) decreases
3. Which of the following is not a type of
(c) remains constant (d) None of these
twisted-pair cable?
(a) UTP (b) FTP 8. Which of the following is not a band?
(c) STP (d) None of these (a) VHF (b) UHF
(c) VLF (d) SLF
4. BNC connectors are used with
(a) satellites (b) fibre-optic cable 9. Unguided signals can propagate in ______
(c) coaxial cables (d) twisted-pair cables ways.
(a) two (b) eight
5. The transmission medium with maximum
(c) three (d) four
error rate is
(a) coaxial cable (b) twisted-pair cable 10. The frequency at which a signal is received by
(c) satellite link (d) optical fibr a satellite is known as its ______ frequency.
(a) downlink (b) microwave
6. Optical fibres transmit a beam of light by
(c) terrestrial (d) uplink
means of

Answers
1. (c) 2. (d) 3. (b) 4. (c) 5. (b) 6. (a) 7. (c) 8. (d) 9. (c) 10. (d)

CN_Chapter 04.indd 82 2/20/2012 6:11:44 PM

 5
Multiplexing and Switching

1. What is meant by bandwidth utilization?

Ans: Bandwidth utilization refers to the wise use of the available bandwidth in order to achieve the
utmost efficien y, privacy and anti-jamming. Bandwidth can be used optimally using two techniques,
namely, multiplexing and spreading. Multiplexing allows sharing the available bandwidth of a commu-
nication link among a number of communicating stations, thus achieving efficienc . On the other hand,
spreading allows expanding the bandwidth of link for achieving privacy and anti-jamming.
2. What is multiplexing? In what situations, it can be used?
Ans: Multiplexing is a technique used for transmitting several signals simultaneously over a single
communication link. An analogy of multiplexing can be made with a multilane highway. Just as a ­multilane
highway can carry increased volumes of traffic in multiple lanes at higher speeds and at relatively low in-
cremental cost per lane, the higher-capacity circuit can carry multiple conversations in multiple channels at
relatively low incremental cost per channel. Multiplexing is done to utilize the available bandwidth properly
and to improve the efficiency during a transmission. In a multiplexed system (Figure 5.1), several devices
share a communication link called common medium. Each part of the communication link being used for
carrying transmission between an individual pair of input and an output line is referred to as a channel.
At the sender’s end, the N input lines are combined into a single stream by a communication device,
called multiplexer (MUX). At the receiver’s end, another communication device, called demultiplexer
(DEMUX), completes the communication process by separating multiplexed signals from a communi-
cation link and distributing them to corresponding N output lines.

MUX DEMUX

N N
inputs 1 Link, N channels outputs

Figure 5.1 A Multiplexed System

CN_Chapter 05.indd 83 2/20/2012 6:20:25 PM

 84 Data Communications and Computer Networks

Multiplexing can be used in situations where the signals to be transmitted through the transmission
medium have lower bandwidth than that of the medium. This is because in such situations, it is possible
to combine the several low-bandwidth signals and transmitting them simultaneously through the trans-
mission medium of larger bandwidth.
3. Why we need multiplexing in a communication channel?
Ans: Multiplexing is needed in a communication channel because of the following reasons:
‰‰ To send several signals simultaneously over a single communication channel.
‰‰ To reduce the cost of transmission.
‰‰ To effectively utilize the available bandwidth of the communication channel.

4. Explain the different multiplexing techniques.

Ans: Multiplexing can be done using three techniques: frequency-division multiplexing (FDM),
wavelength-division multiplexing (WDM) and time-division multiplexing (TDM).

Frequency-Division Multiplexing
FDM is used when the bandwidth of the transmission medium is greater than the total bandwidth require-
ment of the signals to be transmitted. It is often used for baseband analog signals. At the sender’s end, the
signals generated by each sending device are of similar frequency range. Within the multiplexer, these
similar signals modulate carrier waves of different frequencies and then the modulated signals are merged
into a single composite signal, which is sent over the transmission medium. The carrier frequencies are
kept well separated by assigning a different range of bandwidth (channel) that is sufficient to hold the
modulated signal. There may also be some unused bandwidth between successive channels called guard
bands, in order to avoid interference of signals between those channels.
At the receiver’s end, the multiplexed signal is applied to a series of bandpass filters which breaks it
into component signals. These component signals are then passed to a demodulator that separates the
signals from their carriers and distributes them to different output lines (Figure 5.2).
Signal 1 (115 KHz)
Signal 1 (115 KHz)

Signal 2 (180 KHz)

Signal 2 (180 KHz)
Channel

Signal 3 (220 KHz) Signal 3 (220 KHz)

Multiplexer Demultiplexer
(sender) (receiver)

Figure 5.2 Frequency-division Multiplexing

CN_Chapter 05.indd 84 2/20/2012 6:20:48 PM

 Multiplexing and Switching 85

Though FDM is an analog multiplexing technique, it can also be used to multiplex the digital signals.
However, before multiplexing, the digital signals must be converted to analog signals. Some common
­applications of FDM include radio broadcasting and TV networks.

Wavelength-Division Multiplexing
WDM is an analog multiplexing technique designed to utilize the high data rate capability of ­fibre-optic
cables. A fibre-optic cable has a much higher data rate than coaxial and twisted-pair cables and using
it as a single link wastes a lot of precious bandwidth. Using WDM, we can merge many signals into
a single one and hence, utilize the available bandwidth efficientl . Conceptually, FDM and WDM are
same; that is, both combine several signals of different frequencies into one, but the major difference
is that the latter involves fibre-optic cables and optical signals as well as the signals are of very high
frequency.
In WDM, the multiplexing and demultiplexing are done with the help of a prism (Figure 5.3), which
bends the light beams by different amounts depending on their angle of incidence and wavelength. The
prism used at the sender’s end combines the multiple light beams with narrow frequency bands from
different sources to form a single wider band of light which is then passed through fibre-optic cable. At
the receiver’s end, the prism splits the composite signal into individual signals. An application of WDM
is the SONET network.

λ1 λ1

λ2 λ2
Fibre-optic cable

λ3 Mutiplexer Demutiplexer λ3
Figure 5.3 Wavelength-division Multiplexing

Time-Division Multiplexing
TDM is a digital multiplexing technique that allows the high bandwidth of a link to be shared amongst
several signals. Unlike FDM and WDM, in which signals operate at the same time but with different
frequencies, in TDM, signals operate at the same frequency but at different times. In other words, the
link is time-shared instead of sharing parts of bandwidth among several signals.
Figure 5.4 gives a conceptual view of TDM. At the sender’s end, the time-division multiplexer ­allocates
each input signal a period of time or time slot. Each sending device is assigned the ­transmission path for
a predefined time slot. Three sending signals, Signals 1, 2 and 3, occupy the transmission ­sequentially.
As shown in the figure, time slots A, B, P, Q, X and Y follow one after the other to carry signals from the
three sources, which upon reaching the demultiplexer, are sent to the intended receiver.
Though TDM is a digital multiplexing technique, it can also multiplex analog signals. However,
before multiplexing, analog signals must be converted to DSs.
5. Explain different schemes of TDM.
Ans: TDM can be divided into two different schemes, namely, synchronous TDM and statistical TDM.

CN_Chapter 05.indd 85 2/20/2012 6:20:50 PM

 86 Data Communications and Computer Networks

Time division signals Reassembled signals

B A B A
Signal 1 Signal 1

Channel

Q P Q P
Signal 2 Signal 2

Multiplexer Demultiplexer
Y X (sender) (receiver) Y X
Signal 3 Signal 3
Figure 5.4 Time-division Multiplexing

Synchronous TDM
In this technique, data flow of each input source is divided into units where a unit can be a bit, byte or
a combination of bytes. Each input unit is allotted one input time slot. A cycle of input units from each
input source is grouped into a frame. Each frame is divided into time slots and one slot is dedicated for
a unit from each input source. That is, for n connections, we have n time slots in a frame and every slot
has its own respective sender. The duration of input unit and the duration of each frame are same unless
some other information is carried by the frame. However, the duration of each slot is n times shorter
where n is the number of input lines. For example, if duration of input unit is t seconds, then the dura-
tion of each slot will be t/n seconds for n input lines. The data transmission rate for the output link
must be n times the data transmission rate of an input line to ensure the transmission of data. In addition,
one or more synchronization bits are to be included in the beginning of each frame to ensure the data
synchronization between the multiplexer and the demultiplexer.
Figure 5.5 shows a conceptual view of synchronous TDM in which data from the three input sources
has been divided into units (P1, P2, P3), (Q1, Q2) and (R1, R2, R3). As the total number of input
lines is three, each frame also has three time slots. Now, if the duration of an input unit is t seconds, then
after every t seconds, a unit is collected from each source and kept into the corresponding time slot in the
frame. For example, Frame 1 contains (P1, Q1, R1). In our example, the data rate for the transmission
link must be three times the connection rate to ensure the flow of data. In addition, the duration of each
frame is three time slots whereas the duration of each time slot is t/3.
t
t /3
P3 P2 P1
R3 P3 R2 Q2 P2 R1 Q1 P1
Q2 Q1 M Frame 3 Frame 2 Frame 1
U
R3 R2 R1 X Link

Figure 5.5 Synchronous TDM

A major drawback of synchronous TDM is that it is not highly efficient. Since each slot in the frame
is pre-assigned to a fixed source, the empty slots are transmitted in case one or more sources do not have

CN_Chapter 05.indd 86 2/20/2012 6:20:54 PM

 Multiplexing and Switching 87

any data to send. As a result, the capacity of a channel is wasted. For example, in Figure 5.5, an empty
slot is transmitted in Frame 3 because the corresponding input line has no data to send during that time
period.

Statistical TDM
Statistical TDM, also called asynchronous TDM or intelligent TDM, overcomes the disadvantage of
synchronous TDM by assigning the slots in a frame dynamically to input sources. A slot is assigned to
an input source only when the source has some data to send. Thus, no empty slots are transmitted which
in turn result in an efficient utilization of bandwidth. In statistical TDM, generally each frame has less
number of slots than the number of input lines and the capacity of link is also less than the combined
capacity of channels.
Figure 5.6 shows a conceptual view of statistical TDM. At the sender’s end, the multiplexer scans
the input sources one by one in a circular manner and assigns a slot if the source has some data to send;
otherwise, it moves on to the next input source. Hence, slots are filled up as long as an input source has
some data to send. At the receiver’s end, the demultiplexer receives the frame and distributes data in
slots of frame to appropriate output lines.

P3 P2 P1
R3 P3 R2 Q2 P2 R1 Q1 P1
Q2 Q1 M Frame 3 Frame 2 Frame 1
U
R3 R2 R1 X Link

Figure 5.6 Statistical TDM

Unlike synchronous TDM, there is no fixed relationship between the inputs and the outputs in statis-
tical TDM because here slots in a frame are not pre-assigned or reserved. Thus, to ensure the delivery of
data to appropriate output lines, each slot in the frame stores the destination address along with data to
indicate where the data has to be delivered. For n output lines, an m-bit address is required to define each
output line where m = log2 n. Though the inclusion of address information in a slot ensures the proper
delivery, it incurs more overhead per slot. Another difference between synchronous and ­statistical TDMs
is that the latter technique does not require the synchronization among the frames thereby eliminating
the need of including synchronization bits within each frame.
6. Distinguish multilevel, multiple-slot and pulse-stuffed TDMs.
Ans: Multilevel multiplexing, multiple-slot allocation and pulse stuffing are the strategies used in
TDM to handle the different input data rates of the input lines.

Multilevel Multiplexing
Multiple multiplexing is used in situations where the data rate of an input line is an integral multiple
of other input lines. As the name suggests, several levels of multiplexing are used in this technique.
To understand, consider Figure 5.7 where we have three input lines; the first two input lines have a data
rate of 80 kbps each and the last input line has a data rate of 160 kbps. As the data rate of third input
line (160 kbps) is a multiple of that of other two (80 kbps), the first two input lines can be multiplexed

CN_Chapter 05.indd 87 2/20/2012 6:20:55 PM

 88 Data Communications and Computer Networks

to produce a data rate of 160 kbps that is equal to that of the third input line. Then, another level of
multiplexing can be used to combine the output of first two input lines and the third input line thereby
generating an output line with a data rate of 320 kbps.

80 kbps
M 160 kbps
80 kbps U M 320 kbps
X U
160 kbps X

Figure 5.7 Multilevel Multiplexing

Multiple-Slot Allocation
Like multilevel multiplexing, multiple-slot allocation technique is also used in situations where the data
rate of an input line is an integral multiple of other input lines. Generally, one slot per each input source
is reserved in the frame being transmitted. However, sometimes, it is more useful to allocate multiple
slots corresponding to a single input line in the frame. For example, again consider Figure 5.7 where
we can divide one 160-kbps input line into two (each of 80 kbps) with the help of serial-to-parallel
converter and then multiplex the four input lines of 80 kbps each to create an output line of 320 kbps.
However, now there will be total four slots in the transmitting frame with two slots corresponding to
input line with originally 160 kbps data rate (Figure 5.8).

80 kbps
80 kbps M
U
160 kbps 80 kbps 320 kbps
X
80 kbps
Serial-to-parallel
convertor
Figure 5.8 Multiple-slot Allocation

Pulse Stuffing
Pulse stuffing, also known as bit padding or bit stuffin , technique is used in situations where the data
rates of input lines are not the integral multiples of each other. In this technique, the highest input data rate
is determined and then dummy bits are added to input lines with lower data rates to make the data rate of
all the lines equal. For example, consider Figure 5.9 where the first input line has the highest data rate equal
to 80 kbps and other two input lines have data rate of 60 and 70 kbps, respectively. Hence, the data rates
of second and third input lines are pulse-stuffed to increase the rate to 80 kbps.

80 kpbs
60 kpbs 80 kpbs M
Pulse stuffing U
70 kpbs 80 kpbs X 240 kbps
Pulse stuffing

Figure 5.9 Pulse Stuffing

CN_Chapter 05.indd 88 2/20/2012 6:20:58 PM

 Multiplexing and Switching 89

7. What is the purpose of framing bit? Is it used in FDM or TDM?

Ans: Framing bits are used in TDM to ensure the synchronization between multiplexer and ­demultiplexer.
Without synchronization, there may be some error in delivery of bits; that is, a bit may be delivered to the
wrong output line. Thus, one or more synchronization bits, called framing bits, are generally added at the
beginning of each transmitting frame. The framing bits follow a specific pattern in each frame, which allows
the demultiplexer to get in synchronization with the input data and divide the time slots with accuracy.
8. What is the frame format in statistical TDM?
Ans: In statistical TDM, there are two possible frame structures that a multiplexer can use. These
structures are described as follows:
‰‰ One Data Source Per Frame: This frame structure permits to include data from one source per
each frame. The frame consists of two parts: address, which is used to identify the source, and data,
which is of variable length [Figure 5.10(a)]. The end of data field is indicated by the end of overall
frame. This type of frame structure is suitable in environments where the load is less. However, as
the load increases, this frame structure becomes inefficient
‰‰ Multiple Data Sources Per Frame: This frame structure permits to include data from multiple
sources in a single frame. For each source, the frame contains address to identify the source, length
of data field and data [see Figure 5.10(b)]. This frame structure helps to improve efficienc .

Address Data

(a) Subframe with one source per frame

Address Length Data Address Length Data

(b) Subframe with multiple source per frame

Figure 5.10 Statistical TDM Frame Structures

9. Describe the DS hierarchy.

Ans: The DS hierarchy, also called digital signal service, is a hierarchy of digital signals which the
telephone ­companies use for implementing TDM. This hierarchy consists of several levels with each level
supporting a ­different data rate. The digital signals at lower levels in hierarchy are multiplexed into signals
at higher levels in hierarchy (Figure 5.11). Various levels of DS hierarchy are described as follows:

DS-0

T
DS-1
24 D T
M DS-2
D
T
M DS-3
D
T
M DS-4
D
M

Figure 5.11 DS Hierarchy

CN_Chapter 05.indd 89 2/20/2012 6:21:01 PM

 90 Data Communications and Computer Networks

‰‰ DS-0: This service is at the lowest level of hierarchy. It is a single digital channel of just 64 kbps.
‰‰ DS-1: This service has a capacity of 1.544 Mbps which is 24 times of a DS-0 channel with an
­additional overhead of 8 kbps. It can be utilized for multiplexing 24 DS-0 channels, as a single
service for transmissions of 1.544 Mbps or can even be used in different combinations to utilize
capacity up to 1.544 Mbps.
‰‰ DS-2: This service has a capacity of 6.312 Mbps which is 96 times of DS-0 with an additional
overhead of 168 kbps. It can be utilized for multiplexing 96 DS-0 channels, four DS-1 channels or
can also be utilized as a single service for transmission up to 6.312 Mbps.
‰‰ DS-3: This service has a capacity of 44.376 Mbps which is seven times the data rate of DS-2
channels with an additional overhead of 1.368 Mbps. It can be utilized for multiplexing 672 DS-0
channels, 28 DS-1 channels or seven DS-2 channels. In addition, DS-3 can be utilized as a single
transmission line or as a combination of its previous hierarchies.
‰‰ DS-4: This service is at the highest level of the DS hierarchy and has a capacity of 274.176 Mbps. It
can be utilized for multiplexing six DS-3 channels, 42 DS-2 channels, 168 DS-1 channels or 4,032
DS-0 channels. It can also be used as a combination of service types at lower levels of hierarchy.
10. Explain the analog hierarchy used by telephone networks with an example.
Ans: Analog hierarchy is used by telephone companies for maximizing the efficiency of their infra-
structure such as switches and other transmission equipments. In analog hierarchy, the analog signals
from the lower bandwidth channels are multiplexed onto the higher bandwidth lines at ­different levels.
For multiplexing of analog signals at all levels of hierarchy, the FDM technique is used. The standards
for multiplexing vary in countries but the basic principle remains the same. One of such analog hierar-
chies is used by AT&T in the United States. This analog hierarchy comprises voice channels, groups,
super groups, master groups and jumbo groups (Figure 5.12).

4 kHz
4 kHz
12 voice channels

F Group
D
M F Super group
5 groups

4 kHz D
10 super groups

F Master group
M
D Jumbo
6 master groups

M F group
D
M

Figure 5.12 Analog Hierarchy

At the lowest level of hierarchy, 12 voice channels, each having a bandwidth of 4 kHz, are ­multiplexed
onto a higher bandwidth line thereby forming a group of 48 kHz (12 × 4 kHz). At the next level of
­hierarchy, five groups (that is, 60 voice channels) are multiplexed to form a super group that has a band-
width of 240 kHz (5 × 48 kHz). Further, 10 super groups (that is, 600 voice channels) are multiplexed to
form a master group. A master group must have a bandwidth of 2.4 MHz (10 × 240 kHz). However, to
avoid interference between the multiplexed signals, the requirement of guard bands increases the total
bandwidth of a master group to 2.52 MHz. Finally, six master groups are multiplexed to form a jumbo

CN_Chapter 05.indd 90 2/20/2012 6:21:02 PM

 Multiplexing and Switching 91

group that must have a bandwidth of 15.12 MHz (6 × 2.52 MHz). However, the requirement of guard
bands to separate the master groups increases the total bandwidth of a jumbo group to 16.984 MHz.
11. What is inverse multiplexing?
Ans: Inverse multiplexing is the opposite of the multiplexing technique. In this technique, the data
from a single high-speed line or source is split into chunks of data that can be transmitted over low-
speed lines at the same time and without any data loss in the combined data rate.
12. What do you understand by spread spectrum?
Ans: Spread spectrum is a technique used to expand the available bandwidth of a communication link
in order to achieve goals such as privacy and anti-jamming. For achieving these goals, the spread spec-
trum techniques expand the original bandwidth required for every system (adding redundancy) such that
the signals from different sources can together fit into larger bandwidth. For example, if the bandwidth
requirement of each system is B, then the spread spectrum techniques increase it to B´ such that B´ >> B.
To spread the bandwidth, a spreading process is used which is independent of the ­original ­signal. The
spreading process uses a spreading code—a series of numbers following the specific pattern—and
­results in expanded bandwidth.
The spread spectrum technique is generally used in wireless LANs and WANs where the systems
sharing the air transmission medium must be able to communicate without any interception from
­intruders, blocking of message during transmission and so on.
13. List the principles to be followed by the spread spectrum technique?
Ans: The spread spectrum technique uses the following principles to achieve its goals.
‰‰ The spreading process must be independent of the original signal. This implies that spreading process
is used after the source has generated the signal.
‰‰ Each source station is assigned a bandwidth much greater than what it actually needs thereby
­allowing redundancy.
14. Compare frequency hopping spread spectrum (FHSS) and direct sequence spread
­spectrum (DSSS) technique?
Ans: FHSS and DSSS are the techniques to expand the bandwidth of the digital signal. However,
both ­techniques follow a different process to achieve their goal.

Frequency Hopping Spread Spectrum

In this technique, a number of different carrier frequencies (say, N ) are used. The input signal modulates
one of N carrier frequencies in each hopping period (Thop). As shown in Figure 5.13, FHSS uses a pseu-
dorandom code generator, also known as pseudorandom noise (PN), which generates an m-bit pattern
(where m = log2N) for every hopping period Thop. For example, if an application uses 16 ­hopping
­frequencies, then the pseudorandom code generator creates 16 different four-bit patterns one per each
hopping period. During each hopping period, the m-bit pattern generated by pseudorandom code genera-
tor is used by the frequency table to determine the hopping frequency to be used for that hopping period.
The frequency selected from the frequency table is passed to the frequency synthesizer which generates
a carrier signal of the selected frequency. The input signal modulates the carrier signal generated by
the frequency synthesizer and as a result, spread signal with bandwidth (BFHSS) much greater than the
original bandwidth (B) is produced.

CN_Chapter 05.indd 91 2/20/2012 6:21:02 PM

 92 Data Communications and Computer Networks

Original
Modulator Spread signal
signal

Frequency
synthesizer

Pseudorandom
code generator

Frequency table

Figure 5.13 Frequency Hopping Spread Spectrum

Since one out of N hopping frequencies is used by a source station in each hopping period, the remain-
ing N-1 frequencies can be used by other N-1 stations. Thus, N hopping frequencies (N channels) can
be multiplexed into one using the same bandwidth BFHSS. That is, N source stations can share the same
bandwidth BFHSS. In addition, due to randomization in the frequency and the frequent frequency hops,
the intruders may be able to intercept the signal or send noise to jam the signal possibly for one hopping
period but not for the entire period thus, achieving privacy and anti-jamming.

Direct Sequence Spread Spectrum

In this technique, each bit of the original signal is replaced by n bits using a spreading code. As shown in
Figure 5.14, DSSS uses a chip generator that generates a spreading code of n bits (called chips) for each
signal bit. To expand the bandwidth of the original signal, each bit of the original signal is multiplied by
the code generated by the chip generator. After multiplication, the resultant spread signal with a band-
width n times of that of the original signal is produced. DSSS provides privacy between the sender and
the receiver if the intruder does not know the spreading code. It also prevents interference if a different
spreading code is used for each source station.

Original
Modulator Spread signal
signal
Spreading code

Chip generator

Figure 5.14 DSSS Process

Figure 5.15 illustrates the DSSS process. The spreading code used in this example is eight chips
with the pattern 10110100. Each bit of the original signal is multiplied by the eight-bit spreading code
to obtain the new spread signal. Now, if the original signal rate is S, then the new spread signal rate
will be 8S. This implies that the required bandwidth for the spread signal would be eight times of the
original bandwidth.

CN_Chapter 05.indd 92 2/20/2012 6:21:04 PM

 Multiplexing and Switching 93

1 0 1 1

Original signal

10110100101101001011010010110100

Spreading
code

10110100010010111011010010110100

Spread
signal

Figure 5.15 DSSS Example

15. What is meant by the term switching?
Ans: On a network, switching means routing traffic by setting up temporary connections between
two or more network points. This is done by devices located at different locations on the network,
called switches (or exchanges). Switches create temporary connections amongst two or more devices
connected to them. In a switched network, some switches are directly connected to the communicating
devices, while others are used for routing or forwarding information.

III
I VI

IV

II VII
V

Figure 5.16 Switched Network

Figure 5.16 depicts a switched network in which communicating devices are labelled A,  B,  C
and so on and switches are labelled I,  II,  III,  IV and so on. Each switch is connected either

CN_Chapter 05.indd 93 2/20/2012 6:21:06 PM

 94 Data Communications and Computer Networks

to a communicating device or to any other switch for forwarding information. Notice that multiple
switches are used to complete the connection between any two communicating devices at a time,
hence saving the extra links required in case of a point-to-point connection.

16. Explain different types of switching techniques along with their advantages and disad-
vantages.
Ans: There are three different types of switching techniques; namely, circuit switching, message
switching and packet switching.

Circuit Switching
When a device wants to communicate with another device, circuit switching technique creates a fixed
bandwidth channel, called a circuit, between the source and the destination. This circuit is a physical
path that is reserved exclusively for a particular information flow, and no other flow can use it. Other
circuits are isolated from each other, and thus their environment is well controlled. For example, in
Figure 5.17, if device A wants to communicate with device D, sets of resources (switches I,  II and
III) are allocated which act as a circuit for the communication to take place. The path taken by data
between its source and destination is determined by the circuit on which it is flowing, and does not
change during the lifetime of the connection. The circuit is terminated when the connection is closed.
Therefore, this method is called circuit switching.

III
I II

IV

Figure 5.17 Circuit Switching

Some advantages of circuit switching are as follows:

‰‰ It is a simple method and does not require the use of any special facilities.
‰‰ After a circuit is determined and established, data is transmitted with just a minimal propagation
delay and without congestion.
‰‰ It is the preferred choice for the transmission of real-time data and voice.
Some disadvantages of circuit switching are as follows:
‰‰ The time required to establish a dedicated circuit between two stations is 10 s approximately and
this time duration increases more depending upon the distance between the stations. For many
­applications, this is unsuitable and undesirable.

CN_Chapter 05.indd 94 2/20/2012 6:21:07 PM

 Multiplexing and Switching 95

‰‰ The resources are not efficiently utilized during circuit switching in a computer network
‰‰ For communication amongst stations that use costly and high-speed transmission lines, circuit
switching is not cost effective and economical, as communication between stations occurs gener-
ally in fast and small time gaps.

Packet Switching
Packet switching introduces the idea of breaking data into packets, which are discrete units of poten-
tially variable length blocks of data. Apart from data, these packets also contain a header with control
information such as the destination address and the priority of the message. These packets are passed
by the source point to their local packet switching exchange (PSE). When the PSE receives a packet,
it inspects the destination address contained in the packet. Each PSE contains a navigation directory
specifying the outgoing links to be used for each network address. On ­receipt of each packet, the PSE
examines the packet header information and then either removes the header or forwards the packet to
another system. If the communication channel is not free, then the packet is placed in a queue until
the channel becomes free. As each packet is received at each transitional PSE along the route, it is
­forwarded on the appropriate link mixed with other packets. At the destination PSE, the packet is finally
passed to its destination. Note that not all packets of the same message, travelling between the same two
points, will necessarily follow the same route. Therefore, after reaching their destination, each packet is
put into order by a packet assembler and disassembler (PAD).
For example, in Figure 5.18, four packets (1, 2, 3 and 4) once divided on machine A are transmitted
via various routes, which arrive on the destination machine D in an unordered manner. The destination
machine then assembles the arrived packets in order and retrieves the information.

4 3 2 1 1
1
3
3 1

4 3
2 4 3
4 1

4
2 2 4 2341

Figure 5.18 Packet Switching

Some advantages of packet switching are as follows:
‰‰ It is a fault-tolerant technique.
‰‰ It provides a much fairer and cost-efficient sharing of the resources. In addition, if no data is avail-
able to the sender at some point during a communication, then no packet is transmitted over the
network and no resources are wasted.

CN_Chapter 05.indd 95 2/20/2012 6:21:08 PM

 96 Data Communications and Computer Networks

‰‰ It requires lesser storage capacity for stations.

‰‰ The transmission rate in packet switching is fast.
Some disadvantages of packet switching are as follows:
‰‰ As packets do not follow a specific path, they may arrive out of order at the destination
‰‰ The cost incurred per packet is large.

Message Switching
A message is a unit of information which can be of varying length. Message switching is one of the
earliest types of switching techniques, which was common in the 1960s and 1970s. This switching tech-
nique employs two mechanisms; they are store-and-forward mechanism and broadcast mechanism. In
store-and-forward mechanism (Figure 5.19), a special device (usually, a computer system with large
storage capacity) in the network receives the message from a communicating device and stores it into
its memory. Then, it finds a free route and sends the stored information to the intended receiver. In such
kind of switching, a message is always delivered to one intermediate device where it is stored and then
rerouted to its final destination

Store and Forward Store and Forward

D B
D

B D
D

Store and Forward Store and Forward

Figure 5.19 Store-and-Forward Mechanism of Message Switching

In broadcast switching mechanism, the message is broadcasted over a broadcast channel as shown
in Figure 5.20. As the messages pass by the broadcast channel, every station connected to the ­channel
checks the destination address of each message. A station accepts only the message that is ­addressed
to it.
Some advantages of message switching are as follows:
‰‰ In message switching technique, no physical path is established in advance.
‰‰ The transmission channels are used very effectively in message switching, as they are allotted only
when they are required.
Some disadvantages of message switching are as follows:
‰‰ It is a slow process.

CN_Chapter 05.indd 96 2/20/2012 6:21:11 PM

 Multiplexing and Switching 97

Station 1 Station 2 Station 3 Station n

Message

Figure 5.20 Broadcast Mechanism of Message Switching

‰‰ It involves propagation and queuing delay; in addition, a large capacity of data storage is required.
‰‰ With unlimited message length, the switching nodes needed to have a larger storage space to buffer
the message.
17. Compare circuit switching, packet switching and message switching techniques.
Ans: The comparison of circuit switching, packet switching and message switching is summarized
in Table 5.1.

Table 5.1 Comparison of Circuit Switching, Packet Switching and Message Switching

Circuit switching Packet switching Message switching

• A physical circuit is established • No physical circuit is established • No physical circuit is established
before transmission begins. before transmission. before transmission.
• Message to be transmitted is in • Message to be transmitted is in • Message to be transmitted is in
the form of packets. the form of packets. the form of blocks.
• Store-and-forward technique is • Store-and-forward technique is • Store-and-forward technique is
not used. used. used.
• It can be used with real-time • It can be used with real-time • It cannot be used with real-time
applications. applications. applications.
• It is used in telephone network for • It is used for the Internet. • It was used in the transmission of
bi-directional, fast and real-time voice signals and messages.
data transfer.

18. Describe the major components of a telephone network?

Ans: The traditional telephone network, referred to as plain old telephone system (POTS), was purely
an analog system used to transmit voice. However, with advancement in computing technologies, they
were used to transmit other data along with voice. Three major components of a telephone network are
local loops, trunks and switching office (Figure 5.21). These components are described as follows:
‰‰ Local Loops: A twisted-pair cable connection between each subscriber’s telephone and the near-
est end office of telephone company (also called local central offic ) is referred to as a local loop.
Generally, the distance between the subscribers’ telephone and the nearest end office is 1–10 km.

CN_Chapter 05.indd 97 2/20/2012 6:21:12 PM

 98 Data Communications and Computer Networks

Subscriber 1
Local loop
Toll connecting
Intertoll trunks
trunks
Subscriber 2

End offices
Tandem offices
Regional offices
Subscriber n

Figure 5.21 Components of a Telephone Network

Also referred to as the last mile, the local loops follow analog technology and can extend up to
several miles. When a local loop utilized for voice, its bandwidth is 4 kHz. Earlier, uninsulated
copper wires were commonly used for local loops. However, these days, category 3 twisted pairs
are used for local loops.
‰‰ Trunks: The transmission media used for communication between switching offices is referred to
as trunk. Generally, fibre-optic cable or satellite transmission is used for trunks. A single trunk can
carry multiple conversations over a single path by using a multiplexing technique.
‰‰ Switching Offices: In telephone network, no two subscribers are connected by a permanent ­physical
link. Instead, switches are used to provide a connection between the different subscribers. These
switches are located in the switching offices of the telephone company and connect many local
loops or trunks. In telephone network, there are several levels of switching offices including end
offices, toll offices, tandem offices and regional offi
When a subscriber calls to another subscriber connected to same end office, a direct electrical
­connection is established between the local loops with the help of switching mechanism within the end
office. However, if the called subscriber’s telephone is connected to a different end office, the connection is
established by the toll offices or tandem offices. Several end offices are connected to their nearby switching
centre called toll office or tandem office (if end offices are in the same local area) through toll connecting
trunks. If the caller and called subscriber do not have a common toll office, the connection between them
is established by regional offices to which toll offices are co ected via high bandwidth intertoll trunks.
19. Explain dial-up modems along with their standards in detail.
Ans: The word modem is an acronym for modulator-demodulator. The modulator converts ­digital
data into analog signals (that is, modulation) and the demodulator converts the analog signal back into
digital data (that is, demodulation). A dial-up modem uses the traditional telephone line as the trans-
mission media and enables to send digital data over analog telephone lines.
Figure 5.22 depicts the modulation/demodulation process during communication between machines
A and B. At the sender’ side (that is, machine A), the digital signals from machine A are sent to the mo-
dem, which converts them into analog signals. These analog signals are transmitted over the telephone
lines and received by the modem at the receiver’s end (that is, machine B). The modem at machine B
converts analog signals back into digital signals and sends them to computer B.
ITU-T gave V-series standards for effective working of modems. Some modems of this series are
as follows:
‰‰ V.22: This modem transmits data at 1,200 bps and works over two-wire leased line. It provides
full-duplex synchronous transmission.

CN_Chapter 05.indd 98 2/20/2012 6:21:12 PM

 Multiplexing and Switching 99

Digital Digital
signal signal
Analog Analog
signal Telephone signal
Modem network Modem

Figure 5.22 Modulation/Demodulation Process

‰‰ V.32: This modem transmits data at 9,600 bps and designed for full-duplex synchronous trans­
mission. This modem works on two-wire leased line or telephone network. This standard uses the
technique called trellis coding that detects error which gets introduced during transmission. In
this coding, data stream is divided into four-bit sections and one extra bit is added to each quadbit
(four-bit pattern) during data transmission for error detection.
‰‰ V.32bis: This modem transmits data at 14,400 bps and is the enhancement of V.32 standard.
V.32bis includes feature of fall-back and fall-forward which help the modem to adjust ­upstream
and downstream speed automatically. This adjustment of speed depends on the quality of
the signal.
‰‰ V.34: This modem transmits data up to 33.6 kbps and works on a two-wire leased line. It is ­designed
for full-duplex synchronous and asynchronous transmissions and also supports error correcting
feature.
‰‰ V.90: This modem is designed for full-duplex synchronous and asynchronous transmission over
two-wire connection. The V.90 series modems are also called 56-K modems because their data
transmission rate is 56 kbps. They are asymmetric in nature as the speed of upstream and down-
stream varies. They support downloading data rate up to 56 kbps and uploading up to 33.6 kbps.
The main reason for the difference in the speed between uploading and downloading is that in
uploading, signal gets sampled at various switching stations and gets affected by noise which is
introduced here but in downloading, there is no such sampling of signals.
‰‰ V.92: This modem has the capability to upload the data at the rate of 48 kbps while its downloading
data rate is same as that of V.90 standard, that is, 56 kbps. This modem has the advanced feature
of call-waiting service.
20. Write a short note on DSL and its different technologies.
Ans: DSL (stands for digital subscriber line) was developed by the telephone companies to fulfil
the requirement of high-speed data access and the efficient utilization of the bandwidth of the local
loops. DSL is a very promising technology, as it provides the customers with a telephone connection
and high-speed Internet access simultaneously over the existing local loops. To provide a simultaneous
telephone connection and Internet access, DSL systems are installed between the telephone exchange
and the customers’ site.
The DSL technology comprises many different technologies including ADSL, VDSL, HDSL and
SDSL. Often, this group of technologies is referred to as xDSL where x represents A, V, H or S. These
technologies are described as follows:
‰‰ ADSL: It stands for asymmetric digital subscriber line, also called asymmetric DSL. ADSL
uses the existing local loops and provides a higher downstream rate from the telephone ­exchange
to the subscriber than in the reverse direction. This asymmetry in data rates is suitable for the

CN_Chapter 05.indd 99 2/20/2012 6:21:14 PM

 100 Data Communications and Computer Networks

a­ pplications such as video-on-demand and Internet surfing, as the users of these applications need
to download more data in comparison to uploading and that too at a higher speed. Along with the
Internet service, ADSL also provides simultaneous telephone connection over the same copper
­cable. By using FDM, the voice signals are separated from data signals (upstream and down-
stream). ADSL technology is useful for residential customers and not for the business customers,
as business customers often require larger bandwidth for both downloading and u­ ploading.
‰‰ ADSL Lite: It is a newer version of the ADSL technology. Also known as the spliterless
ADSL, this technology can provide a maximum upstream data rate of 512 kbps and a ­maximum
downstream data rate of 1.5 Mbps. This technology differs from ADSL technology in the sense
that ADSL technology requires installing a splitter at the subscribers’ home or business loca-
tion to split voice and data, while if ADSL Lite modem is used, no such splitter is required
at the premises of the subscriber and all the splitting is done at the telephone exchange. The
ADSL Lite modem is directly plugged into the telephone’s jack at the subscribers’ premises and
­connected to the computer.
‰‰ HDSL: It stands for high-bit-rate digital subscriber line, also called high-bit-rate DSL. It was
developed by Bellcore as a substitute for the T-1 (1.544 Mbps) line. In T-1 lines, alternate mark
inversion (AMI) encoding is used. This encoding is subject to attenuation at high frequencies and
repeaters are to be used for covering longer distance. Thus, it was quite expensive. On the other
hand, in HDSL, 2B1Q encoding is used which is less susceptible to attenuation at higher frequen-
cies. Also, a very high data rate of 1.544 Mbps can be achieved without using repeaters to a dis-
tance of approximately 4 km. To achieve full-duplex transmission, HDSL uses two twisted pairs
with one pair for each direction. HDSL2, a variant of HDSL, uses single pair of twisted cables and
is under development.
‰‰ SDSL: It stands for symmetric digital subscriber line, also called symmetric DSL. It was
­developed as a single copper pair version of HDSL. SDSL uses 2B1Q line coding and provides
full-duplex transmission on a single pair of wires. It provides same data rate of 768 kbps for both
upstream and downstream, as it supports symmetric communication.
‰‰ VDSL: It stands for very high-bit-rate digital subscriber line, also called very high-bit-rate
DSL. It is similar to ADSL and offers very high data transfer rates over small distances using
­coaxial, fibre-optic and twisted-pair cables. The downstream data rate for VDSL is 25–55 Mbps
and the upstream data rate is generally 3.2 Mbps.
Table 5.2 summarizes all the DSL technologies along with their characteristics.

Table 5.2 Summary of the DSL Technologies

Number of
Distance in Simultaneous voice twisted
Technology Upstream rate Downstream rate thousands feet capacity pairs
ADSL 16–640 kbps 1.5–6.1 Mbps 12 Yes 1
ADSL Lite 500 kbps 1.5 18 Yes 1
HDSL 1.5–2.0 Mbps 1.5–2.0 Mbps 12 No 2
SDSL 768 kbps 768 kbps 12 Yes 1
VDSL 3.2 Mbps 25–55 Mbps 3–10 Yes 1

CN_Chapter 05.indd 100 2/20/2012 6:21:14 PM

 Multiplexing and Switching 101

21. What are the two types of switch technologies used in circuit switching? Compare them.
Ans: In circuit switching, two types of switch technologies are used, namely, space-division switch
and time-division switch.

Space-Division Switch
In space-division switching, many different connections in the circuit that are active at the same time
use different switching paths which are separated spatially. Initially, this technology was developed for
the analog signals; however, presently, it is being used for both analog signals and DSs. Space-division
switches may be cross-bar switches or multistage switches.
‰‰ Cross-bar Switch: This switch connects p input lines with q ­output lines in the form of a matrix of
the order p×q; the value of p and q may be equal. Each input line is connected to output line with
the help of a transistor at each cross-point—the ­intersection
Output lines
point of input and output lines. Each cross-point is basically
1 2 3 4
an electric switch that can be opened or closed by a control 1
unit ­depending on whether or not communication is required
Input 2
between the connecting input and output lines. Figure 5.23 lines 3
shows a ­schematic diagram of a 4 × 4 cross-bar switch. 4
A cross-bar switch has certain limitations, which are as ­follows:
Figure 5.23 A 4 × 4 Cross-bar Switch
•  To connect m inputs with m outputs, m2 number of cross-
points is required. In addition, the number of cross-points required increases with the increase
in number of lines to be connected. This makes the cross-bar switches expensive and more
complex.
•  The cross-points are not utilized properly. Even when all the devices are connected, only a few
cross-points are active.
• If a cross-point fails, the devices connected via that cross-point cannot connect to each other.

‰‰ Multistage Switch: This switch overcomes the limitations of the cross-bar switch. A multistage
space-division switch consists of several stages, each containing (p×q) cross-bar switches. These
switches can be connected in several ways to build a large multistage network (usually, three-stage
network). In general, for n inputs and n outputs, m = log2n stages are required. Figure 5.24 shows a
three-stage space-division switch. Here, each cross-point in the middle stage can be accessed through
several cross-points in the first or third stage

8 8
7 7
6 4*2 2*2 2*4 6
5 5

4 4
3 3
2 4*2 2*2 2*4 2
1 1

Figure 5.24 A Three-stage Space-division Switch

CN_Chapter 05.indd 101 2/20/2012 6:21:16 PM

 102 Data Communications and Computer Networks

The main advantage of multistage switches is that the cost of having an m × m multistage network is
much lower than that of an m × m cross-bar switch because the former network requires a lesser ­number
of cross-points as compared to the latter network. However, in case of heavy traffic, it suffers from
blocking problem due to which it cannot process every set of requests simultaneously.

Time-division Switch
In time-division switch, TDM is implemented inside a switch. Various active connections inside the
switch can utilize the same switching path in an interleaved manner. One of the commonly used meth-
ods of time-division switching is time slot interchange (TSI), which alters the sequencing of slots
depending on the connection desired. TSI has a random access memory (RAM) and a control unit.
The data is stored in RAM sequentially, however, retrieved selectively based on the information in
control unit. The control unit stores the control information such as which input line to connect to
which output.
Figure 5.25 depicts the operation of TSI. Suppose the input lines want to send data to output lines in
the order 1 to 4, 2 to 3, 3 to 2 and 4 to 1. This information is stored in control unit. At the sender’s end,
the input unit from each of the four input lines is put serially into time slots to build a frame. The data
in the time slots is stored on the RAM in the exact order in which it is received, that is, A,  B,  C and  D.
The data is retrieved from RAM and then filled up in the time slots of the output frame in the order as
determined by the control unit, that is, D,  C,  B and A.

TSI
A RAM D
1 1
B A A B C D C
T D C B A T
2 2
C D B D B
3 M M 3
D C A
4 4
D

1 4
2 3
3 2
4 1
Control unit

Figure 5.25 Time-division Switch

A disadvantage of time-division switching is that it requires a sufficient amount of RAM always to

be able to store and forward data from the incoming frames to the outgoing frames.
22. Explain in brief about the generations of cable TV networks.
Ans: Cable TV networks provide a wide variety of services ranging from distributing video signals
to Internet access. There are mainly two generations of cable networks: traditional cable networks and
hybrid-fib e-coaxial networks.

CN_Chapter 05.indd 102 2/20/2012 6:21:18 PM

 Multiplexing and Switching 103

Traditional Cable Networks

Cable TV network started providing video services in the late 1940s. An antenna mounted on the top of a
building or tower was used to receive the signals from the TV stations and to distribute them to subscribers
with the help of coaxial cables. That’s why it was also referred to as community antenna TV (CATV).
Figure 5.26 shows a conceptual view of traditional cable networks. The broadcasting stations send
video signals to cable TV office called head end. The head end transmits these video signals to subscrib-
ers’ houses with the help of coaxial cable. To deal with the attenuation problem, amplifiers were used.
At the subscribers’ end, splitters were used to split the cable and taps and drop cables were used to make
the connection to the subscriber’s houses. Communication in these cable networks was unidirectional;
signals could be transmitted only in one direction from cable TV office to subscribe ’s premises.

Headend

Coaxial cable
Splitter

Drop
cable

Tap

Figure 5.26 Traditional Cable TV Network

Hybrid Fibre-coaxial (HFC) Network
In this generation of cable TV network, cable operators decided to provide Internet services. This cable
network is so called because it uses both the fibre-optic cable and the coaxial cable, the fibre-optic cable
for long haul runs while coaxial cable to the houses. The interface between the electrical and the optical
parts of the system is provided through electro-optical converters known as fib e nodes.
Figure 5.27 shows an HFC network. The signals transmitted from cable TV office, referred to
as regional cable head (RCH), are passed to the distribution hubs that act as subordinate to RCH.

Coaxial cable

Fibre node

Fibre-optic
cable Switch
Distribution
hub
Fibre node
RCH

Figure 5.27 HFC Network

CN_Chapter 05.indd 103 2/20/2012 6:21:23 PM

 104 Data Communications and Computer Networks

The ­distribution hubs perform modulation of signals and distribute them to fibre nodes over the
­fibre-optic cable. The fibre nodes split the signal to send same signal to each coaxial cable
In comparison to traditional cable TV network, HFC network requires less number of amplifiers as
well as offers bidirectional communication.
23. Which devices are needed for using cable networks for data transmission?
Ans: Traditional cable TV network was used only for providing video signals but nowadays, HFC
network—an extension of traditional cable TV network—is being used to provide high-speed access to
the Internet. However, to use cable TV network for data transmission, two devices are needed, namely,
cable modem (CM ) and CM transmission system (CMTS ).

Cable Modem
This is an external device that is installed at the end user’s premises where the Internet has to be
­accessed. It is connected to the user’s home PC via Ethernet port and it works similar to an ADSL
modem. CM divides the HFC network into two channels, upstream and downstream channels, with
downstream channel having a higher data transmission rate than the upstream channel. The upstream
channel is used for transmission from the home PC to the head end and the downstream channel is used
to handle transmission in the opposite direction. Since HFC is a shared broadcast network, each packet
from the head end is broadcasted on every link to every home; however, vice versa does not happen. As
a result, when several users are downloading contents simultaneously through the downstream channel,
the transmission rate received by each user is much lesser than the actual rate of downstream channel.
In contrast, when only some users are surfing the web, each user will receive a full downstream rate
because it is very rare that two users request the same page at the same time.

Cable Modem Transmission System

This device is installed inside the distribution hub of HFC network by the cable company. The data is
received by the CMTS from both the Internet and the user. When CMTS receives data from Internet, it
passes the data to the combiner. The combiner combines the data with the video signals received from
head end and passes the resultant information to the intended user. On the other hand, the data received
by the CMTS from the end user is simply forwarded to the Internet.
24. Write a short note on ISDN.
Ans: ISDN (stands for integrated services digital network) was developed by ITU-T in 1976. The
idea behind ISDN is to digitize the telephone network, so that all the data including audio, video and text
could be transmitted over existing telephone lines. The main objective of ISDN is to create a wide area
network (WAN) for providing global end-to-end connectivity over digital media. For this, it integrates
different transmission services with no addition of new links or subscriber lines. ISDN provides higher
data rates usually 2 Mbps on a local link and 64 kbps or 128 kbps on a wide area link.
ISDN caters to the need of both voice and non-voice applications of the same network. It also sup-
ports a variety of applications which involve switched (circuit switched or packet switched) as well as
non-switched connections. The new services can be added to ISDN provided they are compatible with
64 kbps switched digital connections. ISDN can use X.25 standards and can be used as local access link
in frame relay and X.25 networks.
Though ISDN is a digital service and offers much more bandwidth than standard telephone service,
still like telephone service, the ISDN users also need to dial an ISDN number (different from standard

CN_Chapter 05.indd 104 2/20/2012 6:21:24 PM

 Multiplexing and Switching 105

telephone number) for establishing a connection and connecting to other sites. A user can establish a digital
­connection by dialling another ISDN number from his/her own ISDN number. However, unlike leased
lines which provide permanent connection, the ISDN users can disconnect ISDN WAN link whenever
desired.
25. What are the two types of ISDN?
Ans: Based on the transmission and switching capabilities, there are two types of ISDN, namely,
narrowband ISDN and broadband ISDN.
‰‰ Narrowband ISDN (N-ISDN): This is the first generation of ISDN and uses circuit-switching
technique. It has smaller bandwidth and supports low bit rates (usually up to 64 kbps). Due to lower
bit rates, N-ISDN cannot cater to the needs of certain applications such as multimedia ­applications.
Four-wire twisted pairs are used for transmission in N-ISDN thereby resulting in poor quality of
service. N-ISDN follows the concept of frame relay.
‰‰ Broadband ISDN (B-ISDN): This is the second generation of ISDN and uses packet-switching tech-
nology. It supports high-resolution video and multimedia applications. It can support data rates of hun-
dreds of Mbps (usually up to 622 Mbps) and thus, is suitable to be used for high-resolution video and
multimedia applications. Optical fibre is used as the transmission media for B-ISDN thereby resulting
in better quality of service as compared to N-ISDN. B-ISDN follows the concept of cell relay.
26. What are the services provided by ISDN?
Ans: ISDN provides various services including data applications, existing voice applications, fac-
simile, videotext and teletext services. These services are grouped under the following three c­ ategories.
‰‰ Bearer or Carrier Services: These services allow the sender and the receiver to exchange informa-
tion such as video, voice and text in real time. A message from the sender can be communicated to the
receiver without any modification in the original content of the message. Bearer services are provided
by ISDN using packet switching, circuit switching, cell relay and frame relay. These services corre-
spond to the physical, data link and network layers of the OSI model.
‰‰ Teleservices: These services not only permit to transport information but include information-
processing function also. To transport information, teleservices make use of bearer services while
to process the information, a set of higher-level functions that correspond to the transport, session,
presentation and application layers of the OSI model are used. Some examples of teleservices are
videotex, teletex and telefax.
‰‰ Supplementary Services: These services are used in combination with at least one teleservice or
bearer service but cannot be used alone. For example, reverse charging, call waiting and message
handling are some of the supplementary services that are provided with the bearer and teleservices.
27. Describe the type of channels provided by ISDN along with their purpose.
Ans: The ISDN standard has defined three types of channels, namely, bearer (B) channel, data (D)
channel and hybrid (H) channel, with each channel having a specific data rate
‰‰ B Channel: This is the basic channel used to carry only the user traffic in digital form such as
digitized voice and digital data at the rate of 64 kbps. The transmission through B channel is
full-duplex as long as the required data transmission rate is up to 64 kbps. Multiple conversations
­destined for a single receiver can be carried over a single B channel with the use of multiplexing.
Since a B channel provides end-to-end transmission, signals can be demultiplexed only at the
­receiver’s end and not in the mid way.

CN_Chapter 05.indd 105 2/20/2012 6:21:24 PM

 106 Data Communications and Computer Networks

‰‰ D Channel: This channel is mainly used to carry the control information that is required for
e­ stablishing and terminating the switched connections on the B channels. For example, the dialled
digits while establishing a telephone connection are passed over the D channel. Under certain
­circumstances, the D channel can be used to carry user data also. The D channel provides data rates
of 16 or 64 kbps depending on the requirements of the user.
‰‰ H Channel: This channel is used for applications having higher data requirements such as video-
conferencing and teleconferencing. There are certain types of H channels including H0 channel
with a data rate of 384 kbps, H11 channel with a data rate of 1,536 kbps and H12 channel with a
data rate of 1,920 kbps.
28. Distinguish BRI and PRI in ISDN.
Ans: In ISDN network, each user is connected to the central office via digital pipes called digital
subscriber loops. The digital subscriber loops are of two types, namely, basic rate interface (BRI) and
primary rate interface (PRI), with each type catering to a different level of customers’ needs. Table 5.3
lists some differences between these two types.

Table 5.3 Differences Between BRI and PRI

BRI PRI
• It comprises two B channels and one D channel, • It comprises up to 23 B channels and one D channel in
thus, referred to as 2B + D channel. North America and Japan. In countries such as Europe,
Australia and other parts of the world, it comprises 30
B channels and one D channel.
• The BRI-D channel operates at 16 kbps. • The PRI-D channel operates at 64 kbps.
• It provides total bit rate of 192 kbps. • In North America and Japan, it provides total bit rate of
1.544 Mbps and in other parts of the world, it provides
total bit rate of 2.048 Mbps.
• It is primarily used at home for connecting to • It is primarily used in business replacing the leased
Internet or business networks. lines that can provide the same bandwidth and signal
quality but with more flexibility.

29. Explain the architecture of ISDN.

Ans: In ISDN network, to enable the users to access the services of BRI and PRI, certain devices are
used. The main devices used in the architecture of ISDN (Figure 5.28) include terminal equipment (types
1 and 2), network termination devices (types 1 and 2) and terminal adapters. All these devices are used
in the subscriber’s premises. In addition, several reference points are used to determine the interfaces be-
tween any two elements of ISDN installation. The description of these equipments is as follows:
‰‰ Terminal Equipment 1 (TE1): It refers to a device that supports ISDN standards. For example,
­computer or ISDN telephone. To connect TE1s to the ISDN network, a four-wire twisted-pair cable
is used.
‰‰ Terminal Equipment 2 (TE2): It refers to a device that is non-specialized ISDN. For example, the
standard telephone is a non-ISDN device.

CN_Chapter 05.indd 106 2/20/2012 6:21:24 PM

 Multiplexing and Switching 107

Subscriber’s
premises Central office
Switched
network
NT2 NT1
S T U
TE1
ISDN Packet ISDN Subscriber
switch network switch or provider
TA NT2 NT1
R S T U Private line
TE2 network

Figure 5.28 Architecture of ISDN

‰‰ Terminal Adapters: It refers to a device that translates the information in a non-ISDN format into
a form usable by an ISDN device. It is generally used with TE2s devices.
‰‰ Network Termination 1 (NT1): It refers to a device that connects the internal system at the sub-
scriber’s end to digital subscriber loop. It organizes the data received from the subscriber’s end into
frames that are to be sent over the network. It also translates the frames received from the network
into a format which can be understood by the subscriber’s devices. It interleaves the bytes of data
from subscriber’s devices but is not a multiplexer; the multiplexing occurs automatically, as NTI
provides synchronization between the data stream and the process of creating a frame.
‰‰ Network Termination 2 (NT2): It refers to a device that performs multiplexing, flow control
and packetizing at the physical, data link and network layers of OSI model, respectively. It acts as
an intermediate device between data-producing devices and NT1. An example of NT2 is private
branch exchange (PBX) that coordinates transmissions from telephone lines and multiplexes them,
so that they can be transmitted by NT1.
‰‰ Reference Points: It refers to a tag that is used to define logical interfaces between various termi-
nation devices. The reference points used in ISDN architecture include R, S, T and U. Reference
point R specifies the link between TE1 and TA. Reference point S specifies the link between TE1
or TA and NT1 or NT2. Reference point T specifies the link between NT1 and NT2. Reference
point U specifies the link between NT1 and ISDN offic
To exchange the control information between end user and the network, protocols are used. ISDN
uses more than one twisted pair to provide the full-duplex communication link between the end user
and the central office. The operation performed by central office includes accommodating multiplexed
access to provide high-speed interface using digital PBX and LAN. With the help of central office, sub-
scriber can access circuit switched network and packet switched network.
30. Five 1-kbps connections are multiplexed together and each unit represents 1 bit. Find:
(a) the duration of 1 bit before multiplexing,
(b) the transmission rate of the link,

CN_Chapter 05.indd 107 2/20/2012 6:21:27 PM

 108 Data Communications and Computer Networks

(c) the duration of a time slot and

(d) the duration of a frame.
Ans: (a) The duration of one bit before multiplexing is 1/1 kbps, that is, 0.001 s or 1 ms.
(b) The transmission rate of the link will be five times the data r te of the connection, that is, 5 kbps.
(c) The duration of a time slot will be one-fifth of the duration of each bit before multiplexing,
that is, 1/5 ms or 200 μs.
(d) Since there are five slots in a frame, the duration of a frame will be equal to five times the
duration of a time slot, that is, 5 × 200 μs or 1 ms.
31. Five channels, each with a 100-KHz bandwidth, are to be multiplexed together. What is
the minimum bandwidth of the link if there is a need for a guard band of 10 KHz between the
channels to prevent interference?
Ans: To multiplex five channels, four guard bands are needed. Thus, the minimum bandwidth (B)
required will be calculated as follows:
B = 5 × (100 KHz) + 4 × (10 KHz) = 540 KHz

Multiple Choice Questions

1. A device that combines n input lines into a 6. Which of the following is not a group in the
single stream is known as ________. analog hierarchy?
(a) Demultiplexer (a) Voice channels
(b) Serial to parallel convertor (b) Master groups
(c) Prism (c) Slave groups
(d) Multiplexer (d) Jumbo groups
2. Multiplexing can be used in situations where 7. A technique used to expand the bandwidth
the signals to be transmitted through the of a communication link to achieve goals
transmission medium have ________ band- such as privacy and anti-jamming is called
width than that of the medium. _________.
(a) Higher (b) Lower (a) Spread spectrum
(c) Equal (d) None of these (b) Multiplexing
(c) Switching
3. Which of the following multiplexing tech-
(d) Pulse stuffin
niques involves signals composed of light
beams? 8. Which of the following is not a switching
(a) FDM (b) TDM technique?
(c) WDM (d) All of these (a) Circuit switching
(b) Data switching
4. Multilevel multiplexing and multiple-slot
(c) Message switching
allocation are the strategies used in
(d) Packet switching
(a) TDM (b) WDM
(c) Both (a) and (b) (d) FDM 9. Store-and-forward technique is used in
(a) Message switching
5. The digital signal service has ________ levels.
(b) Packet switching
(a) Five (b) Seven
(c) Time-division multiplexing
(c) Six (d) Eight
(d) Both (a) and (b)

CN_Chapter 05.indd 108 2/20/2012 6:21:27 PM

 Multiplexing and Switching 109

10. The transmission media used for communica- 12. The broadcasting stations send video signals
tion between switching offices is referred to a to cable TV office, called ______
(a) Switch (a) Fibre node
(b) Local loop (b) Head end
(c) Trunk (c) Toll offic
(d) None of these (d) CATV
11. Which of the following is a DSL technology? 13. Narrow band ISDN supports bandwidth up to
(a) ADSL Lite (a) 128 kbps
(b) VDSL (b) 112 kbps
(c) SDSL (c) 96 kbps
(d) All of these (d) 64 kbps

Answers
1. (d)   2. (b)   3. (c)   4. (a)   5. (a)   6. (c)   7. (a)   8. (b)   9. (d) 10. (c)
11. (d) 12. (b) 13. (d)

CN_Chapter 05.indd 109 2/20/2012 6:21:27 PM

 6
Error Detection and Correction

1. Explain various design issues of the data link layer.

Ans: Data link layer is responsible for efficien and reliable transmission of frames between two
­adjacent nodes connected via a single communication channel. However, sometime, the efficienc of
data transfer is significantl affected due to finit data rate of communication circuits, non-zero propa-
gation delay between the sending and the receiving times of a bit, and errors in connection. Thus, to
achieve efficien data transmission, all these factors must be kept in mind. Some of the important design
issues of data link layer are as follows:
‰‰ Reliable Delivery: Frames sent by a sender must reach the receiver reliably and in the same order
as sent by the sender. For effective transmission, a receiver must let the sender to know which
frames have been received and which frames need to be retransmitted.
‰‰ Flow Control: Each node connected on a link has its own limited frame-buffering capacity. ­Without
a flo control mechanism, if a sender sends frames at a rate that is higher than the ­receiver can re-
ceive, then some frames may be lost. Therefore, the data link layer should provide flow control in
such a way that sender should not send frames at a rate that is faster than the receiver can process it.
‰‰ Acknowledged Delivery: A receiver must send an acknowledgement frame to indicate the sender
that a frame has been received properly. For this, each frame transmitted from the sender is given
a unique sequence number, so that unacknowledged frames that have got damaged or lost during
transmission can be known and the sender can retransmit those frames.
‰‰ Error Detection and Correction: Bit errors may introduce during the transmission of a frame
­because of the signal attenuation and electromagnetic noise. For example, bit 1 may be replaced
with bit 0 or vice versa. Therefore, the sender must add certain error-detection bits in the frame, so
that the receiver can check whether a frame contains some errors or not. The receiver should not
only be able to detect errors but also be able to determine the location of the errors in the frame and
then correct the errors as well.
‰‰ Addressing: The media access control (MAC) address of the sender and receiver must be specifie
in the frames that are transmitted by the data link layer. This would ensure the delivery of frames
to the appropriate receiver.

CN_Chapter 06.indd 110 2/20/2012 6:24:03 PM

 Error Detection and Correction 111

2. What are the different types of services provided by the data link layer to the network layer?
Ans: The data link layer offers various services to the network layer; however, some of the basic
types are as follows.
‰‰ Unacknowledged Connectionless Service: As the name suggests, this is a connectionless service,
that is, no logical connection between the sender and the receiver needs to be established or ­released
before or after the transmission respectively. In addition, the receiver does not send acknowledgement
frame to the sender on receiving a data frame. If a frame is lost or damaged during the transmission,
no attempt is made to discover the loss and recover from it. This type of service is suitable for applica-
tions with low error rate and for real-time traffi such as voice.
‰‰ Acknowledged Connectionless Service: In this service also, no logical connection is established
between the communicating nodes; however, each frame received by the receiver is acknowledged in
order to indicate the sender that the data frame has been received properly. Thus, the frames for which
the sender does not receive any acknowledgement within a stipulated time can be retransmitted. This
service is suitable for wireless systems.
‰‰ Acknowledged Connection-oriented Service: In this type of service, a connection is established
between the communicating nodes before starting data transmission. The data link layer ensures
that each frame is received exactly once by the receiver and all the frames are received in the
­correct order. For this, each frame is numbered before the transmission and the receiver sends back
an ­acknowledgement frame on receiving a data frame. In a connection-oriented ­service, the data
­transfer takes place by following three phases, namely, connection establishment, data ­transmission
and connection release. During the first phase, a connection is made between the sender and the
receiver. For establishing a connection, both sender and receiver initialize some variables and coun-
ters to identify the frames that have been received and that have not. ­After the connection has been
established, the second phase commences during which the actual ­transmission of frames is carried
out between the sender and the receiver. After the frames have been transmitted, the third phase
begins in which the connection is released by freeing up the used variables, counters and buffers.
3. What is framing? Discuss various framing methods.
Ans: The data link layer accepts the bit stream from the physical layer and breaks it into discrete
frames. This process is known as framing. For ensuring that the bit stream is free of errors, the data
link layer computes the checksum of each transmitting frame at the sender’s side. After the frame has
been received, the data link layer on the receiver’s side re-computes the checksum. If the new value of
checksum differs from the checksum contained in the frame, then the data link layer comes to know that
an error has occurred and takes appropriate measures to deal with errors. For example, some bad frames
can be discarded and the request for their retransmission can be made.
In order to make the receiver to detect the frame boundaries, several framing methods have been
developed. These methods are described as follows.

Character Count
This framing method uses a field in the header of each frame which specifies the total number of
characters in that frame. This field helps the data link layer at the destination node in ­knowing the
number of characters that follow and detecting the end of each frame. The disadvantage of this
­method is that while transmitting a frame, the count of total characters in it may alter due to

CN_Chapter 06.indd 111 2/20/2012 6:24:03 PM

 112 Data Communications and Computer Networks

­errors, thereby making the destination node go out of synchronization. Furthermore, the destination
node will not be able to locate the correct start of the new frame and, therefore, asking sender for the
retransmission of the frame, as the destination node does not know the exact count of ­characters that
should be skipped to get to the start of the retransmission.

Byte Stuffing
This framing method uses a byte called the flag byte at the starting and the ending of each frame.
This helps the destination node to find the end of the current frame even if the destination goes out
of synchronization. The presence of two successive flag bytes marks the ending of one frame and the
beginning of the next one. However, a problem occurs when binary data such as object programs or
floating-point numbers are transmitted. In such cases, the bit pattern of the flag byte may also appear
in the data and thus, may cause interference with the framing. To solve this problem, the data link
layer at the sender’s side can insert a special escape byte (ESC) before each fla byte appearing in the
data. At the destination node, the data link layer stripes off the escape byte and passes the data to the
network layer.
Now, it may also happen that the escape byte appears in the data. To prevent the escape byte from
being mixed with the data, each escape byte appearing in data is stuffed with another escape byte. Thus,
a single escape byte in the data indicates that it is a part of escape sequence while the occurrence of two
consecutive escape bytes in the data indicates that one escape byte is the part of data. The disadvan-
tage of this framing method is that it can only be used for eight-bit characters; therefore, this method
cannot be used for the character codes that use more than eight bits. Byte stuffin is also referred to as
character stuffin .

Bit Stuffing
This framing method overcomes the limitation of byte stuffing and can be used with any number of bits
per character and allows a data frame to comprise an arbitrary number of bits. In bit-stuffing method,
a special bit pattern, 01111110, called flag byte is inserted at the start and end of each frame. In order
to prevent the occurrence of fla byte in data to get misunderstood as start or end of frame, when a
sender’s data link layer find fiv consecutive 1s in the data, it automatically stuffs a 0 bit at the end of
these 1s (refer Q13). The data link layer at the destination de-stuffs or deletes a 0 bit followed by five
­consecutive 1s and then passes de-stuffed bits onto the network layer. An advantage of bit stuffing is that
the ­boundary between two consecutive frames can be clearly identified by the flag byte. Therefore, in a
case where receiver misses its track, it can scan the incoming bits for the flag byte, since a flag byte can
be present only at the boundaries of a frame.

Physical Layer Coding Violations

This method can only be used in a network where encoding on physical medium comprises some
r­ edundancy. For example, in some LANs, one bit of data is encoded using two physical bits, with bit 1
is represented as a high-low pair and bit 0 is represented as a low-high pair. Every bit undergoes a
transition during encoding and makes it easier for a receiver to accurately locate the bit boundaries.

CN_Chapter 06.indd 112 2/20/2012 6:24:03 PM

 Error Detection and Correction 113

4. Explain two types of errors with the help of a diagram.

Ans: During data transmission, one or more bits of data being transmitted may get corrupted.
­Depending on the number of bits in data that have been corrupted, there exist two types of errors:
single-bit error and burst error.
‰‰ Single-bit Error: In this type of error, only one bit of the data is altered during transmission, that
is, bit 1 may be received as 0 or bit 0 may be received as 1 (Figure 6.1).

1 0 0 0 1 0 0 0 0 1

Transmitted data Received data

Error
Figure 6.1 Single-bit Error

‰‰ Burst Error: A burst error occurs when two or more bits (consecutive or non-consecutive) are
altered during the transmission (Figure 6.2). The length of burst is measured from the firs altered
bit to the last altered bit. However, it does not necessarily mean that all bits comprising the burst
length have been altered.

1 0 0 0 1 0 0 0 0 1 Transmitted data

Errors

1 0 0 1 1 1 0 0 1 1 Received data

Length of burst error (6 bits)

Figure 6.2 Burst Error

5. Discuss the concept of redundancy in error detection and correction.

Ans: During data transmission, errors may occur because of certain transmission impairments such
as attenuation and electromagnetic noise. Usually, the burst errors are more likely to occur instead of
single-bit ­errors. For the detection and correction of these errors, some extra bits (called redundant bits)
are required to be sent along the data. Redundancy is the central concept in both error detection and error
correction.
At the sender’s end, the redundant bits are added to the data bits using a process that establishes
a ­relationship between the data bits and the redundant bits. At the receiver’s end, this relationship is
­examined by the receiver to detect or correct the errors in the received data. After detecting or ­correcting
the ­errors, the receiver removes the redundant bits and keeps only the data bits.
6. Compare error detection and error correction.
Ans: In error detection, the emphasis is on findin whether or not there is an error(s) in the received
data irrespective of the number of errors. The sender adds only enough extra (redundant) bits to the data
being transmitted that enables the receiver to infer that some error has occurred. Thus, whether there is

CN_Chapter 06.indd 113 2/20/2012 6:24:04 PM

 114 Data Communications and Computer Networks

a single-bit error or burst error, it does not make any difference. Some of the commonly ­error-detecting
codes include simple parity check, two-dimensional (2D) parity check, checksum and cyclic ­redundancy
check (CRC).
Contrastive to error detection, in error correction, the emphasis is on not only to detect the errors
but also to correct them. For this, it becomes important to know the exact number of bits that have
been corrupted as well as their positions in the received data unit. In error correction, the sender adds
enough redundant bits to the data being transmitted that enable the receiver to infer what the original
data was. This helps the receiver to detect as well as correct the errors in the received data. Error
­correction is more difficult than error detection. Hamming code is one of the commonly used categories
of ­error-correcting codes.
Generally, for highly reliable channels such as fibres error-detecting codes are used; however, for
wireless links, the use of error-correcting codes is preferred.
7. Define the following
(a) Codeword
(b) Code Rate
(c) Hamming Weight of a Codeword
Ans: (a) Codeword: While transmitting a block of data (referred to as dataword) containing d bits, a
few parity or redundancy bits (say, r) are added to form an n-bit (where n  =  d  +  r) encoded
block of data what is referred to as codeword. In other words, codeword is formed by
appending redundancy bits to the dataword.
(b) Code Rate: It refers to the ratio of the number of data bits in the codeword to the total number
of bits. That is,
Code rate = d/n
(c) Hamming Weight of a Codeword: It refers to the number of non-zero elements in the code-
word. It can be obtained by performing an XOR operation on the given codeword and a code-
word containing all zeros.
8. What do you understand by Hamming distance? Describe the role of minimum Ham-
ming distance in error detection and error correction.
Ans: Given a pair of codewords, say X and Y, containing same number of bits. The Hamming
distance between X and Y, denoted as d(X, Y  ), is define as the number of positions in which the
­corresponding bits of X and Y differ. For example, if X = 10010010 and Y = 11011001, then the Hamming
distance between X and Y is four, as the bits in X and Y differ in four positions, namely, 1, 2, 4 and 7.
The Hamming distance between two codewords can be calculated by firs performing XOR operation
on them and then checking for the number of 1s in the resulting bit string; the Hamming distance will
be equal to the number of 1s in the result string. For example, the result of XOR operation on X and Y is
01001011. Since there are four 1s in the resulting bit string, the Hamming distance is four.
Generally, during transmission, a large number of binary words are transmitted in a continuous
­sequence from sender to receiver. Accordingly, a block of codewords are received at the receiver; as a
result, many values of Hamming distances can be calculated between each possible pair of codewords.
Therefore, the Hamming distance is not of much significance instead minimum Hamming distance
(dmin) is used for designing a code. The minimum Hamming distance (dmin) of a linear block of code is
define as the smallest of the Hamming distances calculated between each possible pair of codewords.
The minimum Hamming distance (dmin) plays a significant role in error detection and error ­correction.
It is always possible to detect the errors in a received codeword if the total number of errors in the

CN_Chapter 06.indd 114 2/20/2012 6:24:04 PM

 Error Detection and Correction 115

­codeword is less than dmin; otherwise, errors cannot be detected. This is because if the number of errors in
a codeword is equal to or greater than dmin, the codeword may correspond to ­another valid codeword; as
a result, the receiver may not be able to detect errors. The error-detection and error-correction capabilities
of any coding technique largely depend on the minimum Hamming distance as explained below.
‰‰ To enable the receiver to only detect up to n errors in the codeword, dmin must be greater than or
equal to n  +  1. For example, for dmin = 5, up to four errors can be detected in a codeword, as 5  >  =
n  +  1, that is, n  <  =  4.
‰‰ To enable the receiver to correct up to m errors in the codeword, dmin must be greater than or equal to
2m+1. For example, for dmin= 5, up to two errors can be corrected in a codeword, as 5  >  =  2m  +  1,
that is, m  <  =  2.
‰‰ To enable the receiver to correct up to m errors and detect up to n errors (where n>m) in the
codeword, dmin must be greater than or equal to m+n+1.
In case, the value of dmin is even, this scheme is partially ineffective. For instance, if dmin = 4, the value
of m (the number of errors that can be corrected) comes out to be 1.5 that is rounded off to 1. This means
only one error can be corrected for dmin = 4 and thus, efficienc is wasted.
9. Explain various error-detection methods with the help of suitable examples.
Ans: To perform error detection, a few redundant bits must be added to the data. Redundant bits are
also called check bits, as they do not carry any useful information; however, their presence in data block
lets the receiver to detect errors that have occurred during transmission. Some of the commonly used
error-detection methods are discussed here.

Parity Checking
This is the simplest method for detecting errors in the data. An extra bit known as parity bit is added
to each word that is being transmitted. In a word, the parity bit is generally the MSB and remaining bits
are used as data bits. Parity of a word can be either even or odd depending on the total number of 1s in
it. If the total number of 1s in a word including parity bit is even, then it is called even parity while if
the total number of 1s in a word including parity bit is odd, then it is called odd parity.
Depending on the type of parity required, the parity bit can be set to either 0 or 1. For example, if data
is to be sent with odd parity, then the parity bit is set to 0 or 1, such that the number of 1s becomes odd.
Similarly, to obtain even parity, the parity bit is set to 0 or 1, such that the total number of 1s in the whole
word is even. The receiver can detect an error when it receives data in parity different from the expected
parity, that is, if receiver knew that the data will be in even parity but it receives data with an odd parity.
If the receiver detects an error, it discards the received data and requests the sender to retransmit the
same byte. Examples given in Q14 and Q15 illustrate the use of simple parity checking method.
The detection of errors by the parity checking method depends on the number of errors that have
­occurred. If the total number of errors in the received data is odd, then the parity of received data will be
different from that of transmitted data; therefore, error will be detected. However, if the number of errors
in the received data is even, the parity of received data will be the same as that of the transmitted data
and hence, the receiver will not be able to detect any errors. The main advantage of parity method is that
no additional storage or computational overhead is needed for generating parity. The parity checking
method has certain disadvantages too, which are listed below:
‰‰ Parity checking cannot correct error.
‰‰ Even number of errors cannot be detected.
‰‰ It is not possible to fin the location of the erroneous bit.

CN_Chapter 06.indd 115 2/20/2012 6:24:04 PM

 116 Data Communications and Computer Networks

Two-dimensional Parity Check

An improvement over simple parity check is 2D parity check. As the name implies, this technique
o­ rganizes a number of binary words being transmitted or received in succession along two dimensions,
that is, in the form of rows and columns. At the sender’s end, the parity bits are computed for each row
and each column using the simple parity check method. In this way, two sets of parity bits are generated:
one is associated with the rows of data block, referred to as longitudinal redundancy check (LRC),
and the other is associated with the columns of the data block, referred to as vertical redundancy check
(VRC). Each LRC and VRC bits are set such that the parity of corresponding row and column is even.
After generating the parity bits, the parity bits along with data are sent to the receiver. Example given in
Q16 illustrates the generation of LRC and VRC parity bits.
At the receiver’s end, each row and column (including LRC and VRC bits) of the received block of
data are checked to verify whether its parity is still even. If so, the receiver knows that there is no error
in the data block. Otherwise, for any single-bit error in the block of data, one of the LRC bits and one
of the VRC bits are incorrect. The intersection of row and column corresponding to incorrect LRC and
VRC bits respectively determines the position of erroneous bit. Example given in Q17 illustrates the
error detection using 2D parity checking method.
The 2D parity check is also referred to as block parity, as more than one parity bit are used for error
detection. Using 2D parity check, burst errors are more likely to be detected.

Checksum
To overcome the limitation of detecting multiple errors in parity checking, checksum was introduced.
Checksum method can be used for detecting multiple errors within the same word. With the transmis-
sion of a word in a block of data bytes, its contents are added bit by bit to that of its predecessor and
the sum is maintained at the sender’s side. Similarly, the contents of each successive word are added
to the previous sum. After all, the words of a block of data bytes have been transmitted and the total
sum (called checksum byte) up to that time is transmitted. While computing the checksum byte,
the carries of the MSB are discarded. The example described in Q18 illustrates how to compute the
checksum byte.
At the receiver’s end, the checksum is again computed by adding the bits of received bytes. The
­newly computed checksum is compared with the checksum transmitted by the sender. If both are found
to be same, then it implies no error; otherwise, errors are present in the received block of data. Some-
times, the 2’s compliment of the checksum is transmitted instead of the checksum. In such a case, the
receiver adds the data block of received bytes with 2’s complement of the checksum obtained bit by bit
from the sender. If the sum is zero, then there is no error in data; otherwise, the errors are present. The
advantage of checksum method is that it can detect burst errors.
The performance of checksum detection method is better than simple parity checking, as it can detect
all odd number of errors and most of the even number of errors.

Cyclic Redundancy Check (CRC)

This is one of most widely used and most powerful error-detecting techniques developed by IBM. The
CRC code is also referred to as polynomial code, as it treats the bit string to be transmitted as a polyno-
mial (say, M(x)) whose coefficient are the 0 and 1 values in the bit string and operations on bit string
interpreted as polynomial arithmetic. To generate CRC code, the sender and receiver must firs agree

CN_Chapter 06.indd 116 2/20/2012 6:24:04 PM

 Error Detection and Correction 117

on a predetermined pattern of bits known as generator polynomial, denoted by G(x). Notice that the
high- and low-order bits of the generator should always be 1.
CRC is based on binary division. Initially, at the sender’s end, n 0s are appended to the bit string
(say, d bits) to be transmitted where n is one less than the number of bits in the generator (say, m bits)
or we can say equal to the degree of the generator polynomial. Then, the resultant bit string of d  +  n
bits is ­divided by the generator using modulo-2 arithmetic with ignoring the carries and borrows in the
addition and subtraction, respectively. Here, it is important to note that whenever during division, the
leftmost bit of the dividend becomes zero, the regular divisor is not used for division; instead, a ­divisor
containing all zeros is used. After performing the division, the remainder bits (say, r bits), called redun-
dant bits or CRC, are added to the dividend to form the codeword of d  +  r bits. Notice that the number
of bits in CRC must be exactly one less than the number of bits in generator, that is, r  =  m–1. Finally,
the codeword of d  +  r bits is transmitted to the receiver. The example described in Q19 illustrates how
to generate the CRC code.
At the receiver’s end, the received codeword of d  +  r bits is again divided by the same generator as
used by the sender. If the remainder obtained after this division is non-zero, the receiver comes to know
that an error has occurred and thus, rejects the received data unit; otherwise, the receiver knows that
there is no error in the received data and thus, accepts the same. The example described in Q20 illustrates
how to detect errors using CRC method.
10. Reliability of CRC is better than that of simple parity and LRC. Justify this statement.
Ans: CRC is more reliable than simple parity and LRC, because it can detect single-bit errors, burst
errors, double errors and an odd number of errors. On the other hand, simple parity can detect only an
odd number of errors and is considered only 50% efficien as compared to CRC, whereas LRC can
detect only up to three errors. Therefore, both simple parity and LRC are considered less reliable than
CRC and cannot be implemented in all networks.
11. Write a short note on error correction.
Ans: In error correction, the receiver that has received corrupt data needs to discover the original
data, the exact number of error bits and their location. To enable error correction, encoder and decoder
are used at the sender’s and receiver’s ends, respectively. Figure 6.3 shows the structure of encoder and
decoder in error correction. Each m-bit dataword at the sender’s end is converted into an n-bit codeword
(n  >  m) with the help of a generator that applies some encoding on the dataword thereby adding r redun­
dant bits to the dataword. Thus, an n-bit codeword contains m bit of data and r redundant bits, that is,
n  =  m  +  r. The codeword is then transmitted to the receiver. At the receiver’s end, the checker examines
the redundant bits to detect and correct errors in the data. After this, the dataword (without redundant bits)
is passed to the receiver.

Sender Receiver
Encoder Decoder
m-bit dataword m-bit dataword
Correct
Codeword
Generator Checker

n-bit codeword n-bit codeword

Figure 6.3 Structure of Encoder and Decoder in Error Correction

CN_Chapter 06.indd 117 2/20/2012 6:24:05 PM

 118 Data Communications and Computer Networks

The two main methods for error correction are forward error correction (FEC) and ­retransmission.
FEC is the technique in which the receiver makes guesses to find out the actual message using the
­redundant bits. In this technique, no request can be made for the retransmission of the message and
thus, is useful only when the number of errors is small. In retransmission technique, whenever a
receiver detects that there is some error in the received data, it rejects the received data and asks the
sender to retransmit the data. The receiver continues to do so until it receives the data that it considers
free of errors.
12. Explain Hamming codes in detail.
Ans: Hamming code is a linear block code that was developed by a scientist named R.W. Hamming.
It is a single-bit error-correction technique. The basic idea is to insert parity bits in between the data bits
to be transmitted. The parity bits are placed at each 2n bit position where n = 0, 1, 2, and so on. That is,
firs parity bit is placed at firs (20 = 1) position, second parity is placed at second (21=2) position, third
parity bit is placed at fourth (22 = 4) position and so on.
There can be any number of bits (say, n) in the Hamming code including data bits (d) and parity
bits (r). For finding the total number of parity bits in the dataword so as to form Hamming code, the
condition, 2r  >  =  n  +  1, that is, 2r  >  =  d  +  r  +  1 must be satisfied. For example, if the dataword is of
four bits, then three parity bits will be required as r = 3 satisfie the condition 23 > = (4 + 3 + 1). A seven-
bit ­Hamming code is ­commonly used that comprises four data bits and three parity bits. Figure 6.4
shows the structure of ­seven-bit Hamming code.

7 6 5 4 3 2 1 Bit position

D7 D6 D5 P4 D3 P2 P1

Data bits Parity bits

Figure 6.4 Structure of Seven-bit Hamming Code

The value of parity bits can be set to either 1 or 0. The value of P1 is set to 0 or 1, such that the bits
at positions 1, 3, 5 and 7, that is, P1, D3, D5 and D7, make an even parity. The value of P2 is set to 0 or 1,
such that the bits at positions 2, 3, 6 and 7, that is, P2, D3, D6 and D7, together make an even parity.
Similarly, the value of P4 is set to 0 or 1, such that the bits at positions 4, 5, 6 and 7, that is, P4, D5, D6
and D7, together make an even parity.
After Hamming code has been generated at the sender’s side, the code is transmitted to the receiver. At
the receiver’s end, it is decoded back into the original data. The bits at positions (1, 3, 5 and 7), (2, 3, 6 and
7) and (4, 5, 6 and 7) are again checked for even parity. If each group of bits makes an even parity, then the
received codeword is free from errors; otherwise, the error exists. In case of an error, a three-bit number is
made from three parity bits (P4P2P1) and its decimal equivalent is determined, which gives the position of
erroneous bit. After detecting the position of error bit, it is corrected by inverting the error bit (refer Q21).
13. Apply bit stuffing to the given bit st eam 01101111111111111110010.
Ans: In bit stuffing the data link layer stuffs one zero after each group of fiv consecutive 1s. Thus,
the outgoing bit stream after bit stuffin will be:

CN_Chapter 06.indd 118 2/20/2012 6:24:06 PM

 Error Detection and Correction 119

01101111101111101111100010

Stuffed bits

14. Given a dataword 1010011. What codeword will be transmitted if:

(a) an even parity is used.
(b) an odd parity is used.
Ans: (a) As the number of 1s in the given dataword is four (that is, even), the parity bit is zero. Thus,
the codeword that will be transmitted is 01010011 (parity bit is at MSB position).
(b) As the number of 1s in the given dataword is four (that is, even), the parity bit is one. Thus,
the codeword that will be transmitted is 11010011.
15. If the transmitted codeword is 11010011 and the received codeword is 10111011, detect
the error (if any) in following cases.
(a) An even parity is used.
(b) An odd parity is used.
Ans: (a) It is clear from the transmitted and the received codewords that there are three errors in the
received codeword. However, in case of even parity, since the number of 1s in the received codeword is
six (that is, even), the receiver will not be able to detect errors. The received codeword will be assumed
as the correct word.
(b) Since the number of 1s in the received codeword is six (that is, even), there is some error in the
received codeword in case of odd parity.
16. Consider the following bit stream that is to be encoded using VRC, LRC and even ­parity.
What will be the transmitted data?
   1100001   1111001   1011001   0000101   0010101   0010101   1010001
Ans: The given bit stream is organized into rows and columns to compute the LRC and VRC bits,
as shown below:

bytes

bits 1 1 1 0 0 0 1 0 LRC bits

1 1 0 0 0 0 0 0

0 1 1 0 1 1 1 1

0 1 1 0 0 0 0 0

0 0 0 1 1 1 0 1

0 0 0 0 0 0 0 0

1 1 1 1 1 1 1 1

VRC bits 1 1 0 0 1 1 1 1

CN_Chapter 06.indd 119 2/20/2012 6:24:06 PM

 120 Data Communications and Computer Networks

Thus, the bytes that will be transmitted are:

  11000011   11110011   10110010   00001010   00101011   00101011 10100011
   00101011
17. Consider the following bit stream that has been encoded using VRC, LRC and even
­ arity. Locate the error if present.
p
    11000011   11110011   10100010   00001010   00101011   00101011   10100011
00101011
Ans: To locate the error, the received byte stream is organized into rows and columns as shown
below:

bytes

bits 1 1 1 0 0 0 1 0 LRC bits

1 1 0 0 0 0 0 0

0 1 1 0 1 1 1 1
Wrong
0 1 0 0 0 0 0 0
parity
0 0 0 1 1 1 0 1

0 0 0 0 0 0 0 0

1 1 1 1 1 1 1 1

VRC bits 1 1 0 0 1 1 1 1

Wrong parity

Since fourth LRC and third VRC bits make the parity of its respective row and column even, there is an
error in the received data stream. The erroneous bit will be present at the intersecting position of fourth
row and third column, that is, the fourth bit of third byte is incorrect.
18. Compute the checksum byte for the following datawords.
10110011 10101011 01011010 11010101
Ans: The checksum of the given datawords will be computed as shown below:

Carry 1 0100011 1011110 1 1110000

Carry of 10110011 01011110 10111000
MSB +10101011 +01011010 +11010101
ignored 01011110 10111000 10001101 Checksum byte

19. Given message is M(X) = X^5 + X^4 + X + 1 and the generator is G(X) = X^4 + X^3 + 1.
Compute CRC code.
Ans: Given dataword = 110011
  divisor = 11001

CN_Chapter 06.indd 120 2/20/2012 6:24:07 PM

 Error Detection and Correction 121

Since the degree of generator polynomial is four, four zeros will be appended to the given dataword.
Thus, the dividend will be 1100110000. Now, the CRC will be computed as shown below:

100001
11001 1100110000
11001
00001
00000
As MSB = 0,
Zero divisor 00010
00000
used
00100
00000
01000
00000
10000
11001
1001 CRC

The CRC code is obtained by appending CRC (that is, 1001) to the given dataword (that is, 110011).
Thus, the CRC code is 1100111001.
20. The codeword is received as 1100100101011. Check whether or not there are errors in the
received codeword, if the divisor is 10101.
Ans: To check codeword for errors, the received codeword is divided by the given divisor, that is,
10101 as shown below.

111110001
10101 1100100101011
10101
011000
10101
11010
10101
11111
10101
10100
10101
00011
00000
00110
00000
01101
00000
11011
10101
1110 Remainder (CRC)

Since the remainder obtained (CRC) is non-zero, the received codeword contains errors.
21. For the bit sequence 1001110:
(a) Calculate the number of parity bits that will be required to construct Hamming code.
(b) Construct the Hamming code.
Ans: (a) To calculate the number of parity bits, the following condition must be satisfied
2r > = d + r + 1

CN_Chapter 06.indd 121 2/20/2012 6:24:08 PM

 122 Data Communications and Computer Networks

As the number of data bits (d) is 7, so putting d = 7 in above equation, we get

2r > = 7 + r + 1
⇒2r > = 8 + r (I)
Putting r = 4 satisfie the equation (I) since 2 > 12. Thus, four parity bits are required with seven data
4

bits to construct the Hamming code and these parity bits will be inserted at first second, fourth and
eighth positions.
(b) The structure for Hamming code for the data 1001110 is shown below.
D11 D10 D9 P8 D7 D6 D5 P4 D3 P2 P1

1 0 0 --- 1 1 1 --- 0 --- ---

To be decided

As the bits D3D5D7D9D11 are 01101, thus, P1 = 1 to establish the even parity.
As the bits D3D6D7D10D11 are 01101, thus, P2 = 1 to establish the even parity.
As the bits D5D6D7 are 111, thus, P4 = 1 to establish the even parity.
As the bits D9D10D11 are 001, thus, P8 = 1 to establish the even parity.
Therefore, the Hamming code which will be transmitted to the receiver is:
D11 D10 D9 P8 D7 D6 D5 P4 D3 P2 P1
Complete
1 0 0 1 1 1 1 1 0 1 1
codeword

22. A seven-bit Hamming code is received as 1100101. What is the correct code?
Ans: The received codeword is:
D7 D6 D5 P4 D3 P2 P1
Received
1 1 0 0 1 0 1
codeword

As the bits P1D3D5D7 (that is, 1101) together make an odd parity, there is an error. Thus, P1 = 1.
As the bits P2D3D6D7 (that is, 0111) together make an odd parity, there is an error. Thus, P2 = 1.
As the bits P4D5D6D7 (that is, 0011) together make an even parity, there is no error. Thus, P4 = 0.
The erroneous bit position can be determined as follows:
P4 P2 P1
3-bit number, that
0 1 1
is, error word

The decimal equivalent of (P4P2P1), that is, (011) is 3. Hence, third bit in the received codeword
c­ ontains the error, so by inverting the third bit, the correct code can be obtained as shown below.
D7 D6 D5 P4 D3 P2 P1
Correct
1 1 0 0 0 0 1
codeword

Inverted bit

CN_Chapter 06.indd 122 2/20/2012 6:24:10 PM

 Error Detection and Correction 123

Multiple Choice Questions

1. Which of the following service is not pro- 6. Which of the following error-detection
vided by the data link layer? method involves the use of parity bits?
(a) Unacknowledged connectionless (a) LRC
(b) Unacknowledged connection-oriented (b) VRC
(c) Acknowledged connectionless (c) Checksum
(d) Acknowledged connection-oriented (d) Both (a) and (b)
2. The process of breaking stream of bits into 7. VRC parity bit is associated with ________.
frames is known as __________. (a) Rows
(a) Bit stuffin (b) Columns
(b) Byte stuffin (c) Both (a) and (b)
(c) Framing (d) None of these
(d) Character count
8. What is the formula to calculate the number
3. What is the process of adding one extra 0 of parity bits (r) required to be inserted in
whenever there are fiv consecutive 1s in the dataword (d bits) to construct the Hamming
data, so that the receiver does not mistake code?
the data about a flag (a) 2r >  = d + r
(a) Bit stuffing (b) Bit padding (b) 2r >  = d
(c) Byte stuffing (d) Byte padding (c) 2d > = d + r
4. In single-bit error, _____ bit is altered during (d) 2r >  = d + r + 1
the transmission. 9. Hamming code can detect up to ________ if
(a) Zero (b) One minimum Hamming distance is 3.
(c) Two (d) None of these (a) 2 (b) 3
5. CRC computation is based on (c) 4 (d) 5
(a) OR operation 10. If a codeword in Hamming code is of seven
(b) AND operation bits, then how many parity bits it contains?
(c) XOR operation (a) 2 (b) 3
(d) NOR operation (c) 7 (d) 9

Answers
1. (b) 2. (c) 3. (a) 4. (b) 5. (c) 6. (d) 7. (b) 8. (d) 9. (a) 10. (b)

CN_Chapter 06.indd 123 2/20/2012 6:24:10 PM

 7
Flow and Error Control

1. Differentiate between flo control and error control.

Ans: Flow control and error control are the most important functions of data link layer. Flow control
refers to a set of procedures that allows the communicating devices having different speed and memory
characteristics to communicate with each other in an efficient manner. It ensures that a slow receiver is
not being flooded by the fast sender. Thus, flow control restricts the amount of data that can be sent from
the sender before waiting to receive an acknowledgement (ACK) from the receiver. Each receiver has
a block of memory known as buffer for reserving the incoming data before they can get processed. The
flow control mechanism can be achieved in two ways. First, if the receiver’s end buffer begins to fill up,
then the receiver should tell the sender either to stop the transmission completely or send fewer frames.
Second, the receiver must acknowledge the received frames by sending ACK frame to the sender. For
the frames, which are damaged or lost during transmission, the receiver should send negative acknowl-
edgement (NACK) to the sender by transmitting NACK frame. Two basic protocols that can be used to
control the flow of data include stop-and-wait and sliding window protocol
Error control, on the other hand, refers to a set of procedures that ensure that all the frames have
eventually been received at the destination. If any frames are lost or damaged during transmission, the
error control mechanism allows the receiver to inform the sender about it and coordinates the retrans-
mission of those frames by the sender. In data link layer, the error control is based on an automatic
repeat request (ARQ) in which whenever any error is detected, the lost or damaged frames are retrans-
mitted. The basic protocols that can be used for error control include stop-and-wait ARQ, Go-back-N
ARQ and selective-reject ARQ.
2. Explain the working of stop-and-wait flo control protocol.
Ans: Stop-and-Wait is the simplest protocol used to control the flow of data between a sender and a
receiver. As the name implies, the sender sends one frame, then stops and waits for the ACK frame from
the receiver to confirm the receipt of the frame. Only after getting an ACK frame from the receiver, the
sender sends the next frame (Figure 7.1). In stop-and-wait flow control, the receiver can stop the flow
of the data by not sending the ACK frame, thereby keeping the sender in waiting state.

CN_Chapter 07.indd 124 2/20/2012 6:29:08 PM

 Flow and Error Control 125

Sender Receiver
Frame

K
AC
Frame

Sender Receivers
cannot holds the
send next ACK
Frame K
AC

Frame

Figure 7.1 Stop-and-Wait Flow Control

Mostly, the sender breaks the large block of data into smaller blocks called frames because sending
the message in the larger blocks may increase the transmission time and errors are more likely to occur.
In contrast, when message is transmitted in smaller frames, errors may be detected earlier and a smaller
amount of data needs to be transmitted. Another reason of breaking into frames may be the limited ­buffer
size of the receiver due to which entire large frame cannot be processed at a single time, thus causing
long delays. But the disadvantage of sending small frames for a single message is that stop-and-wait
method cannot be used effectively because in this protocol only one frame can be sent at a time (either
from sender to receiver or vice versa) which results in underutilization of the communication link.

Link Utilization
To understand the link utilization in stop-and-wait flow control, consider tframe that denotes the time
taken in transmitting all bits of a frame and tprop denotes the propagation time from the sender to
­receiver. Now, the propagation delay (a) which is defined as the time taken by one bit to reach from the
sender to receiver can be expressed as
a = tprop/tframe
If a < 1 then tprop< tframe, which implies that the frame is longer enough such that the leading bits of
frame reach the receiver before the entire frame is transmitted. That is, link is inefficiently utilized. On
the other hand, if a > 1, then tprop > tframe, which implies that the sender has transmitted the entire
frame and the leading bits of that frame have still not arrived at the receiver. This results in underutiliza-
tion of the link.
3. Derive the expression for the maximum possible utilization of link in stop-and-wait
flo control.
Ans: Suppose a long message is to be transmitted between two stations. The message is divided
into a number of frames F1, F2, F3,…, Fn. The communication between stations proceeds in such a
manner that the frames (from station 1) and the ACKs (from station 2) are sent alternatively. That is,
initially sender (station 1) sends F1, then receiver (station 2) sends an ACK and then sender sends F2
then receiver sends an ACK and so on.

CN_Chapter 07.indd 125 2/20/2012 6:29:08 PM

 126 Data Communications and Computer Networks

Now, if Tf denotes the time taken in sending one frame and receiving an ACK, then the total time (T)
taken to send the message can be expressed as
T = nTF
where n is the total number of frames and TF can be expressed as
Tf = tprop + tframe + tproc + tprop + tack + tproc
where
tprop is the propagation time from station 1 to station 2.
tframe is the time taken in transmitting all bits of the frame.
tproc is the processing time by each station to respond to an incoming event.
tack is the time to transmit an ACK.
tprop + tframe + tproc is the total time taken by the station 1 to send the frame and getting it
­processed at the receiver’s end and tprop + tack + tproc is the total time taken by the station 2 to send
the ACK frame and getting it processed by the sender.
Therefore,      T = n(tprop + tframe + tproc + tprop + tack + tproc)
Assuming that the processing time is negligible and the ACK frame is very small as compared to the
data frame, we can neglect tproc and tack. Thus, T can be written as
T = n(2tprop + tframe)
As n*tframe is the time that is actually involved in sending the frames and rest is considered as
­overhead, the link utilization (U) may be given as
n*tframe tframe
U= =
n*(2tprop +tframe) (2tprop +tframe)
⇒ U = 1/(1 + 2a)
where a = tprop/tframe = propagation delay.
This is the maximum utilization of the link, which can be achieved in situations where propagation
delay (a) is constant and fixed-length frames are often sent. However, due to overhead bits contained in
a frame, the actual utilization of the link is usually lower than this.
4. Derive the expression for the propagation delay showing that it is proportional to the bit
length of the channel.
Ans: We know, propagation delay (a) = propagation time/transmission time.
Now, propagation time              = d/V
where d is the distance of the link and V is velocity of propagation. For guided media, V = 3*108 m/s
and for unguided transmission, V = 0.67 *3*108 (approximately).
Transmission time               = L/R
where L is the length of the frame in bits and R is the data rate of transmission. Therefore,
d/V Rd
a= =
L/R VL
Since, we know that the bit length of channel (B) = Rd/V
⇒ a = B/L
Hence,                  a ∝ B

CN_Chapter 07.indd 126 2/20/2012 6:29:09 PM

 Flow and Error Control 127

5. Explain the mechanism of sliding window flo control.

Ans: The sliding window flow control technique was introduced to remove the disadvantage of the
stop-and-wait protocol in which only one frame can be transmitted at a time, thus, reducing the efficiency
of the link. The efficiency can be greatly improved by using the sliding window flow control protocol as
in this technique multiple frames can be sent at the same time.
The mechanism of sliding window protocol is as follows. Let there be two stations, A and B, ­connected
to each other via a full-duplex link. Station B (destination) has buffer space that can accommodate N
frames, that is, B can accept N frames at a time, and station A (source) is allowed to send N frames to
station B without waiting for any ACKs from B. Each frame is assigned a sequence number that helps to
keep track of the frames which have been acknowledged and which have not yet been received. ­Station B
acknowledges each frame by sending a receive-ready (RR) frame that contains the sequence number of
the next expected frame. An RR frame from station B indicates to the station A that B is ready to ­receive
next N frames starting with the frame having the sequence number specified within the RR frame. In-
stead of sending ACK to each frame, the station B can also acknowledge many frames at the same time
by sending a single RR frame, that is, acknowledgements are cumulative. For example as shown in
­Figure 7.2, after receiving frame 3 (F3), station B can hold to send the ACK of the frames 0, 1, 2 and 3
until frame 4 arrives. After frame 3 has arrived, B can send an ACK with the sequence number 4 indicat-
ing to A that frames 0, 1, 2 and 3 have been received and now it is expecting to receive frame 4.
At any instant, B can also ask A to stop the flow of frames by sending a receive-not-ready (RNR)
message which acknowledges the former frames and indicates the sender not to transmit more frames.
For example, an RNR 3 message from B means that it has received all frames up to number 2 and cannot
accept any further frames. Later, B must send a normal ACK to A to restart the transmission process.
To keep track of the frames to be sent or received, the sender and receiver maintain a list of ­sequence
numbers called the sender’s window and receiver’s window (indicated by the boxes in Figure 7.2),
­respectively. The sender’s window indicates the sequence numbers of the frames that the sender can send
while the receiver’s window indicates the sequence numbers of the frames that the receiver is expecting
to receive. For example, in Figure 7.2, the initial window of A and B indicates that A can transmit frames
Source station A Destination station B
0123456701234567 0123456701234567
F0
F1
F2
F3
0123456701234567
0123456701234567

RR4 0123456701234567
0123456701234567
F4
F5
F6
F7 0123456701234567
0123456701234567
RR5

0123456701234567 0123456701234567

Figure 7.2 Sliding Window Flow Control

CN_Chapter 07.indd 127 2/20/2012 6:29:09 PM

 128 Data Communications and Computer Networks

numbered from 0 to 6 without having to wait for any ACK from B and B can receive frames numbered from
0 to 6 from A, respectively. As the frames are sent or received at the sender’s and receiver’s end respec-
tively, the window size shrinks. For example, after transmitting frames F0, F1, F2 and F3, the size of A’s
window shrinks to three frames and after receiving frames F0, F1, F2 and F3, the size of B’s window
shrinks to three frames (Figure 7.2). Similarly, the window size expands as the ACKs are received or sent
by the sender and receiver, respectively. Therefore, the scheme is named as sliding window flow control.
Notice that at A’s end in Figure 7.2, the frames between the vertical line and the box indicate those
frames that have been sent but not yet acknowledged by B. Similarly, at B’s end, the frames between the
vertical line and the box indicate those frames which have been received but not yet acknowledged.
As each sequence number used occupies some field in the frame, the range of sequence number that
can be used gets limited. In general, for a k-bit field, the range of sequence number will be from 0 to 2k - 1
and frames will be numbered as modulo 2k. That is, after sequence number 2k - 1, the next number
again will be 0. For example, for a 3-bit sequence number, the frames will be numbered sequentially
from 0 to 7 and the maximum window size will be 7 (23 – 1). Thus, seven frames can be accommodated
in the sender’s window and the receiver can also receive a maximum of seven frames at a time.

Link Utilization
In sliding window flow control, the link utilization (U) depends on the window size (N) and propagation
delay (a) and is expressed as
1, if N > = 2a + 1
U = 
N/(2a + 1), if N < 2a + 1
6. Defin piggybacking. What is its usefulness?
Ans: Usually, the link used between the communicating devices is full duplex and both the sender
and receiver need to exchange data. In addition to data, both sender and receiver send ACK for the
received data to each other. To improve the efficiency in a two-way transmission, a technique called
piggybacking can be employed, which permits an ACK from either device (sender or receiver) to be
assembled with the next outgoing data frame from that device instead of sending a separate ACK frame.
In case a device has only ACK but no data to send, it needs to use a separate ACK frame. However, if
the device has to send the ACK followed by data, it can temporarily withhold the ACK and later, include
the ACK in the outgoing data frame.
By temporarily delaying the outgoing ACKs and sending them along with data frames, piggybacking
results in better use of the available channel bandwidth. Moreover, by using piggybacking technique,
less number of frames needs to be sent which further implies less number of interrupts as well as less
number of buffers required at the receiver’s end.
7. What is meant by lost frame and damaged frame? List the common techniques of error
control method.
Ans: A frame that fails to arrive at the destination node due to noise burst is called lost frame. In this case,
the receiver does not know that the frame has been transmitted by the sender. A damaged frame is the frame
in which some bits have been changed or altered during the transmission. In case of lost frame, no frame is
received by the receiver while in case of damaged frame, a frame is received but not in the exact order.
8. Explain the mechanism of stop-and-wait ARQ.
Ans: Stop-and-Wait ARQ is the extended form of stop-and-wait flow control technique where
data frames are retransmitted if the sender does not receive any ACK for the frame. In this method,

CN_Chapter 07.indd 128 2/20/2012 6:29:10 PM

 Flow and Error Control 129

the sender transmits a single frame and waits for the ACK from the receiver before sending the next
frame. The sender cannot transmit a new frame until the ACK is received from the other side. Since
either the frame or the ACK can be transmitted at a time, this technique is also known as one-bit sliding
window protocol.
Notice that the ACK will not be received by the sender if the frame has been lost during the
­transmission, that is, it has not been received by the receiver or if the receiver has received the frame
free of errors but its ACK has been lost or delayed during the transmission. In both cases, the sender
needs to retransmit the frame. For this, the sender maintains a copy of sent data frame in its buffer so
that they can be retransmitted whenever required.
To identify the frames to be retransmitted, sequence numbers are attached to the consecutive data
frames and ACKs. As only one frame or one ACK can be on the transmission link at a given time, only
two numbers 0 and 1 are needed to address the data frames and ACK. Data frames are alternatively
numbered 0 and 1, that is, first data frame will be marked as F0, second will be F1, third again will be
marked as F0 and so on. The numbers that are attached with ACK indicates the sender to send that num-
bered data frame. For example, if ACK1 is sent by the receiver, then the sender comes to know that F0
has been received successfully and now, it has to send F1 to the receiver. This process helps the receiver
to discard the duplicate frames; however, ACKs need to be sent by the receiver for the duplicate frames
also so as to keep the sender in synchronization.

Lost or Damaged Frames

To handle the problem of lost or damaged frames, the sender maintains a timer. Whenever the sender
sends a frame, the timer starts. In case the frame is lost during the transmission, the receiver does not
send any ACK to that frame. If the sender does not receive any ACK before the time-out (timer ­expires),
it retransmits the same frame after the time-out. On the other hand, if the frame is received by the
receiver with errors in it, then the receiver simply discards the frame and sends NACK to the sender
indicating the sender to retransmit the same frame.

Lost or Delayed ACKs

Now, consider the case where the receiver receives the frame with no errors and therefore, sends an
ACK to the sender but the ACK frame is lost or delayed. In both scenarios, the sender retransmits the
same frame after the time-out. This results in the duplication of frames at the receiver’s end. To identify
the duplicate frames, the receiver maintains a buffer at its end hold those frames for which ACK has
been sent.

Link Utilization
The link utilization (U) in the stop-and-wait ARQ is given as
U = (1 - p)/(1 + 2a)
where
p is the probability of receiving a bit in error
a is the propagation delay which is also equal to (tprop*R)/L where tprop, R and L denotes
­propagation time, data rate of the link and number of bits in the frame, respectively.
9. What are the two types of sliding window ARQ error control?
Ans: To provide the efficient utilization of the link, the sliding window error control technique is
­adopted. The sliding window ARQ is also known as continuous ARQ because it allows the sender

CN_Chapter 07.indd 129 2/20/2012 6:29:10 PM

 130 Data Communications and Computer Networks

to send many frames continuously without waiting for an ACK. There are two techniques, namely,
Go-back-N ARQ and selective-reject ARQ that are based on the concept of sliding window.
10. Explain the mechanism of Go-back-N ARQ technique.
Ans: Go-back-N ARQ is the most commonly used form of error control that is based on the sliding
window flow control. In this technique, the sender sends the data frames depending on its window size
but at the receiving end window size is always one. That is, the sender can send many frames without
waiting for an ACK from the receiver but the receiver can receive only one frame at a time. For each
frame that is received without any error, the receiver acknowledges the frames by sending RR message.
However, when it detects error in the frame, it discards that frame and sends a negative acknowledgment
reject (REJ) to the sender. The receiver continues to discard the further transmitted frames until the
frame in error is received correctly. When the sender receives an REJ, it retransmits the frame in error
plus all the succeeding frames that it has sent already.
Figure 7.3 shows the mechanism of Go-back-N ARQ where the sender’s window size is seven. That
is, the sender can transmit frames F0-F6 to the receiver without waiting for an ACK.

Receiver
Sender
0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7
F0

F0
F1
RR1 ∗
0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7
F2

1
F3 REJ F2 and F3
0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 discarded
F1
F1
RR2
F2
0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7
F2 F1, F2 and F3
retransmitted
3 F3
RR

0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 F3

Figure 7.3 Go-Back-N ARQ Mechanism

Lost or Damaged Frames

As shown in Figure 7.4, initially, the sender sends F0, which is received by the receiver. In response the
receiver sends RR1 frame to the sender and slides its window to F1–F7. By the time the sender receives
RR1, it has transmitted F1 which is either lost or damaged during the transmission. On receiving RR1,
the sender transmits F2 and shrinks its window to F3–F7. Though F1 and F2 have been transmitted,

CN_Chapter 07.indd 130 2/20/2012 6:29:10 PM

 Flow and Error Control 131

Sender Receiver

0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7
F0

F1
RR1 ∗
F2 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7
0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7

J1
F3 SRE

F1

RR4
F4
0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 F5

0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7
6
RR

Figure 7.4 Selective-reject ARQ Mechanism

the sender stores them in the buffer until they are acknowledged by the receiver. When the receiver
receives F2, it accepts F2 and stores in its buffer. It also comes to know that F1 has not arrived. Thus, it
sends SREJ1 to the sender to indicate the retransmission of F1. Now, the receiver cannot send RR frame
until it receives F1 because sending RR3 would indicate that all frames up to F2 have been received.
By the time the sender receives SREJ1, it has transmitted F3 which is accepted by the receiver and
kept in the buffer. When sender receives SREJ1, it retransmits frame F1. On receiving F1, the receiver
replies with RR4 indicating to the sender that all frames up to F3 have been received successfully. This
procedure continues until all the frames have been transmitted.

Lost or Delayed RR or REJ

To understand how lost or delayed RR frame is handled, suppose that the receiver receives F2 frame
error free and thus, responds to sender with RR3 which is either lost or delayed. As the sender does not
receive any acknowledgement before the time-out, it retransmits F2 after the timer expires. Now, as the
receiver has already ­received F2, it simply ignores F2 thus, avoiding the duplication. However, as the
acknowledgements in the Go-back-N method are cumulative, it might be possible that before the timer
associated with frame F2 expires, the sender receives next RR frame (that is, RR4) which acknowledges
all the previous data frames. In such a case, the sender remains unaware of the fact that RR3 has lost in
the network and the operation remain unaffected.
To understand how lost or delayed REJ is handled, suppose that the receiver receives F4 frame with
some errors in it. Thus, the receiver discards F4 and transmits REJ4 to the sender, which is either lost

CN_Chapter 07.indd 131 2/20/2012 6:29:11 PM

 132 Data Communications and Computer Networks

or delayed. In both cases, the sender does not receive any further acknowledgements because the
receiver will deny accepting any frame other than F4. Thus, the sender has to retransmit F4.

Link Utilization
The link utilization (U) in Go-back-N ARQ method can be considered in the following two situations:
1. When N > 2a + 1, U = (1 - p)/(1 + 2ap) and
2. When N < 2a + 1, U = N(1 - p)/((2a + 1)*(1 – p + Np)).
11. Explain the mechanism of selective-reject ARQ technique.
Ans: Selective-reject ARQ (also known as selective-repeat ARQ) method offers an improvement
over Go-back-N by retransmitting only the lost or damaged frame but not all the frames succeeding it. In
this method, usually the size of receiver’s window is same as that of sender’s window and the damaged
frames are acknowledged by the receiver by transmitting selective reject (SREJ) frame to the sender. As
this method retransmits only lost or SREJ frames, a large buffer needs to be maintained at the receiver’s
end so that if some frame is not received in proper order then the receiver can accept the rest incoming
frames and buffer them until that valid frame has been received. When all the data frames are reached, the
receiver arranges all frames in proper sequence and sends them to the next upper layer.
Figure 7.4 shows the selective-reject ARQ mechanism where the window size of both sender and
receiver is seven. That is, the sender can transmit frames F0-F6 to the receiver without waiting for an
acknowledgement and the r­ eceiver can also receive frames F0-F6.

Lost or Damaged Frame

As shown in Figure 7.3, initially, the sender sends F0. As the receiver's window size is 1, the receiver ­accepts
F0 and replies with RR1 to the sender and slides its window to F1. By the time the sender receives RR1, it
has transmitted F1 which gets lost in the transmission. At the time the sender receives RR1, it ­transmits F2
and slides its window from F3 to F7. Though F1 and F2 have been transmitted, the sender stores them in
the buffer until they are acknowledged by the receiver. On receiving F2, the receiver finds that F1 is still
not received. Therefore, the receiver discards F2 and sends REJ1 to the sender. Before the sender receives
REJ1, it transmits F3 which is also discarded by the receiver. When the sender receives REJ1, it needs to
retransmit F1, F2 and F3. First it sends F1, which is accepted by the receiver and the receiver responds
with RR2 to the sender. This procedure continues until all the frames have been transmitted.
In case the receiver receives a damaged frame, it simply discards the frame and sends REJ to the
sender indicating the retransmission of that frame.

Lost or Delayed RR
To understand how lost or delayed RR is handled, suppose that the receiver receives frame F2 and in
­response sends RR3 which either gets lost or gets delayed. In both cases, when RR4 reaches the sender, the
sender assumes it as a cumulative ACK of frames up to F3 and thus, the operation remains unaffected.

Lost or Delayed SREJ

To understand how lost or delayed SREJ is handled, suppose that the receiver receives F4 frame with some
errors and thus, sends SREJ4 to the sender, which either gets lost or delayed. To handle this, the receiver
maintains a timer at its end, which is started at the time the receiver sends a SREJ frame. If the receiver does
not receive the frame F4 before the time-out, it retransmits SREJ4 to the sender after the timer ­expires.
Thus, eventually, the sender receives SREJ4 and retransmits F4.

CN_Chapter 07.indd 132 2/20/2012 6:29:11 PM

 Flow and Error Control 133

Link Utilization
The link utilization (U) in selective-reject ARQ can be considered in the following two situations:
1. when N ≥ 2a + 1, U = 1 - p and
2. when N < 2a + 1, U = (N*(1 - p))/(2a + 1).

12. What do you mean by a bit-oriented and byte-oriented protocol?

Ans: In a bit-oriented protocol, the data received from the upper layers whether text, images,
audio or video is carried in the form of bits; the data field of each frame comprises a sequence of bits.
To distinguish between the consecutive frames, a special eight-bit pattern flag 01111110 is used in the
beginning and at the end of each frame. In case the flag bit pattern appears in the data field of frame,
bit stuffin is used at the sender’s end. After each five consecutive 1s in data, an extra bit 0 (a single
bit) is inserted to prevent the bit pattern from looking like a flag. The extra bit stuffed by the sender is
then destuffed at the receiver’s end. An example of bit-oriented protocol is high-level data link control
(HDLC) protocol.
On the other hand, in a byte-oriented (also called character-oriented) protocol, the data received
from upper layers is carried in the form of eight-bit characters (that is, bytes) from a coding system such
as ASCII. Each frame comprises a sequence of bytes. To distinguish between the consecutive frames, a
1-byte (instead of single bit) flag is used in the beginning and at the end of each frame; the flag is usu-
ally composed of special characters depending on the protocol being used. In case the flag byte appears
in the data field of frame, byte stuffin (also called character stuffin ) is used at the sender’s end. An
extra byte, called escape character (ESC), is stuffed in the data when there appears a character with
the same pattern as flag. The extra byte stuffed by the sender is then destuffed at the receiver’s end. An
example of byte-oriented protocol is point-to-point (PPP) protocol.

13. Explain in brief HDLC protocol.

Ans: The HDLC is the most commonly used data link protocol developed by ISO. It supports
both half-duplex and full-duplex communication lines and switched or non-switched channels. That is
why it has been widely implemented throughout the world. It is a bit-oriented protocol that is used to
­provide communication in point-to-point and multipoint networks. Further, HDLC protocol implements
the ARQ mechanism and satisfies a variety of applications. In addition, HDLC defines three types of
­stations, two types of link configuration and three modes of data transfer to work in various possible
network configurations

Types of stations
Three types of stations defined by HDLC are as follows:
‰‰ Primary Station: This station has the responsibility of controlling the operation of the link. The
frames that are sent by a primary station are called commands.
‰‰ Secondary Station: This station works under the control of the primary station and the frames that
are sent by a secondary station are called responses. Each secondary station on the line is linked with
the primary station via a separate logical link. During communication between primary and second-
ary stations, the primary station is responsible for establishing, managing and terminating the link.
‰‰ Combined Station: This station can act as both a primary and a secondary station and thus, can
send both commands and responses. It does not rely on any other stations on the network and thus,
no other station can control a combined station.

CN_Chapter 07.indd 133 2/20/2012 6:29:11 PM

 134 Data Communications and Computer Networks

Types of Link Configurations

For three stations, two link configurations supported by HDLC are as follows
‰‰ Unbalanced Configuration This type of configuration consists of one primary and one or more
secondary stations. Both full-duplex and half-duplex transmissions are possible in unbalanced
­configuration and the primary station polls each secondary station for input
‰‰ Balanced Configuration This type of configuration consists of two or more combined stations
and supports both full-duplex and half-duplex transmissions.

Modes of Data Transfer

Three types of data transfer modes defined by HDLC are as follows
‰‰ Normal Response Mode (NRM): This mode of transfer is used with an unbalanced configur -
tion. In this mode, the data transfer is initiated by the primary station to a secondary station. The
­secondary station is allowed to transmit only in a response to a command from primary station and
not by ­itself. In other words, the secondary station must get an explicit permission from the primary
­station to send responses.
‰‰ Asynchronous Balanced Mode (ABM): This mode of transfer is used with a balanced
­configuration. In this mode, any of the combined stations can start data transmission without the
need of any ­permission from other combined stations. This mode is used widely as it results in an
efficient utilization of full-duplex point-to-point link
‰‰ Asynchronous Response Mode (ARM): This mode of transfer is used with an unbalanced confi -
uration. In this mode, secondary station may initiate data transfer without needing the permission
from the primary station. Despite of this, the primary station is still responsible for link establish-
ment, link disconnection and error recovery. This mode helps to reduce overhead on the link, as no
frames need to be sent to provide explicit permission to secondary stations. However, this mode is
used only in rare situations where it is necessary for a secondary station to initiate transfer.

14. Explain the frame structure of HDLC.

Ans: The HDLC defines three types of frames, namely, information frames (I-frames), supervisory
frames (S-frames) and unnumbered frames (U-frames) to support all modes of data transfer and link
configurations. The I-frames are used to carry user data. The ACK of the received frames can also be
piggybacked on I-frames. The S-frames do not carry any user data and are used to transport only con-
trol information, which cannot be piggybacked on I-frames such as RR, REJ, RNR and SREJ frames.
The U-frames are reserved for carrying information regarding management of the link, mode setting or
other control information.

Frame Structure
The frame structure of I-frames and U-frames is exactly similar (Figure 7.5) and consists of six fields:
starting flag field, an address field, a control field, an information field, a frame check sequence (FCS)
field and an ending flag. Since S-frame cannot carry user data, it consists of rest of the five fields
­excluding the user information field. When multiple frames are transmitted between the communicating
stations, then the ending flag field of one frame may act as the starting flag field of the next frame and
each frame is transmitted from left to right.

CN_Chapter 07.indd 134 2/20/2012 6:29:11 PM

 Flow and Error Control 135

User
Flag Address Control FCS Flag I-frame
information

Flag Address Control FCS Flag S-frame

Management
Flag Address Control FCS Flag U-frame
information

Figure 7.5 Frame Structures in HDLC

The description of the fields in di ferent types of frames in HDLC is as follows:
‰‰ Flag: It is an eight-bit long field having the unique bit pattern of 01111110. This pattern is used to
identify the starting and ending of a frame and the pattern can also be inserted whenever there is
some idle time between the consecutive frames. This field serves as a synchronization pattern for
the secondary station.
‰‰ Address: It is usually an eight-bit (1 byte) long field but can be extended up to several bytes
­depending on the need of the network. This field contains the address of the secondary station
­irrespective of whether the frame is transmitted by the primary station or the secondary station.
If the frame is transmitted by the primary station, then the address field contains a to address and
thus, indicates that the frame is a command. If the frame is transmitted by the secondary station
then the ­address field contains a from address and thus, indicates that frame is a response. The last
bit of ­address field always ends with one whether the address field is of one byte or of many bytes.
­However, if the address field is of more than one byte then the last bit of every intermediate byte
ends with a zero indicating to the receiver that there are more address bytes to come.
‰‰ Control: It is an eight-bit long field that can be extended up to 16 bits. This field identifies the
type of HDLC frame and is used for carrying the sequence number, acknowledgements request for
retransmission and other control information. The implementation of bits in control field depends
on the frame type (discussed in Q15).
‰‰ Information: It is a variable-length field that contains the user’s data bits in I-frame. However, in
U-frames, the information field is used to carry system management information.
‰‰ FCS: It is a 16-bit long field that contains cyclic redundancy check (CRC) for detecting errors in
the address, control and information fields. This field can be extended up to 32 bits
15. Explain the format of control fiel in all types of frames in HDLC.
Ans: The HDLC defines three types of the frames I, S and U and the control field of each frames
differs from each other. This field comprises various subfields (Figure 7.6) that define the functions as-
sociated with the frame.

Control Field for I-frame

The first bit of the control field specifies the type of the frame. For an I-frame, this bit is always set
to zero. The next three bits comprise N(S) subfield that defines the sequence number of the frame. As
three bits are defined for sequence number, the frames can be numbered from 0 to 7. However, if the

CN_Chapter 07.indd 135 2/20/2012 6:29:12 PM

 136 Data Communications and Computer Networks

0 P/F I-frame

N(S) N(R)

1 0 P/F S-frame

Code N(R)

1 1 P/F U-frame

Code Code
Figure 7.6 Control Field Format for HDLC Frame Types

extended frame format in which the control field is of 16-bits is used then N(S) field becomes larger.
The next single bit after N(S) is the P/F bit where P stands for poll and F stands for fina . If P/F bit is
set to one, it means poll, that is, frame has been sent by a primary station to a secondary station. On
the other hand, if P/F bit is zero, it means final, that is, frame has been sent by a secondary station to
the ­primary station. The last three bits in control field comprise the N(R) subfield that defines an ACK
number ­corresponding to received frame, which has been piggybacked on the I-frame.

Control Field for S-frame

In an S-frame, the first two bits of the control field define the type of the frame. For an S-frame, these
bits are always set to 10. Since an S-frame carries control information, the next two bits of the control
field, called code, indicate the type of control information being carried in an S-frame. There can be
four possible combinations of two bits and accordingly, four types of ACKs that an S-frame may carry.
These types are as follows:
‰‰ Receive Ready (RR): If the value of code is 00, then the frame is an RR S-frame. This kind of
frame sends the ACK to the sender for a frame or a group of frames that have been received.
‰‰ Receive not Ready (RNR): If the value of code is 10, then the frame is an RNR S-frame. This
kind of frame is similar to RR frame, however, includes some additional features. In addition to
acknowledging the frames, it also asks the sender to slow down or not to send any more frames as
the receiver is busy.
‰‰ Reject (REJ): If the value of the code is 01, then the frame is an REJ S-frame. This kind of frame
is used in Go-back-N ARQ to inform the sender about the loss or damage of the received frame
before the sender’s timer expires.
‰‰ Selective-Reject (SREJ): If the value of the code is 11, then the frame is an SREJ S-frame. This
kind of frame is used in selective-repeat ARQ to inform the sender about the loss or damage of a
specified frame before the sende ’s timer expires.
The next single bit after code bits is P/F bit having the same purpose as in I-frame and the last three
bits comprise N(R) subfield that defines positive ACK number or NACK depending on the type
of S-frame.

CN_Chapter 07.indd 136 2/20/2012 6:29:12 PM

 Flow and Error Control 137

Control Field for U-Frame

In U-frame, the first two bits specify the type of the frame. For a U-frame, these bits are always set to 11.
The next two bits before the P/F bit and the last three bits following the P/F bit correspond to the code
bits in the control field. These five code bits together can create up to 32 di ferent types of U-frames.
16. Describe the frame structure of SDLC. How HDLC frame format is superior to SDLC
frame format?
Ans: Synchronous Data Link Control (SDLC) is a data link layer protocol that is widely used by
IBM. It is a subset of HDLC protocol with some minor additional features. The frame structure of SDLC
(Figure 7.7) is somewhat similar when compared with the HDLC frame structure. The description of
various fields used in SDLC is as follows:
‰‰ Flag: It is an eight-bit long field having the unique bit pattern of 01111110. The same bit pattern
is also used for the last field in the frame. The flag field in the beginning of the frame is used for
synchronization while the flag field at the end is used to indicate the end of fram
‰‰ Address: It is an eight-bit long field that is used to carry the address of the secondary station
‰‰ Control: It is an eight-bit long field that is used to carry the sequence number of frames. The
implementation of control field depends on the frame type and can be extended up to 32 bits
‰‰ Data: It is a variable-length field that is used to carry use ’s data.
‰‰ Error Checking Field (ECF): It is a 16-bit long field that is used for the error control
The HDLC frame format is superior to SDLC frame format because in HDLC, FCS field can be of
32-bits but in SDLC, it is limited to 16-bit only.

User’s
Flag Address Control ECF Flag
data

Figure 7.7 SDLC Frame Format

17. Describe the services provided by the PPP protocol. Also, list some services which PPP
does not provide.
Ans: The PPP protocol operates over a link, which directly connects two nodes, one on each end
of the link. Thus, PPP is the most commonly used data link layer protocol for PPP access. This protocol
is used by several users who need to connect their home computers with the server of ISP. The PPP
protocol can control and manage the transfer of the data, which a modem was unable to provide. That is
why PPP has become the popular choice among users accessing Internet over a dial-up connection.
PPP provides a wide variety of services, some of which are as follows:
‰‰ It can operate over various types of links such as serial/parallel, synchronous/asynchronous, etc.
‰‰ It defines the format of frames that are exchanged between the communicating nodes
‰‰ It defines how link should be established between two nodes and how data should be exchanged
‰‰ It defines authentication mechanism to be used by the communicating devices to authenticate each
other.
‰‰ It provides a mechanism for the communicating nodes to learn the network layer address of each
other.
‰‰ It supports many network layer protocols such as IP and DECnet.
‰‰ It can detect the errors in the received frames.

CN_Chapter 07.indd 137 2/20/2012 6:29:12 PM

 138 Data Communications and Computer Networks

In order to keep PPP simple, some services have not been implemented in PPP. These services are
as follows:
‰‰ It does not concern about flow control. However, it provides error control to some extent. It enables
to detect errors; however, the errors cannot be corrected.
‰‰ It operates only where there is single sender and single receiver. Thus, cannot be used over multi-
point links.
‰‰ It does not support frame sequencing, thus, frames are delivered to the destination not necessarily
in the same order in which they were sent by the sender.
18. Explain the frame format of the PPP.
Ans: The PPP is a byte-oriented protocol and uses an HDLC like frame format (Figure 7.8). The
description of each field in PP frame is as follows:
‰‰ Flag: It is a one-byte long field having the unique pattern of 01111110. This unique pattern is in-
serted at the start and end of each PPP frame. Though the flag field in PPP uses the same bit pattern
as that of HDLC, the flag is treated as a byte in PP .
‰‰ Address: It is a one-byte long field that is set to a constant value of 11111111, which means a broad-
cast address. This byte can be omitted upon an agreement between two parties during negotiation.
‰‰ Control: It is a one-byte long field that is set to a constant value of 00000011, which indicates an
unnumbered frame. This implies that PPP does not use sequence numbers and ACKs and thus, does
not provide reliable transmission. Though PPP does not provide flow control and error control is
also limited to error detection only, the control field can also be omitted if both parties agree to do
so during negotiation.
‰‰ Protocol: It is a two-byte long field that defines what is being carried in the payload field, that is,
user data or some other information. Only one byte of this field can be used if both parties agree
during negotiation.
‰‰ Payload: It is a variable-length field containing either the user data or other information in the
form of sequence of bytes. The default maximum length of payload field is 1,500 bytes; however,
two parties can negotiate on this maximum length. If the flag byte appears in the data in payload
field, the data is byte stuffed. In case the size of payload field is less than 1,500 bytes or the maxi-
mum negotiated value, padding is required.
‰‰ FCS: It is a two-byte or four-byte long field that is used to detect errors in the transmitted frame.
This two- or four byte is simply a CRC.

Flag Address Control Protocol Payload FCS Flag

Figure 7.8 PPP Frame Format

19. Explain all the transition phases of PPP.

Ans: A PPP connection transits through various phases while configuring, maintaining, and termi-
nating the PPP link (Figure 7.9). These phases are described as follows:
‰‰ Dead: This is the initial phase of protocol, which means the line is quiet and no physical layer con-
nection exists. When an event such as a physical carrier detection occurs, the connection goes into
establishment phase. The PPP protocol always begins and ends with this phase.

CN_Chapter 07.indd 138 2/20/2012 6:29:13 PM

 Flow and Error Control 139

Failed

Carrier
detected
Dead Establish

Carrier Options agreed

dropped by both sides

Failed If authentication
Terminate Authenticate
not needed
Authentication
Done successful

Open Network
Network layer
configuration

Figure 7.9 Transition Phases of PPP

‰‰ Establish: In this phase, one of the nodes starts the communication and negotiations are made
­between the communicating nodes. The link control protocol (LCP) packets and several other
packets are exchanged between the nodes if the negotiation is successful. With this, the connection
transits to authenticate phase. Notice that if authentication is not required, the connection directly
switches to network phase.
‰‰ Authenticate: This is an optional phase that can be skipped upon an agreement between both
nodes during the negotiation in the establishment phase. If both parties agree to undergo this phase
then several authenticate protocols are used by the communicating nodes to verify each other’s
identity. If the authentication becomes successful, then the connection moves to the network phase
else it goes to the terminate phase.
‰‰ Network: In this phase, two nodes negotiate on the network layer protocol that should be used
so that the data can be received. As PPP supports multiple protocols and many protocols may run
simultaneously at the network layer, the receiver node should negotiate with the sender node on a
common protocol to be used for exchanging the data at the network layer. After the protocol has
been decided, the connection switches to open phase.
‰‰ Open: In this phase, data packets are exchanged between the communicating nodes. The ­connection
remains in this phase until one of the nodes wants to close the connection.
‰‰ Terminate: In this phase, the connection link is closed and the nodes now cannot transfer any data
packets between themselves. Certain packets are exchanged between two nodes for house cleaning
the connection and then the link is terminated. After this, the connection goes to dead phase.
20. Write a short note on LCP and NCP.
Ans: The link control protocol (LCP) and network control protocol (NCP) are the PPP protocols
that are used to establish the link, authenticate the communicating nodes and move the network layer data.

Link Control Protocol

This protocol has the responsibility of establishing, configuring, maintaining and terminating the link.
The link can be established when both the communicating nodes agree on certain options. For this, LCP

CN_Chapter 07.indd 139 2/20/2012 6:29:13 PM

 140 Data Communications and Computer Networks

provides the negotiation mechanism. Each LCP packet is encapsulated in the payload field of PPP frame
and a value 0xC021 (hexadecimal) is placed in the protocol field of PPP frame to indicate that PPP is
carrying an LCP frame (Figure 7.10).

Flag Address Control 0xC021 Payload FCS Flag

LCP packet Code ID Length Information

Figure 7.10 LCP Packet Encapsulated into PPP Frame

The frame structure of LCP consists of various subfields, which are described as follows
‰‰ Code: It is a one-byte long field that defines the type of LCP packet. There are 11 types of LCP
packets that are divided into three categories. The first category includes four packets and is used
to configure the link during the establishment phase. The second category includes two packets
that are used to terminate the connection. The third category includes five packets that are used to
debug and monitor the link.
‰‰ ID: It is a one-byte long field that contains a value to match a request with a reply. When one node
sends the data to another node then the sender node inserts a value in the ID field, which is then
copied by the receiver and placed into the reply packet to be transmitted.
‰‰ Length: It is a two-byte long field that defines the length of the whole L packet.
‰‰ Information: It is a variable length field, which carries options like authentication protocol, pay-
load field size, etc. These options can be negotiated between the two communicating nodes.

Network Control Protocol

PPP supports multiple network layer protocols defined by OSI, Novel, DECnet and Xerox. To carry
a network layer data packet from these protocols, PPP has defined an NCP for each network layer
­protocol. The NCP consists of a set of protocols that encapsulate the data received from the network
layer protocols into the PPP frame. The NCP also configures the link at the network layer. Like LCP,
each NCP packet is also included in the payload field of PPP frame and the protocol field of PPP is set
to a value 0x8021 (hexadecimal) which indicates that PPP frame is carrying the NCP packet.
21. Explain two authentication protocols that are used in PPP.
Ans: As PPP is used over point-to-point link, authentication is important between the communicat-
ing nodes for the effective transmission of resources. The PPP uses two protocols, namely, password
authentication protocol (PAP) and challenge handshake authentication protocol (CHAP) for the verifi-
cation of user identity during the authentication phase.

Password Authentication Protocol

PAP uses three types of packets for authentication, namely, authenticate-request, authenticate-ack and
authenticate-nak. A node that wishes to access the system uses authentication-request packet to send
its username and password to the system. After receiving authentication-request packets, the system
checks the validity of username and password. If the requesting node is validated, the system sends
an authenticate-ack packet to the node indicating that access is allowed. However, if the node is not

CN_Chapter 07.indd 140 2/20/2012 6:29:13 PM

 Flow and Error Control 141

v­ alidated, then the system sends an authenticate-nak packet indicating that access is denied. When-
ever a PAP packet is being carried in a PPP frame, the protocol field of PPP frame is set to a value
0xC023 (hexadecimal).

Challenge Handshake Authentication Protocol

This is a three-way handshaking authentication protocol. It uses four packets, namely, challenge,
­response, success and failure. The system sends a challenge packet containing a challenge value to the
user needing access to the system. The user retrieves the challenge value from the challenge packet and
inputs this value along with its password to a predefined function to generate an output. The user encap-
sulates the output obtained in the response packet and sends it to the system. The system also applies
the same function on the challenge value and the user’s password. If the output obtained is same as the
output received in the response packet, the user is validated. The system sends a success packet to the
user indicating that the access is granted. Otherwise, the system sends a failure packet indicating that
the access is denied. Whenever a CHAP packet is being carried in a PPP frame, the protocol field of PPP
frame is set to a value 0xC233 (hexadecimal).
Further, CHAP provides more security as compared to PAP as it prevents the need of sending the
password online. In case an intruder gets access to the challenge value and the output in the response
packet, he/she cannot know the password. That is, password always remains secret.
22. Why CRC is always added in trailer and not in the header of all data link protocols?
Ans: All data link protocols comprise a CRC field that stores the checksum for error detection and
correction. The CRC is calculated over all the bits that are needed to be transmitted to the ­destination
node and added to the trailer of the outgoing data stream as soon as the last bit is transmitted. If CRC is
to be attached in the header part of the data stream, then CRC has to be calculated before the transmis-
sion of the frame. This would result in more overhead as each byte of frame needs to be handled twice,
once for computing CRC and then for transmission. On the other hand, when CRC is put in the trailer,
each byte of frame needs to be handled only once, resulting in less overhead. That is why CRC is added
in the trailer and not in the header of data link protocols.
23. A channel has a bit rate of 4 kbps and propagation delay of 20 msec. For what frame size
does stop and wait gives an efficienc of at least 50%.
Ans: Given, propagation delay, a = 20 msec
Bit rate, R = 4 kbps = 4 × 103 s
Let L denotes the size of frame, then transmission time of a frame (tframe) can be given as
tframe = L/R
= L/(4 × 103)
As a = tprop/tframe
= 20 × 10–3/(L/(4 × 103))
= (20 × 10–3 × 4 × 103)/L
= 80/L
The maximum link utilization (U) can be given as
U = 1/(1 + 2a)
= 1/(1 + 160/L) … (1)
Given U ≥ 50%
⇒ U ≥ 0.5 … (2)

CN_Chapter 07.indd 141 2/20/2012 6:29:13 PM

 142 Data Communications and Computer Networks

From Eqs (1) and (2), we have

  1/(1 + 160/L) ≥ 0.5
⇒ 1/0.5 ≥ 1 + 160/L
⇒ 2 ≥ 1 + 160/L
⇒ L ≥ 160
Thus, for achieving at least 50% efficienc , the frame size should be at least 160 bits.
24. If the window size is 15, give the sequence number of frames in Go-back-N ARQ and
selective-repeat ARQ.
Ans: In Go-back-N ARQ, the maximum window size is equal to 2m – 1 where m is the number of bits
in the frame sequence number and the frame sequence numbers range from 0 to 2m – 1. Here, given that
2m – 1 = 15
  ⇒2m = 16
Thus, the sequence numbers range from 0 to 15.
In selective repeat ARQ, the maximum window size is equal to 2m–1 and the frame sequence numbers
range from 0 to 2m – 1. Here, given that
2m–1 = 15
⇒2m = 30
Thus, the sequence numbers range from 0 to 29.
25. A channel has a bit rate of 20 kbps. The stop-and-wait protocol with a frame size of 4,500 bits
is used. The delay for error detection and sending ACK by the receiver is 0.25 seconds because of a
fault. Find the maximum efficiency of the channel if the destination is 30,000 km away and the speed
of the propagation of the signal is 2.8 × 108 m/s. Find the decrease in efficienc due to the fault.
Ans: Given, bit rate (R) = 20 kbps = 20 × 103 bps
Frame size (L) = 4,500 bits
Distance between sender and receiver (d) = 30,000 km = 30,000 × 103 m
Propagation speed (V) = 2.8 × 108 m/s
Now, the frame transmission time (tframe) can be computed as
tframe = L/R
= 4,500/(20 × 103) = 0.225 s
The propagation delay (a) can be computed as
a = Rd/VL
= (20 × 103 × 30,000 × 103) / (2.8 × 108 × 4500)
= 0.4761
Now, the maximum utilization (with ignoring the fault) can be computed as
Umax = 1/(1 + 2a)
= 1/(1 + 2 × 0.4761)
= 0.512 = 51.2%
If we consider the fault, the time taken in sending each frame and receiving its ACK will be increased
by 0.25 s. Thus, the total time spent in sending all the frames (say, n) can be given as
T = n*(2tprop + tframe + 0.25)

CN_Chapter 07.indd 142 2/20/2012 6:29:14 PM

 Flow and Error Control 143

The link utilization (in case of fault) can be given as

Ufault = (n*tframe)/(n*(2tprop + tframe + 0.25))
= tframe/(2tprop + tframe + 0.25)
= 1/(1 + 2a + 0.25/tframe)
= 1/(1 + 0.9522 + 1.1111)
= 0.326 = 32.6%
The decrease in efficiency (in %) due to fault = ((Umax – Ufault)/Umax)*100
= ((0.512 – 0.326)/0.512)*100
= 36.3%

Multiple Choice Questions

1. Propagation delay is inversely proportional 6. ABM in HDLC stands for _______.
to ________. (a) asynchronous balanced mode
(a) distance of the link (b) asynchronous balanced modem
(b) data rate of transmission (c) asynchronous bisync mode
(c) length of the frame (d) asynchronous bus modem
(d) none of these 7. The shortest frame in HDLC protocol is usu-
2. When data and ACK are sent on the same ally the _________.
frame, this is called ________. (a) information frame
(a) piggybacking (b) supervisory frame
(b) backpacking (c) unnumbered frame
(c) piggypacking (d) control frame
(d) forwarding and backing 8. The maximum number of unconfirmed
3. A frame that fails to arrive at the destination frames that can be outstanding at any one
node due to noise burst is called ___________. time with SDLC is ___________.
(a) damaged frame (a) 4 (b) 7
(b) transmission frame (c) 8 (d) 14
(c) undamaged frame 9. PPP is a ______ -oriented protocol.
(d) lost frame (a) bit
4. The link utilization (U) in stop-and-wait (b) byte
ARQ is expressed as (c) both (a) and (b)
(a) (1 + P)/(1 + a) (b) (1 + P)/(1 + 2a) (d) none of these
(c) (1 – P)/(1 + 2a) (d) (1 – P)/(1 + a) 10. If you connect to the Internet from your home
5. In Go-back-N ARQ, if frames 3, 4 and 5 are computer, chances are that you are using:
received successfully, which ACK number _________.
may be sent to the sender? (a) PPP (b) NCP
(a) 5 (b) 6 (c) DAP (d) FTAM
(c) 7 (d) 8
Answers
1. (c) 2. (a) 3. (d) 4. (c)   5. (b) 6. (a) 7. (b) 8. (a) 9. (b) 10. (a)

CN_Chapter 07.indd 143 2/20/2012 6:29:14 PM

 8
Media Access Control

1. Write a short note on channel allocation.

Ans: In a broadcast (or multipoint) network, multiple users (stations) use a single broadcast channel
to communicate with each other. In case more than one user wish to access the shared channel at the
same time, the channel is allocated to one of the competing users at one time. This is called channel al-
location. The decision to allocate the channel among the competing users can be made using any of the
two schemes, namely, static channel allocation and dynamic channel allocation.

Static Channel Allocation

In this scheme, a single channel can be shared among multiple users using either frequency division
­multiplexing (FDM) or time division multiplexing (TDM) technique. In FDM, for n users, the ­bandwidth
of a channel is divided into n parts of equal sizes and each part is kept fixed for use by a specific user.
With a separate frequency band for each user, there is no interference between competing users. Though
FDM is simple and efficient, it does have some drawbacks. For example, if at a specific point of time,
less than n users communicate through the channel, there will be wastage of frequency spectrum and if
more than n users want to communicate through the channel, some of them will not be permitted due
to bandwidth deficienc . Moreover, when some users are inactive, it results into loss of precious band-
width as no one else is allowed to use the bandwidth allocated to those users.
On the other hand, in TDM, time is divided into n slots for n users and every nth time slot is ­statically
allocated to each user during which it can transmit the frames. However, if at any instance, the user does
not use the allotted slot, it goes wasted.

Dynamic Channel Allocation

The drawbacks of static channel allocation can be overcome by dynamic allocation method. In this
method, neither a fixed frequency nor a fixed slot is allocated to the user in advance; rather, each
user is allocated the channel dynamically at the time it needs using many protocols. Dynamic channel

CN_Chapter 08.indd 144 2/20/2012 6:31:14 PM

 Media Access Control 145

a­ llocation can be used when bursty traffic is present in the network. To implement this method, certain
assumptions are followed.
‰‰ Station Model: The model consists of n stations (sometimes, called terminals) such as ­computers
and telephones that can work independently on a network. Each terminal has the capability to
produce frames for transmission. After a frame has been generated, the station remains idle or is
blocked until the entire frame has been transmitted.
‰‰ Single Channel: A single communication channel is shared by all the stations for transmission.
‰‰ Collision: When two stations transmit frames simultaneously, the reception of frames may overlap
in time resulting in a garbled signal. This phenomenon is known as collision. The collided frames
must be retransmitted by their respective stations.
‰‰ Continuous Time and Slotted Time: Some systems follow continuous time assumption while
­others follow slotted time assumption. If continuous time assumption is followed in a network,
then a station can transmit a frame at any time. However, if slotted time assumption is followed
then time is partitioned into discrete intervals and a frame can be transmitted only at the beginning
of a slot. Thus, for synchronization, some mechanism such as a clock is required.
‰‰ Carrier Sense and no Carrier Sense: A network can have either carrier sensing or not. A station
with a carrier sense can determine whether a channel is in use before attempting to use it. If a chan-
nel is busy, no station attempts to use it until it gets free. However, if the network has no carrier
sense then a station cannot sense the channel before starting the transmission. If the station needs
to send a frame, it just transmits it.

2. What is the need of multiple-access protocol? List the categories of multiple-access

­protocols.
Ans: When multiple stations or nodes connected via a network use a common link or channel (multi-
point or broadcast) to send or receive data, some means is required to coordinate access to the link. This
is what a multiple-access protocol does. It decides which of the stations is going to access the shared
channel next. A number of multiple-access protocols have been devised to handle access to shared
link. All these protocols belong to a sublayer of data link layer called medium access control (MAC)
sublayer. These protocols have been classified into three categories, namely, random access, controlled
access and channelization protocols.

3. What is meant by random access method? Give examples of random access protocols.
Ans: The method, which allows the stations to access the transmission medium (channel) ran-
domly at any time, is known as random access method. No station is under the control of any
other station and there is no specific time for stations to start transmission. Moreover, there are
no rules to be followed for deciding which station should be allowed access to the channel. If
multiple users need to access the channel at the same time, they have to compete with each other.
Hence, this method is also known as contention method. Examples of random access protocols
are ­ALOHA, carrier sense multiple access (CSMA), CSMA/collision detection (CD) and CSMA/­
collision avoidance (CA).

4. Explain pure ALOHA and slotted ALOHA.

Ans: Pure ALOHA and slotted ALOHA are the versions of ALOHA protocol, the earliest contention-
based method that was invented in 1970 at the University of Hawaii.

CN_Chapter 08.indd 145 2/20/2012 6:31:14 PM

 146 Data Communications and Computer Networks

Pure ALOHA
Pure ALOHA is a simple protocol. It was originally developed for packet radio networks but can be
­applied to any system where a single shared channel is being used among multiple competing users. The
basic idea is that each station can send a frame whenever it needs to transmit. When two frames are sent
at the same time, they may collide with each other fully or partially and get corrupted (Figure 8.1). The
sender, therefore, must retransmit the corrupted frames.

S1

S2

S3

Broadcast channel

Complete collision Partial collision

Figure 8.1 Collisions in Pure ALOHA
Detection of collision is easy in pure ALOHA. A station can detect the destruction of a frame
by ­listening to the channel. However, if listening is not possible, the receiving station must send an
­acknowledgement to the sender after receiving a frame. The sending station, after sending a frame,
waits for a random amount of time computed using binary exponential back-off algorithm (discussed
in Q7). If it receives the acknowledgement until that period, it knows that the frame has been received
successfully. Otherwise, it assumes that collision has happened and thus, retransmits the frame.
In pure ALOHA system, the throughput (S), which is defined as the average number of successful
transmissions, is given as
S = Ge-2G
where G = the average number of frames generated by the system during one frame transmission time,
that is, the offered traffic and e = a mathematical constant with the value 2.718.
The offered traffic G) can also be expressed as
G = Ns × Nf × t
where
Ns = total number of stations in the system.
Nf = the number of frames that can be transmitted by a station in one second.
t = frame transmission time.
As shown in Figure 8.2, the maximum throughput (Smax) of pure ALOHA is obtained at G  = 0.5
(50% ordered traffic) and is equal to 1/2e, that is, 0.184 or we can say 18.4%. At lower offered traffic
(G < 0.5), the channel’s capacity is underutilized while on higher-offered traffic (G > 0.5), the throughput
starts decreasing due to increase in number of collisions.

CN_Chapter 08.indd 146 2/20/2012 6:31:15 PM

 Media Access Control 147

0.5

Throughput
Slotted ALOHA
S
Pure ALOH
A

0 0.5 1 2 3

G (offered traffic)
Figure 8.2 Throughput of Pure ALOHA and Slotted ALOHA

Slotted ALOHA
Slotted ALOHA is a modification of ALOHA that has been developed to improve the efficienc . This is
a discrete time system in which time is partitioned into intervals or slots with each slot corresponding
to a frame. Slotted ALOHA demands for the stations to agree on the boundaries of slots and a clock are
needed to synchronize the stations. In this method, a station can begin the transmission of a frame only
at the beginning of a slot. If a station is unable to start the transmission at the beginning of a slot, then it
has to wait until the next slot, thereby reducing the chances of collisions. However, collisions may still
occur when two stations are ready for transmission in same slot resulting in garbled frames, which have
to be retransmitted by the sender (Fig. 8.3).

S1

S2

S3

Broadcast channel

Collisions
Figure 8.3 Collisions in Slotted ALOHA

The throughput (S) of slotted ALOHA is given as

S = Ge-G
As shown in Figure 8.2, the maximum throughput (Smax) of slotted ALOHA is obtained at G = 1
(100% ordered traffic) and is equal to 1/e, that is, 0.368 or we can say 36.8%. Thus, it is clear that the
throughput of slotted ALOHA is twice that of pure ALOHA.

CN_Chapter 08.indd 147 2/20/2012 6:31:15 PM

 148 Data Communications and Computer Networks

5. List some limitations of ALOHA protocol.

Ans: The ALOHA protocol suffers from certain limitations, which are as follows:
‰‰ If any station transmits a frame but the propagation delay between stations is very large in
­comparison to transmission time of a frame, then it will be harder for the stations to know about it.
During that time, another station can transmit a frame, which can result in a collision.
‰‰ It is not possible for stations to sense a channel before utilizing it for transmission. They can only
detect the channel after transmission of a frame.
‰‰ The system needs to maintain queuing buffers to store the transmitted frames until the receiver receives
them successfully. This is required to enable the retransmission of frames in case of a collision.
6. What is the vulnerable period? Show that the vulnerable time period of slotted ALOHA
is half of that of pure ALOHA.
Ans: Consider a system consisting of several stations that share a common link to transmit and
­receive frames. The time interval during which collision may occur in the frame transmission is referred
to as the vulnerable period of the system. To determine the vulnerable period of pure ALOHA and
­slotted ALOHA, consider three stations S1, S2 and S3 which are transmitting frames and the time taken
in transmitting a complete frame by any station is t.
Consider the pure ALOHA protocol. Let station S1 transmits a frame F1 at time t0. This frame will
be completely transmitted by the time t0 + t. Now, suppose station S2 has already transmitted a frame
F2 between the time t0 - t and t0. This will result in a collision as the ending part of frame F2 may
collide with the leading part of frame F1 (Figure 8.4). Further, suppose that station S3 transmits a frame
F3 between the time t0 and t0 + t. This will again result in a collision as the ending part of frame F1
may collide with the leading part of frame F3. Thus, it is clear that if a station transmits a frame at t0,
a collision may occur if any other station transmits frame between the time t0 - t and t0 + t. This
implies that the vulnerable period of pure ALOHA is 2t, that is, twice of the frame transmission time.

F1

Collides with
F2 end of F1

Collides with F3
beginning of F1

t0 − t t0 t0 + t Time
Vulnerable period
Figure 8.4 Vulnerable Period in Pure ALOHA Protocol
In slotted ALOHA, the stations are allowed to transmit frames only in the beginning of each time slot. If
any station misses the chance in one time slot, it must wait for the beginning of next time slot. This implies
that a station, which has started at the beginning of this time slot, has already ­transmitted its frame ­completely.
However, a collision may still occur if two stations attempt to transmit in the beginning of same time slot
(Figure 8.3). Hence, the vulnerable period of slotted ALOHA is t, which is half of that of pure ALOHA.

CN_Chapter 08.indd 148 2/20/2012 6:31:16 PM

 Media Access Control 149

7. Explain the binary exponential back-off algorithm with the help of an example.
Ans: In case of a collision, the stations whose frames have collided are required to retransmit the
frames. However, before attempting to retransmit the frames, each station that was involved in the col-
lision waits for some random amount of time (random delay) to ensure that the collision is not likely
to occur at the time of retransmission. The random delay increases with each repeated collision.
Binary exponential back-off algorithm specifies the random amount of time a station has to wait for
­retransmission of a frame after a collision occurs. After a collision occurs, this algorithm works as follows:
1. Time is divided into discrete slots. For the kth attempt of retransmission, a random number r
between 0 and 2k – 1 is selected. The value of k is incremented by one with every retransmis-
sion attempt and thus, the range of random numbers also increases.
2. A station waits for r number of slots before making an attempt to retransmit the frames.
3. If number of collision or number of attempts is more than 15, transmission is aborted and failure
is reported to the station.
For example, after two collisions, that is, two attempts of retransmission, a station will wait for random
number chosen between 0 and 3 (that is, 22 – 1) of slots. If more than one station selects the same random
number, then collision occurs again. As the number of collisions increases, the range of random number
increases thereby reducing the chances of collision among stations. However, the increase in range of
random numbers also implies increase in average wait time, which results in increased delay.
8. Explain the working of carrier sense multiple-access protocol.
Ans: Carrier Sense Multiple Access (CSMA) is a contention-based protocol that was developed
to overcome the drawbacks of ALOHA system. This protocol tends to reduce the chances of collisions
and thus, improving the performance of the network. This protocol operates on the principle of carrier
­sensing which states that any station, before attempting to use the channel, must sense (listen to) it to
check whether any other station is using the channel or not. A station sends the data only if it finds the
channel free, that is, when there is no other carrier.
There are three variations of CSMA protocol, which are as follows:
‰‰ 1-persistent CSMA: In this method, when a station needs to transmit a frame, it monitors the
­channel, and transmits the frame as soon as it finds the channel idle. However, if it finds the channel
busy, it continuously senses the channel until it becomes free. Despite of the fact that carrier sensing
is used, collisions may still occur if two or more stations find the channel idle at the same time and
thus, transmit frames simultaneously. The performance of this protocol is greatly affected by the
propagation delay. To understand, consider that a station sends a frame and the propagation delay is
greater than the transmission time of a frame. It takes some time for the first bit of the transmitted
frame to reach and be sensed by other stations. In the mean time, any other station that becomes
ready to send, may sense the channel idle and thus, transmits the frame. This leads to a collision.
This protocol is so named as a station transmits with probability one on finding the channel idle
‰‰ Non-persistent CSMA: In this method, when a station needs to send a frame, it first senses the channel.
If the channel is idle, the station transmits the frame instantly. However, if the station finds the channel
busy, it waits for a random amount of time before sensing the channel again. This method reduces the
chances of collision as it is rare that two stations will wait for same amount of time and then will retry
to transmit at the same time. Due to fewer collisions, the utilization of network is also improved.
‰‰ P-persistent CSMA: This method is applicable and used with the channel that has time slots with
each slot being greater than or equal to maximum propagation time. In this method, if the ­channel

CN_Chapter 08.indd 149 2/20/2012 6:31:16 PM

 150 Data Communications and Computer Networks

is found idle, a station is allowed to transmit with a probability p. With probability q (which is
equal to 1 – p), the station waits for the beginning of the next time slot and senses the line again.
If the channel is idle in the next time slot also, the above method is repeated; the station either
­transmits or waits with probabilities p and q, respectively. This process continues and eventually,
either that station transmits the frame or some other station starts transmitting the frame. In the
latter case, the station assumes that a collision has occurred and thus, it waits for a random period
(computed using back-off algorithm) and then starts again.
9. Compare the performance of non-persistent and 1-persistent CSMA protocols.
Ans: The performance of non-persistent and 1-persistent CSMA protocols can be compared with
­respect to two parameters: delay and channel utilization. In case of fewer load on the network, 1-persistent
has shorter delay than non-persistent because 1-persistent method permits a station to begin sending a
frame as soon as the channel is sensed idle. Due to less delay, the channel utilization of 1-persistent method
is better than that of non-persistent method. On the other hand, in case of heavy load on the network,
non-persistent method has shorter delay than 1-persistent method because the number of collisions in
1-persistent method is higher than that of non-persistent method. Due to less probability of collisions,
the channel utilization of non-persistent method is better than that of 1-persistent method. Thus, it can be
concluded that at low loads, 1-persistent method is preferable over non-persistent method while at heavy
loads, non-persistent method is preferable over 1-persistent method.
10. What is CSMA/CD? Explain its working.
Ans: Carrier Sense Multiple Access/Collision Detection (CSMA/CD) is the refinement of CSMA
protocol. It was developed to overcome the inefficiency of CSMA protocol where the stations involved
in the collision do no stop transmitting bits of frames even though the collision is detected. This leads
to inefficient channel utilization. Further, CSMA/CD improves the channel utilization by making the
stations to sense the channel not only before starting the transmission but during the frame ­transmission
also (referred to as collision detection). This implies that in CSMA/CD, both transmission and collision
detection are continuous processes. Once a station starts transmitting bits of a frame, it continuously
senses the ­channel to detect the collision. As soon as the station detects a collision, it aborts the frame
transmission immediately rather than finishing the transmission. Collision can be detected by compar-
ing the power of the received signal with that of transmitted signal.
The CSMA/CD can be in any of the three states, namely, contention, transmission or idle. Conten-
tion refers to the state in which more than one station is ready to transmit and competing to get the
channel. Transmission refers to the state in which data are transmitted. Idle refers to the state in which
channel is idle and no station is transmitting data.
Figure 8.5 depicts the working of CSMA/CD. When a station has a frame to send, a variable k ­denoting
the number of retransmission attempts is initialized with zero. The station applies one of the persistence
methods to sense the channel. When the channel is found idle, the station starts transmitting bits of frame
and detecting for collision as well. If collision does not occur during the entire transmission period, the
station assumes that the frame has been received successfully. However, if a collision takes place during
the transmission, the station aborts the transmission and a jam signal is sent to inform all other stations
on the network about the collision. If the station has not crossed the limit of maximum retransmission
attempts (kmax), the back-off time is computed for the station using the binary exponential algorithm
(­explained in Q7). The station is made to wait for the back-off period and then restart the process.
To better understand CSMA/CD, consider two stations S1 and S2. Suppose when contention period
starts at time t1, station S1 starts transmitting a frame F1. After some time, station S2 senses the channel

CN_Chapter 08.indd 150 2/20/2012 6:31:16 PM

 Media Access Control 151

Start

k=0

Apply any of the Wait for back

persistence methods off time

Idle channel
Apply binary exponential back off
Sendframe algorithm to compute back off time

No

Collision Yes Send jam

k=k+1 Is k > kmax?
detected? signal

No Yes

Success Abort

Figure 8.5 Working of CSMA/CD

and finds it idle as the first bit of frame F1 sent by station S1 has not yet reached. Therefore, station S2
starts transmitting the bits of frame F2 at time t2; the bits are propagated in both forward and backward
direction. Collision occurs between frames F1 and F2, which is detected by the station S2 at time t3.
As soon as the station S2 detects the collision, it discontinues sending further bits of frame F2. Now, let
at time t4, station S1 detects a collision. It also aborts the transmission of frame F1. After detecting a
­collision, stations S1 and S2 immediately abort the transmission rather than finishing the transmission of
frames, which will be garbled anyway (Figure 8.6).

S1 S2

t1 F1
Fi

t2 F2
rs
tb

F
t bit of 2
it o

Firs
fF

t3
1

Collision detection
Collision and abortion by S2
t4 occurs
Collision detection
and abortion by S1

Figure 8.6 CSMA/CD

CN_Chapter 08.indd 151 2/20/2012 6:31:17 PM

 152 Data Communications and Computer Networks

The advantage of this protocol over CSMA is that it saves time and bandwidth as it aborts the transmission
of a frame as soon as the collision is detected. Thus, it offers better performance than CSMA. It is widely
used on LANs in MAC sublayer.
11. Describe CSMA/CA protocol.
Ans: In wireless networks, it is difficult to effectively detect the collision using CSMA/CD. This is
because in wireless networks, most energy of the signal is lost during transmission and as a result, the
signal received by a station is of very less energy. That is why in wireless networks, a need arises to
avoid the collision rather than detect the collision. The CSMA/CA is a protocol developed for wireless
networks to help in avoiding collisions.
Figure 8.7 shows the working of CSMA/CA protocol. When a station is ready to transmit the frames, a
variable k denoting the number of retransmission attempts is initialized with zero. The station uses any of
the persistent methods to sense the channel. If the channel is busy, it again senses the channel. The station
continues to do so until it finds the station idle. After the station has found the channel idle, it does not send
the frame immediately rather it waits for a random amount of time known as interframe space (IFS). The
station is required to wait for IFS time because it may happen that some other distant station has already
started the transmission of a frame but the leading bits of the frame have not reached here until yet.

Start

k=0

Apply any of the

persistence methods

No Is channel
idle?

Yes

Wait for IFS time

No
Yes
No Is channel Is k > k max? Abort
idle?

Yes
Apply binary exponential back off k=k+1
algorithm to choose n between 0 and 2k−1
No

Wait for n time ACK received

Send frame Set the timer before time-
slots
out?
Yes

Success

Figure 8.7 Working of CSMA/CA

CN_Chapter 08.indd 152 2/20/2012 6:31:17 PM

 Media Access Control 153

After waiting for IFS amount of time, the station again senses the channel. If it finds the channel still
idle, it waits for a random amount of time partitioned into slots. This time is referred to as the contention
time and it is computed using the binary exponential back-off algorithm; a random number n is chosen
between 0 and 2k - 1 and the station is made to wait for n time slots. After each time slot, the station
senses the channel. If the channel is busy, the timer is stopped and it is not restarted until the channel is
found idle.
After waiting for n time slots, the station sends the frame and starts waiting for its acknowledgement.
If the station receives the positive acknowledgement before the time-out, it knows that the frame has
been received successfully. However, if acknowledgement is not received before the time-out, the station
needs to retransmit the frame. To check whether the station is eligible for retransmission, the value of k is
increased by one and compared with kmax—the maximum number of retransmissions allowed. If the ­value
of k is less than kmax, the whole process is repeated. Otherwise, the station aborts the transmission.

12. Write a short note on the collision-free protocols.

Ans: Generally, when CSMA/CD protocol is used, collisions do not occur. Still, there are some
chances of occurrence of collisions during the contention period. To resolve the contention problem along
with avoiding the collisions, collision-free protocols were developed. Bit-map and binary ­countdown are
the two commonly used collision-free protocols, which are discussed here.

Bit-Map Protocol
This protocol involves the division of contention period into n number of slots for n number of stations
in the network with one slot corresponding to each station. Both stations and slots are numbered from
0 to n – 1. If a station needs to transmit a frame, it will transmit a 1 bit in its respective slot; no other
station can transmit in this time slot. At the end of n-bit contention period, each station knows which
stations wish to transmit. Then, the stations begin transmitting frames in the numerical order. After the
last ready station has transmitted the frame, the next n-bit contention period starts. In case a station is
ready to send but its slot has passed by, then it must wait for its turn. Since, every station knows which
station has to transmit next, collisions cannot occur.
To understand the bit-map protocol, consider a network
with five stations numbered from 0 to 4. If stations 0, 2 and 3
want to transmit frames, they insert 1 bit into their correspond-
ing slots and they are transmitted in the order 0, 2 and 3 only 1 1 1 0 2 3
(­Figure 8.8).
To analyze the performance of channel in bit-map protocol, 0 1 2 3 4 Frames
let the total number of stations is n and the quantity of data per Contention slots
each frame is d bits. At low loads, the bit-map will be repeated Figure 8.8 Bit-map Protocol
over and over lacking of frames, creating an overhead of n bits
per frame. Thus, the performance of channel at low loads can be
given as
Slow = d/(N + d)
On the other hand, at high loads, almost every station has some data to send. As a result, each n-bit
contention period will have bits per each frame, creating an overhead of one bit per frame. Thus, the
performance of channel at high loads can be given as
Shigh = d/(1 + d)

CN_Chapter 08.indd 153 2/20/2012 6:31:18 PM

 154 Data Communications and Computer Networks

Binary Countdown Protocol

When a large number of stations are present, the bit-map protocol does not prove efficient because of
the overhead of one bit per each station. To overcome this problem, a more efficient protocol named
binary countdown can be used. In this protocol, each station has a binary address that identifies it and
the one out of many competing stations that has the highest binary address is allowed to use the channel
at a given time.
Every station that wants to use the channel, broadcasts its binary address starting with the most
­significant bit. During each bit time, the higher-order bits from the competing stations are ORed ­together.
If the result comes out to be one, then stations with zero high-order bit quit as they come to know that
some station with higher binary address is contending to use the channel. On the other hand, stations
with high-order bit 1 still continue. This process continues until a station with the highest binary address
is determined and thus, allowed access to channel.
To better understand the process, consider three
stations S1, S2 and S3 with binary addresses 0011, Bit time
1000 and 1011, respectively are trying to use the
channel. Each station transmits its higher-order bit 0 1 2 3
in the bit time slot and these bits (0, 1, 1) from all S1 0 0 1 1 0 - - -
the stations are ORed together, which results in 1.
Now, station S1 knows that stations S2 and S3 with a
higher address are contending for the channel, hence,
1 0 0 0 1 0 0 -
it quits. Again, the next high-order bits (0, 0) from S2
stations S2 and S3 are ORed together and the result
comes out to be zero. Both stations continue to trans-
mit next higher-order bit, that is, (0, 1) which are S 1 0 1 1 1 0 1 1
3
ORed together and result obtained is one. The sta-
tion S2 sees this result and quits. Thus, S3 gets the Result of OR operation 1 0 1 1
channel to transmit data. Figure 8.9 depicts the whole
process.
The main limitation of this protocol is that the sta- After seeing this After seeing this
bit, S1 quits bit, S2 quits
tion with higher address will always get the higher
priority. The efficiency of channel in this protocol is Figure 8.9 Binary Countdown Protocol
equal to d/(d + log2n).
13. Explain limited-contention protocol.
Ans: Limited-contention protocol is the combination of contention and collision-free ­protocols.
At high loads, it uses collision-free protocol to achieve better efficiency of network and at low loads, it
uses contention protocols to achieve shorter delay for transmission. Further, limited-contention protocol
is ­asymmetric in nature; different stations attempt to seize the channel with different probabilities.
The basic idea behind the limited-contention protocol is that chances of some station acquiring
the channel can be increased by reducing the number of stations competing for the channel. Thus, in
­limited-contention protocol, stations are divided into groups. For slot 0, stations under group 0 are al-
lowed to compete for the channel. If any of the group 0 stations succeeds, it gets the channel and trans-
mits the frame. However, if some collision occurs then stations under group 1 compete for the channel
in slot 1. This process continues, reducing the amount of competition for each slot.

CN_Chapter 08.indd 154 2/20/2012 6:31:18 PM

 Media Access Control 155

To appropriately divide the stations into groups, adaptive tree-walk algorithm is used, which
c­ onsiders all the stations in the network as the leaves of a binary tree as shown in Figure 8.10. The pro-
cess starts at the root of the tree (node 1). Initially, in the first slot (slot 0), all the stations under node 1
are allowed to compete for the channel. If there is a single station that is ready to transmit, it is allocated
the ­channel. However, if there is a collision, stations under the node 2 try to seize the channel in slot
1. If some ­station under node 2 acquires the channel, the next slot after the successful transmission of
frame is kept reserved for the stations under node 3. However, if collision occurs in slot 1, the stations
under node 4 contend during the slot 2. In essence, in case of collision during slot 0, the search continues
recursively in the left and right subtree of binary tree and it stops when a bit slot is idle or there is only
one station that is ready to transmit the data.
1

2 3

4 5 6 7

Stations
S1 S2 S3 S4 S5 S6 S7 S8
Figure 8.10 Adaptive Tree-walk Algorithm

14. Explain three controlled access methods.

Ans: In controlled access, stations refer each other to know which station has the right to transmit a
frame. A station is allowed to transmit only if other stations authorize it do so. There are three controlled
access methods, namely, reservation, polling and token passing.

Reservation
In this method, any station that wants to transmit a frame has to make a reservation before sending. During
each time interval, the reservation frame precedes the data frames to be transmitted in that interval.
A reservation frame consists of mini-slots with the number of mini-slots being equal to the number of
stations in the network. Whenever, a station wants to transmit a data frame it makes a reservation by
inserting a 1 bit into its corresponding mini-slot. After making reservations, the stations can transmit
their data frames in the numerical order following the reservation frame.

Polling
In this method, two types of stations namely, primary node and secondary node exist. A primary node
refers to the master station through which the exchange of the data takes place irrespective of whether
the destination station is primary or secondary. Secondary nodes are the terminals that follow the com-
mands of a primary node. Primary node decides which secondary node is allowed to use the channel and
transmit at a particular time. Polling involves two operations, namely, poll and select.

CN_Chapter 08.indd 155 2/20/2012 6:31:18 PM

 156 Data Communications and Computer Networks

‰‰ Poll: When the primary node is ready to receive data, it polls or asks all the secondary nodes in a
round-robin fashion to determine if they have any data to transmit. The primary node sends a mes-
sage to a secondary node by specifying the maximum number of frames the secondary node can
send. The secondary node can respond with either negative acknowledgement (NAK) or positive
acknowledgement (ACK). If the response is NAK, the primary node polls the next secondary node.
This process goes on in a cyclic manner until any secondary node responds with an ACK. Once a
positive response is received, the primary node also responds with an ACK to confirm the receipt
‰‰ Select: This operation is performed when primary node has some data to transmit. Before sending
the data, it alerts the receiving secondary node about the transmission and does not transmit until
it gets a ready acknowledgement from the receiving secondary node.
An advantage of polling is that it eliminates collision and results in better efficienc . However, it suffers
from some limitations too. First, this protocol requires large amount of time in notifying secondary nodes and
thus, introduces large amount of delay. Second, if the primary node fails, the entire network goes down.

Token Passing
In this method, all stations in the network are arranged in a logical S4
ring manner (Figure 8.11). To allow stations to access the channel,
a special packet known as token is circulated among the stations
throughout the ring in a fixed order. Whenever a station receives the
token from its predecessor and it has some data to send, it keeps the S S3
1
token with itself and transmits the data. After transmitting the data, T
it releases the token and passes it to its successor in the ring. In case
a station receives the token but has no data to send, it immediately
passes the token to its successor in the ring.
In this method, a priority can be assigned to each station in the S2
ring such that whenever a high priority station wants to transmit, a Figure 8.11 Token Passing
low priority station has to release the token. In addition, type of data
being transmitted from the stations can also be prioritized.
An advantage of token-passing method over polling is that there is no primary node present in the
network, so this method is decentralized. However, it also suffers from some limitations. First, if any
station goes down then entire network goes down. Second, if some problem such as failure of a station
possessing the token occurs then token will vanish from the network.
15. What is the advantage of token-passing protocol over CSMA/CD protocol?
Ans: In CSMA/CD, a frame may be delivered at the destination either after many collisions or not
delivered at all. Thus, CSMA/CD is not suitable for real-time applications. On the other hand, in token
passing, a frame is delivered within the time. Also, the frames can be assigned priorities. This makes
token passing suitable for real-time applications.
16. What is meant by the term channelization? Explain in detail various channelization
­protocols.
Ans: Channelization refers to a multiple-access method in which the bandwidth of a communica-
tion link is partitioned in time, frequency or using code among the stations that want to transmit data.
Three channelization protocols, namely, FDMA, TDMA and CDMA are discussed as follows:
‰‰ Frequency Division Multiple Access (FDMA): This method involves partitioning the bandwidth
of a channel into parts called frequency bands. Each frequency band is allocated to a particular

CN_Chapter 08.indd 156 2/20/2012 6:31:19 PM

 Media Access Control 157

s­ tation for transmitting its data and it remains fixed throughout the entire communication period.
The frequency bands are kept well separated by placing guard bands in between to avoid the station
interference. In case of higher load on the network, the performance of FDMA can be improved by
allocating the frequency bands to the stations dynamically when they demand rather than keeping
the frequency bands reserved for stations.
‰‰ Time Division Multiple Access (TDMA): In this method, the bandwidth of channel is time shared
among the stations. The entire time is partitioned into slots and each slot is allocated to a particular
station in which it can transmit data. The major limitation of TDMA is that synchronization is required
between different stations so that every station could know the beginning as well as position of its
respective slot. To achieve synchronization, preamble bits also known as synchronization bits are
inserted at the start of each time slot.
‰‰ Code Division Multiple Access (CDMA): This method allows all the stations to send data through
the same channel and at the same time. As compared to FDMA, in CDMA, the whole bandwidth
of a link can be utilized by only one channel. Further, CDMA is also different from TDMA in
the sense that in CDMA, all stations can transmit the data at the same time. All transmissions are
­separated using coding theory. Each station is assigned a code—a sequence of numbers called
chips. These sequences are called orthogonal sequences and possess the following properties.
• Each sequence contains a number of elements equal to the number of stations. For example,
(+1 -1 -1 +1) is a sequence with four elements. Here, -1 denotes the 0 bit while +1
denotes the bit 1.
• When a sequence is multiplied with a scalar quantity, every element of the sequence is multi-
plied with it. For example, on multiplying the sequence (-1 -1 +1 +1 -1 -1) with 5,
we get (-5 -5 +5 +5 -5 -5).
• On multiplying two equal sequences, element by element with each other, and further adding
the elements of resulting sequence, the final result obtained is equal to the number of elements
in each sequence. This is also known as inner product of two sequences. For example, when
(-1 -1 +1 +1 -1 -1 +1 +1) is multiplied with (-1 -1 +1 +1 -1 -1 +1 +1),
the resulting sequence is (+1 +1 +1 +1 +1 +1 +1 +1) whose sum of element is 8, equal
to the number of elements in the sequence.
• On multiplying two different sequences, element by element with each other, and then adding
the elements of resulting sequence, the final result obtained is zero. This is called the inner
product of two different sequences. For example when (-1 -1 +1 +1 -1 -1 +1 +1)
is multiplied with (+1 +1 +1 +1 +1 +1 +1 +1), the resulting sequence is (-1 -1 +1
+1 -1 -1 +1 +1) whose sum of elements is zero.
• When two sequences are added, the corresponding elements of both sequences are added to
each other. For example, on adding a sequence (-1 -1 +1 +1 -1 -1 +1 +1) with (-1
-1 +1 +1 -1 -1 -1 -1), the resulting sequence is (-2 -2 +2 +2 -2 -2 0 0).
To understand how CDMA works, consider four stations S1, S2, S3 and S4 with data d1, d2, d3 and
d4 and codes c1, c2, c3 and c4, respectively. Each station multiplies its data by its code to get c1*d1,
c2*d2, c3*d3 and c4*d4, respectively. The data that is sent to the channel is the sum of c1*d1, c2*d2,
c3*d3 and c4*d4. Any station (say S4) that wants to receive data from another station (say S2) mul-
tiplies the data on the channel with the code of the sender (in our case c2) to get the data as shown
here
Data = (c1*d1+c2*d2+c3*d3+c4*d4)*c2
= 4*d2

CN_Chapter 08.indd 157 2/20/2012 6:31:19 PM

 158 Data Communications and Computer Networks

It is clear from the second and third property that c2*c2 will result in four while other products c1*c2,
c3*c2 and c4*c2 will result in zero. Thus, in order to receive data from S2 (which is d2), the station S4
has to divide the result by four.
Further, CDMA is widely used for cellular networks. The main advantage of CDMA is that it uses
a larger bandwidth, so transmission of data is less affected by the transmission impairments. However,
the limitation of CDMA is self-jamming and distance.
17. How FDM differs from FDMA?
Ans: Although FDM and FDMA employ the same concept, there are differences between the two.
The FDM is a physical layer technique, which involves using a multiplexer to combine traffic from low
bandwidth channels and transmitting them on a higher bandwidth channel. In this technique, channels
are statically allocated to different stations, which is inefficie t in case of bursty traffic. In contrast,
FDMA is an access method in the data link layer in which channels can be allocated on demand. In
FDMA, the bandwidth is divided into frequency bands and each band is reserved for a specific ­station.
The efficiency is improved in FDMA by using a dynamic-sharing technique to access a particular
­frequency band. No physical multiplexer is used at the physical layer in FDMA.
18. How TDM differs from TDMA?
Ans: TDM and TDMA have certain differences. The TDM is a physical layer technique, which uses
a physical multiplexer to render data from slower channel, combine them and transmit using faster channel,
whereas TDMA is a multiple-access method in the data link layer where the data link layer of each
station tells the physical layer to use the specific slot allocated for it. No physical multiplexer is used at
the physical layer in TDMA.
19. A slotted ALOHA channel has an average 10% of the slots idle. What is the offered traffic G?
Calculate the throughput and determine whether the channel is overloaded or under loaded?
Ans: According to Poisson’s distribution
P0 = e–G
Also, G = –ln P0
= –ln 0.1
= 2.3
As we know that, throughput (S) of slotted ALOHA = Ge–G
S = 2.3 × 0.1 = 0.23
As G > 1, the channel is overloaded.
20. Consider a slotted ALOHA having fiv stations. If the offered load G1 = 0.1, G2 = 0.15,
G3 = 0.2, G4 = 0.25 and G5 = 0.3 packets, fin the individual throughput of each station and channel
throughput.
Ans: In slotted ALOHA, individual throughput, Si = Gi e–Gi
Let the individual throughput of each station be S1, S2, S3, S4 and S5.
Then, S1 = G1e–G1 = 0.1e–0.1 = 0.0905
S2 = G2e–G2 = 0.15e–0.15 = 0.1291
S3 = G3e–G3 = 0.2e–0.2 = 0.1637
S4 = G4e–G4 = 0.25e–0.25 = 0.1947
S5 = G5e–G5 = 0.3e–0.3 = 0.2222

CN_Chapter 08.indd 158 2/20/2012 6:31:19 PM

 Media Access Control 159

Thus, total channel throughput (S) can be determined as

S = S1 + S2 + S3 + S4 + S5
= 0.0905 + 0.1291 + 0.1637 + 0.1947 + 0.2222
= 0.8002
21. Consider a CSMA/CD LAN running at 1 Gbps over a 1 km long cable with no repeaters.
The signal propagation speed is 200 m/µs. What is the minimum frame size?
Ans: Frame transmission time = 2 × propagation time
Here, propagation time = distance/propagation speed
That is, propagation time = 1000/200 = 1000/2 = 5 µs
So, frame transmission time = 2 × 5 = 10 µs
The minimum size of the frame is given as
data rate × frame transmission time
= 1 Gbps × 10 µs
= 1 × 109 × 10 × 10–6
= 10,000 bits = 1,250 bytes
22. A 4 Mbps token ring has a token holding timer value of 10 ms. What is the longest frame
that can be sent on this ring?
Ans: Here, token holding time = 10 ms
Thus, the longest frame that can be sent on this ring = data rate × token holding time
= 4 × 106 × 10 × 10–3
= 40 × 103 bits = 40,000 bits
= 5 KB (as 1 byte = 8 bits)

Multiple choice questions

1. In which of the following protocols, a station (a) If the line is not idle, it continuously
sends a frame whenever it wants? senses the line until it finds the line idle
(a) pure ALOHA (b) slotted ALOHA (b) If the line is not idle, it waits a random
(c) both (a) and (b) (d) none of these amount of time and then senses the line
again.
2. In which of the following protocols, the
(c) If the station finds the line idle it sends
station senses the channel before trying to
or refrains from sending based on the
use it?
outcome of a random number generator.
(a) ALOHA (b) CSMA
(d) None of these
(c) CDMA (d) none of these
5. In which of the following methods, the stations
3. The throughput of pure ALOHA is expressed
refer each other to know which station has the
as:
right to transmit a frame?
(a) S = Ge – 2G (b) S = Ge – G
(a) random access
(c) S = 2Ge – 2G (d) S = 2Ge–G
(b) controlled access
4. Which of the following is true with reference (c) channelization
to 1-persistent CSMA? (d) none of these

CN_Chapter 08.indd 159 2/20/2012 6:31:20 PM

 160 Data Communications and Computer Networks

6. Which of the following protocol involves divi- 8. In which of the following methods, the stations
sion of time into intervals and at each interval are arranged in logical ring?
reservation frame precedes the data frame? (a) reservation (b) polling
(a) reservation (b) polling (c) token passing (d) none of these
(c) token passing (d) none of these 9. Channelization involves:
7. In which of the following methods, primary (a) FDMA (b) TDMA
station controls the link? (c) CDMA (d) all of these
(a) reservation 10. Which of the following is based on coding
(b) polling theory?
(c) token passing (a) FDMA (b) TDMA
(d) none of these (c) CDMA (d) none of these

Answers
1. (a) 2. (b) 3. (a) 4. (a) 5. (b) 6. (a) 7. (b) 8. (c) 9. (d) 10. (c)

CN_Chapter 08.indd 160 2/20/2012 6:31:20 PM

 9
Ethernet, Virtual Circuit Networks
and SONET

1. Write a short note on IEEE 802 standards.

Ans: IEEE 802 standards were developed by IEEE in 1985 for local area networks (LANs). These
standards are compatible with each other at data link layer and enable intercommunication between
various devices developed by different manufacturers. These standards are classified into the following
categories:
‰‰ IEEE 802.1: This category defines interface primitives for LAN. It deals with internetworking
aspect, which seeks to settle down the conflicts between incompatible devices.
‰‰ IEEE 802.2: This category specifies the upper part of the data link layer and supports the logical
link protocol.
‰‰ IEEE 802.3: This category supports Ethernet. Further, CSMA/CD protocol is used in the Ethernet
to control simultaneous access to the channel by multiple media.
‰‰ IEEE 802.4: This category has been specified for LANs based on Token Bus architecture.
It ­supports token passing access method and bus topology.
‰‰ IEEE 802.5: This category describes standards for LANs based on Token Ring. It supports token
passing access method and ring topology.
‰‰ IEEE 802.6: This category is used for distributed queue dual bus (DQDB) architecture. It has been
developed for usage in metropolitan area networks (MANs).
‰‰ IEEE 802.11: This category applies to wireless Ethernet or Bluetooth. Bluetooth is a technology
that is used for small wireless LANs.
2. Explain IEEE 802 reference model.
Ans: IEEE 802 reference model was developed by IEEE and adopted by all the organizations for
LAN standards. Initially, the American National Standards Institute (ANSI) adopted the standard and
later in 1987, it was approved by the International Organization for Standards (ISO) as an international
standard. The 802 reference model is related to the open systems interconnection (OSI) model as shown
in Figure 9.1.

CN_Chapter 09.indd 161 2/20/2012 6:32:46 PM

 162 Data Communications and Computer Networks

Upper layers Upper layers

LLC
Data link layer
Token Ring Token Bus
Ethernet MAC ...
MAC MAC
Ethernet
Token Ring Token Bus Physical layer
physical layers ...
physical layer physical layer
(several)
(a) IEEE 802 Standard (b) OSI model
Figure 9.1 Relationship Between IEEE 802 Reference Model and OSI Model

Physical Layer
The lowest layer of IEEE 802 reference model is equivalent to the physical layer of the OSI model.
It depends on the type and implementation of transmission media. It defines a specification for the
­medium of transmission and the topology. It deals with the following functions:
‰‰ encoding of signals at the sender’s side and the decoding of signals at the receiver’s side;
‰‰ transmission and reception of bits and
‰‰ preamble generation for synchronization.

Data Link Layer

The layer above the physical layer in the 802 reference model corresponds to the data link layer of the
OSI model. Here, the data link layer is divided into two layers, namely, logical link control (LLC) and
media access control (MAC).
‰‰ LLC: This layer is the upper part of the data link layer that provides an interface to upper layers
and performs functions such as flow control and error control. It also performs a part of framing
function. It provides single data link protocol for all IEEE LANs. In addition, it must support
multi-access channels and multiple user networks.
The LLC layer is concerned with the transmission of link-level protocol data unit (PDU) that is
identical to high-level data link control (HDLC) and referred to as LLC frame. Each LLC frame
consists of a header field followed by a data field. The header of LLC frame contains three sub-
fields, namely, destination service access point (DSAP),
source service access point (SSAP) and control (Figure 9.2) DSAP SSAP Control
and the data field is used to hold data received from upper
­layers. The description of subfields in LLC frame header is Figure 9.2 Format of LLC Frame Header
as follows:
‰‰ DSAP: This field contains a 7-bit address that identifies a user of LLC at the destination. One bit
of the address specifies whether it is a group or individual address. It specifies the upper-layer
protocol at the destination.
‰‰ SSAP: This field contains a 7-bit address that identifies the user of LLC at the source. One bit
of the address specifies whether it is a command or response PDU. It specifies the upper-layer
­protocol at the source.
‰‰ Control Field: This field is used to handle flow and error contro

CN_Chapter 09.indd 162 2/20/2012 6:32:46 PM

 Ethernet, Virtual Circuit Networks and SONET 163

‰‰ MAC: This layer forms the lower part of the data link layer that specifies the access control method
used for each type of LAN. For example, CSMA/CD is specified for Ethernet LANs and the to-
ken passing method is specified for Token Ring and Token Bus LANs. It also handles a part of
framing function for transmitting data. To implement all the specified functions, MAC also uses a
PDU at its layer, which is referred to as MAC frame. The MAC frame consists of various fields
(­Figure 9.3) which are described as follows:
• MAC Control: This field contains protocol control information such as priority level, which
is required for the proper functioning of protocol.
• Destination MAC Address: This field contains the address of the destination on the LAN for
this frame.
• Source MAC Address: This field contains the address of the source on the LAN for th s frame.
• LLC PDU: This field contains the LLC data arriving from the immediate upper laye .
• Cyclic Redundancy Check (CRC): This field contains the CRC code that is used for error
­detection. This field is also referred to as frame check sequence (FCS).

MAC Destination MAC Source MAC

LLC PDU CRC
control address address

Figure 9.3 Format of an MAC Frame

3. Discuss frame format of IEEE 802.3 standards.

Ans: IEEE 802.3 standard supports Ethernet which was developed in 1976 by Xerox Corporation.
It was developed as an improvement over prior networks and is capable of controlling access to channel
in case many stations attempt to transmit simultaneously. To control media access, it uses 1-persistent
CSMA/CD protocol.
In standard Ethernet, the MAC layer is responsible for performing the operation of access method.
Further, IEEE 802.3 has specified a type of an MAC frame for Ethernets. This frame consists of seven
fields (Figure 9.4), which are described as follows
‰‰ Preamble: It is the first field of the Ethernet frame. It contains 7 bytes of alternating 0s and 1s such
as 1010101 that are used to warn the receiver about the incoming frame so that the receiver may
synchronize its timing with the input. In fact, this field is appended at the physical layer and is not
a part of the frame.
‰‰ Start Frame Delimiter (SFD): It is a 1-byte long field that is used to determine the beginning of
the frame. The last two bits of this field are set to 11 to indicate the receiver that the next field is
destination address; also, it is the last chance for the receiver to synchronize its input timing.
‰‰ Destination Address (DA): It is a 6-byte long field that holds the physical address of the next
receiving station(s) to which the frame is to be transmitted.
‰‰ Source Address (SA): It is a 6-byte long field that holds the physical address of the station that
has sent the frame.

Destination Source Length Data and

Preamble SFD CRC
address address or type padding
7 bytes 1 byte 4 bytes
6 bytes 6 bytes 2 bytes 46-1500 bytes

Figure 9.4 Format of IEEE 802.3 MAC Frame

CN_Chapter 09.indd 163 2/20/2012 6:32:47 PM

 164 Data Communications and Computer Networks

‰‰ Length or Type: It is a 2-byte-long field that defines either the length or type of data. The
­Ethernet originally used this field as a type field to define the total length of data in upper-layer
protocols, while the IEEE standard used it as the length field to indicate total number of bytes in
data field
‰‰ Data: This field carries data arriving from upper-layer protocols. The amount of data stored in this
field can range between 46 and 1,500 bytes
‰‰ CRC: It is a 4-byte-long field that contains error detection information. In case of Ethernet
MAC frame, it is the CRC that is computed over all the fields except preamble, SFD and CRC
itself.
4. What are the four generations of Ethernet? Discuss Ethernet cabling in all the ­generations
of Ethernet.
Ans: The Ethernet was developed in 1976 at the Xerox’s Palo Alto Research Center (PARC). Since
its development, the Ethernet has been evolving continuously. This evolution of Ethernet can be cat-
egorized under four generations, which include standard Ethernet, fast Ethernet, gigabit Ethernet and
10-gigabit Ethernet. These generations are discussed here.

Standard Ethernet
Standard Ethernet uses digital signals (baseband) at the data rate of 10 Mbps and follows 1-­persistent
CSMA/CD as access method. A digital signal is encoded/decoded by sender/receiver using the
­Manchester scheme. Standard Ethernet has defined several physical layer implementations, out of
which the following four are commonly used.
‰‰ 10Base5: It uses a thick coaxial cable of 50 ohm and is implemented in bus topology with an
­external transceiver connected to coaxial cable. The transceiver deals with transmitting, receiving
and detecting collisions in the network. The cable used is too firm to bend with the hands. The maxi-
mum length of the cable should not be more than 500 m; otherwise, the signals may deteriorate.
In case greater length of cable is required, the maximum five segments each of 500 m can be used
and repeaters are used to connect the segments. Thus, the length of the cable can be extended up
to 2,500 m. The 10Base5 Ethernet is also referred to by other names including thick Ethernet or
thicknet.
‰‰ 10Base2: It is also implemented in the bus topology but with a thinner coaxial cable. The ­transceiver
in this Ethernet is a part of the NIC. The 10Base2 specification are cheaper than the 10Base5 as
the cost of thin cable is less than that of the 10Base5 specifications. The thinner cable can easily be
bent close to the nodes that results in flexibility and thus, making installation of the 10Base2 speci-
fication easier. The maximum length of the cable segment must not exceed 185 m. The 10Base2
Ethernet is also referred to as thin Ethernet or Cheapernet.
‰‰ 10Base-T: It uses two pairs of twisted cable and is implemented in star topology. All nodes are
connected to the hub via two pairs of cable and thus, creating a separate path for sending and
­receiving the data. The maximum length of the cable should not exceed 100 m; otherwise, the
signals may attenuate. It is also referred to as twisted-pair Ethernet.
‰‰ 10Base-F: It is the most common 10-Mbps Ethernet that is implemented in star topology. It uses
a pair of fibre optic cables to connect the nodes to the central hub. The maximum length of cable
should not exceed 2,000 m. It is also referred to as the fib e Ethernet.

CN_Chapter 09.indd 164 2/20/2012 6:32:47 PM

 Ethernet, Virtual Circuit Networks and SONET 165

Fast Ethernet
The IEEE 802.3 committee developed a set of specifications referred to as the fast Ethernet to provide
low-cost data transfer at the rate of 100 Mbps. It was designed to compete with LAN protocols such as
fibre distributed data interface (FDDI) and it was also compatible with the standard Ethernet. The fast
Ethernet uses a new feature called autonegotiation, which enables two devices to negotiate on certain
features such as data rate or mode of transmission. It also allows a station to determine the capability
of hub and two incompatible devices can also be connected to one another using this feature. Like the
standard Ethernet, various physical-layer implementations of the fast Ethernet have also been specified.
Some of them are as follows:
‰‰ 100Base-TX: It either uses two pairs of either cat5 UTP cable or STP cable. The maximum length
of the cable should not exceed 100 m. This implementation uses MLT-3 line coding scheme due
to its high bandwidth. However, since MLT-3 coding scheme is not self-synchronized, the 4B/5B
block coding scheme is used to prevent long sequences of 0s and 1s. The block coding increases
the data rate from 100 Mbps to 125 Mbps.
‰‰ 100Base-FX: It uses two wires of fibre optic cable that can easily satisfy the high bandwidth
requirements. The implementation uses NRZ-I coding scheme. As NRZ-I scheme suffers from
synchronization problem in case of long sequence of 0s and 1s, 4B/5B block coding is used with
NRZ-I to overcome this problem. The block coding results in increased data rate of 125 Mbps. The
maximum cable length in 100Base-FX must not exceed 100 m.
‰‰ 100Base-T4: It is the new standard that uses four pairs of cat3 or higher UTP cables. For this
implementation, 8B/6T line coding scheme is used. The maximum length of cable must not exceed
100 m.

Gigabit Ethernet
Gigabit Ethernet was developed by the IEEE 802.3 committee to meet the higher data rate requirements.
This standard provides a data rate of 1000 Mbps (1 Gbps). It is backward compatible with traditional
and fast Ethernet and also supports autonegotiation feature. Various physical layer implementations of
gigabit Ethernet are as follows:
‰‰ 1000Base-SX: It is a two-wire implementation that uses short wave fibres. One wire is used for
sending the data and other is used for receiving the data. The NRZ line coding scheme and 8B/10B
block coding scheme is used for this implementation. The length of the cable should not exceed
550 m in the 1000Base-SX specifications
‰‰ 1000 Base-LX: It is also a two-wire implementation that uses long wave fibres. One wire is used
for sending the data and other is used for receiving the data. It is implemented by the NRZ line
coding scheme and the 8B/10B block-coding scheme. The length of the cable should not exceed
5,000 m in the 1000Base-LX specifications
‰‰ 1000 Base-CX: It uses two STP wires where one wire is used for sending the data and other is
used for receiving the data. It is implemented by the NRZ line coding scheme and the 8B/10B
block-­coding scheme. The length of the cable should not exceed 25 m in the 1000Base-CX
­specifications
‰‰ 1000 Base-T: It uses four cat5 UTP wires. It is implemented by the 4D-PAM5 line coding scheme.
In this specification, the length of the cable should not exceed 100 m

CN_Chapter 09.indd 165 2/20/2012 6:32:47 PM

 166 Data Communications and Computer Networks

Ten-Gigabit Ethernet
This standard was named as 802.3ae by the IEEE 802.3 committee. It was designed to increase the data
rate to 1000 Mbps (10 Gbps). It is compatible with standard Ethernet, fast Ethernet and gigabit Ethernet.
It enables the Ethernet to be used with technologies such as Frame Relay and ATM. Various physical-
layer implementations of 10-gigabit Ethernet are as follows:
‰‰ 10GBase-S: It uses short-waves fibres and is designed for 850 nm transmission on multimode
fibre. The maximum length of the cable should not exceed 300 m.
‰‰ 10GBase-L: It uses long-wave fibres and is designed for 1,310 nm transmission on single-mode
fibre. The maximum length of the cable should not exceed 10 km.
‰‰ 10GBase-E: It uses extended waves and is designed for 1,550 nm transmission on single-mode
fibre. The maximum distance that can be achieved using this medium, is up to 40 km.
5. What is Token Ring (IEEE 802.5)? How is it different from Token Bus (IEEE 802.4)?
Ans: The IEEE 802.5 is a specification of standards for LANs S4
that are based on Token Ring architecture. The Token Ring network
was originally developed by IBM in the 1970s. It is the most com-
monly used MAC protocol that uses token passing mechanism with
ring topology. In this protocol, the stations are connected via point-
to-point links with the use of repeaters (Figure 9.5). To control the S1 S3
T
media access, a small frame called token (a 3-byte pattern of 0s and 1s)
is allowed to move around the network and the station possessing
the token can only transmit frames in the allotted time.
Whenever a station wants to transmit a frame, it first needs to grab S2
a token from the network before starting any transmission. Then, it ap-
pends the information with the token and sends it on the network. The Figure 9.5 Token Ring LAN
information frame then circulates around the network and ­eventually,
received by the intended destination. After receiving the information frame, the destination copies the
­information and sends the information frame back on the network with its two bits set to indicate that it is an
acknowledgement. The information frame then moves around the ring and is finall , received by the send-
ing station. The sending station checks the returned frame to determine whether it has been received with
or without errors. If the sending station has now finished the transmission, it creates a new token and inserts
it on the network. Notice that while one station is transmitting the data, no other station can grab a token.
Thus, collisions cannot occur as only one station can transmit at a time. In addition, if a station does not have
a frame to send or the time allotted to it passes away, the token is immediately passed to the next station.
In Token Ring networks, the ring topology is used in which the failure of any one station can bring
the entire network down. Thus, another standard known as Token Bus (IEEE 802.4) was developed as
an improvement over Token Ring networks. Like Token Ring, Token Bus is also based on token-passing
mechanism. In Token Bus, the stations are logically organized into a ring but physically organized into
a bus (Figure 9.6). Thus, each station knows the addresses of its adjacent (left and right) stations. After
the logical ring has been initialized, the highest numbered station may transmit. The sending station
broadcasts the frame in the network. Each station in the network receives the frame and discards if
frame is not addressed to it. When a station finishes the transmission of data or the time allotted to it
passes away, it inserts the address of its next neighbouring station (whether logical or physical) on the
token and passes it to that station.

CN_Chapter 09.indd 166 2/20/2012 6:32:47 PM

 Ethernet, Virtual Circuit Networks and SONET 167

S3
S1 S5

Logical
ring

Broadband
coaxial cable Token

S2 S4 S6
Figure 9.6 Token Bus LAN
6. Compare IEEE 802.3, 802.4 and 802.5.
Ans: There are certain differences among the IEEE 802.3, 802.4 and 802.5 standards that are listed
in ­Table 9.1.
Table 9.1 Comparison among IEEE 802.3, 802.4 and 802.5

IEEE 802.3 IEEE 802.4 IEEE 802.5

• This standard uses 1- persistent • This standard uses Token • This standard uses Token Ring
CSMA/CD medium access protocol. Bus medium access protocol. medium access protocol.
• The stations are logically connected • The stations are logically • The stations are physically
to each other via a broadcast cable connected to each other via connected to each other via point-to-
medium. a broadcast cable medium. point links.
• Frames are broadcasted to the • Frames are broadcasted to • Frames are transmitted to the
destination. the destination. destination using point-to-point links.
• Transmission media used is • Transmission medium used • Transmission medium used is
generally coaxial cable, optical fibre is generally coaxial cable or generally coaxial cable, optical fibre
or twisted pair. twisted pair. It is not well or twisted pair.
suited to fibre cables.
• T here is no prioritization of stations • Stations are prioritized for • S tations are prioritized for
for transmission of data. transmission of data. transmission of data.
• It cannot transmit short frames. • It can handle transmission • It can handle transmission of short
of short frames. frames.
• It cannot be used for real-time • It is used for real-time • It is used for office automation.
applications. applications.
• It applies Manchester encoding. • It applies analog encoding. • It applies differential Manchester
encoding.
• A t high loads, its efficiency is very • A t high loads, its efficiency • At high loads, its efficiency is also
low. However, at low loads, its is very high. However, at high and at low loads, its efficiency
efficiency is high due to less delay. low loads, its efficiency is is low similar to the Token Bus.
low due to more delay.

CN_Chapter 09.indd 167 2/20/2012 6:32:48 PM

 168 Data Communications and Computer Networks

7. Write a short note on FDDI. Explain access method, time registers, timers and station
procedure.
Ans: The fib e distributed data interface (FDDI) refers to the first high speed LAN protocol
­standardized by ANSI and ITU-T. It has also been approved by the ISO and resembles IEEE 802.5
standards. It uses fibre optic cable; thus, packet size, network segment length and number of stations
increase. It offers a speed of 100 Mbps over the distance of up to 200 km and connects up to 1,000
­stations. The distance between any two stations cannot be more than a few kilometers.

Access Method
The FDDI employs the token passing access method. A station possessing the token can transmit
any number of frames within the allotted time. There are two types of frames provided by the FDDI:
­synchronous and asynchronous. Synchronous frame (also called S-frame) is used for real-time ap-
plications such as audio and video. The frame needs to be transmitted within a short period of time
without much delay. Asynchronous frame (also called A-frame) is used for non-real-time applications
(such as data traffic) that can tolerate large delays. If a station has both S-frames and A-frames to send,
it must send S-frames first. After sending the S-frame, if the allotted time is still left then A-frames can
be transmitted.

Time Registers
Three time registers are used to manage the movement of token around the ring, namely, synchronous
allocation (SA), target token rotation time (TTRT) and absolute maximum time (AMT). The SA register
specifies the time for transmitting synchronous data. Each station can have a different value for it. The
TTRT register specifies the average time a token needs to move around the ring exactly once. The AMT
register has a value two times the value of the TTRT. It specifies the maximum time that it can take for
a station to receive a token. However, if a token takes more time than the AMT, then the reinitialization
of the ring has to be done.

Timers
Each station maintains two types of timers to compare the actual time with the value present in time
registers. These timers include token rotation timer (TRT) and token holding timer (THT). The TRT cal-
culates the total time taken by the token to complete one cycle. This timer runs continuously. The THT
starts when the token is received by a station. This timer indicates the time left for sending A-frames
after the S-frames have been sent.

Station Procedure
When a station receives the token, it uses the following procedure:
1. It sets the THT to a value equal to (TTRT – TRT).
2. It sets TRT to zero.
3. It transmits synchronous frame. With each sent unit, the value of TRT is decremented by one.
4. It continues to send asynchronous data as long as the value of THT is positive.

CN_Chapter 09.indd 168 2/20/2012 6:32:48 PM

 Ethernet, Virtual Circuit Networks and SONET 169

8. What are the advantages of FDDI over a basic Token Ring?

Ans: Though FDDI, like Token Ring, is a token-passing protocol, it provides certain advantages
over Token Ring. Some of these advantages are as follows:
‰‰ It provides a data rate of 100 Mbps as compared to the 10–16 Mbps in Token Ring.
‰‰ It supports large number of stations as compared to Token Ring.
‰‰ The access method used in FDDI is timed-token passing. It supports transmission of multiple
frames at the same time after capturing a token that is not possible in Token Ring.
‰‰ In case of Token Ring, the transmitting station releases the token after it receives the acknowledge-
ment of sent frames. On the other hand, in FDDI the station releases the token immediately after it
has finished the transmission. This is known as early token release.
‰‰ It offers higher reliability than Token Ring by using dual counter ring topology. In this, two types
of rings are used, namely, primary and secondary in contrast to Token Ring in which only one ring
is present. Both rings are used to transmit data, however, in opposite directions to provide fault
tolerance. Thus, in case one ring breaks or some station malfunctions, the load can be shifted to
another ring. Moreover, if both rings break at the same point, two rings can be joined together to
form a single ring. Reliability can be further increased by using dual ring of trees or dual homing
that provides multiple paths. Thus, on failure of one path, another path can be chosen for passing
a token or data.
9. What is meant by wireless LAN? Mention its advantages and disadvantages.
Ans: Wireless LAN (WLAN) refers to a network that uses wireless transmission medium. It is used
to connect devices without using cables and introduces a higher flexibility for ad hoc communication.
Some of the advantages of wireless LAN are as follows:
‰‰ It is difficult to design small-size devices such as PDA and laptop with the use of cables and thus,
WLAN becomes an alternative to be used as a transmission media.
‰‰ In case of natural disasters, such as flood and earthquake, noise in the cable increases. Therefore,
WLAN can be used where transmission rate is not affected by such natural calamities.
‰‰ It can cover large area and number of devices. In addition, new devices can be added easily without
affecting the existing network design.
‰‰ The nodes can communicate without any restriction and from anywhere.
Some disadvantages of WLAN are as follows:
‰‰ It is an expensive medium of transmission as compared to cables. Certain wireless devices such as
wireless LAN adaptor and PC card are quite expensive.
‰‰ Installation of WLANs is expensive.
‰‰ Radio waves used in WLAN might interfere with various devices and thus, are not secure
to use.
‰‰ WLANs provide low quality of transmission because of high error rate due to interference.
‰‰ WLANs are restricted to only certain frequency bands in radio transmission.

10. Differentiate between wired and wireless LAN.

Ans: Both wired and wireless LANs are used for establishing a network within an organization but
there are some differences between them. Table 9.2 lists these differences.

CN_Chapter 09.indd 169 2/20/2012 6:32:48 PM

 170 Data Communications and Computer Networks

Table 9.2 Differences Between Wired and Wireless LAN

Wired LAN Wireless LAN

• It is difficult to set up wired LANs, as it requires long and • It is easier to set up WLANs as compared to
large number of wires to connect devices to each other or wired LANs.
to the central device such as hub or switch.
• The total cost required to set up these type of networks • W ireless adapters and access points are
include cost of cable, hubs, switches and various software three to four times expensive as compared to
packages to install the network. components of wired LAN.
• It is more secure because firewalls can be installed on the • It is less secure because radio waves travel
computers. through air and can be easily intercepted.
• The components used in wired LANs including Ethernet • It is less reliable as signals are easily interfered
cable, hub and switch are extremely reliable. from home appliances, causing more problems.
• Wired LANs provide less flexibility, as one need to • The use of WLANs provides flexibility as it is not
separate printers, modems and scanners on every required to separate CD-ROMs, colour printers
computer. and B/W printers.

11. Explain two types of services of IEEE 802.11.

Ans: The IEEE 802.11 is a standard specified for wireless LANs. Two types of services namely,
basic service set (BSS) and extended service set (ESS) has been defined for IEEE 802.11. These two
services are explained as follows:

Basic Service Set

The BSS acts as the main building block of a WLAN.
It consists of many wireless stations (stationary or S4 S3 S4 S3
­mobile) and an optional base station called access
point (AP). If AP is present in the network, then AP
BSS is referred to as the infrastructure network
[­Figure 9.7(a)]; otherwise, BSS is referred to as an S1 S2 S1 S2
adhoc network [Figure 9.7(b)], which cannot transmit (a) Infrastructure network (b) Ad hoc network
data to any other BSS. Figure 9.7 Basic Service Set (BSS)

Extended Service Set

The ESS is formed by the combination of at least two BSSs with APs (Figure 9.8). All BSSs are
­connected via a distribution system that is generally a wired LAN such as Ethernet. The distribution
system connects the APs of each BSS to one another and thus, enables the communication between
BSSs. Each BSS can have two types of stations: mobile and stationary. The stations inside the BSS are
mobile ­stations while AP of each BSS is a stationary station. The stations inside a BSS can communicate
with one another without requiring the use of an AP. However, if communication is required between
two stations of different BSSs, then it occurs via APs. The ESS is analogous to a cellular network in
which each cell can be considered as a BSS and each base station as an AP.

CN_Chapter 09.indd 170 2/20/2012 6:32:48 PM

 Ethernet, Virtual Circuit Networks and SONET 171

Distribution system

BSS Server or
Gateway

AP AP AP

Figure 9.8 Extended Service Set (ESS)

12. What are the types of frames specifie in IEEE 802.11?

Ans: The IEEE 802.11 has specified three types of MAC frames for WLANs, which include manage-
ment frames, control frames and data frames. All these three types are discussed as follows:
‰‰ Management Frame: This frame is used for managing communication between stations and APs.
It manages request, response, reassociation, dissociation and authentication.
‰‰ Control Frame: This frame is used for accessing the channel and acknowledging frames. It ­ensures
reliable delivery of data frames. There are six subtypes of a control frame which are as follows:
• Power Save-Poll (PS-Poll): This frame is used by a station to send a request to an AP for the
transmission of frames buffered by AP for that station while the station was in power saving
mode.
• Request to Send (RTS): Whenever a station wishes to send data to another station, it first
sends an RTS frame to declare to the destination and other stations within its reception range
that it is going to send data to that destination.
• Clear to Send (CTS): This frame is sent by a station in response to an RTS frame. It indicates to
the source station that the destination station has granted permission for sending data frames.
• Acknowledgement: This frame is sent by the destination station to the source station to
­acknowledge the successful receipt of the previous frames.
• Contention-Free (CF)-end: This frame is used to indicate the end of the contention-free period.
• CF-End + ACK: This frame is used to acknowledge the CF-end frame. It ends the period and
all the bound stations are freed from all the restrictions associated with that period.
‰‰ Data Frame: This frame is used for carrying data and control information from a source station
to a destination station. Data frames are divided into eight subtypes, which are further organized
under two groups. One group contains the data frames that are used to carry the user data (received
from upper layers) from the source station to the destination station. The data frames included in
this group are described as follows:
• Data: This is the simplest data frame used to send data in both contention period and
­contention-free period.
• Data + CF-ACK: This frame carries data and also acknowledges the data, which has been
­received previously. It can be used only in the contention-free period.

CN_Chapter 09.indd 171 2/20/2012 6:32:48 PM

 172 Data Communications and Computer Networks

•Data + CF-Poll: This frame is used by a point coordinator to send data to a mobile station. It also re-
quests the mobile station to send a data frame, which may have been buffered by the mobile station.
• Data + CF-ACK + CF-Poll: This frame combines the functionality of two frames Data + CF-
ACK and Data + CF-Poll into a single frame.
Besides, there is another group that contains four more subtypes of data frames, which do not carry any
user data. One of these frames is null-function data frame that carries the power management bit in the
frame control field to the AP. It indicates that the station is moving to a low-power operating state. The
remaining three frames (CF-ACK, CF-Poll, CF-ACK + CF-Poll) function in the same way as that of last
three frames in the first group, the only di ference being that they do not contain any data.
13. Explain the frame format of 802.11 standards.
Ans: The IEEE 802.11 has defined three MAC layer frames for WLANs including control, data, and
management frames. Figure 9.9 shows the format of data frame of IEEE 802.11 that comprises nine
fields. The format of management frames is also similar to data frames except that it does not include
one of the base station addresses. The format of control frames does not include frame body and SC
fields. It also includes one or two address fields
The description of fields included in the IEEE 802. 1 MAC frame is as follows:

FC D Address 1 Address 2 Address 3 SC Address 4 Frame body FCS

2 bytes 2 bytes 6 bytes 6 bytes 6 bytes 2 bytes 6 bytes 0-2312 bytes 4 bytes

Figure 9.9 Frame Format of the IEEE 802.11 Standard

‰‰ Frame Control (FC): It is a 2-byte-long field in which the first byte indicates the type of frame
(control, management, or data) and the second byte contains control information such as fragmen-
tation information and privacy information.
‰‰ Duration (D): It is a 2-byte-long field that defines the channel allocation period for the transmis-
sion of a frame. However, in case of one control frame, this field stores the ID of the frame.
‰‰ Addresses: The IEEE 802.11 MAC frame contains four address fields and each field is 6-byte-
long. In case of a data frame, two of the frame address fields store the MAC address of the original
source and destination of the frame, while the other two store the MAC address of transmitting and
receiving base stations that are transmitting and receiving frames respectively over the WLAN.
‰‰ Sequence Control (SC): It is a 2-byte (that is 16 bits) field, of which 12 bits specify the sequence
number of a frame for flow control. The remaining 4 bits specify the fragment number required for
reassembling at the receiver’s end.
‰‰ Frame Body: It is a field that ranges between 0 and 2,312 bytes and contains payload ­information.
‰‰ FCS: It is a 4-byte-long field that comprises a 32-bit CRC
14. With reference to 802.11 wireless LAN, explain the following:
(a) Hidden terminal problem
(b) Exposed terminal problem
(c) Collision avoidance mechanisms
Ans:
(a) Hidden Terminal Problem The hidden terminal problem occurs during communication
­between two stations in wireless ­networks. It is the problem where a station is unable to detect another

CN_Chapter 09.indd 172 2/20/2012 6:32:49 PM

 Ethernet, Virtual Circuit Networks and SONET 173

station while both of them are competing for the

same data transfer medium. To understand how this
problem occurs, consider three stations X, Y and Z Range Range
situated in a network as shown in Figure 9.10. The of X of Z
transmission range of station X is represented by the X Y Z
left oval and that of Z is represented by the right
oval. The stations falling in either of the ovals can Figure 9.10 Hidden Terminal Problem
hear the signal transmitted by the station situated in
that oval. However, stations X and Z are outside the transmission range of each other, that is, they are
hidden from each other. As the station Y is situated in the area common to X and Z, it can hear the signals
transmitted by both X and Z. Now, suppose, while X is transmitting data to Y, Z also wants to send
data to Y. As Z is unable to hear the transmission from X to Y, it assumes that the transmission medium
is free and starts sending data frames to Y. This results in a collision at station Y, as it is receiving from
both X and Z, which are hidden from each other with respect to Y.
The hidden terminal problem can be solved by making use of the RTS and CTS frames before start-
ing the transmission of data. Initially, station X sends an RTS frame to station Y to request for sending
data. The transmission of RTS frame cannot be detected by Z. In response to RTS frame, station Y sends
a CTS frame, which specifies the duration of transmission from X to Y. As Y in the transmission range
of Z, Z can detect the transmission of CTS frame and knows that Y is busy with any other station and
also for how long. Therefore, it does not initiate the transmission until that duration is finished.
(b) Exposed Terminal Problem This prob-
lem is the reverse of the hidden terminal problem. Range
This problem arises when a station ­restricts itself of P
from using another station that is in fact available
for use. To understand how this problem occurs,
Range
consider four stations P, Q, R and S in a network as of R
shown in Figure 9.11. Suppose that while the station Range
P is sending data to station Q, there arises a need in of Q
Q P R S
station R to send data to S. The transmission from R
to S can be done without disturbing the transmission
from P to Q. However, as the station R is exposed Figure 9.11 Exposed Terminal Problem
to transmission range from P, it stops itself from
transmitting to S after realizing that P is transmitting some data. Such a situation is known as exposed
terminal problem.
(c) Collision Avoidance Mechanisms To avoid the collisions in the wireless networks, a collision
avoidance protocol named as multiple access with collision avoidance (MACA) has been designed. This
protocol requires a sender to make the receiver send a short frame before starting the transmission of data
frames. This frame is used as an announcement to the nearby frames that transmission is going on between
sender and the receiver and no other station should interfere in between, thus, avoiding any collision.
To understand how MACA works, consider five stations P, Q, R, S and T as shown in Figure 9.12.
Suppose P wants to send data to Q. Initially, P sends an RTS frame to Q specifying the length of the
data frame. The station Q then responds with a CTS frame specifying the duration for transmission from
P to Q. After receiving the CTS frame, P begins transmitting data to Q. Now, as the station S is in the
transmission range of Q, it hears the CTS message from Q and thus, remains silent for the duration, Q is
receiving from P. The station R is in the transmission range of P and hears the RTS message from P (but
not the CTS message from Q). Thus, R can transmit to P without avoiding the collision as long as the

CN_Chapter 09.indd 173 2/20/2012 6:32:49 PM

 174 Data Communications and Computer Networks

data from R to P does not interfere with CTS from Q Range of P Range of Q
to P. The station T being in the transmission range
of both P and Q hears RTS and CTS both and thus,
remain silent until the transmission is over. R P Q S
The disadvantage of the MACA protocol is that
collision can still occur in case both Q and R trans- T
mit RTS frames to P simultaneously. The RTS from
Q and R may collide at P; because of collision, nei-
ther Q nor R receives the CTS frame. Then both Q
and R wait for a random amount of time using bi- Figure 9.12 MACA Protocol
nary exponential back off algorithm (explained in
Chapter 8) and again retry to transmit.
To overcome the disadvantage and improve the performance of MACA, it was enhanced in 1994
and renamed as MACA for wireless (MACAW). This newer version of MACA includes several
­enhancements, some of which are as follows:
‰‰ To identify the frames that have been lost during the transmission, the receiver must acknowledge
each successfully received data frame by sending an ACK frame.
‰‰ CSMA protocol is used for carrier sensing so that no two stations could send an RTS frame at the
same time to the same destination.
‰‰ Instead of running the binary back off exponential algorithm for each station, it is run for a pair of
transmitting stations (source and destination).
15. What is meant by Bluetooth? Explain its architecture.
Ans: Bluetooth is a short-range wireless LAN technology through which many devices can be
linked without using wires. It was originally started as a project by the Ericsson Company and then
formalized by a consortium of companies (Ericsson, IBM, Intel, Nokia and Toshiba). Bluetooth is sup-
posed to get its name from 10th century Danish king Harald Bluetooth who united Scandinavian Europe
(Denmark and Norway) during an era when these areas were torn apart due to wars. The Bluetooth tech-
nology operates on the 2.4 GHz industrial, scientific and medical (ISM) band and can connect different
devices such as computer, printer and telephone. The connections can be made up to 10 m or extended up
to 100 m depending upon the Bluetooth version being used.

Bluetooth Architecture
A Bluetooth LAN is an ad hoc network, which means the network is formed by the devices themselves
by detecting each other’s presence. The number of devices connected in Bluetooth LAN should not be
very large as it can support only a small number of devices. The Bluetooth LANs can be classified into
two types of networks, namely, piconet and scatternet (Figure 9.13).
‰‰ Piconet: It refers to a small Bluetooth network in which the number of stations cannot exceed
eight. It consists of only one primary station (known as master) and up to seven secondary stations
(known as slaves). If number of secondary stations exceeds seven, then 8th secondary station is put
in the parked state. A secondary station in the parked state cannot participate in communication in
the network until it leaves the parked state. All stations within a piconet share the common chan-
nel (communication link) and only a master can establish the link. However, once a link has been
established, the other stations (slaves) can also request to become master. Slaves within a piconet
must also synchronize their internal clocks and frequency hops with that of the master.

CN_Chapter 09.indd 174 2/20/2012 6:32:50 PM

 Ethernet, Virtual Circuit Networks and SONET 175

S3 S6

S1
M1 S5 M2

S2 S4 S7
Piconet Piconet

Scatternet
Figure 9.13 Bluetooth Piconets and Scatternets

‰‰ Scatternet: It refers to a Bluetooth network that is formed by the combination of piconets. A ­device
may be a master in one piconet whiles a slave in another piconet or a slave in more than one piconet.
16. Discuss the Bluetooth protocol stack.
Ans: Bluetooth protocol stack is a combination of multiple protocols and layers (Figure 9.14).
­Bluetooth comprises several layers including radio layer, baseband layer, L2CAP layer and other upper
layers. In addition, various protocols are also associated with Bluetooth protocol stack. The description
of these layers and protocols is as follows:

vCard/vCal WAE

OBEX WAP
AT-
TCS BIN SDP
Commands
UDP TCP

IP

PPP

RFCOMM

Audio
L2CAP Layer

Host Controller Interface

LMP

Baseband Layer

Radio Layer

Figure 9.14 Bluetooth Protocol Stack

CN_Chapter 09.indd 175 2/20/2012 6:32:50 PM

 176 Data Communications and Computer Networks

Radio Layer
This is the lowest layer in the Bluetooth protocol stack and is similar to a physical layer of transmission
control protocol/Internet protocol (TCP/IP) model. The Bluetooth devices present in this layer have low
power and a range of 10 m. This layer uses an ISM band of 2.4 GHz that is divided into 79 channels, each
of 1 MHz. To avoid interference from other networks, the Bluetooth applies frequency-hopping spread
spectrum (FHSS) technique. Here, a packet is divided into different parts and each part is transmitted
at a different frequency. The bits are converted to signal using a variant of FSK, known as ­Gaussian
bandwidth filterin shift keying (GFSK). In GFSK, the bit 1 is represented by a frequency deviation
above the carrier frequency used and bit 0 by a frequency deviation below the carrier frequency.

Baseband Layer
This layer is similar to MAC sublayer in LANs. It uses time division multiplexing (TDMA) and the
primary and secondary stations communicate with each other using time slots. Bluetooth uses a form
of TDMA known as TDD-TDMA (time-division duplex TDMA)—a sort of half-duplex communica-
tion, which uses different hops for each direction of communication (from primary to secondary or vice
versa). If there is only one secondary station in the piconet, then the secondary station uses even num-
bered slots while the primary station using odd-numbered slots for communication. That is, in slot 0,
data flows from primary to secondary while in slot 1 data flows from secondary to primary. This ­process
continues until the end of frame transmission. Now, consider the case where there is more than one
­secondary station in the piconet. In this case also, primary station sends in even-numbered slots, how-
ever, only one secondary station (out of many) who had received the data in the previous slot transmits
in the odd-numbered slot. For example, suppose in slot 0, the primary station (P) has sent data intended
for a secondary station (S) then only S can transmit in slot 1.

L2CAP Layer
The logical link control and adaptation protocol
Length Channel ID Data and control
(L2CAP) is similar to LLC sublayer in LANs. This 2 bytes 2 bytes 0-65535 bytes
layer is used for exchanging data packets. Each data
packet comprises three fields (Figure 9.15), which are Figure 9.15 Format of Data Packet of L2CAP Layer
as follows:
‰‰ Length: It is a 2-byte long field that is used to specify the size of data received from upper layers
in bytes.
‰‰ Channel ID (CID): It is a 2-byte long field that uniquely identifies the virtual channel made at
this level.
‰‰ Data and Control: This field contains data that can be up to 65,535 bytes as well as other control
information.
The L2CAP layer performs many functions that are discussed as follows:
‰‰ Segmentation and Reassembly: Application layer sometimes delivers a packet that is very large
in size, however, baseband layer supports only up to 2,774 bits or 343 bytes of data only in the
payload field. Thus, the L2CAP layer divides large packets into segments at the source and these
packets are reassembled again at the destination.

CN_Chapter 09.indd 176 2/20/2012 6:32:51 PM

 Ethernet, Virtual Circuit Networks and SONET 177

‰‰ Multiplexing: The L2CAP deals with multiplexing. At the sender’s side, it acquires data from
the upper layers, frames them and gives them to the baseband layer. At the receiver’s station,
it ­acquires frames from the baseband layer, extracts the data and gives them to the appropriate
­protocol layer.

Link Manager Protocol

The link manager protocol (LMP) helps a Bluetooth device to discover other devices when they come
across within the radio range of each other. It uses peer-to-peer message exchange in order to perform
various security functions such as authentication and encryption. The LMP layer performs the following
functions:
‰‰ Generation and exchange of encryption keys.
‰‰ Link setup and negotiation of baseband packet size.
‰‰ Controlling the power modes, connection state and duty cycles of Bluetooth devices in a piconet.

Host Controller Interface

The host controller interface (HCI) provides command line access to LMP and the baseband layer in
order to control and receive the status information. It consists of the following three parts:
‰‰ The HCI firmwa e, which is a part of the actual Bluetooth hardware.
‰‰ The HCI driver, which is present in the Bluetooth device software.
‰‰ The host controller transport layer, which is used to connect the firmware with the drive .

Radio Frequency Communication

Radio frequency communication (RFCOMM) is a serial line communication protocol. It communicates
with other upper layer protocols and tells them that the current Bluetooth devices are working over a
RS232 wired serial interface.

Service Discovery Protocol (SDP)

Service discovery protocol (SDP) allows a Bluetooth device to join a piconet. It also tells about the
available services, their types and the mechanism to access these services.

Telephony Control Protocol Binary (TCS BIN)

Telephony control protocol binary (TCS BIN) is a bit-oriented protocol that helps to setup speech and data
calls between Bluetooth devices by defining all essential call control signalling protocols. It also defines mo-
bility management procedures to handle a group of Bluetooth telephony control services (TCS) devices.

AT-Commands
This protocol consists of a set of AT-commands (attention commands) which are used to configure and
control a mobile phone to act as a modem for fax and data transfers.

CN_Chapter 09.indd 177 2/20/2012 6:32:51 PM

 178 Data Communications and Computer Networks

Point-to-Point Bluetooth
This is a point-to-point protocol (PPP) that takes IP packets to/from the PPP layer and places them onto
the LAN.

TCP/IP
This protocol is used for Internet communication.

Object Exchange Protocol

The Object Exchange (OBEX) is a session protocol, which is used to exchange objects. It works like the
hypertext transfer protocol but with a much lighter fashion. It helps to browse the contents of a folder
on some remote device.

vCard/vCal
These are the content formats supported by OBEX protocol. A vCard specifies the format for electronic
business card while vCal specifies the format for entries in personal calendar, which are maintained by
Internet mail consortium.
17. Write a short note on virtual circuit networks.
Ans: A virtual circuit network includes the characteristics of both circuit-switched and a datagram net-
work and performs switching at the data link layer. Like circuit-switched networks, it requires a virtual
connection to be established between the communicating nodes before any data can be transmitted. Data
transmission in virtual circuit networks involves three phases: connection setup, data transfer and con-
nection teardown phase. In connection setup phase, the resources are allocated and each switch creates
an entry for a virtual circuit in its table. After establishment of virtual connection, data transfer phase
begins in which packets are transmitted from source to destination; all packets of a single message take
the same route to reach the destination. In connection teardown phase, the communicating nodes inform
the switches to delete the corresponding entry.
In virtual circuit networks, data is transmitted in the form of packets, where each packet contains an
­address in its header. Each packet to be transmitted contains a virtual circuit identifier (VCI) along with
the data. The VCI is a small number, which is used as an identifier of packets between two switches.
When a packet arrives at a switch, its existing VCI is replaced with a new VCI when the frame leaves
from the switch.
The main characteristic of virtual circuit networks is that nodes need not make any routing decision
for the packets, which are to be transferred over the network. Decisions are made only once for all the
packets using a specific virtual circuit. At any instance of time, each node can be connected via more
than one virtual circuit to any other node. Thus, transmitted packets of a single message are buffered at
each node and are queued for output while packets using another virtual circuit on the same node are
using the line. Some of the advantages associated with virtual circuit approach are as follows:
‰‰ All packets belonging to the same message arrive in the same order to the destination as sent by the
sender. This is because every packet follows the same route to reach the receiver.
‰‰ It ensures that all packets arriving at the destination are free from errors. For example, if any node
receives a frame with an error, then a receiving node can request for retransmission of that frame.
‰‰ Packets transmit through the virtual circuit network more rapidly.

CN_Chapter 09.indd 178 2/20/2012 6:32:51 PM

 Ethernet, Virtual Circuit Networks and SONET 179

18. Write a short note on datagram networks.

Ans: Datagram networks are the connectionless networks used for packet switching at the network
layer. Here, the packets are referred to as datagrams. No virtual connection exists between the source
and the destination and each arriving datagram is treated independently by the switch regardless of
the source and destination address provided in the datagram. Thus, the different datagram even if they
belong to the same message may be forwarded through different paths to reach the destination. This
results in the unordered arrival of datagrams at the receiver and with varying delay times. As switches
are involved in processing datagrams belonging to other messages also, it might be possible that some
datagrams are lost or dropped because of the unavailability of resources.
In datagram networks, there is no need of any connection setup or teardown phase. Each switch
maintains a dynamic routing table that helps to deliver the datagrams to the intended receiver. This rout-
ing table contains the destination address of every node connected to the network and the corresponding
forwarding output port. This approach provides better efficiency as compared to other networks such as
circuit-switched networks because resources are allocated only when datagrams need to be transferred
instead of setting a connection and reserving the resources in advance. However, datagrams may have
to experience more delay as compared to packets in virtual circuit networks. This is because each data-
gram of a message can be forwarded through different switches and thus, may have to wait at a switch
depending on the resources available at the switch at that instance of time.
19. Differentiate virtual circuits and datagram networks.
Ans: There are certain differences between virtual circuits and datagram networks that are listed in
Table 9.3.
Table 9.3 Differences between Virtual Circuit Network and Datagram Networks

Virtual Circuit Networks Datagram Networks

• It is a connection-oriented service. Thus, it requires • It is a connectionless service.
setting up a circuit between the sender and receiver
before transmission.
• Each frame that is to be transmitted contains a • E ach frame that is to be transmitted contains source
virtual circuit identifier. and destination address.
• All frames belonging to the same message follow the • The frames belonging to the same message can follow
same route to a destination. The route is selected at different routes to reach the destination.
the time when the virtual circuit is set up.
• If a router fails then all virtual circuits passing • T here is no effect on the datagram networks when the
through it are terminated. router fails. Only the loss of some packets takes place.
• Congestion control is easy as the virtual circuit is set • Congestion control is difficult.
up depending upon the available buffers.
• The delay associated with each packet is less. • T he delay associated with each packet is more.
• It provides less efficiency because resources remain • It provides more efficiency as resources are allocated
allocated to stream of packets even if the connection only when required. If a packet of the message being
is not been used. transmitted fails due to some reason, the resources
can be allocated to a packet of another message.

CN_Chapter 09.indd 179 2/20/2012 6:32:51 PM

 180 Data Communications and Computer Networks

20. What is X.25? With reference to X.25, explain the following:

(a) Switched virtual circuit and permanent virtual circuit
(b) Protocols used at the link level
(c) State diagrams to explain call setup and call clearing.
Ans: X.25 is the first public network that was developed in 1976 by ITU-T. It specifies an interface
for exchanging data packets between the packet mode end system called data terminal equipment
(DTE) and the access node of switched packet data network called data circuit terminating equip-
ment (DCE). The DTE is operated by the user, while the DCE is operated by the service provider.
X.25 is a virtual circuit-switching network, which requires the prior establishment of virtual con-
nection between sender and receiver. Each connection is assigned a unique connection number that is
included in each packet to be transmitted. Each packet comprises a 3-byte header followed by data up
to 128 bytes. The header consists of 12-bit connection number packet sequence number and many other
fields. X.25 provides flow and error control at both data link and network lay .
(a) Switched Virtual Circuit and Permanent Virtual Circuit X.25 offers end-to-end virtual
communication path through the network for communication between two DTEs. This virtual path can
be of two types: switched virtual circuit (SVC) and permanent virtual circuit (PVC). An SVC is a tem-
porary switched connection that is established upon the request of a DTE and is terminated when data
transmission is over. It involves three phases, namely, call setup, data transfer and call clearing. In the
call setup phase, an entry for the virtual circuit (connection between a source and a destination) is made
in each switch. The network resources are allocated for the entire duration of transmission. In data
transfer phase, the data packets are exchanged between the communicating DTEs. The communication
between DTEs is made via local and remote DCEs. The calling DTE sends the data packets to its local
DCE, which forwards the packets to remote DCE through the virtual circuit, established between them.
The remote DCE finally hands over the packets to the called DTE. In call clearing phase, the virtual
connection is terminated and resources are deallocated.
On the other hand, PVC is a constant (fixed) connection established between two DTEs. It need not
be established or terminated for every instance of communication between the DTEs. Thus, it does not
require call setup and call clearing phases and always remains in the data transfer phase.
(b) Protocols used at the Link Level The interface of X.25 has been defined at three levels
including level 1, level 2 and level 3 which correspond to physical, data link and network layer of OSI
model, respectively. Various protocols are used at each level. The link level (level 2) uses data link
­protocols whose functionality is same as that of the HDLC. These protocols are as follows:
‰‰ Link Access Protocol, Balanced (LAPB): This protocol is the most common protocol and
has been derived from HDLC. It supports all the characteristics of HDLC and can also form
a logical link connection.
‰‰ Link Access Protocol (LAP): This protocol is the earlier version of the LAPB protocol
and it is rarely used today.
‰‰ Link Access Procedure, D Channel (LAPD): This protocol has been derived from
the LAPB protocol. It is mainly used in integrated services digital networks (ISDNs),
­supporting data transmission between DTE and ISDN node. The transmission is mainly
done through channel D.
‰‰ Logical Link Control (LLC): This is an IEEE 802 protocol used in LANs. It allows trans-
mission of X.25 packets through a LAN channel.

CN_Chapter 09.indd 180 2/20/2012 6:32:51 PM

 Ethernet, Virtual Circuit Networks and SONET 181

(c) State Diagrams to explain Call Setup and Call Clearing The communication between
two DTEs initiates through the call setup phase. In this phase, initially, the calling DTE sends a Call
Request packet to its local DCE. After receiving a Call Request packet, the local DCE forwards this
packet to the next node thus, establishing the virtual connection up to the remote DCE, which serves
the required DTE. The remote DCE then sends an Incoming Call packet to the called DTE to indicate
the ­willingness of calling DTE to communicate with it. If the called DTE is ready to communicate, it
sends a Call ­Accepted packet to the remote DCE, which then forwards this packet to the local DCE via
the same ­virtual connection. After receiving the Call Accepted packet, the local DCE sends a Call Con-
nected packet to the calling DTE to indicate the successful establishment of connection. Figure 9.16
depicts the whole process of call setup phase.

Local DCE Remote DCE

Calling DTE Call Incoming Called DTE
request call

Call Call
Virtual circuit
connected accepted

Figure 9.16 Call Setup Phase

Generally, the call-clearing phase is initiated after the completion of data transfer between calling
and called DTEs. However, in certain situations, such as when call is not accepted by the called DTE or
when a virtual circuit cannot be established, the call-clearing procedure is also initiated. The call can be
terminated by either of the communicating parties or by the network. For example, if the calling DTE
wants to clear the connection, it sends a Clear Request packet to the local DCE which forwards this
packet to the remote DCE. To forward the call-clearing request to called DTE, the remote DCE sends a
Clear Indication packet to it. In response, the called DTE sends a Clear Confir packet to the remote
DCE, which then forwards this packet to local DCE. The local DCE passes this packet to the calling
DTE, thus terminating the connection. Figure 9.17 depicts the whole process of call-clearing phase
initiated by DTE.

Local DCE Remote DCE

Calling DTE Clear Clear Called DTE
request indication

Clear Clear
Virtual circuit
confirm confirm

Figure 9.17 Call-clearing Phase

Now, consider the case where the call-clearing phase is initiated by the network. In this case, both
the local and remote DCE send a Clear Indication packet to the calling and called DTE, respectively.
On receiving the packets, the calling and called DTEs respond with a Clear Confir packet to the local
and remote DCE respectively, thus, terminating the connection. The call clearing by the network may
result in the loss of some data packets.

CN_Chapter 09.indd 181 2/20/2012 6:32:51 PM

 182 Data Communications and Computer Networks

21. List some drawbacks of X.25.

Ans: X.25 is a virtual circuit network that was first developed in 1976 by ITU-T. It has some
­drawbacks, which are as follows:
‰‰ It provides a low data rate only up to 64 kbps. Thus, it cannot be used to transmit bursty data.
‰‰ The error and flow control is performed at both data link and network layer. This results in great
overhead and reduced speed of transmission.
‰‰ It has its own network layer as it was designed for the private use. Thus, if X.25 is to be used with
some network that has its own network layer such as Internet, the network layer packets of Internet
have to be delivered to X.25, which then encapsulates them into X.25 packets. This increases the
overhead.
22. What are T1/T3 lease lines? List some of their drawbacks.
Ans: Some organizations started working in separation from the X.25. They started using their
own private wide area networks (WANs), where a line (T1 or T3) was leased from the private service
­providers. These lines (T1 or T3) are known as leased lines. Like X.25, the leased lines also have certain
drawbacks, which are as follows:
‰‰ It was too costly, as organizations have to pay for them. The payment has to be made even if the
lines are not in use.
‰‰ Only fixed rate data can be transmitted with T1/T3 lines. It was not possible to send different
frames at different bandwidths.
23. Discuss Frame Relay in detail.
Ans: Frame Relay is a virtual circuit WAN that came into existence in the late 1980s to meet the
demands of a new WAN with faster transmission capability. Prior to Frame Relay, some organizations
were using virtual circuit network X.25 and some organizations were having their own private WANs
using lease lines (T1 or T3) from public service providers. Both these technologies suffered from severe
limitations and thus, were replaced by Frame Relay.
Frame Relay provides a higher transmission speed of 44.376 Mbps and allows sending a frame of
size up to 9,000 bytes. It operates in physical and data link layer only and thus, can be easily used with
networks having their own network layer such as Internet. It allows bursty data to send through it and is
less expensive as compared to other WANs. It does not provide flow control; however, error detection
is supported and that too only at the data link layer.

Architecture
Frame Relay is a virtual circuit network in which each virtual circuit is uniquely identified by a number
known as data link connection identifie (DLCI). It provides two types of virtual circuits, which are
as follows:
‰‰ Permanent Virtual Circuit (PVC): In this circuit, a permanent connection is created ­between
a source and a destination and the administrator makes a corresponding entry for all the
­switches in a table. An outgoing DLCI is given to the source and an incoming DLCI is given
to the ­destination. Using PVC connections is costly, as both source and destination have to
pay for the connection even if it is not in use. Moreover, it connects a single source to a single
destination. Thus, if source needs connection with another destination, then separate PVC is
to be set up.

CN_Chapter 09.indd 182 2/20/2012 6:32:51 PM

 Ethernet, Virtual Circuit Networks and SONET 183

‰‰ Switched Virtual Circuit (SVC): In this circuit, a short temporary connection is created and that
connection exists as long as data transmission is taking place between source and the destination.
After the transmission of the data, the connection is terminated.
In a Frame Relay network, the frames are routed with the help of a table associated with each switch
in the network. Each table contains an entry for every virtual circuit that has already been set up. The
table contains four fields for each entry: incoming port, incoming DLCI, outgoing port and outgoing
DLCI. Whenever a packet arrives in a switch, it searches the incoming port DLCI combination in the
table to match with an entry. After a match has been found, the DLCI of the arrived packet is replaced
with outgoing DLCI (found in table) and the packet is routed to the outgoing port. This way the packet
travels from switch to switch and eventually, reaches the destination. Figure 9.18 depicts how data is
transferred from a source to a destination in a Frame Relay network.
Incoming Outgoing
Port DLCI Port DLCI
2 48 1 50

Switch
2 1
48

Da

a
at
ta

D
50

Data 24 4 3
Data 70
Switch 3 Switch
1 4
2 S2
S1 2 1

Incoming Outgoing Incoming Outgoing

Port DLCI Port DLCI Port DLCI Port DLCI
1 24 4 48 3 50 4 70

Figure 9.18 Data Transfer in a Frame Relay Network

Frame Relay Layers

Frame Relay comprises two layers: physical and data link layer. There is no specific protocol defined for
the physical layer; the implementer can use any of ANSI-recognized protocols at this layer. In contrast,
at the data link layer, a simple protocol is used, which does not offer any flow or error control. However,
error detection is supported by this protocol. Frame Relay uses a frame at the data layer provides whose
format is shown in Figure 9.19.

Flag Address Information FCS Flag

DLCI C/R EA DLCI FECN BECN DE EA

6 bits 1 bit 1 bit 4 bits 1 bit 1 bit 1 bit 1 bit

Figure 9.19 Format of Frame Relay Frame

CN_Chapter 09.indd 183 2/20/2012 6:32:52 PM

 184 Data Communications and Computer Networks

The Frame Relay frame consists of five fields, which are described as follow
‰‰ Flag: It is an 8-bit long field, which is used at the start and end of the frame. The starting flag
­indicates the start of the frame, whereas the ending flag indicates the end of the frame.
‰‰ Address: This field has a default length of 16 bits (2 bytes), and may be extended up to 4 bytes.
The address field is further divided into various subfields, which are described as follow
• DLCI: This field is specified in the two parts in the frame as shown in Figure 9.19. The first
part is 6 bits long while the second part is 4 bits long. These 10 bits together identify the data
link ­connection defined by the standard
• Command/Response (C/R): It is 1-bit long field that enables the upper layers to recognize
­whether a frame is a command or a response.
• Extended Address (EA): This field is also specified in two parts in the frame, with each part of
1 bit. It indicates whether the current byte is final byte of the address. If the value is zero, then it
indicates that another address byte is to follow; else, it means that the current byte is the final byte
• Forward Explicit Congestion Notificatio (FECN): It is a 1-bit long field that informs the
­destination that congestion has occurred. It may lead to loss of data or delay.
• Backward Explicit Congestion Notificatio (BECN): It is a 1-bit long field that informs the
sender that congestion has occurred. The sender then slows down the speed of the transmission
in order to prevent data.
• Discard Eligibility (DE): It is a 1-bit long field that indicates the priority level of the frame. If
its value is set to one, it indicates the network to discard the packet in case of congestion.
‰‰ Information: It is a variable-length field that carries highe -level data.
‰‰ FCS: It is a 16-bit long field, which is used for error detection.

24. Discuss congestion control in Frame Relay networks.

Ans: Frame Relay was designed to provide higher data rate and to achieve that congestion must be
avoided. However, congestion control is difficult for a Frame Relay networks as it supports variable
data rate. In addition, it does not have a network layer for flow control. To avoid congestion in the net-
work Frame Relay uses two notification bits in the frame, namely, forward explicit congestion notific -
tion (FECN) and backward explicit congestion notification (BECN). The FECN bit is used to signal the
­receiver about the congestion in the network so that the receiver could control the transmitting of frames
by delaying the acknowledgement to the higher layers. The BECN bit is used to signal the sender about
the congestion in the network. To inform the source, the switch can either send response frame from the
receiver or use a special connection having DLCI 1023 to send special frames. The sender then reduces
the rate of data transmission to control congestion.
25. What is the limitation of using Frame Relay?
Ans: In Frame Relay, the length of frame is not fixed, that is, a user may transmit frames of ­different
sizes. As all the frames are stored in the same queue, a small frame after a long frame in the queue
­experiences different delay than a small frame before the large frame in the queue. That is, delay varies
from frame to frame. This makes the Frame Relay unsuitable for real-time applications such as audio
and video as these applications are time sensitive.
26. What is ATM? Explain the architecture of ATM network?
Ans: Asynchronous transfer mode (ATM) is the standard specified by ITU-T for cell relay in
which multiple service types including video and data are transferred in the form of fixed-size cells.
A cell is the basic unit of data exchange in an ATM. Each ATM cell is 53 bytes long including 5 bytes

CN_Chapter 09.indd 184 2/20/2012 6:32:52 PM

 Ethernet, Virtual Circuit Networks and SONET 185

of header and 48 bytes of payload or data. Further, ATM networks are connection-oriented though they
employ packet-switching technique. They also allow bursty traffic to pass through as well as devices
with different speeds can communicate with each other via ATM network. Thus, it combines the advan-
tages of both packet switching and circuit switching.

Architecture of ATM Network

An ATM network is composed of ATM switches and ATM endpoints. An ATM switch deals with trans-
mission of a cell through the network. It takes the cell from an ATM switch or ATM endpoint, reads the
cell header information and updates it. After this, it switches the cell to an output interface towards its
intended destination. An ATM endpoint is a user access device such as router, workstation and digital
service unit (DSU) that consists of a network interface adapter.
Two types of interfaces, namely, user-to-network interface (UNI) and network-to-network interface
(NNI) are used in an ATM network. The UNI connects the ATM endpoints to the ATM switches inside
the network, whereas the NNI connects the switches within the network. Each UNI and NNI can also
be classified into public or private UNIs and NNI. A private UNI connects an ATM endpoint and a pri-
vate switch. However, the public UNI connects an ATM endpoint or private switch to a public switch.
A ­private NNI connects two ATM switches located in the same private organization. However, the
public NNI connects two ATM switches located in the same public organization. The architecture of an
ATM network is shown in Figure 9.20.

Endpoint

UNI UNI
NNI
Switch Switch

NNI NNI
Switch

UNI

Figure 9.20 Architecture of an ATM Network

Two ATM endpoints are connected through transmission path (TP), virtual path (VP) and virtual
circuits (VC). A transmission path such as wire or cable that links an ATM endpoint with an ATM
switch or two ATM switches with one another. It consists of a set of virtual paths. A virtual path refers
to the link (or a group of links) between two ATM switches. Each virtual path is a combination of virtual
circuits having the similar path. A virtual circuit refers to the logical path that connects two points. All
the cells corresponding to the same message pass through the same virtual circuit and in the same order
until they reach the destination.
In order to route cells from one end ATM end point to another, the virtual connections must be
uniquely identified. Each virtual connection is identified by the combination of virtual path identifier
(VPI) and virtual circuit identifier (VCI). Further, VPI uniquely identifies the virtual path while VCI
uniquely identifies the virtual circuit; both VPI and VCI are included in the ATM cell header. Notice that
all the virtual circuits belonging to the same virtual path possess the same VPI. The length of VPI is differ-
ent in UNI and NNI. It is of 8 bits in UNI but of 12 bits in NNI. On the other hand, the length of VCI is

CN_Chapter 09.indd 185 2/20/2012 6:32:52 PM

 186 Data Communications and Computer Networks

same in both UNI and NNI and is 16 bits. Thus, to identify a VC, a total of 24 bits are required in UNI
while 28 bits are required in NNI.
The ATM also uses PVC and SVC connections like Frame Relay. However, the difference is that
ATM was developed from the starting to support audio and video applications using SVCs while in a
Frame Relay, PVC and SVC were added later.
27. How are ATM cells multiplexed?
Ans: In an ATM, asynchronous time division multiplexing technique is followed to multiplex the cells
from different input sources. The size of each slot is fixed and is equal to the size of a cell. Each input source
is assigned a slot when it has some cell to send. In case none of the sources has cell to send, the slots remain
empty. Whenever any channel has a cell to send, the ATM multiplexer puts any of these cells into a particu-
lar slot. However, if all the cells have been multiplexed, the empty time slots are sent in the network.
Figure 9.21 shows the cell multiplexing from four input sources P, Q, R and S. At the first clock tick,
as the input source Q has no cell to send, the multiplexer takes a cell from S and puts it into the slot.
Similarly, at the second clock tick, Q has no data to send. Thus, the multiplexer fills the slot with a cell
from R. This process continues until all the cells have been multiplexed. After all the cells from all the
sources have been multiplexed, the output slots are empty.

Empty slots

P3 P2 P1
P S4 Q3 P3 S3 R2 Q2 S2 R1 P2 S1 Q1 P1
Q3 Q2 Q1
Q MUX
R2 R1
R
S4 S3 S2 S1
S

Figure 9.21 Multiplexing in ATM

28. Discuss ATM layers.
Ans: The ATM standard has defined three layers, namely, physical layer, ATM layer and application
adaptation layer (AAL) as shown in Figure 9.22.

AAL Layer

AAL1 AAL2 AAL3/4 AAL5

ATM Layer

Physical Layer

Figure 9.22 ATM Layers

CN_Chapter 09.indd 186 2/20/2012 6:32:53 PM

 Ethernet, Virtual Circuit Networks and SONET 187

Physical Layer
The physical layer is responsible for managing the medium-dependent transmission. It carries the ATM
cells by converting them into bit streams. It is responsible for controlling the transmission and receipt
of bits as well as maintaining the boundaries of an ATM cell. Originally, ATM was designed to use
­synchronous optical network (SONET) as the physical carrier. However, other physical technologies
can also be used with ATM.

ATM Layer
An ATM layer is similar to the data link layer of the OSI model. It is responsible for cell multiplexing
and passing cells through ATM network (called cell relay). Other services provided by the ATM layer
include routing, traffic management and switching. It accepts a 48-byte segment from the AAL layer
and adds a 5-byte header, transforming it into a 53-byte cell. Further, ATM uses a separate header for
UNI and NNI cell (Figure 9.23). The header format of the UNI cell is similar to the NNI cell except the
GFC field that is included in the UNI cell, but not in the NNI cell

GFC VPI VCI PT CLP HEC Payload

4 bits 8 bits 16 bits 3 bits 1 bit 8 bits data
(a) UNI cell

VPI VCI PT CLP HEC Payload

12 bits 16 bits 3 bits 1 bit 8 bits data
(b) NNI cell
Figure 9.23 Format of an ATM Cell Header

The description of fields included in cell header is as follows

‰‰ Generic Flow Control (GFC): It is a 4-bit long field that is used for flow control in UNI. In NNI,
this field is not required. Thus, all the bits are assigned to VPI. The longer is the VPI, the more
number of virtual paths will be specified at NNI.
‰‰ Virtual Path Identifie (VPI): It is an 8-bit long field in UNI cell and 12-bit long field in NNI cell
that identifies a specific virtual pat
‰‰ Virtual Circuit Identifie (VCI): It is 16-bit long field that identifies a specific virtual circuit
inside the virtual path.
‰‰ Payload Type (PT): It is a 3-bit long field in which the first bit is used to define the type of information,
that is, data or managerial information while the context of other two bits are based on the first bit
‰‰ Cell Lost Priority (CLP): It is a 1-bit long field that is used for congestion control
‰‰ Header Error Correction (HEC): It is the cyclic redundant code that is used for error correction.
It is calculated over the first 4 bytes of header using a divisor x8 + x2 + x + 1 to correct single
and multiple-bit errors.

Application Adaptation Layer

Application adaptation layer (AAL) accepts data frames or stream of bits from upper layers and divides
them into fixed-size segments of 48 bytes. At the receiver’s side, these data frames or stream of bits
are again reassembled into their original form. An AAL layer is partitioned into two sublayers, namely,

CN_Chapter 09.indd 187 2/20/2012 6:32:53 PM

 188 Data Communications and Computer Networks

segmentation and reassembly sublayer (SAR) and convergence sublayer (CS). The SAR sublayer is re-
sponsible for segmentation of payload at the sender’s side and reassembling the segments to create the
original payload at the receiver’s side. The CS sublayer is responsible for ensuring the integrity of data
and preparing it for segmentation by the SAR sublayer. There are various types of AAL including AAL1,
AAL2, AAL3/4 and AAL5. Out of these four versions, only AAL1 and AAL5 are commonly used.
29. Explain the structure of ATM adaptation layer.
Ans: The ATM standard has defined four versions of AAL, which include AAL1, AAL2, AAL3/4
and AAL5. All these versions are discussed as follows:

AAL1
It is a connection-oriented service that supports applications needing to transfer information at constant
bit rates such as voice and video conferencing. The bit stream received from upper layer is divided into
47-byte segments by the CS sublayer and then segments are passed to SAR sublayers below it. The SAR
sublayer appends a 1-byte header to each 47-byte segment and sends the 48-byte segments to the ATM
layer below it. The header added by SAR sublayer consists of two fields (Figure 9.24) namely, sequence
number (SN) and sequence number protection (SNP). The SN is a 4-bit field
that specifies a sequence number for ordering the bits. Further, SNP is a 4-bit SN SNP
field that is used to protect the sequence number. It corrects the SN field by 4 bits 4 bits
using first three bits and the last bit is used as a parity bit to discover error in Figure 9.24 SAR Header
all eight bits.

AAL2
Initially, it was designed to support applications that require variable-data rate. However, now it has
been redesigned to support low bit rate traffic and short-frame traffic such as audio, video or fax. The
AAL2 multiplexes short frames into a single cell. Here, the CS sublayer appends a 3-byte header to the
short packets received from the upper layers and then passes them to the SAR layer. The SAR layer
combines the short frames to form 47-byte frames and adds a 1-byte header to each frame. Then, it
passes the 48-byte frames to the ATM layer.
The header added by CS sublayer consists of five fields (Figure 9.25(a)) which are described as
­follows:
‰‰ Channel Identifie (CID): It is an 8-bit long field that specifies the channel user of the packe
‰‰ Length Indicator (LI): It is a 6-bit long field that indicates the length of data in a packet
‰‰ Packet Payload Type (PPT): It is a 2-bit long field that specifies the type of a packe
‰‰ User-to-User Indicator (UUI): It is a 3-bit long field that can be used by end-to-end use .
‰‰ Header Error Control (HEC): It is a 5-bit long field that is used to correct errors in the heade .
The header added by SAR consists of only one field [Figure 9.25(b)], start fiel (SF) that specifies
an offset from the beginning of the packet.

CID LI PPT UUI HEC SF

8 bits 6 bits 2 bits 3 bits 5 bits 8 bits

(a) CS header (b) SAR header

Figure 9.25 CS and SAR Headers

CN_Chapter 09.indd 188 2/20/2012 6:32:54 PM

 Ethernet, Virtual Circuit Networks and SONET 189

AAL3/4
Originally, AAL3 and AAL4 were defined separately to support connection-oriented and connectionless
services, respectively. However, later they were combined to form a single format AAL3/4. Thus, it supports
both connection-oriented and connectionless services. Here, the CS sublayer forms a PDU by inserting a
header at the beginning of a frame or appending a trailer. It passes the PDU to SAR sublayer, which partitions
the PDU into segments and adds a 2-byte header to each segment. It also adds a trailer to each segment.
The header and trailer added by the CS layer together consist of six fields (Figure 9.26) that are
­described as follows:
‰‰ Common Part Identifie (CPI): It is an 8-bit long field that helps to interpret the subseque t fields
‰‰ Begin Tag (Btag): It is an 8-bit long field that indicates the beginning of a message. The value of
this field is same for all the cells that correspond to a single message
‰‰ Buffer Allocation Size (BAsize): It is a 16-bit long field that specifies to the receiver buffer size
needed to hold the incoming data that is to be transmitted.
‰‰ Alignment (AL): It is an 8-bit long field that is added to make the trailer 4 bytes long
‰‰ Ending Tag (Etag): It is a 16-bit long field that indicates the end of the message. It has the same
value as that of Btag.
‰‰ Length (L): It is a 16-bit long field that specifies the length of the data uni

CPI Btag BAsize

8 bits 8 bits 16 bits

(a) CS header

AL Etag L
8 bits 8 bits 16 bits

(b) CS trailer
Figure 9.26 CS Header and Trailer

The header and trailer added by SAR layer together consists of five fields (Figure 9.27) that are
­described as follows:
‰‰ Segment Type (ST): It is a 2-bit long field that specifies the position of a segment corresponding
to a message.
‰‰ Sequence Number (SN): It is a 4-bit long field that specifies the sequence numb .
‰‰ Multiplexing Identifie (MID): It is a 10-bit long field that identifies the flow of data to which the
incoming cells belong.
‰‰ Length Identifie (LI): It is a 6-bit long field in the trailer that specifies the length of a data in the
packet excluding padding.
‰‰ CRC: It is a 10-bit long field that contains CRC computed over the entire data unit

ST SN MID LI CRC
2 bits 4 bits 10 bits 6 bits 10 bits

(a) SAR header (b) SAR trailer

Figure 9.27 SAR Header and Trailer

CN_Chapter 09.indd 189 2/20/2012 6:32:55 PM

 190 Data Communications and Computer Networks

AAL5
This layer supports both connection-oriented and connectionless data services. It assumes that all cells
corresponding to single message follow one another in a sequential order and the upper layers of the
application provide the control functions. This layer is also known as simple and efficien algorithm
layer (SEAL). Here, the CS sublayer appends a trailer to the packet taken from upper layers and then
passes it to the SAR layer. The SAR layer forms 48-bytes frames from it and then passes them to the
ATM layer.
The trailer added by CS layer consists of four fields (Figure 9.28) that are described as follows
‰‰ User-to-User (UU): It is an 8-bit-long field that is used by users
‰‰ Common Part Identifie (CPI): It is an 8-bit-long field that is used for the similar function as that
in the CS layer of AAL3/4.
‰‰ Length (L): It is a 16-bit-long field that specifies the length of dat
‰‰ CRC: It is a 32-bit-long field, which is used for error correction

UU CPI L CRC
8 bits 8 bits 16 bits 32 bits

Figure 9.28 CS Trailer

30. List some benefit of ATM.

Ans: An ATM is a cell-switched network that provides several benefits over its counterpart, the
Frame Relay. Some of these benefits are as follows:
‰‰ It provides high bandwidth for applications that require bursty traffic. For example, applications
such as video involve bursty data in which amount of motion in a video is not fixed and also in audio,
conversation does not go the same way all the time. Thus, ATM can be used for these applications.
‰‰ Technologies such as Frame Relay use frames of different sizes. Thus, it is difficult to manage the
traffic. In contrast, ATM was developed to carry audio and video using a cell-switched technology
in which fixed-size frames are used. This improves the efficienc , as it is easier to quantify, predict
and manage the network traffic.
‰‰ Further, ATM is a WAN technology but it can be used as both LAN and WAN technology. It can
be used to cover large distances to link LANs or WANs.
31. What is the relationship between SONET and SDH?
Ans: Both Synchronous optical network (SONET) and synchronous digital hierarchy (SDH) are WANs
that are used as transport networks to carry data at high rate from other WANs because of higher band-
widths of fibre optic cables as well as carrying vast amount of lower-rate data. The only difference is that
ANSI of the United States has provided SONET while SDH has been provided by ITU-T of Europe. Both
SONET and SDH are synchronous networks that use synchronous TDM multiplexing. A master clock con-
trols all the clocks present in the system. Both SONET and SDH standards are independent of each other;
however, they are functionally similar and compatible. It can be said that they are nearly identical.
32. How is an STS multiplexer different from an add/drop multiplexer since both can add
signals together?
Ans: Both synchronous transport signal (STS) and add/drop multiplexer are the devices used in
SONET. Further, STS is used as an interface between electrical and optical networks. At the sender’s

CN_Chapter 09.indd 190 2/20/2012 6:32:55 PM

 Ethernet, Virtual Circuit Networks and SONET 191

end, STS multiplexer is used that multiplexes the signals coming from various electrical sources into
the corresponding optical carrier (OC) signal. This optical signal passes through the SONET link and
finall , reaches the receiver. At the receiver’s end, STS demultiplexer is used that demultiplexes the OC
signals into the corresponding electrical signals.
Add/drop multiplexer is used in the SONET link to insert or remove signals. It can combine the STSs
from several sources into a single path or extract some desired signal and send it to some other path
without demultiplexing. In SONET, the signals multiplexed by the STS multiplexer (optical signals) are
passed through regenerator, which regenerates the weak signals. The regenerated signals are then passed
to add/drop multiplexer that transmits them in the directions as per the information available in data
frames (Figure 9.29). The main difference between add/drop multiplexer and STS multiplexer is that it
does not demultiplex the signals before delivering them.
Terminal

Regenerator

STS STS
MUX ADM DEMUX

Add Drop

Figure 9.29 A Simple Network using SONET Equipment

33. What is the function of a SONET regenerator?

Ans: The SONET regenerator is a repeater that regenerates the weak signals. Sometimes, because
of the long distances travelled from one multiplexer to another, the signals becomes weak and need to be
regenerated. The regenerator receives an OC signal and demodulates it into the corresponding ­electrical
signals. These electrical signals are then again regenerated and finall , modulated into OC signals.
The regenerator functions at the data link layer.
34. What are the four SONET layers? Discuss the functions of each layer.
Ans: There are four functional layers included in the SONET standard, namely, photonic, section,
line and path (Figure 9.30).
‰‰ Photonic Layer: It is the lowest layer whose function-
Upper layers
alities are similar to that of the physical layer of the OSI
model. This layer includes the physical specifications Path layer
­related to optical fibre, the multiplexing functionalities
Line layer Data link layer
and the sensitivity of the receiver. Further, NRZ encoding
is used by SONET where the presence of light shows bit 1 Section layer
while absence of light shows 0.
Photonic layer Physical layer
‰‰ Section Layer: The function of section layer includes
handling framing, scrambling and error control. It has (a) SONET (b) OSI model
the responsibility of moving a signal across a physical
section. In addition, the section layer header is added to Figure 9.30 Relationship of SONET Layers
the frame at this layer. with the OSI Model

CN_Chapter 09.indd 191 2/20/2012 6:32:56 PM

 192 Data Communications and Computer Networks

‰‰ Line Layer: The line layer takes care of the movement of signal across a physical line. At this
layer, the line layer overhead is added to the frame. The line layer functions are provided by STS
multiplexers and add/drop multiplexers.
‰‰ Path Layer: The movement of a signal from the optical source to the optical destination is the
responsibility of the path layer. The electronic signal is changed into the optical form from the elec-
tronic form at the optical source and then multiplexed with other signals, finally being encapsulated
into a frame. The received frame is demultiplexed and the individual optical signals are changed
into their electronic form at the optical destination. The STS multiplexers are used to provide path
layer functionalities. At this layer, the path layer overhead is added to the signal.
35. What are virtual tributaries?
Ans: SONET was originally introduced to hold the broadband payloads. However, the data rates of
the current digital hierarchy ranging from DS-1 to DS-3 are lower than STS-1. Thus, virtual tributary
(VT) was introduced to make the SONET compatible with the present digital hierarchy. A VT is a partial
­payload, which can be filled into an STS-1 frame and combined with many other partial payloads to
cover the entire frame. The VTs filled in the STS-1 frame are organized in form of rows and columns.
There are four types of VTs, which have been defined to make SONET compatible with the existing
digital hierarchies. These four categories are as follows:
‰‰ VT1.5: This VT adapts to the US DS-1 service and provides a bit rate of 1.544 Mbps. It gets three
columns and nine rows.
‰‰ VT2: This VT adapts to the European CEPT-1 service and provides a bit rate of 2.048 Mbps. It gets
four columns and nine rows.
‰‰ VT-3: This VT adapts to the DS-1C service and provides a bit rate of 3.152 Mbps. It gets six
­columns and nine rows.
‰‰ VT 6: This VT adapts to the DS-2 service and provides a bit rate of 6.312 Mbps. It gets 12 columns
and nine rows.

Multiple Choice Questions

1. Fast Ethernet operates at 5. Extended service set in IEEE 802.11 consists of
(a) 10 Mbps (b) 100 Mbps (a) only one basic service set with AP
(c) 1000 Mbps (d) none of these (b) only one basic service set without AP
(c) at least two basic service sets without APs
2. 10-Base F uses
(d) at least two basic service sets with APs
(a) optical fibr (b) coaxial cable
(c) twisted pair (d) none of these 6. An example of wireless LAN is
(a) Bluetooth (b) Ethernet
3. IEEE 802.4 is used to describe
(c) both (a) and (b) (d) none of these
(a) Token Ring (b) Token Bus
(c) CSMA/CD (d) Ethernet 7. Which of the following statements is false?
(a) Virtual circuit network provides less
4. The access method used in FDDI is
efficienc .
(a) CSMA/CD (b) token passing
(b) Congestion control is easy in virtual circuit
(c) timed token passing (d) none of these
networks.

CN_Chapter 09.indd 192 2/20/2012 6:32:56 PM

 Ethernet, Virtual Circuit Networks and SONET 193

(c) Virtual circuit does not require set up of a 9. Which of the following is a type of interface
connection before transmission. in an ATM network?
(d) all of these (a) user-to-network (b) network-to-network
(c) user-to-user (d) both (a) and (b)
8. Which of the following statements is false?
(a) In Frame Relay, frame length is not fixed. 10. AAL1 is a
(b) Frame Relay was developed for real-time (a) connection-less service
applications. (b) connection-oriented service
(c) ATM uses fixed-size cells (c) both (a) and (b)
(d) none of these (d) none of these

Answers
1. (b) 2. (a) 3. (b) 4. (c) 5. (d) 6. (a) 7. (c) 8. (b) 9. (d) 10. (b)

CN_Chapter 09.indd 193 2/20/2012 6:32:56 PM

 10
Routing and Congestion Control

1. Explain the various design issues of the network layer.

Ans: The network layer provides an end-to-end communication between two hosts. There are mainly
four design issues in the network layer, which are discussed as follows:
‰‰ Network l ayer i nterface: The network layer can follow two methods, namely, virtual circuit
(connection-oriented network service) and datagram (connectionless network service) for sending
packets from node to node. In the virtual circuit method, first a logical connection is established
between the two communicating users and then the quality of service to be used is decided. All the
packets travel through the established path to reach the destination. On the other hand, the data-
gram method does not require to set up any fixed path before the transmission. Thus, each packet
should contain sufficient information for routing from source to destination so that it need not ­depend
upon previous exchanges. Each packet may follow a different path to reach the destination.
‰‰ Routing issues: It is the responsibility of the network layer to route packets from source to desti-
nation. For this, a piece of software, known as routing algorithm, is used which decides where the
packets are to be sent. The routing algorithm has to deal with the following issues:
• Correctness and simplicity: The networks should never go down; the links or routers may
fail but the whole network should still function.
• S tability: When a router fails, the time elapsed before other routers realize this change should
be stable.
• Fairness and optimality: Some aspects such as whether to maximize channel usage or mini-
mize the average delay should be considered while making the routing decisions.
‰‰ Congestion: The network layer also deals with congestion issues, which are as follows:
• The network should not be flooded with packets more than its capacity; otherwise, the delay
­increases and the performance degrades. If the situation does not improve then the packets
have to be discarded.

CN_Chapter 10.indd 194 2/20/2012 6:34:22 PM

 Routing and Congestion Control 195

• I f the delay increases, packets may be incorrectly transmitted by the sender making the situa-
tion worse.
• If congestion is not controlled, the overall performance of the network degrades as the network
resources would be used for processing the packets that have actually been discarded.
‰‰ I nternetworking: Internetworking is used to connect different network technologies together.
Certain issues related to internetworking are as follows:
• The packets should be able to pass through many different networks.
• Different networks may have different frame formats. Thus, the network layer should support
multiple frame formats.
• The network layer should support both connectionless and connection-oriented networks.

2. Write a short note on I P addresses.

Ans: Each system (computer or router) that connects to the Internet is assigned a specific Internet
address. This Internet address, referred to as I P address, uniquely identifies the connection of the sys-
tem to the Internet. Here, uniquely means that no two systems can be assigned the same IP address at
the same time. Moreover, IP addresses are universal which means each system wishing to connect to
Internet must accept the addressing system.
Each protocol that defines the IP addresses has an address space—the total number of addresses that
the protocol can use. If a protocol assigns n-bit addresses to machines, the address space of protocol
will be 2n. For example, traditional IP protocol, which is IPv4, defines 32-bit IP addresses and thus,
its address space is 232. On the other hand, IPv6 protocol—the new generation of IP—defines 128-bit
­addresses and thus, its address space is 2128.
3. Briefl describe the notations used to represent IPv4 addresses.
Ans: An IPv4 address is of 32 bits in length and can be represented using two notations, namely,
binary notation and dotted-decimal notation.
In binary notation, the IP address is represented as 32 bits grouped into four octets. As each octet
is one byte, an IPv4 address is sometimes also referred to as a 4-byte address. The problem with this
notation is that the binary address is too lengthy and difficult to read. To overcome this problem, dotted-
decimal notation is used.
In dotted–decimal notation, each byte of an IPv4 address is represented by a decimal value ranging
between 0 and 255 and different bytes are separated with a decimal point (.). Figure 10.1 shows a 32-bit
IPv4 address written in binary as well as dotted decimal notation.

Binary notation 11011101 00111000 01001110 01001110

Dotted decimal
221.56.7.78
notation
Figure 10.1 An IPv4 Address in Binary and Dotted–Decimal Notation

4. Explain classful and classless addressing.

Ans: IP addressing can be either classful or classless. Both classful and classless addressing are
discussed as follows:

CN_Chapter 10.indd 195 2/20/2012 6:34:23 PM

 196 Data communications and Computer Networks

Classful Addressing
The traditional IP addressing used the concept of classes thereby named as classful addressing. IPv4
addressing is classful in which all the IP addresses (address space) are divided into five classes, namely,
A, B, C, D and E with each class covering a specific portion of the address space. Each class is further
divided into a fixed number of blocks with each block of fixed size. Given an IPv4 address, its class can
be identified by looking at either few bits of first byte in case IP address is represented in binary notation
or the first byte in case the I address is represented in dotted-decimal notation (Table 10.1).

Table 10.1 Determining Class of IP Address

Class Binary notation (few bits of first byte) Dotted-Decimal notation (first byte)
A 0   0–127
B 10 128–191
C 110 192–223
D 1110 224–239
E 1111 240–255

The blocks of class A, B and C addresses were granted to different organizations depending on their
size. The large organizations using a vast number of hosts or routers were assigned class A addresses.
The class B addresses were designed to be used by mid-size organizations having tens of thousands of
hosts or routers while the class C addresses were designed for small organizations having small number
of hosts or routers. The class D addresses were projected to be used for multicasting where each address
identifies a specific group of hosts on the Internet. Only a few of class E addresses were used while
rest were kept reserved for the future use. The main problem with classful addressing was that a large
number of available addresses were left unused resulting in a lot of wastage.
Each IP address in class A, B and C is divided into two parts: net I D that identifies the network on
the Internet and host I D that identifies the host on that network. The size of each part is different for the
different classes. In a class A address, net ID is identified by one byte and host ID is identified by three
bytes. In a class B address, net ID is identified by two bytes and host ID is also identified by two bytes
and in a class C address, three bytes specify the net ID and one byte specifies the host ID

Classless Addressing
There were certain problems with classful addressing such as address depletion and less organization
­access to Internet. To overcome these problems, classful addressing is replaced with classless addressing.
As the name of the addressing scheme implies, the addresses are not divided into classes; however, they
are divided into blocks and the size of blocks varies according to the size of entity to which the addresses
are to be allocated. For instance, only a few addresses may be allocated to a very small organization while
a larger organization may obtain thousands of addresses. IPv6 addressing is a classless addressing.
The Internet authorities have enforced certain limitations on classless address blocks to make the
handling of addresses easier. These limitations are as follows:
‰‰ The addresses of a block must be contiguous.
‰‰ Each block must have a power of 2 (that is, 1, 2, 4, 8…) number of addresses.
‰‰ The first address in a block must be evenly divisible by the total number of addresses in that block.

CN_Chapter 10.indd 196 2/20/2012 6:34:23 PM

 Routing and Congestion Control 197

5. Describe the notation used to represent I Pv6 addresses.

Ans: An IPv6 address is of 128 bits (16 bytes) and is represented using hexadecimal colon notation.
In this notation, the 128-bit address is divided into eight sections of two bytes each and each byte is repre-
sented by four hexadecimal digits. Thus, the address consists of 32 hexadecimal digits with each group of
four hexadecimal digits separated by a colon. Figure 10.2 shows an IPv6 address written in hexadecimal
colon notation.

FABD : 0037 : 0000 : 0000 : 0000 : ABCF : 0000 : FFFF

Figure 10.2 An IPv6 Address in Hexadecimal Colon Notation

Since many of the digits in the IP address represented in Figure 10.2 are zeros, the address even in
hexadecimal format is still too long. However, it can be abbreviated by omitting leading (but not trail-
ing) zeros of a section. For example, 0037 in the second section can be written as 37 and 0000 in third,
fourth, fifth and seventh section can be written as 0. Figure 10.3 shows the abbreviated form of a hexa-
decimal address after omitting leading zeros of a section.
As still there are many consecutive sections containing
zero only in the address shown in Figure 10.3, the address FABD : 37 : 0 : 0 : 0 : ABCF : 0 : FFFF
can be more abbreviated. The consecutive section of zeros
Figure 10.3 Abbreviated Form of IPv6 Address
can be eliminated and replaced with double semicolon (::)
as shown in Figure 10.4. However, there is one limitation with this
type of abbreviation that it can be applied only once per address. FABD : 37 : : ABCF : 0 : FFFF
That is, if there are two or more runs of consecutive sections of
zeros in a single address, only one run can be replaced with double Figure 10.4 More Abbreviated Form
semicolon but not others. of IPv6 Address

6. Defin network address. Is it different from net ID?

Ans: The address that defines the network itself is referred to as the network address. It cannot be
assigned to a host. It is also different from a net ID. It comprises both net ID and host ID with all bits of
host ID set to zero. The network address is required to route the packets to the destination. Table 10.2
lists examples of network addresses for classes A, B and C.

Table 10.2 Examples of Network Addresses

Class Network address
A 112.0.0.0
B 155.7.0.0
C 229.48.57.0

7. L ist some special I P addresses.

Ans: Depending on their contents, some IP addresses have been designated as special IP addresses.
These addresses are listed as follows:
‰‰ An IP address with all 0s (that is, 0.0.0.0) identifies this host. This address is used by the hosts at
the time when they are being booted up but not afterwards.

CN_Chapter 10.indd 197 2/20/2012 6:34:23 PM

 198 Data communications and Computer Networks

‰‰ An IP address with all 1s (that is, 255.255.255.255) indicates broadcast on this network. This
­address is used for forwarding the packet to all the hosts on the local network.
‰‰ An IP address with net ID all 0s and a proper host ID (see diagram below) identifies a specific host
on the local network.
000 ..................... 0 Host ID

‰‰ An IP address with a proper net ID, and host ID containing all 1s (see diagram below) indicates
broadcast to some distant LAN in the Internet.

Net ID 111 ........... 1

‰‰ An IP address of the form 127.aa.bb.cc (see diagram below) indicates reserved address used for
loopback testing.

127 Anything

8. Write a short note on address masks.

Ans: An address mask, also referred to as default mask, is a string made up of contiguous 1s
­followed by contiguous 0s. Like IP address, it is of 32 bits, that is, four octets where each octet is equiva-
lent to a decimal value in the range of 0–255. Table 10.3 lists the default masks used for classes A, B and
C IP addresses.

Table 10.3 Default Masks for Classes A, B and C

Address mask Address mask
Class (in binary) (in dotted–decimal)
A 11111111 00000000 00000000 00000000 255.0.0.0
B 11111111 11111111 00000000 00000000 255.255.0.0
C 11111111 11111111 11111111 00000000 225.255.255.0

The address mask (also simply called mask) identifies which part of an IP address defines the net ID
and which part defines the host ID. To determine this, the IP address and mask are compared bit by bit.
If a given mask bit is 1, its corresponding bit in IP address will form a part of net ID. On the other hand,
if mask bit is 0, its corresponding bit in IP address will form a part of host ID. For example, consider an
IP address 132.7.21.84 (that is, 10000100 00000111 00010101 01010100) that belongs to class B. It is
clear from Table 10.3 that the mask for class B addresses is 255.255.0.0 (that is, 11111111 11111111
00000000 00000000). Now, if we compare given IP address with its corresponding mask, we can easily
determine that 132.7 define the net ID and 21.84 define the host I

9. Explain the concept of subnetting in IP. What is subnet mask?

Ans: The subnetting was introduced at the time when classful addresses were being used. It means
breaking large blocks of addresses into many contiguous groups, which are further assigned to smaller
networks known as subnets. It came in use when a single network had to be subdivided internally to
accommodate more computers, increasing the size of the whole network. It was impossible to acquire
new network addresses every time a network is divided. Therefore, in such a situation, subnetting was

CN_Chapter 10.indd 198 2/20/2012 6:34:24 PM

 Routing and Congestion Control 199

used to split the network into many subnets. When subnetting is done, the network is divided internally,
however, it appears as a single network to the outside world. For example, in a college campus network
the main router may be connected to an Internet service provider (ISP) and different departments may
use their own routers, which are connected to the main router. Here, each department with a router and
some communication lines form a subnet, however to the outside world, the whole campus appears as a
single network (Figure 10.5).

PC Router
ISP

Main
Subnets router Subnets

Figure 10.5 A Campus Network Divided into Subnets

Now, a problem arises when a packet comes to the main router; how does it know to which subnet
the packet is to be delivered. One way is to maintain a table in the main router having an entry corre-
sponding to each host in the network along with the router being used for that host. Though this scheme
seems fine, it is not feasible as it requires a large table and many manual operations as hosts are added or
deleted. Thus, a more feasible scheme was introduced, where a part of the host address is used to define
a subnet number. For example, in a class B address where a host ID is of 16 bits, the six bits can be used
to define the subnet number and the rest 10 bits can be used to define the host ID. This would lead to up
to 64 subnetworks, each with a maximum of 1,022 hosts.
For the implementation of subnetting, the main router requires subnet mask that shows the division of
addresses between network plus subnet number and host ID. For example, Figure 10.6 shows a subnet
mask written in binary notation for a subnetted class B address, where six bits have been used to repre-
sent the subnet number and 10 bits for host ID. The subnet mask can also be written in dotted-decimal
notation. For example, for the IP address shown in Figure 10.6, the subnet mask in dotted-decimal
notation can be written as 255.255.192.0. Alternatively, slash (/) notation, also called CI DR (classless
interdomain routing) notation can also be used to represent the subnet mask. In this notation, a slash
­followed by the number of bits in the network plus subnet number defines the subnet mask. For example,
the IP address shown in Figure 10.6 can be defined in slash notation as /22 where 22 is the size of subnet
mask. Notice that /22 indicates that the first 22 bits of subne mask are 1s while rest 10 bits are 0s.

Subnet
10 Network Subnet Host
mask
111111111111111111 1 1 1 10000000000
Figure 10.6 A Class B Network Address Subnetted into 64 Subnets
10. What is supernetting? Why it is needed?
Ans: Supernetting is just the opposite of subnetting. There was a time when all the class A and B
addresses were used up but still there was a great demand of address blocks for mid-size organiza-
tions. The class C address blocks were also not suitable for serving needs of those organizations; more

CN_Chapter 10.indd 199 2/20/2012 6:34:25 PM

 200 Data communications and Computer Networks

a­ ddresses were required. As a solution to this problem, supernetting was used. In this method, several
class C blocks can be combined by an organization to create a large address space, that is, smaller
­networks can be combined to create a super-network or supernet. For example, an organization having
the requirement of 1,000 addresses can be allocated four contiguous blocks of class C addresses.
11. What is IP spoofing
Ans: IP spoofin is a mechanism, which is used to hide the original IP address by replacing it with
an arbitrary IP address. The IP datagrams, which are transferred over a network, carry the sender’s IP
address apart from the data from the upper layers. Any user having control over the operating system
can place an arbitrary address into a datagram’s source address field by modifying the device protocols.
Thus, an IP packet may be made to appear as though it is originating from an arbitrary host. IP spoofing
is generally used in denial of service attacks to hide the originator of attack.
If IP spoofing is destructive, it can be prevented through a mechanism known as ingress filtering.
When it is implemented, the routers receiving the IP datagrams see their source addresses to check
whether these addresses are in the range of network addresses known to the routers. This check is per-
formed at the edge of a network such as a corporate gateway or a firewall

12. What is routing? Differentiate adaptive and non-adaptive routing algorithms.

Ans: Generally, in a network there exist multiple paths available for going from a source to destina-
tion. To forward the packets, one of the available routes is to be chosen by the network layer. This is
called routing.
The part of the network layer software which is responsible for deciding which output line should
be used to transmit an incoming packet is referred to as routing algorithm. The routing algorithms can
be divided into two categories; namely, non-adaptive and adaptive. Non-adaptive algorithms are the
algorithms, which make routing decisions regardless of the current status of the network such as ­current
traffic or current topology. If a packet has to be transferred from A to B, then the route to get from A to B
is traced in advance and is entered in the routing table. This is done at the time of booting of the network.
This is also sometimes called as static routing. Some examples of non-adaptive routing ­algorithms
­include flooding, shortest path routing and flow-based routin
On the other hand, the adaptive algorithms change their decisions dynamically in coordination with
the changes in the network attributes as topology, traffic and time delay. Whenever changes occur in the
network attributes, the routes of the routers are also changed accordingly. That is why it is also referred
to as dynamic routing. Some examples of the adaptive routing algorithms are distance vector routing
and link state routing.
13. What is optimality principle?
Ans: O ptimality principle refers to a general statement about the optimal routes without taking into
account the topology and the traffic load in the network. According to the principle, if an optimal path
(say d1) from router A to G goes through a router F, then the path from F to G (say, d2) is also optimal.
To prove this principle, let us assume that d2 is not optimal. This means there exist some other path
(say, d3 ) from F to G that is more optimal than d2. Thus, d3 can be combined with d1 to improve the path
from A to G. This implies that d1d2 is not optimal which contradicts our statement that d1d2 is optimal.

14. What is routing table? What are its categories?

Ans: Each host or router in a network maintains a data structure, known as routing table, which
contains an entry corresponding to each destination or a group of destinations to route the IP packets.

CN_Chapter 10.indd 200 2/20/2012 6:34:25 PM

 Routing and Congestion Control 201

Whenever an IP packet arrives at a router, it sees the destination address of the packet. Then, it looks
up in the routing table to check whether the destination address matches some existing entry. If so,
the IP packet is forwarded through the interface specified in the routing table corresponding to that
­matching entry.
Routing tables are of two types: static and dynamic routing table. Both these types are discussed as
follows:
‰‰ S tatic routing t able: This routing table is created manually. The route of each of the destination
is entered into the table by the administrator. If there is any change in the network, it is also done
manually by the administrator. Any broken link in the network would require the routing tables to
be manually reconfigured immediately so that alternate paths could be used. It is not possible to
automatically update the table. Static routing tables are well suited for small networks. However, in
large networks such as Internet, maintenance of static routing tables can be very tiresome.
‰‰ Dynamic routing t able: This routing table is updated automatically by the dynamic routing
protocols such as RIP, BGP and OSPF whenever there is a change in the network. Generally, each
router periodically shares its routing information with adjacent or all routers using the routing
­protocols so that the information about the network could be obtained. If some change is found in
the network, the routing protocols automatically update all the routing tables.
15. Differentiate between intradomain and interdomain routing.
Ans: Due to the ever-growing size of Internet, using only one routing protocol to update the ­routing
tables of all routers is not sufficient. Therefore, the network is divided into various autonomous ­systems.
An autonomous system (AS ) is a group consisting of networks and routers, which is controlled by
a ­single administrator. Thus, a network can be seen as a large collection of autonomous system. The
routing, which is done inside an autonomous system, is known as intradomain routing. One or more
intradomain routing protocols may be used to handle the traffic inside an autonomous system. Two most
commonly used intradomain routing protocols include distance vector routing and link state routing. On
the other hand, routing which is done between different autonomous systems is known as ­interdomain
routing. Only one interdomain routing protocol is used to handle the routing between autonomous
­systems. The most popular interdomain routing protocol is the path vector routing.
16. Explain the following algorithms in brief.
(a) Flooding
(b) Flow-based routing
Ans: (a) Flooding: Flooding is a static routing algorithm that works based on forwarding of packets.
Here, every packet arriving in a router is forwarded (flooded) to all the outgoing lines from the router,
except the one through which it has arrived. Due to bulk forwarding of packets, a large number of dupli-
cate packets are generated. The flooding can be controlled by ignoring the flooded packets rather than
resending them. Some preventive measures that can be used to control flooding are as follows
‰‰ A hop counter can be included in the header of each packet. Initially, the hop counter can be set to
a value equal to the total distance from source to destination. If the distance is not known, then the
counter can be initialized to the whole length of the subnet. As the packet reaches at every hop, the
counter is decremented by one and finall , when it becomes zero the packet is discarded.
‰‰ Another technique to prevent duplication is to keep track of all those packets, which have already
been flooded so that they could not be sent for the second time. This technique requires a sequence

CN_Chapter 10.indd 201 2/20/2012 6:34:25 PM

 202 Data communications and Computer Networks

number to be included in every packet that is to be forwarded. This sequence number is entered by
the source router whenever it receives a packet from a host in its network. Every router maintains
a list per each source router indicating the sequence numbers that have already been seen by that
source router. The packet is not flooded if it is already on the list
‰‰ A variation of flooding known as selective floodin can be used to overcome the problem. In this
algorithm, not every incoming packet is sent to every outgoing line, but only to those lines, which
are going in the right direction.
Few applications where flooding is useful are as follows
‰‰ It is used in distributed database applications where the need arises to update all the databases at
the same time and to choose the shortest path to the destination resulting in shorter delay.
‰‰ It is used in wireless networks where one station can transmit a message, which can be received by
all other stations within the range of that network.
‰‰ It can be used as a metric in routing algorithms. This is because of the reason that flooding always
gives a shorter delay which no other algorithm can.

(b) Flow-based Routing: This is also a static routing ­algorithm

in which packets are ­forwarded by taking into account the traffic K
condition and the structure of the ­network. To understand, consider a
network shown in Figure 10.7. Suppose there always remains a great
J L
traffic from node M to L. If node M has to send packets to node K,
routing through node L would not be efficient because of the heavy Heavy load
load on path ML. Thus, another path MNJK is used even though it is a
N M
longer than the MLK path. This type of routing is known as flow-base
­routing. The optimal routes can be found by analyzing the flow of data Figure 10.7 Flow-based Routing
mathematically. However, this is possible only if average traffic from
a node to any other node is already known and it always remains constant. Once the average traffic and
capacity of a line has been known, the mean delay for each packet can be calculated in advance.

17. What is shortest path routing? Explain with the help of a suitable example.
Ans: The shortest path routing is a static routing algorithm, which is based on the concept of
graphs. Here, the whole subnet is depicted as a graph where the nodes of the graph represent the routers
and the edges represent the links between two routers. The optimal route between two routers is deter-
mined by finding the shortest path between them. The metric used in shortest path routing can be num-
ber of hops, distance or the transmission delay. Any of the metrics used is represented as the weights
assigned to the edges in the graph.
There are many algorithms for computing the shortest path between two nodes of a graph represent-
ing a network. However, the Dijkstra’s algorithm is the most often used to calculate the shortest path.
In this algorithm, each node is labelled with a value equal to its distance from the source node along the
optimal path. Since no paths are known initially, all nodes are labelled with infinit . As the algorithm
proceeds, paths are found and thus, label may change. The labels on the nodes are divided into two
­categories: tentative and permanent. Initially, and each label is tentative. However, as it is assured that
the label presents the shortest distance, it is made permanent.
To understand the Dijkstra’s algorithm, consider a sample subnet graph shown in Figure 10.8 where
the weights on the edges represent the distance between nodes connected by that edge. Suppose the
shortest path has to be calculated from node A to D.

CN_Chapter 10.indd 202 2/20/2012 6:34:25 PM

 Routing and Congestion Control 203

B 14 C
4 4 6 6
4 D
A
E F 4
2 4
12
G 8 H
Figure 10.8 A Sample Subnet Graph

The steps involved in calculating the shortest path from A to D are as follows:
1. The node A is marked permanent (shown by the filled circle in Figure 10.9). Each adjacent neigh-
bour of A is examined and is relabelled with distance from A. For example, B is relabelled as
(4, A) which indicates that distance of B is 4 from A. Similarly, G is relabelled as (12, A). Rest of
the nodes still remain labelled with infinit . Thus, the new tentative nodes are B and G.

B(4, A) C(∞, −)

A E(∞, −) F(∞, −) D(∞, −)

G(12, A) H(∞, −)
Figure 10.9 Node A Marked as Permanent

2. As out of nodes B and G, B has the shorter distance from A, it is selected and marked permanent
as shown in Figure 10.10. Thus, now each adjacent neighbour of B is examined and relabelled
with distance from B. Notice that a node is relabelled only if the existing label on it is greater than
the sum of the label on the source node and the distance between that node and source node. For
­example, while examining the node C with respect to node B, we find that the sum of label on B
(that is, 4) and the distance between B and C (that is, 14) is 18, which is far less than the existing
­label on C (that is, ∞). Thus, node C is relabelled as (18,B). Similarly, E is relabelled as (8,B).

B(4, A) C(18, B)

A E(8, B) F(∞, −) D(∞, −)

G(12, A) H(∞, −)
Figure 10.10 Node B Marked as Permanent

3. As out of nodes C and E,E has the shorter distance from B, it is selected and marked permanent
as shown in Figure 10.11. Thus, now each adjacent neighbour of E is examined and relabelled with
distance from E. As the node G can be reached via E with a shorter distance (that is, 10) than its
previous distance (that is, 12), G is relabelled as (10,E). Similarly, F is relabelled as (12,E).

CN_Chapter 10.indd 203 2/20/2012 6:34:26 PM

 204 Data communications and Computer Networks

B(4, A) C(18, B)

A E(8, B) F(12, E) D(∞, −)

G(10, E) H(∞, −)
Figure 10.11 Node E Marked as Permanent

4. Though out of nodes G and F,G has the shorter distance from E, it cannot be selected as it will
result in a loop. Therefore, node F is selected and marked permanent as shown in Figure 10.12 and
its adjacent neighbours are examined. While examining the neighbouring nodes, the node E is not
considered because we have reached F via E and thus, cannot go back. In addition, as the existing
label on node C (that is, 18) is equal to the sum of distance between F and C (that is, 6) and label
on F (that is, 12), node C is not relabelled. Contrastive to this, node H is relabelled as (16,F).

B(4, A) C(18, B)

A E(8, B) F(12, E) D(∞, −)

G(10, E) H(16, F)
Figure 10.12 Node F Marked as Permanent

5. The node H is marked permanent as shown in Figure 10.13 and its adjacent neighbour D is
­examined. The node D is relabelled as D(20,H). Since our intended destination D has been
reached, the node D is also marked permanent. Thus, the shortest path from A to D is ABEFHD as
shown in Figure 10.13 and the shortest distance from A to D is 20.

B(4, A) C(18, B)

A E(8, B) D(20,H)
F(12, E)

G(10, A) H(16, F)
Figure 10.13 Shortest Path from A to D

18. Explain distance vector routing algorithm with the help of a suitable example.
Ans: The distance vector routing (also known as Bellman–Ford routing) is a dynamic routing
­algorithm. In this method, each router maintains a table (known as vector) in which routing information
about the distance and route to each destination is stored. The table contains an entry per each router in

CN_Chapter 10.indd 204 2/20/2012 6:34:27 PM

 Routing and Congestion Control 205

the network, where the preferred outgoing line and the distance to that router are specified. Each router
periodically exchanges its routing information with their neighbours and updates its table accordingly.
Thus, each router knows the distance to each of its neighbours. To make the routing decisions, the
metric used maybe the total number of packets queued along the path, number of hops, time delay in
milliseconds, etc. For example, if delay is used as the metric for making routing decisions then after
every t milliseconds, every router updates its neighbour with the estimated delays needed to reach each
destination in the network. Suppose, one such table has just arrived to router J showing that its neighbour
A can reach the router I in estimated time Ai. If now the router J knows that its delay time to A is n
milliseconds, then it can reach to router I via A in Ai + n milliseconds. By doing such calculation for
each neighbour, the optimum estimate is found.
To understand how calculations are performed in distance vector routing, consider a subnet consist-
ing of routers and networks as shown in Figure 10.14(a). Suppose at any instant the router D receives the
delay vectors from its neighbours A,C, and E as shown in Figure 10.14(b). Further, assume that the
estimated delay from D to its neighbours A,C, and E are 5, 8 and 6, respectively. The router D can now
calculate the delays to all the other routers in the network. Suppose D wants to compute the route to B,
using the available delay information. Now, D has a delay time of 5 ms to reach A while A takes 10 ms
to reach B [Figure 10.14(b)], so D would take a total of 15 ms to reach B via A. In the same way, D can
calculate the delays to reach B using the lines C and E as 18(11 + 7), 31(9 + 12 + 10) ms, respectively.
Since, the optimum value is 15 ms, the router D updates its routing table mentioning that the delay to B
is 15 ms via the route A. The same procedure can be followed by D to calculate the delays to all the other
routers and then making entries in its routing table as shown in Figure 10.14(c).

E Delays received by D New routing table for D

A C E line
A 0 20 10 A 5 A
B 10 7 22 B 15 A
A D C 17 0 20 C 8 C
D 9 11 9 D 0 –
E 12 20 0 E 6 E
B C DA DC DE
5 8 6
(a) (b) (c)
Figure 10.14 Distance Vector Routing

19. What is the count-to-infinit problem in the distance vector routing algorithm?
Ans: Theoretically, the distance vector routing algorithm works efficiently but practically, it has
serious loopholes. Even though a correct answer is derived, it takes time. In practice, this algorithm acts
quickly to good news but acts very late to bad news. For example, let us consider a router whose best
route to destination X is large. If on the next neighbour exchange, one of its neighbours (say, A) declares
a short delay to X, the router starts using the line through A to reach X. The good news is thus processed
by B within a single vector exchange. The news spreads even faster as it is received by its neighbours.
To understand how fast good news spreads, consider a linear subnet consisting of five nodes
­[Figure 10.15(a)]. Suppose, the metric used in the subnet is the number of hops. In the beginning, the
node A is down and therefore, all the routers have recorded their delay to A as infinit . When A becomes
functional, the routers perceive it through the vector exchanges. At the first vector exchange, B realizes
that its left neighbour has zero delay to A, so it makes an entry 1 in its routing table as A is just one
hop away from it. At the second vector exchange, the router C comes to know that B is just one hop

CN_Chapter 10.indd 205 2/20/2012 6:34:27 PM

 206 Data communications and Computer Networks

A B C D E A B C D E

∞ ∞ ∞ ∞ Initially 1 2 3 4 Initially
1 ∞ ∞ ∞ after 1 exchanges 3 2 3 4 after 1 exchange
1 2 ∞ ∞ after 2 exchanges 3 4 3 4 after 2 exchanges
1 2 3 ∞ after 3 exchanges 5 4 5 4 after 3 exchanges
1 2 3 4 after 4 exchanges 5 6 5 6 after 4 exchanges
• 7 6 7 6 after 5 exchanges
•
• 7 8 7 8 after 6 exchanges
•
•
∞ ∞• ∞ ∞

(a) (b)
Figure 10.15 The Count-to-Infinity Problem

away from A, so C updates its table making an entry of two hops to A. Similarly, D and E update their
­routing tables after the third and fourth exchanges, respectively. Thus, the news about the recovery of A
­propagates at a rate of one hop per exchange.
However, the situation is very different in the case of bad news propagation. To understand the
propagation of bad news, consider the situation shown in Figure 10.15(b). Here, initially all routers
are properly functioning and the routers B,C,D and E are 1, 2, 3 and 4 hops away from A, respec-
tively. Now, if A goes down all of a sudden, the routers still think A to be functional. At the first vector
­exchange, no vector is received from A but the node B sees in the vector received from C that C has a
path of length 2 to A. As B does not know that path from C to A goes through it, it updates its routing
table with distance to A as 3 and via C. Now, at the second vector exchange, as C sees the distance of
B to A as 3, it updates its distance to A as 4 and via B. Similarly, all the routers continue to update their
routing tables and ­increase their distance to A after every vector exchange. Eventually, all routers update
their routing tables with distance to A as infinity where infinity denotes the longest path plus one. This
problem is known as the count-to-infinit problem.
20. Explain the link state routing protocol.
Ans: The link state routing is a dynamic routing algorithm that had been designed to replace its
­previous counterpart distance vector routing. The distance vector routing has certain limitations, as it
does not take into account the line bandwidth while choosing among the routes. It also suffers from
count-to-infinity problem. To overcome these limitations, link state routing algorithm was devised.
In link state routing, the information such as network topology, metric used and type and ­condition
of links is available to each router (node). Therefore, each router can use the Dijkstra’s algorithm to
­compute the shortest path to all the available routers and then build their routing tables. Link state
­routing involves the following phases:
‰‰ Learning about the neighbours: Whenever a router boots up, the first step for it is to identify all
its neighbours and get their network addresses. For this, the router sends a HELLO packet on each
point-to-point line. Each router receiving the HELLO packet responds with its name (or network
address) to the sending router. Similarly, all the routers in the network discover their neighbours.
Notice that the name of the routers must be globally unique in order to avoid any ambiguity.
‰‰ Measuring the l ine cost: After discovering all the neighbours, the next step for a router is to have
an estimate of delay to reach every other router in the network. For this, the simplest method that a
­router can adopt is to send a special ECHO packet over the line and then start a timer. As soon as a router

CN_Chapter 10.indd 206 2/20/2012 6:34:28 PM

 Routing and Congestion Control 207

r­ eceives an ECHO packet, it immediately sends the packet back to the sending router. After receiving
the ECHO packet back, the sending router can determine the round-trip time of packet and then divide it
by two to calculate the delay. To have better estimate of delay, the router can send ECHO packet several
times and then use the average time as delay. While computing the delay, the traffic load in the network
may or may not be taken into account. If the load is considered, the sending router starts the timer at
the time the packet is queued up. On the other hand, if load is not taken into account, the sending router
starts the timer at the time the packet reaches at the front of the queue.
‰‰ Building l ink s tate packets: T he next step for a router is to package the collected information to
form a link state packet (LSP). Each LSP comprises four fields: the identity of the sending router,
sequence number, age and list of its neighbouring links. The first and fourth fields of LSP together
determine the network topology. The second field indicates the sequence number assigned to the
LSP packet; the sequence number is incremented by one each time the packet is created by the
router. The third field indicates the duration for which the LSP has been residing in the domain.
The LSP also includes delay to reach each adjacent neighbour. For example, Figure 10.16(b)
shows the LSPs for each of the five routers of the subnet shown in Figure 10.16(a).

A
A B C D E
5 7 SEQ SEQ SEQ SEQ SEQ
8 AGE AGE AGE AGE AGE
B C
B 5 A 5 A 7 B 4 D 3
4 1 C 7 B 8
C 8 E 3 C 1
D E D 4 E 1
3
(a) (b)
Figure 10.16 Subnet and Link State Packets
The LSPs are created by the routers periodically at regular intervals of time or at certain occasions such
as when some changes occur in the topology or either when some router goes down or comes up.
‰‰ Flooding of LS Ps: After each router has created the LSP, it needs to distribute its LSP to its neigh-
bours in the network. This process is referred to as floodin . The router forwards a copy of its LSP
through each of its neighbouring interface. For example, the router A sends its LSP through lines
AB and AC [Figure 10.16(a)]. Each receiving router compares the received LSP against the list
of LSPs that it already has. If the sequence number of newly arrived LSP is found lower than the
highest sequence number in the list, the packet is simply discarded. Otherwise, the receiving router
stores the new LSP and also forwards a copy of it through its neighbouring interfaces except the
one through which the LSP has arrived.
‰‰ Computing new routes: After receiving all LSPs, the next step for a router is to compute the short-
est path to every possible destination using the Dijkstra’s algorithm (explained in Q 17) and build the
routing table. The routing table of a router lists all the routers in the network, the minimum cost of
reaching them and the next router in the path to which the packet should be forwarded.

21. Differentiate link state and distance routing vector algorithms.

Ans: The link state routing algorithm came after the distance vector routing algorithm and it offers
many advantages over the previous one. Some of the differences between the two are as follows:
‰‰ The traffic generated due to routing process is less in link state routing as compared to the distance
vector routing. The reasons for this difference are as follows:

CN_Chapter 10.indd 207 2/20/2012 6:34:28 PM

 208 Data communications and Computer Networks

• The HELLO messages exchanged between adjacent routers are much smaller in size than the
vectors in the distance vector routing. The LSPs of the link state routing contain information
only about the neighbours, while the distance vectors include the net IDs of all the routers in
the network.
• In link state routing, the LSPs are exchanged between neighbours after every 30 min, while in
distance vector routing, the vectors are exchanged after a comparatively very small period (for
example, 30 s in RIP).
‰‰ In link state routing, the optimal paths are calculated by each router independently. However, in
distance vector routing each router depends on its neighbour for updating its distance vector result-
ing in slow processing.
‰‰ Alternate routes are also possible in link state routing but in distance vector routing only specific
routes are used.
‰‰ Multiple cost metrics can be used in link state routing and the optimal paths can be computed with
respect to each metric separately. Further, packets can be forwarded based on any one metric.
22. Explain the path vector routing protocol.
Ans: Path vector routing is an interdomain routing protocol, which is used between various autono-
mous systems (ASs). In every AS, there is one node (can be more, but only one is considered here) which
acts on behalf of the whole system. This node is known as the speaker node. It creates the routing table
for its AS and announces it with the speaker nodes residing in the neighbouring ASs. Path vector routing is
similar to that of distance vector routing; however, here only the speaker nodes in ASs can ­communicate
with each other. In the advertised tables, only the path is shown excluding the metric of the nodes.
To understand path vector routing, consider three ASs, AS1, AS2 and AS3 with K, L and M as the speaker
nodes, respectively (Figure 10.17). The path vector routing works in various phases, which are as follows:

Dest Path
K AS1
K1 K1 AS1 L1
K2 AS1
K2 K3 AS1
K L L2
AS1 K3
AS2
Dest Path
M L AS2
Dest Path L1 AS2
M AS3 M2 M1 L2 AS2
M1 AS3 AS3
M2 AS3

Figure 10.17 Initial Routing Tables in Path Vector Routing

‰‰ I nitialization: Initially, the speaker nodes know only how to reach the nodes inside their own
­autonomous system. Thus, each speaker node initializes its routing table with an entry per each
node that resides in its own network. For example, the routing table of node K shows that the nodes
K1, K2 and K3 are located in AS1 and can be reached through K. The routing tables of L and M are
also initialized in the same way.
‰‰ S haring: After the routing tables have been initialized, they are shared between the speaker nodes
of all ASs. For example, the node K shares its table with L and M providing the routing information
inside AS1. Similarly, node L and M also share their routing tables.

CN_Chapter 10.indd 208 2/20/2012 6:34:28 PM

 Routing and Congestion Control 209

‰‰ Updating: When a routing table is received by a speaker node, it updates its routing table by adding
the nodes that are not present in its own routing table along with the path to reach them (Figure 10.18).
After updating the table, each speaker node knows how to reach every node in other ASs. For
example, if node K receives a packet for node K2, it knows that the path is in AS1. However, if it
receives a packet for M2, it knows that it has to follow the path AS1-AS3.

K’s routing table L’s routing table M’s routing table

Dest Path Dest Path Dest Path
K AS1 K AS2-AS1 K AS3-AS1
K1 AS1 K1 AS2-AS1 K1 AS3-AS1
K2 AS1 K2 AS2-AS1 K2 AS3-AS1
K3 AS1 K3 AS2-AS1 K3 AS3-AS1
L AS1-AS2 L AS2 L AS3-AS2
L1 AS1-AS2 L1 AS2 L1 AS3-AS2
L2 AS1-AS2 L2 AS2 L2 AS3-AS2
M AS1-AS3 M AS2-AS3 M AS3
M1 AS1-AS3 M1 AS2-AS3 M1 AS3
M2 AS1-AS3 M2 AS2-AS3 M2 AS3

Figure 10.18 Routing Tables After Updating

23. L ist some advantages of path vector routing.

Ans: The path vector routing is an interdomain routing protocol that offers certain advantages. Some
of these advantages are as follows:
‰‰ L oop prevention: Path vector routing avoids the creation of loops in a path. Whenever a router
receives a message, it determines whether or not its own AS is in the path list to destination. If so,
looping tends to occur in which case the message is ignored.
‰‰ Policy routing: The path vector routing can implement the policy routing easily. Whenever a
router receives a message, the path of its AS is checked. If any of the ASs listed in the path is
against its policy then that path and destination are ignored. This path is also not included in the
routing table and the message is not forwarded.
‰‰ O ptimum path: The optimum path is the path that is best for the organization. Path vector routing
only shows the path and does not include the metrics involved in the path as different ASs may use
different metrics. Optimal path does not mean only the shortest path, several other measures like
security, safety and reliability are also taken into account while determining optimal path.
24. Explain in the following routing protocols.
(a) RI P
(b) OS PF
(c) BGP
Ans: (a) RI P: Routing information protocol (RI P) is an intradomain routing protocol (also known
as interior gateway protocol) which is used inside an autonomous system. It is a simple protocol
whose ­operating procedures are similar to the widely used distance vector routing. The operation of RIP
is based on c­ ertain considerations, which are as follows:
‰‰ RIP is used in autonomous systems, which consists of routers and networks (links). Routing tables
are created for the routers but the networks do not implement any such table.

CN_Chapter 10.indd 209 2/20/2012 6:34:28 PM

 210 Data communications and Computer Networks

‰‰ As the first column of a routing table specifies a destination and in case of RIP, the destination is the
network; therefore, a network address is defined in the first column of the R routing table.
‰‰ The metric used by RIP is the hop count where the term hop defines the number of subnets ­traversed
from the source router to the destination subnet with including the destination subnet.
‰‰ The maximum cost of a path in RIP cannot be more than 15, that is, any route in an autonomous
system cannot have more than 15 hops.
‰‰ The next-node column in the routing table of RIP contains the address of the router to which the
packet is to be forwarded (the first hop to be done)
Consider an autonomous system consisting of six networks and three routers as shown in Figure 10.19.
Each router has its own routing table showing how to reach each network existing in the AS. Let us
study the routing table of one of the router, say R1. The networks 120.10.0.0 and 120.11.0.0 are directly
connected to the router R1; therefore, R1 need not make any hop count entries for these networks. The
packets destined for both these networks can be sent directly by the router; however, such forwarding
cannot be done in the case of the other networks. For example, if the router R1 needs to send packets
to the two networks to its left, then it has to forward the packets to the router R3. Thus, for these two
networks, the next-node column of the routing table stores the interface of router R2 with the IP ad-
dress 120.10.0.1. Similarly, if R1 has to send packets to two networks to its right, then it has to forward
­packets to the router R2. Thus, for these two networks, the next-node entry of routing table is the inter-
face of router R2 with IP address 120.11.0.1. The routing tables of other routers can also be explained
in the same way.

R3’s routing table 225.2.3.0 120.10.0.0 120.11.0.0 225.6.6.0

Hop R2’s routing table
Dest Next 225.2.3.1
R1 225.6.6.1
225.2.3.0 1 − Dest Hop Next
R3 120.10.0.2 120.11.0.2 R2
225.2.6.0 1 − 225.6.6.0 1 −
120.10.0.0 1 − 120.10.0.1 120.11.0.1 225.6.7.0 1 −
225.2.6.1 225.6.7.1 120.11.0.0
120.11.0.0 2 120.10.0.2 1 −
225.6.6.0 3 120.10.0.2 R1’s routing table 120.10.0.0 2 120.11.0.2
225.2.6.0 225.6.7.0 225.2.3.0
225.6.7.0 3 120.10.0.2 Dest Hop Next 3 120.11.0.2
225.2.6.0 3 120.11.0.2
120.10.0.0 1 −
120.11.0.0 1 −
225.2.3.0 2 120.10.0.1
225.2.6.0 2 120.10.0.1
225.6.6.0 2 120.11.0.1
225.6.7.0 2 120.11.0.1

Figure 10.19 Example of a Domain Using RIP

(b) OS PF: O pen S hortest Path First (OS PF) is an intradomain routing protocol which works
­ ithin an ­autonomous system. It is based on the link state routing algorithm. In OSPF, the autonomous
w
system is divided into many areas so that the routing traffic can be handled easily. Each area is formed by
the combination of networks, hosts and routers within an autonomous system and is assigned a unique
identification number. The networks residing in an area must be connected to each other. The routers
inside an area function normally, that is, periodically updating the area with routing information.
In OSPF, there are two types of routers, namely, area border router and backbone router. The area
border router is situated at the boundary of each area and it shares the information about its area with
the other areas. The backbone router is situated in a special area (called the backbone area) among all
the areas inside an autonomous system. The backbone is considered as a primary area and all the other

CN_Chapter 10.indd 210 2/20/2012 6:34:29 PM

 Routing and Congestion Control 211

Autonomous system

net net Area 0 (back bone)

Area border To other

router ASs

net Area 1 net

net Area border

router
net net net

net Back bone

Area 2
router

Figure 10.20 Areas in an Autonomous System

areas are connected to it. The backbone routers can also function as area border routers. Figure 10.20
depicts the idea of an OSPF protocol.
Like every other routing protocol, in OSPF also, there is a cost associated with each route known as
metric. This metric may depend upon the type of service such as minimum delay, minimum hop count
or maximum throughput and there may be multiple routing tables with each based on a different type
of service.
The connections in OSPF are known as links. In OSPF, four types of links have been identified:
point-to-point, transient, stub and virtual.
‰‰ Point-to-point link: This link defines a direct connection
­between the routers. It can be represented as a graph where
the nodes of the graph represent routers while the bidirectional R1 R2
­arrows connecting the nodes represent the link between the Figure 10.21 Point-to-Point Link
routers (Figure 10.21).
‰‰ T ransient link: This link defines a network to which multiple routers are attached. In this type
of network, each router has many neighbours all linked through a common network. The data
can enter the network through any of the routers and leave the network through any of the rout-
ers. For example, consider an Ethernet network consisting of six routers R1, R2, R3, R4, R5 and R6
[Figure 10.22(a)]. To represent transient link graphically, the network is represented as a node,
which is connected directly to all other nodes [Figure 10.22(b)].

R1 R2 R3 R4 R2
R3
R1
R4

Designated
router
R5 R6
R5 R6
(a) (b)
Figure 10.22 Transient Link and its Graphical Representation

CN_Chapter 10.indd 211 2/20/2012 6:34:30 PM

 212 Data communications and Computer Networks

‰‰ S tub link: This link is a special case of transient link, which defines a network with only one
router connected to it [Figure 10.23(a)]. The packets enter and leave the network using this single
router. Graphically, it can be represented using the designated router for the network and the router
as a node [Figure 10.23(b)]. The link is only unidirectional from the router to the network.

R1
R1

Designated
router
(a) (b)
Figure 10.23 Stub Link and its Graphical Representation

‰‰ Virtual link: When the link between two routers is broken, a new link has to be established by
the administrator, which is known as virtual link. This link may use a longer path covering more
number of routers.
(c) BGP: Border Gateway Protocol (BGP) is an interdomain routing protocol, which came into
existence in 1989. It is implemented using the concept of path vector routing. As BGP works between
the different ASs, it is an exterior gateway routing protocol. It follows certain policies to transfer packets
between various types of ASs and these policies are to be configured manually in each BG router.
In BGP, the ASs are divided into three categories, namely, stub AS, multihomed AS and transit AS.
The stub AS has a single connection with any other AS. The data traffic can either originate or terminate
at a stub AS, that is, a stub AS is either a source or a sink; however, data traffic cannot pass through it.
The multihomed AS can have multiple connections with other ASs, that is, it can send data to more
than one AS and receive also from many. However, still it is a sink or source as it does not allow data
traffic to pass through it. The transit AS is also a multihomed AS except that it allows third-party traffic
to pass through it.
Each BGP router maintains a routing table that keeps track of the path being used by the BGP router.
The routing information is exchanged periodically between the routers in a session, referred to as BGP
session. Whenever a BGP router needs to exchange the routing information with its neighbour, a trans-
mission control protocol (TCP) connection is established between the two, which indicates the start
of session. Using TCP connection provides a reliable communication between the routers. However,
the TCP connection used for BGP is not permanent and it is terminated as soon as some unusual event
­occurs.
To understand the BGP operation, consider a set of BGP routers shown in Figure 10.24. Let the
router H uses the path HDE to travel to E. When the neighbours of H provide routing information,
they give their complete paths to reach E. After receiving all the paths, H examines which one will be
the optimal path for going to E. The path originating from A is discarded as it passes through H. The
choice now has to be made between G, C and D. To select from G, C and D, BGP uses a scoring func-
tion that is responsible for examining the available routes to a destination and scoring them, returning
a number indicating the distance of each route to that destination. After the paths have been scored,
the one with the shortest distance is selected. As the path DE has the shortest distance, it is selected.

CN_Chapter 10.indd 212 2/20/2012 6:34:30 PM

 Routing and Congestion Control 213

A G F

Information received by H:
H
E From C: “CDE”
From D: “DE”
From A: “AHDE”
B C D From G: “GFE”

Figure 10.24 A Set of BGP Routers Along with Routing Information for H

BGP solves the count-to-infini y problem previously encountered in the distance vector routing
a­ lgorithm. To understand how it happens, suppose D crashes and the line HD goes down. When H
­receives routing information from the remaining working neighbours A,C and G, it finds the new routes
to E as GFE, AHGFE and CHGFE, respectively. The route GFE is selected as the other two pass through
H itself. Thus, the new shortest path for H to reach E is HGFE via G.
25. Explain unicast and multicast routing.
Ans: Unicast and multicast routing are the techniques of routing based on the network structure.
These are explained as follows:

Unicast Routing
In unicast communication, there is only one source and one destination, that is, one-to-one connection.
The data packets are transferred between the source and the destination using their unicast addresses.
The packets go through various routers to arrive at their destination. Whenever a packet arrives at the
router, the shortest path to the intended destination has to be found. For this, the router checks its routing
table to find out the next hop to which the packet should be forwarded in order to reach the destination
through the shortest path. If a path is not found in the table, the packet is discarded; otherwise, the packet
is forwarded only through the interface specified in the routing table

Multicast Routing
In multicast communication, there is one source and many destinations. The communication is one-to-
many communication, that is, the packets are forwarded to one or more destination addresses ­defined
altogether by a group address. Whenever a router receives a packet destined for a group address,
it ­forwards the packet through its many interfaces so that the packet could reach to all destinations
­belonging to that group. To forward a single packet to a group address, the router needs to create a
shortest path tree per each group. This makes the multicast routing complex, as N shortest path tree need
to be created for N groups. To resolve the complexity, two approaches including source-based tree and
group-shared tree can be used to construct the shortest path tree.
‰‰ S ource-based t ree: In this approach, each router constructs a shortest path tree for each group,
which defines the next hop for each network containing the members for that group. For example,
consider Figure 10.25 in which there are five groups G1, G2, G3, G4 and G5 and four routers R1, R2,
R3 and R4. The group G1, G2 and G5 has its members in four networks while the group G2 and
G3 in two networks. Each router is required to construct five shortest path trees, one per each group
and having the shortest path entries in its routing table. Figure 10.25 shows a multicast routing table
maintained by the router R3. Now, suppose the router R3 receives a packet destined for group G5.
Then, it forwards the copy of a packet to router R2 and R4. Further, router R2 forwards the copy of

CN_Chapter 10.indd 213 2/20/2012 6:34:30 PM

 214 Data communications and Computer Networks

a packet to router R1. This process continues and eventually, the packet reaches each member of
G5 in the whole network. The problem with this approach is the complexity of routing table, which
may have thousand of entries in case of large number of groups.

G1,G2

G1,G4,G5 G1,G2,G3,G5 G1,G2

R1 R2 R3

Routing table of R3 G3,G5

Dest Next hop
G1 –R2
R4
G2 –R4,R2
G3 –R2 G2,G4,G5
G4 –R4
G5 –R2,R4

Figure 10.25 Source-based Tree Approach

‰‰ Group-shared t ree: This approach overcomes the problem of source-based approach, as each
router is not required to construct shortest path for each group. Rather, one router is designated as
the core (or rendezvous) and only it is responsible for having N shortest paths in its routing table
for N groups, one per each group. Whenever a router receives a packet containing a group address,
it encapsulates the packet into a unicast packet and forwards it to the core router. The core router
then extracts the multicast packet and checks its routing table to determine the route for the packet.
Figure 10.26 shows the core router and its routing table.

G1,G2

G1,G4,G5 G1,G2,G3,G5 G1,G2

R1 R2 R3

Routing table of core router G3,G5

Dest Next hop
G1 –R1,R3 R4
Core
G2 –R4,R3
router G2,G4,G5
G3 –
G4 –R1,R4
G5 –R1,R4

Figure 10.26 Group-shared Tree Approach

26. Write a short note on internetworking.
Ans: I nternetworking refers to the interconnection of many different networks. There exist many
­different networks like LAN, MAN and WAN using different protocols and design. When users ­existing
in these different networks need to communicate with each other, internetworking has to be ­implemented.

CN_Chapter 10.indd 214 2/20/2012 6:34:31 PM

 Routing and Congestion Control 215

Internetworking not only allows these users to communicate with each other, but to access each other’s
data also. The communication between users is carried through transfer of packets. To transfer packets
from one network to another, internetworking devices such as routers, transport gateways and appli-
cation gateways are used at the junction between two networks. These devices help in converting the
packets in a format accepted by the network to which the packets are to be transferred.
27. Explain the concept of tunnelling with respect to internetworking.
Ans: Consider two hosts A and B residing in the same type of networks (say, TCP/IP-based Ethernet
LAN) and connected via a different type of network (say, a non-IP WAN) in between the two similar
networks. Now, if A and B wish to communicate with each other, they can do so using the technique
named as tunnelling, which implements internetworking.
To send a packet to host B, the host A creates an IP packet containing the IP address of the host B.
The IP packet is then put in an Ethernet frame addressed to the multiprotocol router residing in A’s
network and eventually, put on the Ethernet. When the multiprotocol router on A’s network receives
the frame, it removes the IP packet and inserts it in the payload field of the WAN network layer packet.
The packet is now addressed to the WAN address of the multiprotocol router residing in B’s network.
After the packet reaches to the multiprotocol router on B’s network, it removes the IP packet from the
Ethernet frame and sends to the host B in the local network inside an Ethernet frame. The host B then
extracts the IP packet from the Ethernet frame and uses it. Figure 10.27 shows the communication
between host A and B.

Ethernet frame WAN packet Ethernet frame

IP IP IP
Multiprotocol Multiprotocol
router router
A B

WAN
Ethernet LAN Tunnel Ethernet LAN
Figure 10.27 Tunnelling

Here, the WAN situated in between acts as a big tunnel connecting both the multiprotocol ­routers.
The IP packet and the communicating hosts know nothing about the WAN architecture. Only the
multiprotocol routers need to understand the IP and WAN packets. In effect, the distance from
the middle of one multiprotocol router to another is like a tunnel that carries IP packets without any
obstruction.
28. What is congestion? Why do we need congestion control?
Ans: Congestion is a situation that occurs in a network when the number of packets sent to the
network is far more than its capacity. It results in increased traffic in the network layer and as a conse-
quence, the TCP segments start getting lost as they pass through the network layer. Since TCP follows
retransmission policy in case of lost or delayed segments, the congestion problem is further aggregated
due to retransmission of segments. The situation becomes even worse and the performance of network
degrades rapidly. That is why congestion needs to be controlled.

CN_Chapter 10.indd 215 2/20/2012 6:34:31 PM

 216 Data communications and Computer Networks

29. What is congestion control? What are its categories?

Ans: Congestion control refers to the techniques and mechanisms used to prevent or remove
­congestion from the network. There are various ways that can be used to control the congestion all
these ways are divided into two categories, namely, open-loop and closed-loop congestion control. The
open-loop congestion control aims to prevent the congestion from occurring by using some policies.
Either the source or destination handles the congestion. On the other hand, the closed-loop congestion
­control attempts to remove congestion after it has occurred. The closed-loop congestion control is fur-
ther ­divided into two categories, namely, implicit feedback closed-loop congestion control and explicit
feedback closed-loop congestion control. In implicit feedback algorithms, the source is required to
detect the congestion by making certain observations such as monitoring the time needed for acknowl-
edgement to come back. In contrast, in explicit feedback algorithms, whenever congestion occurs, the
packets are sent back to the source from the point of congestion to indicate the congestion.
30. What are the prevention policies used by the congestion?
Ans: The open-loop congestion control method adopts certain policies to prevent the congestion.
Some of these policies are described as follows:
‰‰ Retransmission policy: When the sent packets are damaged or lost then they need to be
­retransmitted; however, the retransmission usually results in more congestion in the network. Thus,
a reliable retransmission policy needs to be implemented to prevent the congestion. The retrans-
mission timers are introduced to check, how fast the sender sends the data at a particular time. Both
the retransmission policy and timers are designed in such a way that efficiency is optimized with
preventing the congestion.
‰‰ Acknowledgement policy: Congestion may be affected by the acknowledgement policy used
by the receiver, as acknowledgements are also a part of the network. Thus, if the receiver does
not send an acknowledgement for every packet, it helps to reduce the congestion. Moreover, the
­receiver may decide to send acknowledgement only for specific packets or only when it is having
packets to send or when the timer expires. All these factors result in less traffic in the network and
thus, prevent congestion.
‰‰ Discarding policy: A reliable discarding policy can be implemented to prevent the congestion. For
instance, the routers may start to discard packets when the network is not able to handle more packets,
thus, preserving the integrity of transmission. If discarding policy is implemented correctly, then it
is the best method to prevent the congestion; otherwise, it will make the situation worse.
‰‰ Admission policy: This policy is used to prevent congestion in virtual-circuit networks. According
to this policy, whenever a request for virtual connection arrives to a router/switch, it first deter-
mines the flow of data and other resources in the network before admitting the incoming data into
the network. If there is congestion or even the possibility of congestion in the network, the router/
switch denies the connection request.
31. What are the different mechanisms carried out to remove congestion?
Ans: After congestion has occurred in the network, it can be removed by implementing various
mechanisms, some of which are described here.

Backpressure
Backpressure is a node-to-node congestion control technique where each node knows its immediate
upstream node from which it is receiving the flow of data. When congestion occurs, the congested

CN_Chapter 10.indd 216 2/20/2012 6:34:31 PM

 Routing and Congestion Control 217

node rejects to receive any more data from its upstream node. This makes the upstream node to become
­congested and thus, it also starts rejecting the data coming from its upstream node. Therefore, they drop
all the data coming from the upper node. This procedure continues and eventually, the congestion infor-
mation reaches the original source of flo , which may then slow down the flow of data. As backpressure
begins with the node where congestion is detected first and propagates back to the source, this technique
is a kind of explicit feedback algorithm.
Backpressure technique can be applied only in virtual circuit networks that support node-to-node
connection. It was introduced in X.25, the first virtual-circuit network. However, this technique
is not applicable for IP-based networks, as in such networks, a node does not know its upstream
node.

Choke Packet
Choke packet is a control packet that is created by the congested node to inform the source about
the congestion in the network. Unlike backpressure method, the choke packet is sent by the congested
node directly to the source of flow rather than to the intermediate nodes. The source receiving the
choke packet is required to reduce the rate of flow towards the router from where the choke packet
has come.
An example of choke packet is the source quench message used in ICMP (discussed in Chapter 11).
This message is sent by the router or destination node to ask the source to slow down the rate of ­sending
traffic. The router sends a source quench message for every datagram it discards due to overloaded
­buffer.

Implicit Congestion Signalling

In this method, there is no communication between the source and the nodes where congestion has
­occurred. Rather, the source is required to determine the congestion in the network by observing certain
things. For example, if the source after sending many packets still does not receive any ­acknowledgement,
it may assume that there is congestion in the network and thus, reduce its rate of sending flo . This
is called implicit congestion signalling. This method can be used in connection-oriented networks
such as frame relay networks; however, it is more effective in connectionless networks such as packet-
­switching networks and IP-based networks.

Explicit Congestion Signalling

The explicit signalling method is used by a congested node to send information to the source or
­destination. It is different from the choke packet method. In the choke packet method, when ­congestion
occurs, a separate packet signalling the congestion is sent to the source whereas in explicit ­signalling
method, the signals are included in the packets carrying data. The explicit congestion signalling can
work in two directions, namely, backward signalling and forward signalling. In backward signalling,
a bit is set in the packet that is moving in a direction opposite to the congestion. This bit informs the
source that there is congestion in the link and the traffic needs to be slow down to avoid the discard-
ing of packets. In forward signalling, a bit is set in the packet that is moving in the direction of
­congestion. This bit informs the destination that there is congestion in the link and thus, the destination
needs to adopt some policy such as reducing the speed of sending acknowledgements to remove the
­congestion.

CN_Chapter 10.indd 217 2/20/2012 6:34:31 PM

 218 Data communications and Computer Networks

32. Write a short note on load shedding.

Ans: It is one of the simple and effective closed-loop congestion control techniques. Whenever
a router discovers congestion in the network, it starts dropping out the packets until the congestion
­disappears. To decide which packet to drop, it uses certain policies. The simplest but less efficient policy
is to randomly pick up the packets to be dropped. Another complex but effective policy requires some
cooperation from the sender side. This policy uses some kind of priorities among the packets. As some
packets are more important than others, different priority classes are used to distinguish them from less
important ones. At the time of congestion in the network, the packets from the lowest priority class are
discarded first, then from the next lower class and so on. There is one more policy, which allows a host
to break the agreement made when the network was set up. According to this policy, the host can use
resources beyond its specified limit in order to deal with congestion. For example, host can increase its
bandwidth more than the specified limit.

33. List the differences between congestion control and flo control.
Ans: The congestion control and flow control are distinct in some ways that are as follows:
‰‰ In congestion control, the whole network traffic (containing many nodes) is checked for ­congestion,
while in flow control, the point-to-point traffic is checked for smooth functionin
‰‰ The congestion control involves many hosts, routers, connections while transferring the data,
whereas in flow control only sender and receiver interact with each othe .
‰‰ In congestion control, when the network is unable to handle the load, packets are dropped to
­remove the congestion. However, in flow control, when the receiver is unable to absorb more data,
it informs the sender to slow down the speed of sending data.

34. Write a short note on hop-by-hop choke packets.

Ans: The hop-by-hop choke packet method is an improvement over choke packet method.
In choke packet method, at the time of congestion, the choke packets are sent directly to the source
from congested node. However, until the time choke packets reach the source, many data packets will
have already been sent out from the source and still have to be dealt with. Thus, the choke packet
method is not desirable at high speeds or over long distances. In such cases, hop-by-hop choke
packet method is preferred. In hop-by-hop choke packet method, not only the source node but also
each intermediate node receiving the choke packet gets affected, thus, reducing the traffic between the
intermediate nodes as well as the source.
To understand how hop-by-hop choke packet method works, consider the subnet shown in
­Figure 10.28(a). Here, the data is being sent from the node P to node R via path PTSR [Figure 10.28(b)].
Now, suppose the node R runs out of buffers and thus, sends a hop-by-hop choke packet to P via
path RSTP. As the packet reaches from R to S, it informs the node S to slow down the speed of
sending data packets to R. However, as S is still receiving the data packets, it has to include more
buffers to accommodate the packets. This puts more demand on S but makes R congestion free
[Figure 10.28(c)]. Thus, S forwards the choke packet to T, which informs T to reduce its speed of
sending data packets to S. This puts more demand on T but makes S congestion free [Figure 10.28(d)].
Finally, the choke packet reaches P and the traffic flow is reduced [Figure 10.28(e)]. This mecha-
nism makes all the nodes congestion free. Thus, the congestion is controlled and that too without
losing any packets.

CN_Chapter 10.indd 218 2/20/2012 6:34:32 PM

 Routing and Congestion Control 219

Q Q Q

P R Heavy P R P R
traffic
Choke Reduced
packet flow
T S T S T S
(a) (b) (c)
Q Q

P R P R
Reduced
flow

T S T S
Reduced flow
(d) (e)
Figure 10.28 Hop-by-Hop Choke Packet Method

35. Find the class of each of the following address.

(a) 00000001 00001011 00001011 11101111
(b) 11000001 10000011 00011011 11111111
(c) 14.23.120.8
(d) 252.5.15.111
Ans: (a) As the first bit of the address is 0, it is a class A address.
(b) As the first two bits of the address are one and the third bit is zero, it is a class B address
(c) As the first byte of the address is 14 (between 0 and 127), it is a class A address.
(d) As the first byte of the address is 252 (between 240 and 255), it is a class E address
36. Why the largest octet in an Internet address is 255?
Ans: Each octet in an IP address consists of 8 bits, each of which can be either 0 or 1. The largest
value that can be expressed in 8 bits is 11111111, that is, 255. That is why the largest octet in an IP
address is 255.
37. Given the following I P addresses,
(a) 32.46.7.3
(b) 200.132.110.35
(c) 140.75.8.92

determine (i) class, (ii) network address, (iii) mask and (iv) broadcast address for each.
Ans: (a) Given IP address = 32.46.7.3
(i) As the first byte of this address is 32 (between 0 and 127), it is class A address.

CN_Chapter 10.indd 219 2/20/2012 6:34:32 PM

 220 Data communications and Computer Networks

(ii) Since it is class A address, its first byte (that is, 32) denotes the net ID. To obtain the
­network address, the host ID bits are made zero. Thus, the network address is 32.0.0.0
(iii) Being a class A address, the mask for this address is 255.0.0.0.
(iv) The broadcast address is obtained by keeping the net ID of the address same and replacing
each byte of host ID with 255. Thus, the broadcast address is 32.255.255.255.
(b) Given IP address = 200.132.110.35
(i) As the first byte of this address is 200 (between 192 and 223), it a Class C address
(ii) Since it is Class C address, its first three bytes (that is, 200.132.110) denote the net ID. To
obtain the network address, the host ID bits are made zero. Thus, the network address is
200.132.110.0.
(iii) Being a Class C address, the mask for this address is 255.255.255.0.
(iv) The broadcast address is obtained by keeping the net ID of the address same and replacing
each byte of host ID with 255. Thus, the broadcast address is 200.132.110.255.
(c) Given IP address = 140.75.8.92
(i) As the first byte of this address is 140 (between 128 and 191), it a class B address
(ii) Since it is class B address, its first two bytes (that is, 140.75) denote the net ID. To
­obtain the network address, the host ID bits are made zero. Thus, the network address is
140.75.0.0.
(iii) Being a class B address, the mask for this address is 255.255.0.0.
(iv) The broadcast address is obtained by keeping the net ID of the address same and replacing
each byte of host ID with 255. Thus, the broadcast address is 140.75.255.255.

38. T
 he IP network 192.168.130.0 is using the subnet mask 255.255.255.224. What subnets
are the following hosts on?
(a) 192.168.130.10
(b) 192.168.130.67
(c) 192.168.130.93
(d) 192.168.130.199
(e) 192.168.130.222
(f ) 192.168.130.250
Ans: To determine which host is lying on which subnet, we need to determine the total number of
subnets in the network, number of hosts in each subnet and the range of IP addresses assigned to each
subnet. For finding the number of subnets and the number of hosts in each subnet, we need to know the
number of masked and unmasked bits in the subnet mask, which can be found as follows:
Given, subnet mask = 255.255.255.224
The binary equivalent of host ID byte (that is, 224) is 11100000. Here, three bits are 1s while five
bits are 0s. Thus, the number of masked bits (m) is 3 and the number of unmasked bits (n) is 5. Now,
the number of subnets and the number of hosts in each subnet can be determined using the following
formulas.

CN_Chapter 10.indd 220 2/20/2012 6:34:32 PM

 Routing and Congestion Control 221

Number of subnets = 2m
⇒ 23
⇒8
Number of hosts in each subnet = 2n – 2
⇒ 25 – 2
⇒ 30
Thus, there are eight subnets in the network with each subnet comprising 30 hosts. As the ­given
IP­ ­network is 192.168.130.0, the range of IP addresses assigned to each subnet can be given as
­follows:
Range of first subne = 192.168.130.0 – 192.168.130.31
Range of second subnet = 192.168.130.32 – 192.168.130.63
Range of third subnet = 192.168.130.64 – 192.168.130.95
Range of fourth subnet = 192.168.130.96 – 192.168.130.127
Range of fifth subne = 192.168.130.128 – 192.168.130.159
Range of sixth subnet = 192.168.130.160 – 192.168.130.191
Range of seventh subnet = 192.168.130.192 – 192.168.130.223
Range of eighth subnet = 192.168.130.224 – 192.168.130.255
Notice that the first and last address of each range cannot be assigned to hosts. For example, in the
first subnet, the addresses 192.168.30.0 and 192.168.130.31 cannot be used for the hosts
Now, we can find which host lies on which network as given below
(a) The host 192.168.130.10 lies on the first subnet
(b) The host 192.168.130.67 lies on the third subnet.
(c) The host 192.168.130.93 lies on the third subnet.
(d) The host 192.168.130.199 lies on the seventh subnet.
(e) The host 192.168.130.222 lies on the seventh subnet.
(f) The host 192.168.130.250 lies on the eighth subnet.
39. W hat will be the subnet address if the destination address is 200.45.34.56 and subnet
mask is 255.255.240.0?
Ans: Given, destination address = 200.45.34.56
  Subnet mask = 255.255.240.0
Now, the subnet address can be determined by performing AND of destination address and the sub-
net mask.
The destination address and subnet mask can be written in binary notation as shown below:
11001000 00101101 00100010 00111000
11111111 11111111 11110000 00000000
On ANDing the both addresses, we get the following subnet address:
11001000 00101101 00100000 00000000
This subnet address can be written in dotted decimal notation as:
200.45.32.0.

CN_Chapter 10.indd 221 2/20/2012 6:34:32 PM

 222 Data communications and Computer Networks

40. G iven an IP address 156.36.2.58 with default mask 255.255.0.0/23. Determine its subnet
mask in the event of subnetting.
Ans: Given IP address = 156.36.2.58
Default mask = 255.255.0.0/22
Here, /22 indicates that the first 22 bits of the subnet mask are 1s while the rest are 0s. Thus, the
subnet mask is:
11111111 11111111 11111100 00000000
This subnet mask can be written in dotted decimal notation as 255.255.252.0.

Multiple choice questions

1. How many bits are allocated for the host ID in 7. Which protocol uses backbone router?
an IP address of class A? (a) OSPF   
(a) 8      (b) 24   (b) RIP
(c) 48     (d) 16 (c) BGP   
(d) all of these
2. Which class of IP address is used for multi-
casting? 8. ________ is a node-to-node congestion
(a) class E   (b) class B ­control technique where the congested node
(c) class D   (d) class M stops receiving data from the upstream
nodes.
3. Which class address is used to implement su-
(a) backpressure
pernetting?
(b) choke packet
(a) class D    (b) class A
(c) implicit congestion
(c) class C    (d) class B
(d) explicit congestion
4. What is the maximum limit for hop count in
9. Which of the following belongs to explicit
RIP protocol?
feedback closed-loop congestion control?
(a) 15     (b) 12  
(a) choke packet
(c) 25     (d) 10
(b) hop-by-hop choke packet
5. Which of the following metrics is used in (c) backpressure
RIP? (d) all of these
(a) delay      (b) hop count
10. Which of the following is not a prevention
(c) throughput     (d) bandwidth
policy used for congestion?
6. Which of the following is an interdomain (a) retransmission policy
routing protocol? (b) discarding policy
(a) RIP     (b) OSPF   (c) acknowledgement policy
(c) BGP     (d) none of these (d) error control policy

Answers

1. (b) 2. (c) 3. (c) 4. (a) 5. (b) 6. (c) 7. (a)  8. (a) 9. (d) 10. (d)

CN_Chapter 10.indd 222 2/20/2012 6:34:32 PM

 11
Quality of Service and Protocols

1. What is meant by quality of service? What are its characteristics?

Ans: A stream of packets being transmitted from the source node to destination node is referred to as
the flo . Each flow is characterized by a certain set of performance parameters such as reliability, jitter,
delay and bandwidth. The ability of a network to deliver the flow of packets to the destination node with
the defined set of performance parameters is defined as its quality of service (QoS). The characteristics
that a flow seeks to attain are as follows
‰‰ Reliability: This characteristic ensures that no packet is damaged or lost during the transmission and all
packets are received at the destination correctly. If the flow does not attain reliability then retransmis-
sion needs to be done. Reliability can be achieved by applying checksum to each packet at the sender’s
end and then verifying it at the receiver’s end. Some applications such as e-mail, remote login require
high reliability while others such as telephony and video-on-demand are less sensitive to reliability.
‰‰ Delay: This characteristic slows down the flow of transmitted packets that are needed to be re-
ceived by the destination node. Like reliability, delay requirements also depend on the type of the
application used. In case of file transfer applications such as e-mail, if packets are delayed by a few
seconds then it cannot cause any harm. However, if the delay occurs in real-time applications such
as telephony, then the users will not be able to understand each other’s conversation. Thus, for such
applications, delay should be the least.
‰‰ Jitter: This characteristic defines the variation in delay of packets that correspond to the same
flo . If the delay between successive packets is constant, then there is no harm but if the packets
are received with irregular time intervals between them, the result may be unacceptable. Video and
audio applications are highly sensitive to jitter while applications such as e-mail and file transfer
have less stringent requirements with respect to jitter.
‰‰ Bandwidth: This characteristic defines the number of bits that can be delivered per second. Dif-
ferent applications have different bandwidth requirements. For example, in e-mail applications,
bandwidth consumption is less as number of bits transmitted is not much but in the case of video
conferencing, more bandwidth is utilized as number of bits to be transmitted is in millions.

CN_Chapter 11.indd 223 2/20/2012 6:35:54 PM

 224 Data Communications and Computer Networks

2. What are the four general techniques to improve the QoS?

Ans: The QoS can be improved by using the four techniques, namely, scheduling, traffic shaping,
resource reservation and admission control.
‰‰ Scheduling: This technique schedules the arrival of packets from different flows so that packets
can be processed in an appropriate manner by the router or switch. Some scheduling techniques to
improve the QoS are as follows:
• FIFO Queuing: In this technique, packets are scheduled based on First-In-First-Out (FIFO)
method. A queue is maintained by the router or switch in which packets wait for their turn to
get processed. Each incoming packet is appended to the queue if there is space in the queue.
The packet, which enters the queue, first, is processed first. If the queue is filled, then the new
packets are discarded until some space is freed up in the queue.
• Priority Queuing: In this scheduling, each packet is assigned a priority class and a separate queue
is maintained for each priority class. The packets in the queue of the highest priority class are
processed first while packets in the queue of the lowest priority class are processed last. Thus,
this technique processes the higher-priority packets with less delay. However, the demerit of this
technique is that low-priority queue packets will not be processed if there is a continuous flow of
packets in high-priority queues. This situation is referred to as starvation of low-priority packets.
• Weighted Fair Queuing: In this technique, weights are assigned to different priority classes
such that higher-priority queues get higher weight and the lower-priority queues get the lower
weight. The weight assigned to a queue indicates the number of packets that will be processed
from the queue. For example, consider three queues Q1, Q2 and Q3 with priorities 1, 2 and 3 and
weights assigned are 4, 1 and 2, respectively. Initially, four packets will be processed from the
Q1; then, one packet will be processed from Q2 and finall , two packets from Q3.
‰‰ Traffi Shaping: It is a technique that improves the QoS by managing the amount of traffic, which
is to be sent to the network. This technique tries to make the traffic flow at the uniform data rate,
resulting in less congestion and improved QoS. The flow of traffic is regulated by monitoring the
traffic flow during the connection period what is referred to as traffi policing. If a packet in a
stream does not obey the policy, then it is penalized by either discarding it or by assigning low pri-
ority. There are two traffic-shaping techniques, namely, leaky bucket and token bucket (discussed
later) used to improve the QoS.
‰‰ Resource Reservation: This technique reserves the resources required for a flow beforehand when
a specific route to reach the destination node has been decided. The three kinds of resources that
are normally reserved for the flow include bandwidth, buffer space and central processor unit
(CPU) cycles. The reservation of bandwidth enables the flow to reach the destination node more
effectively. For example, if a flow requires 2 Mbps and the capacity of outgoing line is 5 Mbps then
reserving 2 Mbps for one flow will effectively work but trying to direct three flows through the
line will lead to more congestion. The buffer space must be reserved at the destination side so that
a specific flow does not compete with other flows for being placed in the queue. This is because
if the buffer space is not available, then the packets have to be discarded which may degrade the
QoS. The CPU cycles also need to be reserved, as the router takes some CPU time to process each
packet and also, it can process only a few numbers of packets per second. Thus, for effective and
timely processing of each packet, the CPU must not be overloaded.
‰‰ Admission Control: This technique is used by the router or switch to decide whether to accept or
reject the incoming flow. The decision to accept or reject the flow is made based on certain parameters

CN_Chapter 11.indd 224 2/20/2012 6:35:54 PM

 Quality of Service and Protocols 225

such as bandwidth, buffer cycles, CPU cycles and packet size. The set of these parameters is referred
to as flo specificatio . Whenever a flow comes to a router, the router checks the flow specific -
tion to determine whether it can handle the incoming flow. To determine this, the router checks its
current buffer size, bandwidth and CPU usage. It also checks its prior commitments made to other
flows. The flow is accepted only after the router becomes sure that it can handle the flo .
3. Explain the leaky bucket algorithm.
Ans: The leaky bucket algorithm is a traffic-shaping technique that is used to control the congestion
in a network. It was proposed by Turner in 1986 that used the concept of a leaky bucket—a bucket with
a hole at the bottom. The water is pouring into the bucket and it is leaking continuously. However, the
rate of leakage is always constant irrespective of the rate of water pouring into the bucket. This process
continues until the bucket is empty. If the bucket is overflowed, then additional water falls through
the sides of bucket but leakage rate will always be constant. Turner applied the same idea to packets
transmitted over the network and therefore, the algorithm was named as leaky bucket algorithm. This
algorithm smooths out the bursty traffic by storing bursty chunks of packets in the leaky bucket so that
they can be sent out at a constant rate.
To understand the leaky bucket algorithm, let us assume that network has allowed the hosts to trans-
mit data at the rate of 5 Mbps and each host is connected to the network through an interface containing
a leaky bucket. The leaky bucket enables to shape the incoming traffic in accordance with the data rate
committed by the network. Suppose the source host transmits the bursts of data at the rate of 10 Mbps
for the first three seconds, that is, a total of 30 Mbits of data. After resting for 3 s, the source host again
transmits the bursts of data at the rate of 5 Mbps for 4 s, that is, a total of 20 Mbits. Thus, the total
amount of data transmitted by the source host is 50 Mbits in 10 s. The leaky bucket sends the whole data
at the constant rate of 5 Mbps (that is, within the bandwidth commitment of the network for that host)
regardless of the rate arrived from the source node. If the concept of leaky bucket was not used, then
more bandwidth would be consumed by the starting burst of data, leading to more congestion.
The leaky bucket algorithm is implemented by maintaining a FIFO queue to hold the arriving pack-
ets. The queue has a finite capacity. Whenever a packet arrives, it is appended at the end of queue if there
is some space in queue; otherwise, the packet is discarded (Figure 11.1). If each packet is of fixed size,
then a fixed number of packets are removed from the queue per each clock tick. However, if packets are
of variable sizes, a fixed number of bytes (say, p) are removed from the queue per each clock tick. The
algorithm for variable size packets is as follows:
1. A counter is initialized to p at the tick of the clock.
2. If packet size is smaller than or equal to p, the packet is sent and value of counter is decremented
by the packet size.
3. Repeat step 2 until p becomes smaller than the packet size.
4. Reset the counter and go back to step 1.

N Remove packets
Arrival Full?
at a constant rate
Queue

Discard Y

Figure 11.1 Leaky Bucket Implementation

CN_Chapter 11.indd 225 2/20/2012 6:35:54 PM

 226 Data Communications and Computer Networks

4. Explain token bucket algorithm.

Ans: The token bucket algorithm is a traffic-shaping technique used to control the congestion in the
network. Though it is a variation of leaky bucket algorithm, it is more flexible as it allows the output
to go at the higher rate when a large burst of traffic arrives in the bucket rather than persisting with the
constant output rate. Moreover, it takes into account the time for which a host remains idle so that the
idle hosts could take benefit in the future. For this, the system generates many tokens (say, n) per each
clock tick and these tokens are fed into the bucket. The token are added to the bucket as long as there
is a space in the bucket. For transmitting each packet, one token is removed from the bucket and is de-
stroyed. Now, if some host remains idle for 50 clock ticks and the system sends 20 tokens per each clock
tick then there will be 1,000 (50*20) tokens in the bucket of idle host. When host becomes active, either
it can transmit 1,000 packets (one per each token) in a single clock tick or it can transmit 100 packets
per each clock tick up to 10 clock ticks or any other way thereby sending the bursty traffic
The token bucket algorithm is implemented by the counter to count tokens. The counter variable
is incremented by one each time a token is added in the leaky bucket and is decremented by one each
time a packet is transmitted. If the counter value becomes zero then the host cannot send any further
packets.
The performance of token bucket algorithm is measured by the burst time (S) which is given as
S = C/(M - P)
where C is the token bucket capacity in bytes, M is the maximum output rate in bytes/s and P is the token
arrival rate in bytes/s.

5. Distinguish between leaky bucket algorithm and token bucket algorithm.

Ans: Though both leaky bucket and token bucket are traffic-shaping techniques that smooth out the
traffic between routers and regulate output from nodes, still there are some differences between the two.
These differences are as follows:
‰‰ Token bucket algorithm stores tokens up to the maximum size of the bucket if the node is not send-
ing packet for some period of time, whereas in leaky bucket algorithm no such storing of tokens
happens.
‰‰ Token bucket algorithm can send large bursts of packets at once if needed but in leaky bucket
­algorithm, the output rate always remains constant.
‰‰ Token bucket algorithm never discards the packet but throws away the tokens not the packets if the
bucket overflows, whereas in leaky bucket algorithm the packets are discarded when the bucket
fills up
‰‰ Token bucket algorithm does not result in loss of data as node with the token bucket stops sending
the packets if the router defines such a rule but in leaky bucket algorithm there is no such provision
to prevent any data loss.

6. Write a short note on ARP.

Ans: Address resolution protocol (ARP) is a network layer protocol used to map an Internet pro-
tocol (IP) address (logical address) to its corresponding media access control (MAC) address (physical
address). The mapping from logical to physical address can be of two types, namely, static mapping and
dynamic mapping. In static mapping, each node on the network maintains a table that contains entries
of IP addresses of all other nodes along with their corresponding MAC addresses. If any node knows the
IP address of a particular node but does not know its MAC address then it can find the corresponding

CN_Chapter 11.indd 226 2/20/2012 6:35:55 PM

 Quality of Service and Protocols 227

entry from the table. The disadvantage of static mapping is that as the physical address of a node may
change, the table must be updated at regular intervals of time and thus, causing more overhead. On the
other hand, in dynamic mapping, a protocol can be used by the node to find the other address if one is
known. The ARP is based on dynamic mapping.
Whenever a host wishes to send IP packets to another host or router, it knows only the IP address of
the receiver and needs to know the MAC address as the packet has to be passed through the physical
network. For this, the host or router broadcasts an ARP request packet over the network. This packet
consists of IP address and MAC address of the source node and the IP address of the receiver node. As
the packet travels through the network, each node in between receives and processes the ARP request
packet. If a node does not find its IP address in the request, it simply discards the packets. However,
when an intended recipient recognizes its IP address in the ARP request packet, it sends back an ARP
response packet. This packet contains the IP and MAC address of the receiver node and is delivered
only to the source node, that is, ARP response packet is unicast instead of broadcast.
The performance of ARP decreases if every time the source node or router has to broadcast an ARP
request packet to know the MAC address of the same destination node. Thus, to improve the efficienc ,
ARP response packets are stored in the cache memory of the source system. Before sending any ARP
request packet, the system first checks its cache memory and if the system finds the desired mapping in
it then the packet is unicasted to the intended receiver instead of broadcasting it over the network.
The format of ARP packet is shown in Figure 11.2.
The ARP packet comprises various fields, which are described as follows
‰‰ Hardware Type: It is a 16-bit long field that defines the type of the network on which ARP is run-
ning. For example, if ARP is running on Ethernet then the value of this field will be one. ARP can
be used on physical network.
‰‰ Protocol Type: It is a 16-bit long field that defines the protocol used by ARP. For example, if
ARP is using IPv4 protocol then the value of this field will be (0800)16. ARP can be used with any
protocol.
‰‰ Hardware Length: It is an 8-bit long field that defines the length of MAC address in byte

Hardware type Protocol type

16 bits 16 bits

Hardware length Protocol length Operation

8 bits 8 bits 16 bits

Sender hardware address

Sender protocol address

Target hardware address

Target protocol address

Figure 11.2 ARP Packet Format

CN_Chapter 11.indd 227 2/20/2012 6:35:55 PM

 228 Data Communications and Computer Networks

‰‰ Protocol Length: It is an 8-bit long field that defines the length of address in bytes.
‰‰ Operation: It is a 16-bit long field that defines the type of packet being carried out. For ARP
request packet, the value of this field will be one and for ARP response packet, the value will be
two.
‰‰ Sender Hardware Address: It is a variable-length field that defines the MAC address of the
sender node.
‰‰ Sender Protocol Address: It is a of variable-length field that defines the IP address of the sender
node.
‰‰ Target Hardware Address: It is a variable-length field that defines the MAC address of the desti-
nation node. In case of an ARP request packet, the value of this field is 0s as the MAC address of
the receiver node is not known to the sender node.
‰‰ Target Protocol Address: It is a variable-length field that defines the IP address of the destination
node.
7. Write a short note on the following.
(a) RARP
(b) BOOTP
(c) DHCP
Ans:
(a) RARP: Reverse address resolution protocol, as the name implies, performs the opposite of
ARP. That is, it helps a machine that knows only its MAC address (logical address) to find the IP ad-
dress (logical address). This protocol is used in situations when a diskless machine is booted from read-
only-memory (ROM). As ROM is installed by the manufacturer, it does not include the IP address in its
booting information because IP addresses are assigned by the network administrator. However, MAC
address of the machine can be identified by reading its NIC. Now, to get the IP address of the ­machine
in a network, RARP request packet is broadcast to all machines on the local network. The RARP
request packet contains the MAC address of the inquiring machine. The RARP server on the network
that knows all the IP addresses sees this request and responds with a RARP reply packet containing the
corresponding IP address to the sender machine.
The problem in using RARP protocol is that if there is more than one network or subnets then RARP
server needs to be configured on each network as RARP requests are not forwarded by the routers and,
thus, cannot go beyond the boundaries of a network.
(b) BOOTP: Bootstrap protocol is an application layer protocol designed to run in a ­client/
server environment. The BOOTP client and BOOTP server can be on the same or different network. The
BOOTP uses user datagram protocol (UDP) packets, which are encapsulated in an IP packet. A BOOTP
request packet from a BOOTP client to BOOTP server is broadcast to all the nodes on the network. In
case the BOOTP client is in one network and BOOTP server is on another network and two networks
are separated by many other networks, the broadcast BOOTP request packet cannot be forwarded by the
router. To solve this problem, one intermediary node or router, which is operational at the application
layer, is used as a relay agent. The relay agent knows the IP address of the BOOTP server and when
it receives a BOOTP request packet, it unicasts the BOOTP request packet to the BOOTP server by
including the IP address of the server and of itself. On receiving the packet, the BOOTP server sends
BOOTP reply packet to the relay agent, which further sends it to the BOOTP client.

CN_Chapter 11.indd 228 2/20/2012 6:35:55 PM

 Quality of Service and Protocols 229

A problem associated with BOOTP is that it is a static configuration protocol. The mapping table
containing MAC and IP addresses is configured manually by the network administrator. Thus, a new
node cannot use BOOTP until its IP and MAC address have been entered manually by the network
administrator in the table.
(c) DHCP: Dynamic host control protocol supports both static and dynamic address allocation
that can be done manually or automatically. Thus, it maintains two databases one for each allocation. In
static address allocation, DHCP acts like BOOTP, which means that a BOOTP client can request for a
permanent IP address from a DHCP server. In this type of allocation, DHCP server statically maps MAC
address to IP address using its database. On the other hand, in dynamic address allocation, dynamic
database is maintained by DHCP that contains the unused IP addresses. Whenever a request comes for
an IP address, DHCP server assigns a temporary IP address to the node from the dynamic database using
a technique called leasing. In this technique, the node requests the DHCP server to renew the IP address
just before the time of using this IP address expires. If the request is denied by the DHCP server, then
the node cannot use the same IP address that was assigned to it earlier.
Like BOOTP, DHCP also uses relay agent on each network to forward the DHCP requests. When
a DHCP node requests for an IP address, the relay agent on network helps the request to unicast to the
DHCP server. On receiving the request, the server checks its static database to find the entry of the
requested physical address. If it finds the address, it returns the permanent IP address to the node; oth-
erwise, it selects some temporary address from the available pool, returns this address to the host and
also, adds this entry to the dynamic database.
8. Draw and discuss the IP datagram frame format. Discuss in detail the various fields
Ans: The Internet protocol version 4 (IPv4) is the most widely used internetworking protocol. In
IPv4, the packets are termed as datagrams (a variable length packet). Further, IPv4 is a connectionless
and unreliable datagram protocol; connectionless means each datagram is handled independently and
can follow different paths to reach the destination; unreliable means it does not guarantee about the
successfully delivery of the message. In addition, IPv4 does not provide flow control and error control
except the error detection in the header of the datagram. To achieve reliability, the IP is paired with
transmission control protocol (TCP), which is a reliable protocol. Thus, it is considered as a part of
TCP/IP suite.
An IPv4 datagram consists of a header field followed by a data field. The header field is 20–60
bytes long and contains routing and delivery information. The header comprises various subfields
(­Figure 11.3), which are described as follows:
‰‰ Version (VER): It is a 4-bit long field, which indicates the version being used. The current version
of IP is 4.
‰‰ Header Length (HLEN): It is a 4-bit long field, which defines the IPv4 header length in 32-bit
words. The minimum length of IPv4 header is five 32-bit words
‰‰ Service: It is an 8-bit long field, which provides an indication of the desired QoS such as prece-
dence, delay, throughput and reliability. This field is also called type of service (ToS) field
‰‰ Total Length: It is a 16-bit long field, which defines the total length (in bytes) of the datagram in-
cluding header and data. The maximum permitted length is 65,535 (216–1) bytes with 20–60 bytes
for header and rest for data.
‰‰ Identification It is a 16-bit long field, which uniquely identifies the datagram. The datagrams
can be fragmented at the sender’s end for transmission and then reassembled at the receiver’s end.

CN_Chapter 11.indd 229 2/20/2012 6:35:55 PM

 230 Data Communications and Computer Networks

VER HLEN Service Total length

4 bits 4 bits 8 bits 16 bits

Identification Flags Fragmentation

16 bits 3 bits offset 13 bits

Time to live Protocol Header checksum

8 bits 8 bits 16 bits

Source IP address
32 bits

Destination IP address
32 bits

Options

Figure 11.3 IPv4 Datagram Header Format

When a datagram is fragmented into multiple datagrams, all datagrams belonging to the same da-
tagram are labelled with same identification number and the datagrams having same identification
number are reassembled at the receiving side.
‰‰ Flags: It is a 3-bit long field in which the first bit is reserved and always zero, the second bit is do
not fragment (DF) and the third bit is more fragment (MF). If DF bit is set to 1 then it means the
datagram must not be fragmented while a value of zero indicates the fragmentation of the data-
gram. If MF bit is set to zero, then it means this fragment is the last fragment. However, if MF bit
is zero then it means there are more fragments after this fragment.
‰‰ Fragmentation Offset: It is a 13-bit long field that indicates the relative position of this fragment
with respect to the entire fragment. The fragmentation offset indicate the actual datagram to which
the given fragment belongs. It is measured in units of eight octets (64 bits). The first fragment is
set to the offset zero.
‰‰ Time-to-Live (TTL): It is an 8-bit long field, which indicates the total time (in seconds) or
number of hops (routers) that an IPv4 datagram can survive before being discarded. As a router
receives a datagram, it decrements the TTL value of datagram by one and then forwards that da-
tagram to the next hop and so on. When the TTL value becomes zero, it is discarded. Generally,
when a datagram is sent by the source node, its TTL value is set to the two times of the maximum
number of routes between the sending and the receiving hosts. This field is needed to limit the
lifetime of datagram because it may travel between two or more routers for a long time without
ever being delivered to the destination host. Therefore, to avoid this, we discard the datagram
when its TTL value becomes zero.
‰‰ Protocol: It is an 8-bit long field, which specifies the higher-level protocols that uses the services
of IPv4.
‰‰ Header Checksum: It is a 16-bit long field that is used to verify the validity of the header and is
recomputed each time when the TTL value is decremented.
‰‰ Source IP Address: It is a 32-bit long field, which holds the IPv4 address of the sending host. This
field remains unchanged during the lifetime of the datagram

CN_Chapter 11.indd 230 2/20/2012 6:35:55 PM

 Quality of Service and Protocols 231

‰‰ Destination IP Address: It is a 32-bit long field, which holds the IPv4 address of the receiving
host. This field remains unchanged during the lifetime of the datagram
‰‰ Options: These are included in the variable part of header and can be of maximum 40 bytes. These
are optional fields, which are used for debugging and network testing. However, if they are present
in the header then all implementations must able to handle these options. Options can be of single
or multiple bytes. Some examples of single byte options are no-operation and end of option and
of multiple bytes are record route and strict source route.
9. Explain in detail about IPv6.
Ans: The IP is the foundation for most Internet communications. Further, IPv6 is a version of IP that
has been designed to overcome the deficiencies in IPv4 design. Some of the important issues that reflect
IPv4 inadequacies include:
‰‰ The IPv4 has a two-level address structure (network number, host number), which is inconvenient
and inadequate to meet the network prefixes requirements. In addition, IPv4 addresses need extra
overhead like subnetting, classless addressing, NAT, address depletion, etc., which are still a big
issue for efficient implementation
‰‰ Internet also deals with real-time audio and video transmission, which requires high speed with
minimum delays and requires reservation schemes of resources. There is no such a procedure in
IPv4 to deal with such kind of transmission.
‰‰ Some confident al applications need authentication and encryption to be performed during data
transmission, However, IPv4 does not provide any authentication and encryption of the packets.
Thus, an Internetworking Protocol, version 6 (IPv6), also known as Internetworking Protocol Next
Generation (Ipng) with enhanced functionality is proposed by the Internet Engineering Task Force
(IETF) to accommodate the future growth of Internet.

Packet Format of Ip v 6
Figure 11.4 shows the format of IPv6 packet. An IPv6 packet consists of two fields: base header field of
40 bytes and a payload field of length up to 65,535 bytes. The payload field further consists of optional
extension headers and data packets from upper layer.

40 bytes Up to 65,535 bytes

Base header Payload

Extension header
Data packet from upper layer
(optional)

Figure 11.4 IPv6 Datagram Format

CN_Chapter 11.indd 231 2/20/2012 6:35:55 PM

 232 Data Communications and Computer Networks

The format of IPv6 datagram header is shown in Figure 11.5.

VER PRI Flow label

4 bits 4 bits 24 bits
Payload length Next header Hop limit
16 bits 8 bits 8 bits
Source address
128 bits
Destination address
128 bits
Next header Header length

Next header Header length

Next header Header length

Figure 11.5 IPv6 Datagram Header Format

The description of various fields included in IPv6 header is as follows
‰‰ Version (VER): It is a 4-bit long field, which indicates the version being used. The current version
of IPv6 is 6.
‰‰ Priority (PRI): It is a 4-bit long field, which indicates the transmission priority of the packet in
accordance to traffic congestion
‰‰ Flow Label: It is a 24-bit (3-byte) long field which is used to provide special handling to some
particular data packet flow like audio and video packets
‰‰ Payload Length: It is a 16-bit (2-byte) long field which defines the length of the remainder of the
IPv6 packet following the header, in octets.
‰‰ Next Header: It is an 8-bit long field which identifies the first extension header (if extension
header is available) following the base header or defines the protocol in the upper layer such as
TCP, UDP or ICMPv6.
‰‰ Hop Limit: It is an 8-bit long field, which defines the maximum number of routers the IPv6 packet
can travel during its lifetime. The purpose of this field is same as of TTL field in IPv4
‰‰ Source Address: It is a 128-bit long field, which holds the IPv6 address of the sending host. This
field remains unchanged during the lifetime of the datagram
‰‰ Destination Address: It is a 128-bit long field, which holds the IPv6 address of the receiving host.
It remains unchanged during the lifetime of the datagram.
10. Discuss the advantages of IPv6 over IPv4.
Ans: The IPv6 has the following advantages over IPv4:
‰‰ Bigger Address Space: IPv6 has 128-bit address, which can make 2128 addresses.
‰‰ Improved Header Format: In IPv6 header format, options are kept distinct from the base header
and can be inserted when needed. These optional headers are not examined by any router during
the datagram path thereby simplifying and speeding up the routing process.

CN_Chapter 11.indd 232 2/20/2012 6:35:56 PM

 Quality of Service and Protocols 233

‰‰ Auto Configuratio of Addresses: IPv6 protocol can provide dynamic assignment of IPv6 addresses.
‰‰ Allow Future Extensions: IPv6 design allows any future extensions, if needed, to meet the
­requirements of future technologies or applications.
‰‰ Support for Resource Allocation: In IPv6, a new mechanism called flo label has been intro-
duced to replace TOS field. With this mechanism, a source can request special handling of the
packet. This mechanism supports real-time audio and video transmission.
‰‰ Enhanced Security: Ipv6 includes encryption and authentication options, which provide integrity
and confidentiality of the packet
11. Compare IPv4 header with IPv6 header.
Ans: Both IPv4 and IPv6 are the variants of IP, which have certain differences between them. These
differences are listed in Table 11.1.

Table 11.1 Differences Between IPv4 and IPv6 Header

IPv4 Header IPv6 Header
• IPv4 has header length (HLEN) field and it is variable • IPv6 does not have header length field because the
in size. length of the header is fixed.
• It has service field to provide desired QoS such as • The priority and flow label fields together provide the
precedence, delay, throughput and reliability. functionality provided by the service field.
• It has total length field, which shows the total length • It has payload field that indicates the length of data
(header plus data) of the datagram. but not the header.
• It has identification, flag, and offset fields in its base • T hese fields have been removed from the base header
header. and are included in the fragmentation extension header.
• In this, the TTL field specifies time to live in seconds. • IPv6 uses a hop limit field, which serves the same
purpose as TTL field in IPv4.
• There is a protocol field. • T here is a next header field in place of protocol field in
IPv4.
• There is a header checksum field in IPv4, which is used • There is no header checksum field needed because it
to detect errors in the header of an IP packet and results is provided by the protocol in upper layer.
into an extra overhead while processing an IP packet.
• There is an options field. • There is an extension header field.
• The source and destination address fields are of 32 bits. • The source and destination fields are of 128 bits.

12. Discuss some deficiencie of IP.

Ans: The IP is the best effort service to deliver a datagram from the original source to the final des-
tination. However, it has certain deficiencies, which are as follows
‰‰ It does not support error control and assistance mechanism. Sometimes, unexpected errors may oc-
cur, such as a router may have to discard a datagram because it cannot find a route to the destination
or TTL field has a zero value. In such a situation, the I protocol is unable to inform the source host.
‰‰ It does not provide any support for host and management queries. Sometimes, a host wants to
know whether the router or another host is alive and sometimes, a network administrator may need
information about another host or router.

CN_Chapter 11.indd 233 2/20/2012 6:35:56 PM

 234 Data Communications and Computer Networks

13. Explain message types associated with ICMP.

Ans: Internet control message protocol (ICMP) has been designed to overcome the problems
with IP. It provides a mechanism for error reporting and host and management queries. It addresses and
reports these problems through ICMP messages. The ICMP messages are of two types, namely, error-
reporting messages and query messages.
Error-reporting Messages
The ICMP facilitates for reporting error messages to the original source. It does not correct errors but
only reports them. The error-reporting messages report problems that a router faces while processing an
IP packet. The ICMP handles five types of errors, which are as follows
‰‰ Destination Unreachable: The destination-unreachable messages are used in case of routing or
delivery error. It may happen that routers are unable to route the datagrams or a host is unable to
deliver it. In such situations, destination-unreachable messages are sent back to the source host. It
is to be noted that these messages cannot be created by the routers or destination host.
‰‰ Time Exceeded: The time-exceeded message is sent in two cases: first, when the packet is dropped
because the value of TTL field reaches zero and second, when a destination does not receive all the
fragments belonging to the same packet within a certain time limit.
‰‰ Parameter Problem: When a router or the destination host finds an illegal or missing value in any
field of the datagram then it discards and sends a parameter problem message to the source
‰‰ Source Quench: If a router or host discards the datagram due to congestion in the network, then it
sends a source quench message to the source host. There are two purposes of sending this message.
First is to tell the source that the sending datagram has been discarded and second is to slow down
the process of sending datagrams to prevent the congestion.
‰‰ Redirection: When a router finds that the packet has been sent to the wrong route because of some
sudden change in the routing table or loss of a routing table update, then it sends a redirect message
to the source host to notify the problem.
Query Messages
The ICMP allows sending query messages in addition to error-reporting messages. The query messages
help a host or a network manager to receive specific information from a router or another host. For
example, nodes can search their neighbours, hosts can search and know about routers on their network,
and routers can help a node to redirect its messages. The query messages always occur in pairs. Differ-
ent types of query messages are as follows:
‰‰ Echo Request and Echo Reply: The echo-request and echo-reply messages are used to determine
whether there is communication between two systems at the IP level. If the machine that has sent
the echo-request message, receives the echo-reply message, it is proved that these two machines
are communicating with each other through IP datagrams. A ping command is used to determine
the communication in which a series of echo-request and echo-reply messages are transferred
­between two hosts to check the connectivity.
‰‰ Timestamp Request and Reply: The timestamp request and timestamp reply messages are used
by the hosts or routers to find out the round-trip time required for an IP datagram to reach from one
to another. It is also used to synchronize the clocks.

CN_Chapter 11.indd 234 2/20/2012 6:35:56 PM

 Quality of Service and Protocols 235

‰‰ Address Mask Request and Reply: The address mask request is sent by a host to a router on a ­local
area network (LAN) that knows its IP address but wants to know its corresponding mask. In response,
the address mask reply messages are sent by the router to the host by providing it necessary mask.
‰‰ Router Solicitation and Advertisement: To send data to a host on another network, the source
host needs to know the router address of its own network which connects it to other networks.
Along with this, it needs to know whether neighbour’s routers are alive and functioning. In these
­situations, the router solicitation and advertisement message helps the host to find out such
­information. The source host broadcasts a router-solicitation message. The router, which receives the
router-­solicitation ­message, broadcast their routing table information through router-­advertisement
­message. The ­router-advertisement message includes an announcement about its presence as well
as of other routers on the network.
14. In a leaky bucket, what should be the capacity of the bucket if the output rate is 5 gal/min
and there is an input burst of 100 gal/min for 12 s and there is no input for 48 s?
Ans: Total input = 100*(12/60) + 0*48 = 20 gal/min
Given, output rate = 5 gal/min
Thus, the capacity of bucket = 20 – 5 = 15 gal.
15. Imagine a flo specificatio that has a maximum packet size of 1,000 bytes, a token ­bucket
rate of 10 million bytes/s, a token bucket size of 1 million bytes and a maximum transmission rate
of 50 million bytes/s. How long can a burst at maximum speed last?
Ans: Given that
Bucket size, C = 1 million bytes
Maximum transmission rate, M = 50 million bytes/s
Token bucket rate, P = 10 million bytes/s
Burst time, S = ?
We know that, S = C/(M – P)
Putting the respective values in the above formula, we get
S = 1/(50 – 10)
⇒ S = 1/40
⇒ S = 0.025 s
Therefore, burst will last for 0.025 s.

Multiple Choice Questions

1. ________ defines the variation in the packet (c) priority queuing
delay. (d) weighted fair queuing
(a) jitter (b) bandwidth 3. Which of the following ICMP messages is
(c) reliability (d) none of these sent by the router to source host to inform
2. Which of the following are not the ­scheduling that the packet has been discarded due to
techniques used to improve the QoS? congestion?
(a) FIFO queuing (a) redirection (b) parameter problem
(b) LIFO queuing (c) time exceeded (d) source quench

CN_Chapter 11.indd 235 2/20/2012 6:35:56 PM

 236 Data Communications and Computer Networks

4. In which of the following algorithms, the (a) 16 (b) 8

output rate of burst packets can be of (c) variable (d) 32
­variable rate? 6. Mapping from MAC address to IP address is
(a) token bucket (b) leaky bucket done by
(c) both (a) and (b) (d) none of these (a) RARP (b) BOOTP
5. The protocol field in the ARP packet ­format (c) DHCP (d) All of these
is of ________bits.

Answers
1. (a) 2. (b) 3. (d) 4. (a) 5. (b) 6. (d)

CN_Chapter 11.indd 236 2/20/2012 6:35:56 PM

 12
Internet Transport Protocols

1. Explain the duties of transport layer.

Ans: The transport layer ensures process-to-process delivery of message. As multiple processes
may be running at the same time on the communicating hosts, it is the duty of the transport layer to
-communicate message from a specifi process on sending host to the desired process on the receiving
host. It also ensures that the receiving process receives the whole segment in the exact order as it was
sent by the sending process. Some other responsibilities of transport layer are as follows:
‰‰ Segmentation and Reassembly: A message is divided into segments by the transport layer with
each segment being given a sequence number. These sequence numbers enable the destination
transport layer to reassemble the segments in exact order as they were sent by the sender and thus,
help in identifying the segments that have been lost during the transmission.
‰‰ Addressing: As many processes may be running on the communicating hosts at the same time, it is
necessary to identify the desired process out of many processes. For this, the transport layer header
must include a service point address (port address) in each segment.
‰‰ Connection Control: The transport layer provides both connection-oriented and connectionless
-services. In connection-oriented service, a connection must be established firs between two commu-
nicating processes before transmitting any segments. After transmitting the whole data, the connection
is released. On the other hand, in connectionless service, no such connection is established and each
segment takes different route to reach destination process.
‰‰ Flow Control: The transport layer is responsible for controlling the flo of data such that no
sending process should send segments at a rate faster than the receiving process can process. The
transport layer provides end-to-end flo control rather than across a single channel as provided by
the data link layer.
‰‰ Error Control: The transport layer functions in such a way that the receiving process not only can
detect the errors but also can determine the location of errors in the segment. The transport layer provides
process-to-process error control rather than across a single link as provided by the data link layer.

CN_Chapter 12.indd 237 2/20/2012 6:45:19 PM

 238 Data communications and Computer Networks

2. In OSI model, both data link layer and transport layer are involved in error control. Why
same activity twice? Justify.
Ans: Both data link layer and transport layer provide error control, but where the transport layer
­provides end-to-end error control and the data link control provides error control across a single link.
The data link layer makes the physical link reliable by adding the mechanism for detecting the errors and
retransmitting frames in case of lost and damaged frames. On the other hand, the transport layer ensures
that the entire message is delivered to the receiving process without any error and in the same order as
sent by the sending process. Retransmission is used in the transport layer to perform error correction.
3. Why do we need port addresses?
Ans: Usually, a machine provides a variety of services such as electronic mail, TELNET and FTP.
To differentiate among these services, each service is assigned with a unique port number. To avail some
specifi service on a machine, firs it is required to connect to machine and then connect to the port
­assigned for that service. The port numbers less than 1,024 are considered well-known and are reserved
for standard services. For example, the port number used for TELNET is 23.
4. What is socket address? Explain socket addressing.
Ans: In transport layer, two processes communicate with each other via sockets. A socket acts as an
end-point of the communication path between the processes. To ensure the process-to-process delivery,
the transport layer needs the IP address and the port number at each communicating end. The IP ­address
is used to identify the machine on the network and the port number is used to identify the specifi
-process on that machine. The IP address and port address together defin the socket address.
To enable the communication, each of the communicating processes (client and server) creates its
own socket and these sockets are to be connected. The client socket address uniquely identifie the client
process while the server socket address uniquely identifie the server process.
The server listens to a socket bound to a specifi port for a client to make connection request. When-
ever a client process requests for a connection, it is assigned a port number (greater than 1,024) by the
host computer (say, M). Using this port number and the IP address of host M, the client socket is ­created.
For example, if the client on host M having IP address (125.61.15.7) wants to connect to TELNET server
(listening to port number 23) having IP address (112.56.71.8), it may be assigned a port number 1,345.
Thus, the client socket and server socket used for the communication will be (125.61.15.7:1345) and
(112.56.71.8:23), respectively as shown in Figure 12.1.
Each connection between client and server employs a unique pair of sockets. That is, if another client
on host M wants to connect to TELNET server, it must be assigned a port number different from 1,345
(but greater than 1,024).

Socket Socket
(125.61.15.7:1345) (112.56.71.8:23)

Host M TELNET server

Figure 12.1 Communication Between Sockets
5. Write a short note on Berkeley sockets.
Ans: The transport layer offers certain operations called transport service primitives that enable
the application programmes to use the transport services. Each transport service has its own access prim-
itives. The set of transport service primitives used in Berkeley UNIX for transmission control protocol

CN_Chapter 12.indd 238 2/20/2012 6:45:20 PM

 Internet Transport Protocols 239

(TCP) are referred to as socket primitives. These socket primitives are commonly used for Internet
programming and also provide more flexibility. Various socket primitives used in Berkeley UNIX for
TCP are described as follows:
‰‰ SOCKET: This primitive is used to create a new communication end point. It also allocates table
space to the end point within the transport entity. This primitive is executed both by the client and
by the server.
‰‰ BIND: This primitive is used to assign network addresses to newly created sockets. After the addresses
have been attached with sockets, the remote systems can connect to them. This primitive is executed
only at the server side and always after the SOCKET primitive. At the client side, there is no need to
execute BIND primitive after the SOCKET primitive. This is because the server has nothing to do with
the address used by the client.
‰‰ LISTEN: This primitive is used to indicate the willingness of a server to accept the incoming
­connection requests. If a number of clients attempt to make a connection with the server, it allo-
cates space to queue up all the incoming requests. This primitive is executed only at the server side
and always after the BIND primitive.
‰‰ ACCEPT: This primitive is used to make some incoming connection wait as long as a ­connection
request does not arrive. This primitive is executed only at the server side and always after the
LISTEN primitive.
‰‰ CONNECT: This primitive is executed by the client to attempt to establish a connection with the
server. As the client executes the CONNECT primitive, it gets blocked. It remains blocked until
it receives a transport packet data unit (TPDU) from the server, which indicates the completion of
CONNECT primitive. Then, the client gets unblocked and a full-duplex connection is established
between the client and the server.
‰‰ SEND: This primitive is used to send data over the full-duplex connection. Both client and server
can execute this primitive.
‰‰ RECEIVE: This primitive is used to receive data from the full-duplex connection. Both client and
server can execute this primitive.
‰‰ CLOSE: This primitive is used to release the connection between the client and the server. The con-
nection is terminated only after both the communicating parties have executed the CLOSE primitive.
6. Explain various schemes used by the transport layer to fin the transport service access
point (TSAP) at a server.
Ans: Whenever an application process (client) wishes to establish a connection with some -application
process running on the remote system (server), it needs to specify which one to connect to. For this, the
transport layer specifie transport addresses to which the process can listen for connection requests. In
transport layer, these end points are termed as TSAP. Both client and server processes get attached to
a TSAP for establishing a connection with remote TSAP. However, now the problem arises, how the
client knows which TSAP a specifi process on the server is listening to? To identify this, some scheme
is needed.
One such scheme is the initial connection protocol scheme. In this scheme, not all the server ­processes
are required to listen to well-known TSAPs; rather, each machine that wants to serve to ­remote users has
a proxy server called process server. The process server can listen to a number of ports ­simultaneously.
When a CONNECT request from a user specifying the TSAP address of a specifi server process arrives,
it gets connected to the process server in case no server process is waiting for it. After the desired server
process gets available, the process server creates the requested server which then inherits the existing

CN_Chapter 12.indd 239 2/20/2012 6:45:20 PM

 240 Data communications and Computer Networks

connection with the user. This way, the user gets connected to the desired server process. The newly
created server then performs the requested job while the process server gets back to listening to ports.
Though this scheme seems fine, it cannot be used for servers that cannot be created when required. To
overcome this limitation, another scheme is used.
In the alternative scheme, there exists a process known as a name server or a directory server which
listens to a well-known TSAP. The name server contains an internal database in which all the service
names along with their TSAP addresses are stored. This scheme requires every newly created service
to register itself with the name server. When a client needs to fin the TSAP address corresponding to
a service, it establishes a connection with the name server. After the connection has been established,
the client sends a message including the requested service name to the name server. The name server
searches through its database and sends the requested TSAP address back to the client. After receiving
the TSAP address by the client, the connection between client and name server is released and a new
connection is established between the client and the service requested by it.
7. What is meant by upward and downward multiplexings?
Ans: Multiplexing is a technique that allows the simultaneous transmission of multiple signals
across a single data link. In transport layer, we may need multiplexing in the following two ways, which
are as follows:
‰‰ Upward Multiplexing: When there is only one network address available on a host, then all
the transport connections on that host use the same network connection. Whenever a TPDU is
­received, some means is required to indicate which process is to give the TPDU. This situation is
known as upward multiplexing.
‰‰ Downward Multiplexing: The subnet uses virtual circuits with each virtual circuit having a fixed data
rate. Now, if a user needs higher bandwidth than that of a single virtual circuit to transport the data,
then the traffic from a single transport connection can be multiplexed to multiple network connections
(virtual circuits) thereby increasing the bandwidth. This is what is called downward multiplexing.
8. Explain in detail user datagram protocol (UDP). Also list its uses.
Ans: UDP is a connectionless and unreliable transport protocol that offers process-to-process
­delivery with limited error checking. By connectionless, we mean that the segments are sent to the
destination host without any prior establishment of the connection between communicating hosts. By
unreliable, we mean that UDP does not perform error and flo control and thus, does not guarantee
about the proper delivery of segments at the destination. As segments in UDP are not numbered, there
is no means to identify the frames that have lost during the transmission.
Though UDP is considered powerless due to unreliable transport, it is a simple protocol that incurs
a minimal overhead. It is generally used by a process that has only a small message to send and is not
much concerned about the reliability. The applications such as speech and video for which instant
­delivery is more important than accurate delivery, prefer to adopt UDP for transport.
The UDP packets, known as user datagrams, have a fixed-size header of eight bytes, which is -followed
by the data. The header of UDP packet contains four field of 16 bits each as shown in Figure 12.2.

Source port number Destination port number

16 bits 16 bits
Total length Checksum
16 bits 16 bits

Figure 12.2 UDP Packet Header Format

CN_Chapter 12.indd 240 2/20/2012 6:45:20 PM

 Internet Transport Protocols 241

The description of various field of a UDP packet header is as follows:

‰‰ Source Port Number: It is a 16-bit long fiel that define the process running on the source host.
If the UDP packet is being sent from the client, that is, if the source host is client, then the source
port number will be chosen randomly by the UDP software running on the client. However, if the
UDP packet is sent by the server, the source port number will be a well-known port used with UDP.
Some of the well-known ports used with UDP are listed in Table 12.1.

Table 12.1 Some Well-known Ports Used with UDP

Port Protocol Description

13 Daytime Returns the date and the time
53 Name server Domain name service
111 RPC Remote procedure call
123 NTP Network time protocol

‰‰ Destination Port Number: It is a 16-bit long fiel that define the process running on the desti-
nation host. If the destination host is the client, the destination port number will be an ephemeral
port number. However, if the destination host is the server, the destination port number will be a
well-known port used with UDP.
‰‰ Total Length: It is a 16-bit long fiel that specifie the total length of the UDP packet including
header as well as data. The size of this fiel can range from 0 to 65,535 (that is, 216–1) bytes but the
size of UDP packet must be much less, as it is to be encapsulated in an IP datagram having a total
length of 65,535 bytes.
‰‰ Checksum: It is a 16-bit long fiel that is used for error detection over both the header and
the data.

Uses of UDP
‰‰ It is suitable for multicasting.
‰‰ It is used for management processes such as SNMP.
‰‰ It is used for the route updating protocols such as routing information protocol (RIP).
9. What is transmission control protocol (TCP)? What are the services provided by it?
Ans: TCP is a connection-oriented and reliable transport layer protocol. By connection-­oriented,
we mean that a virtual connection must be established between the sending and the receiving pro-
cesses before any data can be transferred. Whenever a process on source host wishes to communicate
with a specific process on destination host, first a virtual connection is established between the TCPs
of the sending and receiving processes. Then, the data can be transferred in both directions. After the
completion of data transfer, the connection is released. TCP accepts a stream of bytes from the upper
layer and divides it into a sequence of segments which are then sent to the destination. By reliable,
we mean that TCP provides error and flow control and thus, ensures the delivery of segments to
the destination. Each segment sent from the source needs to be acknowledged by the receiver on
its receipt.

CN_Chapter 12.indd 241 2/20/2012 6:45:20 PM

 242 Data communications and Computer Networks

TCP provides a variety of services to the processes at the application layer. Some of these services
are described as follows:
‰‰ Process-to-Process Communication: Like UDP, TCP is also a process-to-process protocol that
connects a process on a source host to a specifi process on the destination host using the port
numbers. Some of the well-known ports used with TCP are listed in Table 12.2.

Table 12.2 Some Well-known Ports Used with TCP

Port Protocol Description

9 Discard Discards any datagram that is received
13 Daytime Returns the date and the time
19 Charg en Returns a string of characters
23 SMTP Simple mail transfer protocol

‰‰ Stream Delivery Service: TCP is a byte-oriented protocol that allows the sending process to send
the stream of bytes and the receiving process to receive the stream of bytes. Since the speed of
sending and receiving processes may differ, buffers need to be maintained at both ends to store
the bytes. The buffer at the sender’s end is divided into three sections: empty slots to hold the
bytes produced by the sending process, slots containing bytes that have been sent but not yet
­acknowledged and the slots containing bytes that are to be sent. On the other hand, the buffer at the
receiver’s end is divided into two sections: empty slots to hold the bytes received from the sending
process and the slots containing bytes that are to be read by the receiving process.
‰‰ Segments: The communication between sending and receiving TCPs takes places through the IP
layer, which supports data in packets rather than a stream of bytes. Therefore, before sending the
bytes (stored in the sending buffer), the sending TCP groups multiple bytes together into a packet
called segment; different segments may or may not contain equal number of bytes. TCP header is
attached with each segment and then the segment is delivered to IP layer. The IP layer encapsulates
the segment into IP datagram and sends the IP datagrams. The IP layer at the receiving machine
processes the header of IP datagrams and passes the segments to the receiving TCP. The receiving
TCP stores the bytes of segments in the receiving buffer.
‰‰ Full-Duplex Communication: TCP provides full-duplex connection, that is, both the sender and
the receiver processes can simultaneously transmit and receive the data.
‰‰ Reliable Service: TCP provides a reliable service, that is, every received byte of data is
­acknowledged to the sending process. This helps in detecting the lost data. To ensure reliability,
TCP uses byte number, sequence number and acknowledgement number.
• Byte Number: Each of the bytes that are transmitted in a connection is numbered by TCP. The
numbering does not start necessarily with zero. As TCP receives the first byte from the sending
process, it chooses a random number between 0 and 232 –1 and assigns that number to the first
byte. The subsequent bytes are numbered accordingly. For example, if the first byte is numbered
as 330 and total 1,000 bytes are to be transmitted, then byte numbers will be from 330 to 1,330.
• Sequence Number: After numbering the individual bytes, the groups of bytes, that is, seg-
ments are numbered. Each segment is assigned a sequence number by TCP, which is same as
the number of firs byte carried by that segment.

CN_Chapter 12.indd 242 2/20/2012 6:45:20 PM

 Internet Transport Protocols 243

• Acknowledgement Number: The communicating parties send acknowledgement to each

-other to confir the receipt of bytes. TCP numbers acknowledgement and the acknowledge-
ment number ­indicate the number of byte expected to be received next. Moreover, the acknowl-
edgement number is cumulative. For example, if one of the communicating devices sends an
acknowledgement number 540 to the other, then it means that the bytes from the beginning up
to 539 have been received and the byte number 540 is expected to be received next.
‰‰ Flow Control: TCP implements a byte-oriented flo control mechanism to prevent the receiving
TCP from being overloaded with data from the sending TCP.
‰‰ Error Control: To ensure reliability, byte-oriented error control mechanism is implemented by TCP.
10. Give the segment format of TCP.
Ans: TCP allows exchange of data in the form of segments, where each segment consists of header
and data. The size of segment header varies from 20 to 60 bytes, depending on whether or not the header
contains Options field If the header does not contain Options field then it is of 20 bytes else it can be
up to 60 bytes. Figure 12.3 shows the format of a TCP segment header.

Source port address Destination port address

16 bits 16 bits
Sequence number
32 bits
Acknowledgement number
32 bits
HLEN Reserved Control Window size
4 bits 6 bits 6 bits 16 bits
Checksum Urgent pointer
16 bits 16 bits

Options and padding

Figure 12.3 TCP Segment Header Format

The TCP segment header comprises various fields which are described as follows:
‰‰ Source Port Address: It is a 16-bit long fiel that define the port number of the application
­programme running on the source host.
‰‰ Destination Port Address: It is a 16-bit long fiel that define the port number of the application
programme running on the destination host.
‰‰ Sequence Number: It is a 32-bit long fiel that indicates to the receiving process the number of
firs byte contained in the received segment. While the connection is established, both the commu-
nicating processes make use of random number generator to generate the initial sequence number
(ISN), which is different in each direction.
‰‰ Acknowledgement Number: It is a 32-bit long field that defines the byte number the receiving
process is expecting to receive from the sending process. If the sending process sends the byte
number x to the receiver, then the receiving process adds one to that byte number to create an
acknowledgement number that is, x+1. The acknowledgement number x+1 means that the re-
ceiving process has successfully received all bytes up to x and now, it is expecting to receive byte
number x+1.

CN_Chapter 12.indd 243 2/20/2012 6:45:21 PM

 244 Data communications and Computer Networks

‰‰ HLEN: It is a 4-bit long fiel that define the length of segment header in terms of 32-bit words.
This fiel can take value between 5 (for 20-byte header) and 15 (for 60-byte header).
‰‰ Reserved: It is a 6-bit long fiel which has been kept reserved for the future use.
‰‰ Control: It is a 6-bit long fiel consisting of six flag each of one bit. These fla bits help in flo
control, establishment of connection, termination of connection and the mode of transferring data
in TCP. The description of fla bits are as follows:
• URG: This bit is set to 1 if the Urgent pointer fiel is in use else it is set to 0.
• ACK: This bit is set to 1 to indicate the valid acknowledgement number. If ACK bit is set to 0, then
it indicates that the segment does not carry the acknowledgement and thus, the ­Acknowledgement
number fiel is ignored.
• PSH: This bit indicates the pushed data. The receiver is asked to deliver the data to the applica-
tion immediately as it arrives and not buffer the data until a full buffer has been received.
• RST: This bit is used to reset the data connection in case the connection has been distorted. It also
rejects an invalid segment and denies making another connection.
• SYN: This bit is used to synchronize the sequence numbers during the connection establishment.
If the piggyback Acknowledgement number fiel is not in use, then SYN bit is set to 1 and ACK
bit is set to 0. If the connection uses an acknowledgement, then SYN bit is set to 1 and ACK bit
is also set to 1.
• FIN: This bit is used to terminate the connection. If this bit is set to 1, then it means the sending
process has transmitted all data and has no more data to transmit.
‰‰ Window Size: It is a 16-bit long fiel that describes the size of the window (in bytes) the receiving
process should maintain. The maximum size of window is 65,535 bytes. This value is regulated by
the receiver and is usually known as the receiving window.
‰‰ Checksum: It is a16-bit long fiel that is used to detect the errors. The checksum in TCP is neces-
sary, unlike UDP. The same psuedoheader as that of UDP is added to the segment and for the TCP
psuedoheader, the value of Protocol fiel is set to 6.
‰‰ Urgent Pointer: It is a 16-bit long fiel which is used only when segment contains the urgent data.
‰‰ Options: This fiel is up to 40 bytes that is used to contain additional or optional information in
the TCP header.
11. Describe three phases of connection-oriented transmission in TCP.
Ans: The connection-oriented transmission in TCP needs three phases, which are described as follows:
‰‰ Connection Establishment: This is the firs phase of connection-oriented transmission in TCP. In
this phase, the TCPs in the machines that wish to communicate need to be firs connected. Each of
the communicating parties must initialize communication and take permission from the other party
before transferring any data. In TCP, connection is established using three-way ­handshaking
(discussed in Q12).
‰‰ Data Transfer: After the TCPs of the communicating machines are connected, the data transfer
phase begins. In this phase, both the parties can send segments to each other at the same time as the
connection established between TCPs is full-duplex. After receiving a segment, the receiving party
is also required to send an acknowledgement number to the sending party to confir the receipt
of segment. An acknowledgement from either side (client to server or server to client) can also be
piggybacked on the segment (containing data) that is travelling in the same direction. That is, a
single segment may contain both data and acknowledgement.

CN_Chapter 12.indd 244 2/20/2012 6:45:21 PM

 Internet Transport Protocols 245

‰‰ Connection Termination: This is the last phase of connection-oriented transmission that ­commences
after the data have been transferred. Though either of the communicating parties can close the
­connection, generally the client initiates the connection close command. In TCP, connection is termi-
nated using three-way handshaking mechanism (discussed in Q14).
12. Explain how connection is established in TCP using three-way handshaking mechanism.
Ans: TCP is a connection-oriented protocol that allows full-duplex transmission. In TCP, the
­connection is established using three-way handshaking mechanism.

Three-Way Handshaking
The server starts the mechanism. The server process informs its TCP that it is ready to accept an
­incoming connection by executing the LISTEN and ACCEPT primitives. This is called a request for a
passive open. The client process then sends the request for an active open to its TCP by executing the
CONNECT primitive. This primitive specifie the IP address and port number that the TCP on client
could identify the specifi server process to which the client process wants to connect. Now, TCP starts
the three-way handshaking process, which involves the following steps (Figure 12.4).

Client Server

Active SYN
segm
open SEQ ent
=m
ment Passive
A C K seg +1 open
+ K=m
SYN
Q = n, AC
SE A
CK s
SEQ egme
= m, nt
ACK
=n+
1

Figure 12.4 Connection Establishment Using Three-way Handshaking

1. The client sends a TCP segment with SYN bit set to 1 and ACK bit set to 0. This SYN segment is
used for synchronization of sequence numbers between client and server. The SYN segment does
not actually contain any data; however, it consumes one byte space and is assigned a sequence
number (say, m), so that it can be acknowledged unambiguously.
2. If the server accepts the connection, it sends a TCP segment with both SYN and ACK bits set to 1.
This SYN+ACK segment serves the purpose of acknowledgement to the SYN segment sent
by the client and does not carry any data. It also consumes a sequence number (say, n) and the
acknowledgement number of this segment is set to sequence number of SYN segment plus one
(that is, m+1).
3. The client sends a TCP segment with ACK bit set to 1 and a valid Acknowledgement number.
This ACK segment does not carry any data; it just confirm the receipt of SYN+ACK segment
from the server. The sequence number of ACK segment is same as that of the firs segment (SYN
segment) sent by the client while the acknowledgement number is set to sequence number of
(SYN+ACK) segment plus one (that is, n+1).

CN_Chapter 12.indd 245 2/20/2012 6:45:21 PM

 246 Data communications and Computer Networks

13. Write a short note on SYN floodin attack.

Ans: The SYN floodin attack is a type of security attack that can occur during the connection estab-
lishment process in TCP. In this attack, an attacker sends a large number of SYN segments to the server
with each segment carrying a fake source IP address. As the server believes that an active open request
is being issued from different clients, it allocates the required resources to each client. It then sends
SYN + ACK segment to each fake client, which are lost. This results in a lot of server resources ­allocated
but not used. Now, during this time, if more requests arrive to server, then it is quite possible that the
server may not have enough resources to grant their requests and thus, server may crash.
To handle SYN floodin attacks, some strategies have been imposed by TCP. First, the number of
-requests during a specifie period of time can be limited. Second, filterin strategy can be implemented
in which datagrams are filtere out if they arrive from unwanted source addresses. Third, some other
transport layer protocols such as SCTP have adopted a new strategy called cookie in which the alloca-
tion of resources to clients is delayed until the entire connection has been set up.
14. Discuss three-way handshaking mechanism for connection termination in TCP.
Ans: The connection termination in TCP is easier than the connection establishment. After the com-
municating processes have finished with exchanging data, any one of them can terminate the connection.
In TCP, the connection is terminated with the help of three-way handshaking mechanism which involves
the following steps (Figure 12.5).

Client Server

Active FIN s
close SEQ egme
= m, nt
ACK
=n
nt Passive
egme
+ A CK s +1 close
FIN K=m
Q = n, AC
SE A
CK s
SEQ egme
= m, nt
ACK
=n+
1

Figure 12.5 Connection Termination Using Three-way Handshaking

1. The client TCP initiates the termination process by sending a TCP segment with FIN bit set to 1
to the server. This FIN segment may also include the last chunk of data byte that needs to be sent
to the server process. If the FIN segment does not contain any data, then it consumes only one
sequence number (say, m). The FIN segment may also contain a valid acknowledgement number
(say, n) to confir the receipt of the latest bytes from the server.
2. On receiving the FIN segment, the server TCP sends the FIN+ACK segment to the client. This
segment indicates the receipt of FIN segment by the server as well as closing of connection from
the server side. This segment may also include last chunk of data byte to be sent by the server.
However, if it does not contain data, it is assigned only one sequence number which is same as
the acknowledgement number of FIN segment (that is, n). The acknowledgement number of this
segment is equal to the sequence number of FIN segment plus one (that is, m+1).
3. The client TCP sends the last segment, an ACK segment, to acknowledge the FIN+ACK
segment from the server TCP. This segment does not carry any data and thus, takes no new

CN_Chapter 12.indd 246 2/20/2012 6:45:21 PM

 Internet Transport Protocols 247

s- equence number. The sequence number of this segment is same as that of firs FIN segment
sent by the client (that is, m). It includes an acknowledgement number equal to sequence number
received in FIN+ACK segment plus one (that is, n+1).
15. Compare UDP with TCP.
Ans: Both UDP and TCP are transport layer protocols that provide process-to-process delivery of
packets. Some differences between UDP and TCP are listed in Table 12.3.
Table 12.3 Comparison Between UDP and TCP

UDP TCP
• It is simple and unreliable protocol. • It is high featured and reliable protocol.
• It is a connectionless protocol, which means an es- • TCP is a connection-oriented protocol, which means
tablishment of connection between client and server a virtual connection needs to be established between
is not required before starting the transfer of data. client and server before initiating the data transfer.
• UDP application sends message-based data in • TCP application sends stream-based data with no
discrete packages. specific structure.
• No acknowledgement of data takes place. • Each received byte of data needs to be acknowledged.
• Retransmission is not performed automatically. • The lost data is retransmitted automatically by TCP.
Applications must detect the lost data by themselves
and retransmit if required.
• It does not provide any flow control mechanism. • It offers flow control the sliding window mechanism.
• Its transmission speed is very high. • Its transmission speed is high but lower than UDP.
• It is suitable for transferring small amount of data up • It is suitable for small as well as large amount of
to hundreds of bytes. data up to a few gigabytes.
• Some protocols and well-known applications of UDP • Some protocols and well-known applications of TCP
include DNS, BOOTP, TFTP and SNMP. include FTP, NNTP, DNS, SMTP, IRC and HTTP.
• Its header is shorter than that of TCP and thus it • It incurs more overhead than UDP.
incurs less overhead.

16. What is meant by remote procedure call (RPC)? Explain its mechanism.
Ans: RPC, as the name implies, is a communication mechanism that allows a process to call a proce-
dure on a remote system connected via network. It was introduced by Birrell and Nelson in 1984. This
method is implemented to allow programmes to call procedures located on remote host. The calling
process (client) can call the procedure on the remote host (server) in the same way as it would call the
local procedure. The syntax of RPC call is very similar to conventional procedure call as given below:
Call <Procedure_id>(<List of parameters>);
The RPC system facilitates the communication between client and server by providing a stub on
both client and server. For each remote procedure, the RPC system provides a separate stub on the client
side. When the client process wants to invoke a remote procedure, the RPC call is implemented in the
following steps.
1. The RPC system invokes the stub for the remote procedure on the client, passing to it the parameters
that are to be passed further to the remote procedure. The client process is suspended from execution
until completion of the call.

CN_Chapter 12.indd 247 2/20/2012 6:45:21 PM

 248 Data communications and Computer Networks

2. The client stub performs parameter marshalling, which involves packaging the parameters into
a machine-independent form, so that they can be transmitted over the network. It now prepares a
message containing the identifie of the procedure to be executed and the marshalled parameters.
3. The client stub sends the message to the server. After the message has been sent, the client stub
blocks until it gets reply to its message.
4. The corresponding stub on the server side receives the message and converts the parameters into
a machine-specifi form suitable for the server.
5. The server stub invokes the desired procedure, passing parameters to it. The server stub is suspended
from execution until completion of the call.
6. The procedure executes and the results are returned to the server stub.
7. The server stub converts the results into a machine-independent form and prepares a message.
8. The server stub sends the message containing the results to the client stub.
9. The client stub converts the results into machine-specifi form suitable for client.
10. The client stub forwards the results to the client process. With this, the execution of RPC is
­completed, and now, the client process can continue its execution.
Figure 12.6 depicts all the steps involved in execution of RPC.

Client Server

2 4
1 3 5
Process Stub Stub Procedure
10 8 6
9 7

Figure 12.6 Implementation of RPC

Multiple Choice Questions

1. The main function of transport layer is: (c) Node-to-node, Host-to-host

(a) Node-to-node delivery (d) Host-to-host, Process-to-process
(b) Process-to-process delivery 3. Which of the following is true for UDP?
(c) Synchronization (a) It is connection oriented.
(d) Source-to-destination delivery. (b) It provides flo control.
2. IP is responsible for _______communica- (c) It offers reliable service.
tion while TCP is responsible for_____ (d) It performs error control.
communication. 4. TCP exchanges data in the form of:
(a) Process-to-process, Host-to-host (a) Datagrams (b) Packets
(b) Node-to-node, Process-to-process (c) Segments (d) Frames

CN_Chapter 12.indd 248 2/20/2012 6:45:22 PM

 Internet Transport Protocols 249

5. In TCP, control fiel consists of: (c) PSH and URG bits
(a) Six flag  
(b) Three flag (d) None of these
(c) Five flag  
(d) Seven flag 7. During connection establishment in TCP,
6. In TCP, the connection establishment uses: the mode of data transmission is:
(a) FIN and SYN bits (a) Full-duplex (b) Half-duplex
(b) SYN and ACK bits (c) Simplex (d) None of these

Answers
1. (b) 2. (d) 3. (c) 4. (c) 5. (a) 6. (b) 7. (a)

CN_Chapter 12.indd 249 2/20/2012 6:45:22 PM

 13
Application Layer Protocols

1. Explain how TELNET helps in remote login.

Ans: The word “TELNET” is derived from telecommunications and network and is a protocol
that allows a user to log on to a remote computer. Further, TELNET is also known as remote login,
which means connecting one machine to another in such a way that a person may interact with another
­machine as if it is being used locally. It means that someone in New Delhi could connect to a computer
in the New York City Public Library and search the card catalogue the same way as someone located at
a terminal in the library. Once connected, the user’s computer emulates the remote computer. When the
user types in commands, they are executed on the remote computer. The user’s monitor displays what
is taking place on the remote computer during the TELNET session.
The user’s computer, which initiates the connection, is referred to as the local computer or TELNET
client, and the machine being connected to, which accepts the connection, is referred to as the remote
computer or TELNET server. The transmission control protocol/Internet protocol (TCP/IP) protocol is
used to transmit information between the TELNET client and the TELNET server. In addition, ­TELNET
is text based and only the keyboard can be used for navigation. For this reason, it is widely used in
UNIX-based systems. Further, TELNET creates a standard and a fictional terminal called the network
virtual terminal (NVT) that is used for communication by all the computers on the network. The
follow­ing steps are performed while conducting a TELNET session.
� The inputs from the user are taken and translated by the TELNET client to the NVT form.
� This input is send to a TELNET server running on a remote computer.
� The server translates the NVT form to whatever representation the computer being accessed
­requires.
The same steps are repeated when data are sent from the remote computer back to the user. This system
allows clients and servers to communicate even if they use entirely different hardware and internal data
representations.

CN_Chapter 13.indd 250 2/20/2012 6:38:25 PM

 Application Layer Protocols 251

2. What is a domain name system?

Ans: The TCP/IP protocol identifies entities that are connected on the Internet using their unique IP
addresses. These addresses being numeric are difficult to remember for the users as compared to their
names, therefore, a system was needed that can change a name to an address and vice versa. Domain
name system (DNS) is such a system that is used to map the user friendly host names to their IP ­addresses
and vice versa. This system is used extensively over the Internet and other private businesses.
3. What is a resolver? What is its role in DNS?
Ans: A resolver is a library procedure used by the DNS during mapping of a host name to its corre-
sponding IP address. Whenever an application program requires mapping of a host name to IP ­address, or
IP address to host name it invokes the resolver, passing the host name or IP address to it as a parameter. The
resolver passes this parameter to a local DNS server, which searches the name or address in its database.
After the name or address has been found, the DNS server returns the result back to the resolver. Then, the
resolver passes on the result to the calling application program so that a connection could be established
with the destination.
4. Explain the DNS in terms of name space, resource record and name server.
Ans: The DNS is used to map host names to their corresponding IP addresses or vice versa. Various
components of DNS include name space, resource record and name server.

Name Space
It is a representation of the domains of the Internet as a tree structure. Each domain can be subdivided
into many other domains, which are further partitioned thereby creating a hierarchy. The leaf nodes of
the tree cannot be subdivided and each leaf node may contain only a single host or several hosts. The
top-level domains are classified into two groups, namely, generic and country. The generic group con-
tains domain names such as com (commercial), edu (educational institutions), gov (governments), int
(international organizations), mil (armed forces), net (network providers) and org (organizations). The
country domains contain one entry for every country, as per ISO 3166 specification. The tree structure
of the name space is shown in Figure 13.1.

Generic Countries

com edu net mil ... ip nl ...

Sun Yale ac co oce vu

CS Iinda keio nec

Figure 13.1 A Portion of Internet Domain Name Space

A label is included in each node of the tree (Figure 13.2). The label is a string, which has a maximum
size of 63 characters. The label at the root node is just an empty string. The child nodes, which have the
same parent, are not allowed to have the same label name as it may cause ambiguity. Each node in the

CN_Chapter 13.indd 251 2/20/2012 6:38:25 PM

 252 Data Communications and Computer Networks

root(.)

edu.
edu

flag.edu. Domain
flag names

Labels
tc.flag.edu.
tc

terminator.tc.flag.edu.
terminator
Figure 13.2 Labels and Domain Names

tree has a domain name, which is formed by a sequence of labels separated by dots (.). The domain
names are again divided into two types, namely, fully qualifie domain name (FQDN ) and partially
qualified domain name (PQDN ).
‰‰ FQDN: If the domain name ends with a dot (.), that is, null string, it is said to be an FQDN. It is
the full name of a host, which includes all labels, starting from the host label toward the root label,
which is a dot (.). For example, in Figure 13.2, the FQDN of a host named terminator installed at
the technology centre tc is terminator.tc.flag.ed .
‰‰ PQDN: If a domain name does not end with a null string, then it is said to be a PQDN. This means
a PQDN starts with the host label but does not end with the root node. A partial domain name ad-
dress is used when the domain name and the client reside in the same site. The PQDN can take
help of the resolver to convert it to an FQDN. For example, if a person at the flag.ed site wants
to get the IP address of the terminator computer, he/she defines only the partial name terminator.
The rest of the part (called suffi ), that is, tc.flag.ed , is added by the resolver and then the address
is passed to the DNS server.

Resource Records
Every domain is associated with a set of information known as resource records (RRs). The most com-
mon RR is the IP address for a single host, although many other kinds of resource records are also found.
A RR is a five-tuple set, which is mostly represented as ASCII text, but for better efficienc , it can also
be encoded in binary form. The different fields of the five-tuple set are described as follow
‰‰ Domain_Name: This field identifies the domain with which this record is associated. Usually, many
records related to a single domain exist in the databases and the database contains RRs for multiple
domains. Thus, this field is used for search operations so that queries an be executed efficientl .
‰‰ Time_to_Live: This field indicates the stability of the record. It specifies the time interval with-
in which a RR may be cached by the receiver so that the server need not be consulted for RR
­repeatedly. A zero value in this field indicates that the RR is to be used for a single transaction and
therefore, should not be cached. The highly stable records are given large values while less stable
records are given small values.

CN_Chapter 13.indd 252 2/20/2012 6:38:25 PM

 Application Layer Protocols 253

‰‰ Class: This field identifies the protocol family. It is set to IN for Internet information; otherwise,
some other codes are used.
‰‰ Type: This field specifies the type of the recor
‰‰ Value: This field depends on the Type field and can be a domain name, a number or an ASCII
character.

Name Servers
As DNS database is very vast, it is impossible for a single server to hold information about the complete
database and respond to all queries. Even if it is done, then a failure in the single name server would
bring the whole network down. To avoid such a situation, the DNS name space is divided into many
non-overlapping zones (represented by dotted areas in Figure 13.3) with each zone containing name
servers holding information about that zone. Each zone covers some nodes of the tree and contains a
group of one or more subdomains and their associated RRs that exist within that domain. The name
server creates a zone file, which holds all the information of the nodes under it. Each zone has a primary
name server and one or more secondary name servers. The primary name server of a zone obtains the
information from a file on the disk, whereas the secondary name servers obtain information from the
primary name server. Some servers can also be located outside the zone to improve the reliability.

Generic Country

com edu net mil ... ip nl ...

S

Sun Yale ac co oce vu

CS cs
Iinda keio nec

john
texas

Figure 13.3 Division of DNS Name Space into Zones

Whenever a query from a resolver for a domain name arrives at a local name server, it checks ­whether
that information falls under its zone. If so, then the domain name is provided; otherwise, the query is for-
warded to the upper-level name servers. For example, consider a resolver john.cs.vu.nl wants to query
about the IP address of the host texas.cs.yale.edu (Figure 13.3). To handle this query, firstly the resolver
sends a query to the local name server cs.vu.nl to check whether it already has the IP address. If the
local name server has the IP address in its cache, it returns the address to the resolver; else, it forwards
the query to the root server. The root server then forwards the UDP packet to yale.edu, which is further
forwarded to cs.yale.edu because this name server has the IP address of the host texas.cs.yale.edu. The
whole process is repeated in the reverse order in order to reach to the client who had sent the original
request. Once the client has been reached, the local name server stores this IP address in its cache so that
it may be used later when required.

CN_Chapter 13.indd 253 2/20/2012 6:38:26 PM

 254 Data Communications and Computer Networks

5. Why was there a need of dynamic domain name system (DDNS)?

Ans: The DNS database contains large number of addresses that need to be changed very often
due to addition of a new host, removal of some existing host or change in the IP address of some
host. All these changes must be updated in the DNS master file, which leads to lot of manual work
consuming a lot of time. To overcome this problem, DDNS was devised that updates the master fil
automatically. In DDNS, whenever a binding of name and address is found, the information is passed
to a primary DNS server by dynamic host control protocol (DHCP). After receiving the information,
the primary server updates its zone and the secondary servers are informed about the change using
one of the two modes: active and passive. In the active mode, the primary name server itself informs
about the change to the secondary servers by sending the message while in the passive mode, the
secondary name server checks for any change at regular intervals of time. However, in both modes,
the secondary server requests information about the entire zone after being notified about the change.
Further, DDNS also secures DNS database from unauthorized changes by using an authentication
mechanism.
6. What is e-mail? State its advantages and disadvantages.
Ans: E-mail (or electronic mail) is the process of exchanging messages electronically via a commu­
nications network using the computer. E-mail allows users to communicate with each other in less time
and at nominal cost as compared to traditional phone or mail services. Apart from the textual ­message,
e-mails can also consist of other data formats such as pictures, sound and video. E-mails can be sent
anywhere in the world using the computer and a modem. Its delivery is almost instant and is very eco-
nomical to use. One may send many messages at a time or just one to a designated location.
In order to use e-mail, one must have access to the Internet and an e-mail account. An e-mail account is
a service that allows the user to send and receive e-mails through the Internet. The e-mail account pro-
vides a unique e-mail address and a mailbox where the user can save all his/her mails. The e-mail address
is made up of two parts, namely, the logon identity and the identity of the e-mail server. Both these parts
are separated by the symbol @ (­pronounced as at
the rate). A typical e-mail address is username@ Name:
website.com. The first part of the address indicates Street:
Pin:
name of the user. The symbol @ in the address is Envelope
Priority:
used to separate the user name from the rest of the Encryption:
address. Next comes the host name (website.com),
also called the domain name. From:
E-mail consists of two fields: envelope and mes- To:
sage (Figure 13.4). The envelope field defines the Address:
Header Location:
address of the sender and the receiver. The mes-
Date:
sage field is again subdivided into parts, namely, Subject:
header and the body. The header of the message Message
specifies the sender and the receiver while the Dear X,
body contains the information that needs to be
transmitted. Body
Though e-mail is quite popular and efficient Yours friend
XYZ.
but it has some disadvantages along with its ad-
vantages. The advantages and disadvantages are
discussed in Table 13.1. Figure 13.4 Format of an E-mail

CN_Chapter 13.indd 254 2/20/2012 6:38:26 PM

 Application Layer Protocols 255

Table 13.1 Advantages and Disadvantages of E-mail

Advantages Disadvantages
• The delivery of messages is very fast, sometimes • Although e-mail is delivered instantly, the recipient may
almost instantaneous, even though the message is or may not read his/her mail on time. That defeats the
meant for overseas or just to a friend next door. quickness of electronic mailing.
• The cost of e-mailing is almost free as it involves • The user must stay online to read and write more than
negligible amount of telephone and ISP charges. one mail. In addition, most webmail either display
advertisements during use or append them to mails
sent. It results in increased size of the original mail,
which brings a significant decrease in speed of use.
• M
 ultiple copies of the same message can be sent • Since e-mail passes through a network, therefore, it
to a group of people at the same time and can be may be intercepted in between. Moreover, viruses can
sent as easily to a single person. enter the system while downloading the e-mails.
• Pictures, documents and other files can also be • The slightest error in the address or a failure in one
attached to messages. of the links between sender and receiver is enough to
prevent a delivery.

7. Describe the architecture of e-mail.

Ans: The architecture of e-mail includes three components, namely, the user agent (UA), message
transfer agent (MTA) and message access agent (MAA). The UA is a program that helps the user in
reading, writing, replying and forwarding message. It also handles the user’s mailboxes. The MTA is a
client/server program where the MTA client can push messages to the MTA server. The MAA is a client/
server program where the MAA client pulls (retrieves) messages from the MAA server.
To understand how e-mail system works, consider two users A and B on two different systems.
Further, assume that both A and B are connected to their respective mail servers by local area network
(LAN) or wide area network (WAN). When A wishes to send a message to B, it executes the UA pro-
gram to prepare the message. Now, as the message is to be sent to A’s mail server through LAN/WAN, a
pair of MTAs (MTA client and MTA server) is used. The UA after preparing the message sends it to an
MTA client, which then sets up a connection with MTA server on the mail server. At A’s mail server, the
message waiting to be sent is kept in a queue maintained by the mail server. Then, the mail server calls
its MTA client to send the message to MTA server at B’s mail server connected via Internet. At B’s mail
server, the received message for B is kept in his/her mailbox. To retrieve the message from mailbox at
mail server, B calls the MAA client, which then establishes a connection with MAA server at the mail
server. This way B can receive the message sent by A. Figure 13.5 depicts the whole mechanism.
8. What are the services offered by user agent? Also, discuss its types.
Ans: The UA offers various services to the user to make the process of sending and receiving mes-
sages easier. These services are discussed as follows:
‰‰ Composing Messages: This service helps the user in composing the messages that are to be sent.
A template is provided by UA, which can be filled by the user to create the messages. Some UAs
provide users with built-in editor to perform functions like spelling and grammar checking, empha-
sizing text by making it bold, italic, etc.

CN_Chapter 13.indd 255 2/20/2012 6:38:26 PM

 256 Data Communications and Computer Networks

MTA MAA
UA A B UA
client client

LAN/WAN LAN/WAN

MTA MAA
server server

Queue B's mail

box

Message Message
MTA MTA
Internet
client server

A's mail B's mail

server server
Figure 13.5 Mechanism of E-mail Transfer

‰‰Reading Messages: This service helps the user to read the messages, which are in its inbox. Most
user agents show a one-line description of each received mail.
‰‰ Replying to Messages: This service is used to reply to the messages that have been received by
the user. While replying, a user can send a new reply or may include the original message sent by
the sender along with the new one. Moreover, the user can reply either to the original sender or to
all the recipients of message.
‰‰ Forwarding of Messages: This service helps the user to forward the message to the third party
instead of sending it to the original sender. The user can also add some more content in the message
to be forwarded.
‰‰ Handling Mailboxes: The user agent is responsible for maintaining all the mailboxes in e-mail
system. Basically, it creates two types of mailboxes, namely, inbox and outbox. The inbox contains
all the messages received by a user and the outbox contains all the messages sent by the user. The
messages are kept in both mailboxes until the user deletes them.
There are two types of UAs namely, command-driven and graphical-user-interface (GUI )-based
UAs. These types are described as follows:
‰‰ Command-driven UA: This UA was used in the early days in e-mail. In this type, the user can
type one character at a time at the command prompt while replying to the sender. A few command-
driven UAs include pine, elm and mail.
‰‰ GUI-based UA: This UA being used nowadays allows the user to use both mouse and keyboard
to interact with the software. As the name of this UA suggests, it provides GUI components such
as menus and icons that help the users to access the services more easily. Thus, GUI-based UAs
are more user friendly.

CN_Chapter 13.indd 256 2/20/2012 6:38:26 PM

 Application Layer Protocols 257

9. Write a short note on MIME.

Ans: Multipurpose Internet mail extensions (MIME) is a protocol that enables the transfer of
non-ASCII data through e-mails and thus, overcome the limitation of simple e-mail format. It converts
non-ASCII messages to a 7-bit NVT ASCII format at the sender’s side. The converted message is then
forwarded to the MTA client so that it can be sent over the Internet to the receiver. At receiver’s side, the
message is converted to its original format. Further, MIME can also be used to send messages in differ-
ent languages such as French, German, Chinese, etc. The structure of MIME defines five new headers
that were included in the original e-mail header section. These headers are described as follows:
‰‰ MIME Version: This header specifies the MIME version and tells the receiver that the sender is
using MIME message format. The version number 1.1 is being used nowadays.
‰‰ Content Type: This header defines the type and subtype of the data used in the message body. The
type of the data is followed by its subtype, separated by a slash, that is, type/subtype. Some of the
types and their subtypes used by MIME are listed in Table 13.2.
Table 13.2 Data Types and Subtypes in MIME

Type Subtype Description

Text Plain Unformatted
HTML HTML format
Image JPEG Image is in JPEG format
GIF Image is in GIF format
Video MPEG Video is in MPEG format
Audio Basic Single-channel encoding of voice at 8 kHz

‰‰ Content Transfer Encoding: This header defines the different methods used for encoding the
messages into various formats, so that it can be transmitted over the network. Some schemes used
for encoding the message body are listed in Table 13.3.

Table 13.3 Content Transfer Encoding

Type Description
7 bit NVT ASCII characters and short lines
8 bit Non-ASCII characters and short lines
Binary Non-ASCII characters with unlimited length

‰‰ Content Id: This header uniquely identifies the message content

‰‰ Content Description: This header tells what the body of message contains, that is, whether it
contains picture, audio or video. It is an ASCII string that helps the receiver decide whether the
­message needs to be decoded.
10. Explain SMTP.
Ans: Simple mail transfer protocol (SMTP) is a TCP/IP application protocol that supports e-mail
service. It handles the transfer of messages between the sender and receiver. It is based on the client-
server model and defines the MTA client and server in the Internet. During the exchange of message

CN_Chapter 13.indd 257 2/20/2012 6:38:26 PM

 258 Data Communications and Computer Networks

between the sender and the receiver, SMTP is used twice. Once, it is used to transfer the mail from
sender’s end to sender’s mail server, and then to transfer the mail from sender’s mail server to receiver’s
mail server. To retrieve the mail from receiver’s mail server at the receiver’s end, a different mail pro-
tocol such as POP3 and IMAP (discussed in the next question) is used. While transferring mails, SMTP
uses commands and responses between MTA client and MTA server.
‰‰ Commands: They are sent from the client machine to the server machine. The syntax of a com-
mand consists of a keyword followed by zero or more arguments. There are total 14 commands
defined by SMT some of which are listed in Table 13.4.
Table 13.4 SMTP Commands

Keyword Arguments Description

HELO<domain> Sender’s host name It is used for sending sender’s identification.
MAIL FROM: <...> Sender of the message It specifies the sender’s name.
RCPT TO: <...> Intended recipient of the message It specifies the receiver’s name.
DATA Body of the mail It indicates the beginning of mail transmission.
RSET No arguments It ends the current mail transaction.
SEND FROM: <...> Intended recipient of the message It specifies that this mail should be sent directly
to the user’s terminal.
SOML FROM: <...> Intended recipient of the message It is used to specify that the mail should be
sent to user’s terminal if possible; otherwise to
mailbox.
VRFY Name of the recipient to be verified It is used to confirm the user name.

‰‰ Responses: They are just the opposite of commands, that is, they are sent from a server machine
to a client machine. A response consists of a three-digit code, which may be followed by additional
textual information. Some of the SMTP responses are shown in Table 13.5.

Table 13.5 SMTP Responses

Code Information
221 Service closing transmission channel
354 Start mail input
500 Syntax error, unrecognized command
503 Bad sequence or commands

11. Explain in brief the following:

(a) POP3
(b) IMAP4
Ans:
(a) POP3: It stands for post office protocol and the number 3 is its version number. It is a simple
MAA protocol that has the limited functionality. To use POP3, the client POP3 software and server

CN_Chapter 13.indd 258 2/20/2012 6:38:26 PM

 Application Layer Protocols 259

POP3 software must be installed on the recipient’s machine and on its mail server, respectively. Further,
POP3 works in two modes: delete mode and keep mode. In the delete mode, as a message has been
pulled from the mail server, it is deleted from the mailbox on the mail server. On the other hand, in the
keep mode the message remains in the mailbox even after it has been pulled from the mail server. This
mail can be read later from any other computer or location.
Whenever a recipient (client) needs to retrieve mails from the mail server, it establishes a TCP con-
nection to the server on the port 110. Then, it passes its username as well as the password to the mail
server to get access to the mailbox on the mail server. After the server has verified the client, the client
can list and download the messages one at a time.
POP3 has some disadvantages, which are as follows:
‰‰ POP3 does not support mail organization on the server, that is, a user cannot have different folders
on the mail server.
‰‰ POP3 does not allow the contents of the mail to be checked in parts while the mail is being down-
loaded. The mail can be checked after it has been completely downloaded.
(b) IMAP4: It stands for Internet mail access protocol and the number 4 denotes its version number.
Like POP3, it is also an MAA protocol but it provides more functionality and is more complex than
POP3. Some of the additional features provided by IMAP4 are as follows:
‰‰ A user can create folders on the mail server and can delete or rename the mailboxes.
‰‰ IMAP enables the user to partially download the mails. This is especially useful in cases where a
message contains large audio and video files, which may take a lot of time to download because
of slow Internet connection. In such cases, the user can download only the text part of message if
required using IMAP4.
‰‰ A user can search through the contents of the messages while the messages are still on the mail
server.
‰‰ IMAP allows the user to check the contents of the e-mail before it has been downloaded.
‰‰ A user can selectively retrieve the attributes of messages such as body, header, etc.

12. What is FTP? How the files a e transferred using FTP?

Ans: File transfer protocol (FTP) is a mechanism provided by TCP/IP to transfer the files between
hosts connected via Internet. Further, FTP allows access to the files stored in the directory of a remote
computer that is connected to the Internet. In order to access a remote system by FTP, you need to know
either the uniform resource locator (URL) or the IP address of the FTP site such as ftp: ftp.­microsoft.com.
Unlike other client-server applications, FTP establishes two TCP connections between the hosts,
namely, control connection and data connection. The control connection is established over well-
known TCP port 21 and it remains active throughout the session. This connection is used for transfer-
ring control information such as commands and responses. Only one line of command or response can
be transferred at one time through the control connection. On the other hand, the data connection is
­established over the well-known TCP port 20 and it is established only after the control connection has
established. In addition, it needs to be established and released for each file to be transferred while the
control ­connection is open.
Figure 13.6 shows the FTP file transfer procedure. At the client side, there are three components,
namely, user interface, control process and the data transfer process while the server side uses only
two components, namely, control process and the data transfer process. The control processes of client

CN_Chapter 13.indd 259 2/20/2012 6:38:26 PM

 260 Data Communications and Computer Networks

User

User
interface

Control
connection
Control Control
process process

TCP/IP

Data Data
Disk transfer transfer Disk
process Data connection process

Client Server
Figure 13.6 Mechanism of File Transfer in FTP

and server are connected via control connection, whereas the data transfer process of client and server
are connected via data connection. The control processes of client and server communicate using NVT
format. They are responsible for converting from their local syntax such as DOS or UNIX to NVT
­format and vice versa. The data transfer processes of client and server communicate under the control
of commands transferred through the control connection.
13. Explain the following with respect to FTP:
‰‰ File Type
‰‰ Data Structure
‰‰ Transmission Mode
Ans: To transfer a file through the data connection in FTP, the user (client) has to specify certain
attributes to the server including type of file to be transferred, the data structure and the transmission
mode so that the control connection could be prepared accordingly. These attributes are described as
follows:
‰‰ File Type: FTP supports three types of files for transmission over the data connection, namely, an
ASCII fil , EBCDIC fil or image file. The ASCII fil is the default format used for text files. It
uses the 7-bit ASCII format to encode each character of text file. The sender converts the file from
its original form to ASCII characters, while the receiver converts the ASCII characters back to the
original form. If EBCDIC encoding (file format used by IBM) is supported at the sender or receiver
side, then files can be transmitted using the EBCDIC encoding. The image fil is the default format
used in the transmission of binary files. Binary files are sent as continuous stream of bits without
using any encoding method. Usually, the compiled programs are transferred using the image file
‰‰ Data Structure: FTP uses three data structures to transfer a file, namely, file structure, record struc-
ture and page structure. When file structure format is used, the file is sent as a continuous stream of
bytes. The record structure can be used only with text files and the file is divided into many records.

CN_Chapter 13.indd 260 2/20/2012 6:38:27 PM

 Application Layer Protocols 261

In page structure, each file is divided into a number of pages where each page contains a page
­number and a page header. These pages can be accessed sequentially as well as randomly.
‰‰ Transmission Mode: FTP uses three types of transmission modes, namely, stream mode, block
mode and compressed mode. The default mode of transmission is the stream mode, which sends
the data as a continuous pattern of bytes. In case the data contains only a stream of bytes, then no
end-of-file (closing of data connection) is required; the end of file is simply indicated by closing
of connection by the sender. In the block mode, data is sent in blocks, where each block is pre-
ceded by a 3-byte header. The first byte is just a description about the block and the next two bytes
­define the size of the block in bytes. The compressed mode is used in case of large files to reduce
their size so that they can be transmitted conveniently. The size of file is reduced by replacing
multiple consecutive occurrences of characters with a single character or reducing the number of
­repetitions. For example, in text files blank spaces can be compressed

14. What is anonymous FTP access?

Ans: Generally, in FTP a user needs an account (username) and password to login to a particular site
for accessing the file. However, some sites provide public access to a set of files; the user can access
these files without having any account or password. The users can type “anonymous” in the username
part and “guest” in the password part. Using FTP in this way is known as anonymous FTP access. With
an anonymous FTP, users have restricted access rights and usually, can only list, view or copy files to
and from a public directory on the remote system.
15. Define WWW. Explain its architecture.
Ans: The World Wide Web (abbreviated as the Web or WWW) is a collection of linked documents
or pages, stored on millions of computers and distributed across the world. The concept of the Web
began at CERN (the European Center for Nuclear Research) Geneva, Switzerland in the year 1989.
Since then, WWW is the most popularly used Internet subnetwork. One of the main reasons that led to
its popularity is that it provides information in multimedia form, that is, in more than one medium such
as, text, graphics, video and audio. Further, it provides a simple and consistent way of accessing the in-
formation available on the Internet by using hypertext system. In the hypertext system, the documents
are connected to other related documents on the Internet through links. The Web uses a specific Internet
protocol called hypertext transfer protocol (HTTP) to support hypertext documents.
Further, WWW is based on a distributed client/server architecture in which the services provided by
the server are distributed over various locations termed as sites and the client can access these services.
Each site contains one or more documents, which are known as web pages. A web page contains hy-
perlinks that enable the user to jump on other web pages on the same site or on the different sites. The
client can access the web pages through the browser—the software that enables a user to read/view web
pages. Whenever a user needs to access some web page stored on some site, it sends a request includ-
ing the address of the site and the web page (known as the URL) through its browser to the server. On
receiving the request, the server searches the required document and returns it to the client.
The WWW includes many components, which are discussed as follows:

Client (Browser)
Browser is a program which accesses and displays the web pages. It consists of three components, namely,
controller, interpreter and client protocol. The user provides inputs (request for a web ­document)
to the controller through a keyboard or a mouse. After receiving the input, the controller uses client

CN_Chapter 13.indd 261 2/20/2012 6:38:27 PM

 262 Data Communications and Computer Networks

p­ rotocols such as FTP or HTTP to access the web document. Once the controller has accessed the
­desired web document, it selects an appropriate interpreter such as hypertext markup language (HTML)
or JavaScript depending on the type of the web document accessed. The interpreters help the ­controller
to display the web document. A few of the web browsers used today include Microsoft Internet ­Explorer,
Opera and Google chrome.
To understand how a browser works, consider a user who wants to access the link http://www.
ipl.com/home/teams.html. When the user provides this link (URL) to the browser, the browser goes
through the following steps:
1. The browser determines the given URL and sends a query to the DNS server asking for the IP
address of www.ipl.com.
2. The DNS sends a reply to the browser, providing the desired IP address.
3. A TCP connection to port 80 on the received IP address is made by the browser.
4. The browser then sends a request for the file home/teams.html.
5. The file home/teams.html is sent by the www.ipl.com server.
6. The TCP connection is ended.
7. The browser displays the text in the file /home/teams.html. It also fetches and displays the images
in the file

Server
Server is the place where web pages are stored. On a request from the client, the server searches the
desired document from the disk and returns the document to the browser through a TCP connection. The
steps performed by a server are as follows:
1. The server accepts the TCP connection request arriving from the client.
2. It then acquires the name of the file requested by the client
3. The server retrieves the file from the disk
4. The file is sent back to the client
5. The TCP connection is released.
The efficiency of a server can be improved by caching the recently accessed pages so that those pages
could be directly accessed from memory and need not be accessed from the disk. Moreover, server can
support multithreading, that is, serving multiple clients at the same time to increase the efficienc .

Uniform Resource Locator

Each web page has a unique address, called the URL that identifies its location on the Internet. This ad-
dress can be used to locate that page all over the world by millions of people. Usually, the format of an
URL consists of four parts: protocol, name of the web server (or domain name), path and filenam . Here
is an example: http://www.xyz.com/tutor/start/main.htm. The structure of this URL is:
‰‰ Protocol: http
‰‰ Web server name/domain name: www.xyz.com
‰‰ Path: tutor/start/
‰‰ File name: main.htm
The first part of the address, the part before the colon, is the protocol. Most of the time http is used for
accessing a web page. After the protocol, comes the domain name. Colons and slashes respectively separate

CN_Chapter 13.indd 262 2/20/2012 6:38:27 PM

 Application Layer Protocols 263

the protocol and the domain name. Then comes the last part of a URL, namely, the path and the file name.
The path name specifies the hierarchical location of the said file on the computer. For instance, in http://
www.xyz.com/tutor/start/main.htm, the file main.htm is located in start, which is a subdirectory of tutor.

Cookies
Cookies are the small files or strings, which are used to store information about the users. This stored
information may be later used by the server while responding to the requests of the client(s). For some
particular sites, only registered users are permitted to access the information. In such a case, the server
stores the user’s registration information in the form of cookies on the client’s machine. The size of a
cookie file cannot exceed 4 KB. A user can disable the cookies in the browser or can even delete them.
16. What is HTTP? Describe the format of HTTP request and response message.
Ans: Hypertext transfer protocol (HTTP) is the most common protocol that is used to access infor-
mation from the Web. It manages the transfer of data between the client and the server. The older version
of HTTP was 1.0, in which TCP connection was released after serving a single request. This was not ad-
equate as every time a new connection had to be established. This led to the development of HTTP ­version
1.1 that supports persistent connection, that is, it is meant for multiple request–response operations.
Further, HTTP is a stateless protocol and all the transactions between the server and client are carried
out in the form of messages. The client sends a request message to the server and the server replies with
a response message. The HTTP request and response messages have a similar format (Figure 13.7) ex-
cept that in request message, the first line is the request line while in response message, and the first line
is the status line. The remaining part of both the messages consists of a header and sometimes, a body.

Request line Status line

Headers Headers

Blank line Blank line

Body Body

Request message Response message

Figure 13.7 Format of HTTP Request and Response Message

The HTTP request messages are of different types, which are categorized into various methods as
shown in Table 13.6.
For each HTTP request message, the server sends an HTTP response that consists of status line and
some additional information. The status line comprises a three digit status code, similar to the response
message of FTP and SMTP. The status code indicates whether the client request is satisfied or if there is
some error. The first digit of the status code can be 1, 2, 3, 4 or 5 and it indicates one of the five groups

CN_Chapter 13.indd 263 2/20/2012 6:38:27 PM

 264 Data Communications and Computer Networks

Table 13.6 The Built-in HTTP Request Methods

Method Description
GET Request to access a web page from the server.
HEAD Request to get the header of a web page
PUT Request to store a web page
POST Append to a named resource
DELETE Remove the web page
TRACE Echo the incoming request
CONNECT Reserved for future use
OPTIONS Enquire about certain options

into which response messages have been divided. Codes falling in the 100 range are only informational
and thus, rarely used. The codes falling in the 200 range indicate a successful request, codes in the 300
range redirect the client to some other site, the codes in the 400 range indicate an error in the client side
and the codes in the 500 range indicate an error at the server site.
Further, HTTP also contains various headers, which are
used to transfer additional information other than the normal
Header name Header value
message between the client and the server. For example, the
request header can ask for a message to be delivered in some
particular format while a response header can contain a de-
scription of the message. The additional information can be Space
included in one or more header lines within the header. Each Figure 13.8 HTTP Header Format
header line has a format as shown in Figure 13.8.
Each header line may belong to one of the four types of HTTP
headers, which are discussed as f­ ollows:
‰‰ General Header: This header can be included in both request and response message. It contains
general information about the sent or received messages. An example of a general header is Date
that is used to display the current date.
‰‰ Request Header: This header can be used only in the request messages from the client. The details
about the client setup and the preference of the client for any particular format are included in this
header. An example of a request header is the From header, which shows the e-mail address of
the user.
‰‰ Response Header: This header is part of the response messages only. It contains the server’s setup infor­
mation. An example of a response header is the Age header, which shows the age of the d­ ocument.
‰‰ Entity Header: This header includes information about the body of a document. It is mostly pres-
ent in the request or response messages. An example of entity header is the Allow header, which
lists the valid methods that can be used with a URL.

17. Compare HTTP with FTP and SMTP.

Ans: Functionally, HTTP works as a combination of FTP and SMTP. It is similar to FTP as it is also
used to transfer files over the TCP connection. However, it differs from FTP in the respect that it can
transfer only data and thus, no separate control connection is needed. It uses only one TCP connection
over port number 80.

CN_Chapter 13.indd 264 2/20/2012 6:38:27 PM

 Application Layer Protocols 265

Further, HTTP is similar to SMTP because in both protocols, the client initiates a request and the
server responds to that particular request. However, HTTP messages can only be read by HTTP server
and HTTP client, whereas in SMTP, messages can be read by humans also. Moreover, HTTP messages
are forwarded immediately unlike in SMTP, where messages are first stored and then forwarded
18. What is a proxy server and how it is related to HTTP?
Ans: A proxy server is a computer, which stores the copies of responses to most recently requests in
its cache so that further requests for these pages need not be sent to the original server. The proxy server
helps to reduce the load on the original server as well as decrease the network traffic thereby improving
the latency. To use the proxy server, the client must be configured to send HTTP requests to proxy server
instead of the original server. Whenever a HTTP client wishes to access a page, it sends HTTP request to
a proxy server. On receiving a request, the proxy server looks up in the cache for the desired web page.
If the page is found, the stored copy of response is sent back to the client; otherwise, the HTTP request
is forwarded to the target server. Similarly, the proxy server receives responses from the target server.
The proxy server stores these responses in the cache and then sends it to the client.
19. Explain SNMP and mention the two protocols used by it for managing tasks.
Ans: Simple network management protocol (SNMP) is an operational framework, which helps
in the maintenance of the devices used in the Internet. In addition, SNMP comprises two components,
namely, manager and agent. A station (host) which manages another station is called the manager.
It runs the SNMP client program. The agent is the station (router) which runs the server program and
is managed by the manager. Both the stations interact with each other to carry out the management. The
manager can ­access all the information stored in the agent station. This information can be used by
the manager to check the overall performance. The manager can also perform some remote operation on the
agent station such as rebooting the remote station. On the other hand, the agent can also perform some
management operations, such as informing the manager about the occurrence of any unusual situation,
which may hamper the performance.
In order to manage tasks efficie tly, SNMP uses two protocols: structure of management information
(SMI) and management information base (MIB).These two protocols function together with the SNMP.
The role of SMI and MIB protocols is discussed as follows:
‰‰ SMI: It generally deals with the naming of objects and defining their range and length. However,
it does not specify the number of objects maintained by an entity, the relationship between objects,
and their corresponding values.
‰‰ MIB: MIB basically performs the work that SMI has left behind. It defines the object name as per
the conventions specified by SMI. It also states the number of objects and their types

Multiple Choice Questions

1. Which of the following is/are an application (a) 54 (b) 53
layer service? (c) 51 (d) 50
(a) remote login 3. The client TELNET translates characters
(b) file transfer and acces that come from the local terminal into_____
(c) domain name service character form and delivers them to the
(d) all of these network.
2. Which port number is used by the DNS (a) ASCII (b) BCD
server for both TCP and UDP connections? (c) NVT (d) EBCDIC

CN_Chapter 13.indd 265 2/20/2012 6:38:27 PM

 266 Data Communications and Computer Networks

4. The well-known port _____ is used for the (a) SMTP (b) MIME
control connection and the well-known port (c) IMAP4 (d) HTTP
______ is used for the data connection. 8. Which of the following supports a persistent
(a) 20, 21 (b) 21, 20 connection?
(c) 21, 22 (d) 20, 22 (a) FTP (b) SMTP
5. Which is the default format used by FTP for (c) HTTP (d) MIME
transferring text files 9. Which of the following are MAA protocols?
(a) EBCDIC (b) binary fil (a) SMTP and MIME
(c) ASCII fil (d) bytes (b) FTP and HTTP
6. Which port is used by SMTP for TCP con- (c) SMTP and FTP
nection? (d) POP3 and IMAP4
(a) 25 (b) 26 10. Which of the following is the protocol used
(c) 21 (d) 20 for network management?
7. Which protocol uses the GET and POST (a) SNMP (b) POP3
methods? (c) FTP (d) IMAP4

Answers
1. (d) 2. (b) 3. (c) 4. (b) 5. (c) 6. (a) 7. (d) 8. (c) 9. (d) 10. (a)

CN_Chapter 13.indd 266 2/20/2012 6:38:27 PM

 14
Multimedia

1. Defin multimedia.
Ans: The word multimedia is made up of two separate words, multi means many and media means
the ways through which information may be transmitted. Therefore, multimedia can be described as
an integration of multiple media elements (such as text, graphics, audio, video and animation) together
to influence the given information, so that it can be presented in an attractive and interactive manner.
In simple words, multimedia means being able to communicate in more than one way.
2. What are the different categories of Internet audio/video services?
Ans: Earlier, the use of Internet was limited for only sending/receiving text and images; however,
nowadays, it is vastly being used for audio and video services. These audio and video services are
broadly classified into three categories, namely, streaming stored audio/video, streaming live audio/video
and real-time interactive audio/video. Here, the term streaming implies that the user can listen to or
view the audio/video file after its downloading has begun
‰‰ Streaming Stored Audio/Video: In this category, the audio/video files are kept stored on a server
in a compressed form. The users can download these stored files whenever required using the
­Internet; that is why this category is also termed as on-demand audio/video. Some examples of
stored audio/video include songs, movies and video clips.
‰‰ Streaming Live Audio/Video: In this category, as the term live implies, the users can listen to
or view audio/video that are broadcast through the Internet. Some examples of live audio/video
­applications include Internet radio and Internet TV.
‰‰ Real Time (Interactive) Audio/Video: In this category, users can communicate with each other in
an interactive manner using the Internet. This category of Internet audio/video is used for real-time
interactive audio/video applications such as Internet telephony, online video chatting and Internet
teleconferencing.

CN_Chapter 14.indd 267 2/20/2012 6:39:52 PM

 268 Data communications and Computer Networks

3. How does streaming live audio/video differ from streaming stored audio/video?
Ans: The major difference between streaming live audio/video and streaming stored audio/video is
that in the former, the communication is meant for a single user (unicast) and occurs on demand when
the user downloads the file whereas in the latter, the communication is live instead of on demand and
file is broadcast to multiple users at the same time
4. Explain the approaches that can be used to download streaming stored audio/video files
Ans: The streaming stored audio/video files are stored on a server in compressed form. In order to
listen to or view these files, one needs to download them. There are mainly four approaches that can be
used to download streaming audio/video files. These approaches are discussed as follows:

Using a Web Server

This is the simplest approach that allows the compressed audio/video file stored on a Web server to be
downloaded as a text file. Following steps are involved in downloading file from th Web server.
1. The browser on the client machine establishes a TCP connection with the Web server on which
the desired audio/video file is stored
2. After the connection has been established, the browser requests for desired file by sending a GET
message to the Web server using the HTTP services.
3. The Web server in response retrieves the desired compressed file from the disk and sends it to the
browser on the client machine.
4. The browser writes the received audio/video file to the disk
5. The browser uses some helper application such as Windows Media Player that retrieves the
­contents of file from the disk block by block and plays back the fil
This approach does not require streaming. Thus, the file cannot be played black on the media player
until it gets downloaded entirely.

Using a Web Server with Metafile

This approach overcomes the limitation of first approach by allowing the media player on client ­machine
to directly connect with the Web server, so that file can be played back as its downloading starts. There
are two files stored on the Web server, one is the original audio/video file and the other is a metafil that
contains the link (URL) to the desired file. Following are the steps for this approach. Following steps
are involved in this approach.
1. The browser on the client machine sends the GET message using the HTTP services to access the
Web server.
2. The Web server in response sends back the metafile
3. The browser passes the metafile to the media playe .
4. The media player reads the URL in the metafile and sends GET message using the HTTP services
to access the audio/video file
5. The Web server responds now to the media player and not to the browser.
6. The file is played back on the media playe , as it is being fetched by the Web server.

CN_Chapter 14.indd 268 2/20/2012 6:39:52 PM

 Multimedia 269

Using a Media Server

In the second approach, both the browser and the media player access the Web server using the HTTP
services. However, in certain cases, it may happen that the server mentioned in URL stored in metafile is
different from the Web server. Moreover, it may not be even an HTTP server, but some specialized media
server. In such cases, the audio/video files cannot be downloaded using HTTP; rather any protocol that
runs over UDP can be used to download the file. Following are the steps invo ved in this approach.
1. The browser on the client machine uses the HTTP services and sends the GET message to access
the Web server.
2. The Web server in response sends back the metafile
3. The browser passes the metafile to the media playe .
4. The media player reads the URL in the metafile and sends GET message to the media server to
download the audio/video file from it
5. The media server responds to the media player.
6. The file is played back on the media playe , as it is being fetched by the media server.

Using a Media Server and Real Time Streaming Protocol (RTSP)

In this approach, the media server uses RTSP that offers additional functionalities to the streaming of
media files such as pla , record and pause. The steps involved in this approach are as follows:
1. The browser on the client machine uses the HTTP services and sends the GET message to access
the Web server.
2. The Web server in response sends back the metafile
3. The browser passes the metafile to the media playe .
4. The media player reads the URL in the metafile and sends a SETUP message to the media server
to establish a connection between them.
5. The media sever responds to the media player.
6. The media player forwards a PLAY message to the media server to start downloading of file
7. The downloading of file gets started using a protocol that runs over UDP. The file is played back
on the media player, as it is being fetched by the media server. Notice that while the file is being
downloaded, the media player can send a PAUSE message to the media server to temporarily
stop the downloading which can later be resumed by sending PLAY message.
8. After the file has been downloaded completely, the media player sends a TEARDOWN message
to the media server to release the connection between them.
9. The media server in response releases the connection.
5. Write a short note on real-time transport protocol (RTP)?
Ans: RTP is a transport layer protocol designed for interactive (real-time) multimedia applica-
tions on the Internet such as Internet telephony, video-on-demand and video conferencing. These
applications involve multiple streams of multimedia data including text, audio, video and graphics,
being transmitted in real time over the Internet and RTP is responsible for managing these streams.
RTP has been placed above UDP in the protocol hierarchy and is implemented in the user space
(application layer); however, it must be used with UDP.

CN_Chapter 14.indd 269 2/20/2012 6:39:52 PM

 270 Data communications and Computer Networks

The basic function of RTP is to multiplex multiple real-time data streams of multimedia data
onto single stream of UDP packets. To do this, different streams of multimedia data are first sent to
RTP library that is implemented in the user space along with the multimedia application. The RTP
library multiplexes these streams and encodes them in RTP packets, which are then stuffed into a
socket. At the other end of the socket, the UDP packets are created and embedded in IP packets.
Now, depending on the physical media say Ethernet, the IP packets are further embedded in Ethernet
frames for transmission. Besides, some other functions of RTP include sequencing, time-stamping
and mixing.
6. Explain the RTP packet header format.
Ans: The RTP is a transport layer protocol that has been designed to handle real-time traffic on the
Internet. Figure 14.1 shows the format of an RTP packet header, which consists of various fields. Each
field of TP packer header is described as follows:
‰‰ Ver.: It is a 2-bit long field that indicates the version number of RTP. The current version of
RTP is 2.
‰‰ P: It is a 1-bit long field that indicates the presence of padding (if set to 1) or absence of padding (if
set to 0) at the end of the packet. Usually, the packet is padded if it has been encrypted. The number
of padding bytes added to the data is indicated by the last byte in padding.
‰‰ X: It is a 1-bit long field that is set to 1 if an extra extension header is present between the basic
header and the data; otherwise, its value is 0.
‰‰ Contributor Count (CC): It is a 4-bit long field that signifies the total number of contributing
sources participating in the session. Since this field is of four bits, its value can range from 0 to 15,
that is, the maximum of 15 contributing sources can participate in the same session.
‰‰ M: It is a 1-bit marker field that is specific to the application being used. The application uses
this marker to indicate certain things such as the start of a video frame and the end of a video
frame.
‰‰ Payload Type: It is a 7-bit long field that indicates the type of encoding algorithm used such as
MP3, delta encoding and predictive encoding.
‰‰ Sequence Number: It is a 16-bit long field that indicates the sequence number of the RTP packet.
Each packet that is sent in an RTP stream during a session is assigned a specific sequence number.
The sequence number for the first packet during a session is selected randomly and it is increased
by one for every subsequent packet. The purpose of assigning sequence number to packets is to
enable the receiver to detect the missing packets (if any).
‰‰ Timestamp: It is a 32-bit long field that is set by the source of stream to indicate when the first
sample in the packet was produced. Generally, the first packet is assigned a timestamp value at
random while the timestamp value of each subsequent packet is assigned relative to that of the
first (or previous) packet. For example, consider there are four packets with each packet contain-
ing 15 s of information. This implies, if the first packet starts at t = 0, then second packet should
start at t = 15, the third packet at t = 30 and the fourth at t = 45. This time relationship between
the packets must be preserved at the playback time on the receiver to avoid the jitter problem. For
this, timestamp values are assigned to receiver. Suppose the first packet is assigned a timestamp
value 0, then the timestamp value for the second, third and fourth packets should be 15, 30 and 45,
respectively. Now, the receiver on receiving a packet can add its timestamp to the time at which it
starts playback. Thus, by separating the playback time of packets from their arrival time, the jitter
problem is prevented at the receiver.

CN_Chapter 14.indd 270 2/20/2012 6:39:52 PM

 Multimedia 271

‰‰ Synchronization Source Identifier It is a 32-bit long field that indicates the source stream of
the packet. In case of a single source, this field identifies that source. However, in case of multiple
sources, this field identifies the mixer—the synchronization source—and the rest of the sources are
the contributors identified by the contributing source identifiers. The role of mixer is to combine
the streams of RTP packets from multiple sources and forward a new RTP packet stream to one or
more destinations.
‰‰ Contributing Source Identifier It is a 32-bit long field that identifies a contributing source in the
session. There can be maximum 15 contributing sources in a session and accordingly, 15 contribut-
ing source identifiers
32 bits
Ver. P X CC M Payload type Sequence number

Timestamp

Synchronization source identifier

Contributing source identifier

Contributing source identifier

Figure 14.1 RTP Packet Header Format

7. What is RTCP? Describe the packet types define by it?
Ans: RTCP (stands for real time transport control protocol) is called a sibling protocol of RTP.
RTP enables to transport messages containing only data; however, at certain times, some different types
of messages that control the flow and quality of data are required to be sent. For example, a receiver may
send a feedback on received quality of service to the source as well as other participants in the session,
so that better quality could be achieved in future. To send such messages, RTCP is used. RTCP trans-
mission contains five types of packets (containing messages) encapsulated into a single-UDP datagram.
These packet types are described as follows:
‰‰ Sender Report (SR): This packet is used by RTCP receivers to provide feedback about the quality
aspects of data transmission and reception from active senders. It may contain information regard-
ing jitter, delay, congestion, bandwidth and other network properties. Absolute time stamp, that is,
the number of seconds elapsed since midnight on January 1, 1970 is included in the sender’s report.
This helps the receiver in synchronizing multiple RTP packets.
‰‰ Receiver Report (RR): This packet is used for passive participants that do not transmit any RTP
packets. The RR packet communicates the information related to quality of service to the senders
as well as other receivers.
‰‰ Source Description (SDES): This packet is transmitted periodically by a source to give more informa-
tion about itself such as its user name, e-mail address, geographical location and telephone number.
‰‰ BYE: This packet is a direct announcement sent by a source to indicate its exit from a session.
Whenever a mixer receives a Bye message, it forwards this message along with list of sources still
participating in the session to all the stations in session. A Bye packet can also contain the reason
for exit as a description in text format.

CN_Chapter 14.indd 271 2/20/2012 6:39:52 PM

 272 Data communications and Computer Networks

‰‰ Application-define Packet: This packet is an experimental packet for applications that wish to use
new applications that have not been defined in the RTCP standard. Eventually, if an experimental packet
type is found useful, it may be assigned a packet type number and included in RTCP standards.
8. How is SIP used in the transmission of multimedia?
Ans: SIP, which stands for session initiation protocol, is designed by IETF to handle communica-
tion in real-time (interactive) audio/video applications such as Internet telephony, also called voice over IP
(VoIP). It is a text-based application layer protocol that is used for establishing, managing and terminating
a multimedia session. This protocol can be used for establishing two-party, multiparty or multicast sessions
during which audio, video or text data may be exchanged between the parties. The sender and receiver
participating in the session can be identified through various means such as e-mail addresses, telephone
numbers or IP addresses provided all these are in SIP format. That is why SIP is considered very flexible
Some of the services provided by SIP include defining telephone numbers as URLs in Web pages,
initiating a telephone call by clicking a link in a Web page, establishing a session from a caller to a callee
and locating the callee. Some other features of SIP include call waiting, encryption and authentication.
SIP defines six messages that are used while establishing a session, communicating and terminating
a session. Each of these messages is described as follows:
‰‰ INVITE: This message is used by the caller to start a session.
‰‰ ACK: This message is sent by the caller to callee to indicate that the caller has received callee’s reply.
‰‰ BYE: This message is used by either caller or callee to request for terminating a session.
‰‰ OPTIONS: This message can be sent to any system to query about its capabilities.
‰‰ CANCEL: This message is used to cancel the session initialization process that has already started.
‰‰ REGISTER: This message is sent by a caller to the registrar server to track the callee in case the
callee is not available at its terminal, so that the caller can establish a connection with the callee.
SIP designates some of the servers on the network as registrar server that knows the IP addresses
of the terminals registered with it. At any moment, each user (terminal) must be registered with at
least one registrar server on the network.
Before starting transmission of audio/video between the caller and callee, a session needs to be started.
A simple SIP session has three phases (see Figure 14.2), which are as follows:
Caller Callee
INVITE

OK
Session establishment

ACK

Audio exchange
Communication

BYE
Session termination

Figure 14.2 Phases in Session Using SIP

CN_Chapter 14.indd 272 2/20/2012 6:39:53 PM

 Multimedia 273

1. Session Establishment: The session between the caller and callee is established using the three-
way handshake. Initially, the caller invites the callee to begin a session by sending it an INVITE
message. If the callee is ready for communication, it responds to the caller with a reply message
(OK). On receiving the reply message, the caller sends the ACK message to callee to confirm the
session initialization.
2. Communication: Once the session has been established, the communication phase commences
during which the caller and callee exchange audio data using temporary port numbers.
3. Session Termination: After the data has been exchanged, either the caller or the callee can ­request
for the termination of session by sending a BYE message. Once the other side ­acknowledges the
BYE message, the session is terminated.
9. Explain the H.323 standard.
Ans: H.323 is an ITU standard that allows communication between the telephones connected to
a public telephone network and the computers (called terminals) connected to the Internet. Like SIP,
H.323 also allows two-party and multiparty calls using a telephone and a computer. The general archi-
tecture of this standard is shown in Figure 14.3.

Gateway
Gatekeeper

Telephone
Internet
network

Terminal
Figure 14.3 H.323 Architecture

As shown in Figure 14.3, the two networks, Internet and telephone networks, are interconnected via
a gateway between them. As we know that a gateway is a five-layer device that translates messages
from a given protocol stack to different protocol stack. In H.323 architecture, the role of gateway is to
convert between the H.323 protocols on the Internet side and PSTN protocols on the telephone side.
The gatekeeper on the local area network serves as the registrar server and knows the IP addresses of
the terminals registered with it.
H.323 comprises many protocols which are used to initiate and manage the audio/video communi-
cation between the caller and the callee. The G.71 or G.723.1 protocols are used for compression and
encoding/decoding speech. The H.245 protocol is used between the caller and the callee to negotiate on
a common compression algorithm that will be used by the terminals. The H.225 or RAS (Registration/
Administration/Status) protocol is used for communicating and registering with the gatekeeper. The
Q.931 protocol is used for performing functions of the standard telephone system such as providing dial
tones and establishing and releasing connections.

CN_Chapter 14.indd 273 2/20/2012 6:40:02 PM

 274 Data communications and Computer Networks

Following are the steps involved in communication between a terminal and a telephone using H.323.
1. The terminal on a LAN that wishes to communicate with a remote telephone broadcasts a UDP
packet to discover the gatekeeper. In response, the gatekeeper sends its IP address to the terminal.
2. The terminal sends a RAS message in a UDP packet to the gatekeeper to register itself with the
gatekeeper.
3. The terminal communicates with the gatekeeper using the H.225 protocol to negotiate on band-
width allocation.
4. After the bandwidth has been allocated, the process of call setup begins. The terminal sends a
SETUP message to the gatekeeper, which describes the telephone number of the callee or the IP
address of a terminal if a computer is to be called. In response to receipt of SETUP message, the
gatekeeper sends CALL PROCEEDING message to the terminal. The SETUP message is then
forwarded to the gateway, which then makes a call to the desired telephone. As the telephone starts
ringing up, an ALERT message is sent to the calling terminal by the end office to which the desired
telephone is connected. After someone picks up the telephone, a CONNECT message is sent by
the end office to the calling terminal to indicate the establishment of connection. Notice that dur-
ing call setup, all entities including terminal, gatekeeper, gateway and telephone communicate us-
ing the Q.931 protocol. After the connection establishment, the gatekeeper is no longer involved.
5. The terminal, gateway and telephone communicate using the H.245 protocol to negotiate on the
compression method.
6. The audio/video in form of RTP packets is exchanged between the terminal and telephone via
gateway. For controlling the transmission, RTCP is used.
7. Once either of the communicating parties hangs up, the connection is to be terminated. For this,
the terminal, gatekeeper, gateway and telephone communicate using Q.931 protocol.
8. The terminal communicates with the gatekeeper using the H.225 protocol to release the allocated
bandwidth.
10. Defin compression. What is the difference between lossy and lossless compression?
Ans: The components of multimedia such as audio and video cannot be transmitted over the Internet
until they are compressed. Compression of a file refers to the process of cutting down the size of the file by
using special compression algorithms. There are two types of compression techniques: lossy and lossless.
In lossy compression technique, some data is deliberately discarded in order to achieve massive reduc-
tions in the size of the compressed file. In this compression format, we cannot recover all of its original
data from the compressed version. JPEG image files and MPEG video files are the examples of lossy
compressed files. On the other hand, in lossless compression technique, the size of the file is reduced
without permanently discarding any information of the original data. If an image that has undergone loss-
less compression is decompressed, the original data can be reconstructed exactly, bit-for-bit, that is, it will
be identical to the digital image before compression. PNG image file formats use lossless compression
11. Write a short note on audio compression.
Ans: Before the audio data can be transmitted over the Internet, it needs to be compressed. Audio
compression can be applied on speech or music. There are two categories of techniques that can be used
to compress audio, namely, predictive encoding and perceptual encoding.
‰‰ Predictive Encoding: In digital audio or video, successive samples are usually similar to each
other. Considering this fact, the initial frame and the difference values in the successive samples for
all the samples are stored in the compressed form. As the size of the difference values between two

CN_Chapter 14.indd 274 2/20/2012 6:40:02 PM

 Multimedia 275

samples is much smaller than the size of sample itself, this encoding technique saves much space.
While decompressing, the previous sample and the difference value are used to reproduce the next
sample. Predictive encoding is generally used for speech.
‰‰ Perceptual Encoding: The human auditory system suffers from certain flaws. Exploiting this fact,
the perceptual encoding technique encodes the audio signals in such a manner that they sound
similar to human listeners, even they are different. This technique is generally used for compress-
ing music, as it can create CD-quality audio. MP3, a part of MPEG standard, is the most common
compression technique based on perceptual encoding.
12. How does frequency masking differ from temporal masking?
Ans: Effective audio compression takes into account the physiology of human hearing. The com-
pression algorithm used is based on the phenomenon named simultaneous auditory masking—an
­effect that is produced due to the way the nervous system of human beings perceives sound. Masking
can occur in frequency or time, accordingly named as frequency masking and temporal masking.
In frequency masking, a loud sound in a frequency range can partially or completely mask (hide) a
low or softer sound in another frequency band. For example, in a room with loud noise, we are unable
to hear properly the sound of a person who is talking to us.
In temporal masking, a loud sound can make our ears insensitive to any other sound for a few
­milliseconds. For example, on hearing a loud noise such as a gunshot or explosion, it makes our ears
numb for a very short time before we are actually able to start hearing again properly.
13. Explain the JPEG process.
Ans: JPEG, which stands for joint photographic experts group, is the standard compression tech-
nique used to compress still images. It can compress images in lossy and lossless modes and produces
high-quality compressed images. Following are the steps involved in the JPEG image compression
(lossy) process.
1. Colour Sub-Sampling: This step is performed only if the image to be compressed is coloured;
for gray-scale images, this step is not required. The RGB colour space of the image is changed to
YUV colour space and its chrominance component is down-sampled.
2. Blocking: The image is divided into a series of 8 × 8-pixel blocks. Blocking also reduces the
number of calculations needed for an image.
3. Discrete Cosine Transformation (DCT): Each block of 8 × 8 pixels goes through the DCT
transformation to identify the spatial redundancy in an image. The result of DCT transformation
for each 8 × 8 block of pixels is 8 × 8 block of DCT coefficients (that is, 64 frequency compo-
nents in each block).
4. Quantization: In this phase, the DCT coefficients in each block are scalar quantized with the
help of a quantization table (Q-table) in order to wipe out the less important DCT coefficients.
Each value in the block is divided by a weight taken from the corresponding position in Q-table
by ignoring the fractional part. The changes made in this phase cannot be undone. That is why
this JPEG method is considered lossy.
5. Ordering: The output of quantization is then ordered in a zigzag manner to distinguish the low-
frequency components (usually, non-zero) from the high-frequency components (usually, zero).
Ordering results in bit stream in which zero-frequency components are placed close to the end.
6. Run-Length Encoding: The run-length encoding is applied to zeros of zigzag sequence to elimi-
nate the redundancy. This encoding replaces each repeated symbol in a given sequence with the

CN_Chapter 14.indd 275 2/20/2012 6:40:02 PM

 276 Data communications and Computer Networks

symbol itself and the number of times it is repeated. For example, the text “cccccccbbbbuffffff”
is compressed as “c7b4u1f6”. Thus, redundant zeros are removed after this phase.
7. Variable-length Encoding: The variable-length encoding is applied on the output of the
­previous phase to get the compressed JPEG bit stream. In variable-length encoding, a vari-
able number of bits are used to represent each character rather than a fixed number of bits for
each character. Fewer bits are used to represent the more frequently used character; the most
­frequently used character can be represented by one bit only. This helps in reducing the length
of compressed data.
14. What is MPEG? Describe spatial and temporal compressions?
Ans: MPEG, which stands for moving picture experts group, is a method devised for the
­compression of a wide range of video and motion pictures. It is available in two versions: MPEG1 and
MPEG2. The former version has been designed for CD-ROM with a data arête of 1.5 Mbps while the
latter version has been designed for DVD with a data rate of 3–6 Mbps.
Each video is composed of a set of frames where each frame is actually a still image. The frames in
a video flow so rapidly (for example, 50 frames per second in TV) that the human eye cannot notice the
discrete images. This property of human eye forms the basis of motion pictures. Video compression us-
ing MPEG involves spatial compression of each frame and temporal compression of a set of frames.
‰‰ Spatial Compression: Each frame in the video is spatially compressed with JPEG. Since each
frame is an image, it can be separately compressed. Spatial compression is used for the purposes
such as video editing where frames need to be randomly accessed.
‰‰ Temporal Compression: In this compression, the redundancy is removed among the consecutive
frames that are almost similar. For example, in a movie, there are certain scenes where the back-
ground is same and stationary and only some portion such as hand movement is changing. In such
cases, the most consecutive frames will be almost similar except the portion of frame covering the
hand movements. That is, the consecutive frames will be containing redundant information. The
temporal redundancy can be eliminated using the differential encoding approach, which encodes
the differences between adjacent frames and sends them. An alternative approach is motion com-
pensation that compares each frame with its predecessor and records the changes in the coordinate
values due to motion as well as the differences in pixels after motion.
15. Differentiate among the different types of encoded frames used in MPEG video compression.
Ans: In MPEG video compression, the encoded frames fall under three categories, namely, intracoded
(I ) frames, predicted (P) frames and bidirectional (B) frames. These frames are described as follows:
‰‰ I-frame: This frame is not associated with any other frame (previous or next). It is encoded
­independently of other frames in the video and contains all the necessary information that is needed
to recreate the entire frame. Thus, I-frames cannot be constructed from any other frames. I-frames
must appear in a movie at regular intervals to indicate the sudden changes in the frame.
‰‰ P-frame: This frame relates to the frame preceding to it whether it is an I-frame or a P-frame.
It contains small differences related to its preceding I-frame or P-frame; however, it is not
useful for recording major modifications; for example, in case of fast-moving objects. Thus,
P-frames carry only a small amount of information as compared to other frames and even more
less number of bits after compression. Unlike I-frames, a P-frame can be constructed only from
its preceding frame.

CN_Chapter 14.indd 276 2/20/2012 6:40:02 PM

 Multimedia 277

‰‰ B-frame: This frame, as the name implies, relates to its preceding as well as succeeding I-frame
or P-frame. However, a B-frame cannot relate to any other B-frame. B-frames provide improved
motion compensation and the best compression.

Multiple Choice Questions

1. Which of the following services is provided (c) Blocking
by RTP? (d) Discrete cosine transformation
(a) Time-stamping (b) Sequencing 5. Which of the following approaches that is
(c) Mixing (d) All of these used to download streaming stored audio/
2. In __________, the user can listen to or video does not involve streaming?
watch the file only after the file has bee (a) Using a Web server
downloaded. (b) Using a media server
(a) Streaming stored audio/video (c) Using a Web server with a metafil
(b) Streaming live audio/video (d) Using a media server and RTSP
(c) Interactive audio/video 6. Which of the following is a characteristic of
(d) None of these real-time interactive audio/video?
3. MP3 audio compression uses two phenom- (a) Time relationship (b) Mixing
ena, namely, __________. (c) Ordering (d) All of these
(a) Spatial compression and temporal com- 7. Which of the following message types is
pression provided by both RTCP and SIP?
(b) DCT and quantization (a) INVITE (b) BYE
(c) Frequency masking and temporal masking (c) Sender report (d) None of these
(d) None of these
8. In H.323 architecture, __________ serves as
4. Which step of the JPEG image compression the registrar server.
is not needed for grey-scale images? (a) Gateway (b) Terminal
(a) Quantization (c) Gatekeeper (d) Telephone network
(b) Colour sub-sampling

Answers
1. (d) 2. (a) 3. (c) 4. (b) 5. (a) 6. (d) 7. (b) 8. (c)

CN_Chapter 14.indd 277 2/20/2012 6:40:02 PM

 15
Network Security

1. What are the desirable requirements for a secure communication?

Ans: Nowadays, the computer network is used by most people for performing their tasks such as
shopping, bill payment and banking. Thus, it becomes important to secure the network, so that unau-
thorized people could not access the information. For secured communication, there are some basic
requirements that must be met. These are as follows:
‰‰ Confidentiality It refers to maintaining secrecy of the message being transmitted over the network.
Only the sender and the intended receiver should be able to understand and read the message and no
eavesdropper should be able to read or modify the contents of the message. Therefore, the users want
their message to be transmitted over network in encrypted form.
‰‰ Authentication: It is concerned with determining whom you are communicating with. Authenti-
cation is must to ensure the receiver that the message has been received from the actual sender and
not from the attacker. That is, the receiver should be able to authenticate the sender, which can be
achieved by sharing a common secret code word, by sending the digital signatures or by the use of
digital certificates
‰‰ Integrity: Any message sent over the network must reach to its intended receiver without any
modification made to it. If any changes have been made, the receiver must be able to detect that
alteration has happened. Integrity can be achieved by attaching a checksum to the message.
This checksum ensures that an attacker cannot alter the message; therefore, integrity can be
preserved.
‰‰ Non-repudiation: After a message has been sent and received, the sender and the receiver should
not be able to deny about the sending and receiving of the message. The receiver should be able to
prove that the message has come from the intended sender and not from anyone else. In addition,
the receiver should be able to prove that the contents of the received messages are same as sent by
the sender.

CN_Chapter 15.indd 278 2/20/2012 6:41:11 PM

 Network Security 279

2. What do you understand by network security attack? Describe active and passive attacks.
Ans: A network security attack refers to an act of breaching the security provisions of a network.
Such an act is a threat to the basic goals of secure communication such as confidentialit , integrity and
authenticity. Network security attacks can be classified under two categories, namely, passive attack and
active attack.
‰‰ Passive Attack: In a passive attack, an opponent is indulged in eavesdropping, that is, listening to
and monitoring the message contents over the communication channel. The term passive indicates
that the main goal of the opponent is just to get the information and not to do any alteration in the
message or harm the system resources. A passive attack is hard to recognize, as the message is not
tampered or altered; therefore, the sender or receiver remains unaware of the message contents been
read by some other party. However, some measures such as encryption are available to prevent their
success. Two types of passive attacks are as follows:
‰‰ Release of Message Contents: This type of passive attack involves the learning of the sensitive
information that is sent via e-mail or tapping a conversation being carried over a telephone line.
‰‰ Traffi Analysis: In this type of attack, an opponent observes the frequency and the length of
messages being exchanged between the communicating nodes. This type of passive attack is
more elusive, as location and identity of communicating nodes can be determined.
‰‰ Active Attack: In active attack, an opponent either alters the original message or creates a fake
message. This attack tries to affect the operation of system resources. It is easier to recognize an
active attack but hard to prevent it. Active attacks can be classified into four different categories
which are as follows:
‰‰ Masquerade: In computer terms, masquerading is said to happen when an entity impersonates
another entity. In such an attack, an unauthorized entity tries to gain more privileges than it is
authorized for. Masquerading is generally done by using stolen IDs and passwords or through
bypassing authentication mechanisms.
‰‰ Replay: This active attack involves capturing a copy of message sent by the original sender
and retransmitting it later to bring out an unauthorized result.
‰‰ Modificatio of Messages: This attack involves making certain modifications in the captured
­message or delaying or reordering the messages to cause an unauthorized effect.
‰‰ Denial of Service (DoS): This attack prevents the normal functioning or proper management
of communication facilities. For example, network server can be overloaded by unwanted
packets, thus, resulting in performance degradation. DoS attack can interrupt and slow down
the services of a network or may completely jam a network.
3. What is meant by cryptography?
Ans: The term cryptography is derived from a Greek word kryptos which means “secret writing”. In
simple terms, cryptography is the process of altering messages to hide their meaning from ­adversaries
who might intercept them. In data and telecommunications, cryptography is an essential technique
­required for communicating over any untrusted medium, which includes any network, such as Internet.
Cryptography provides an important tool for protecting information and is used in many aspects of
computer security. By using cryptography techniques, the sender can first encrypt a message and then
transmit it through the network. The receiver on the other hand, must be able to decrypt the message and
recover the original contents of message.

CN_Chapter 15.indd 279 2/20/2012 6:41:11 PM

 280 Data Communications and Computer Networks

Sender Key Key Receiver

Encryption Decryption
Plaintext Ciphertext Plaintext

Figure 15.1 Cryptography Components

4. What are the various cryptography components? Show with the help of a diagram.
Ans: Cryptography allows a sender to disguise a message to prevent it from being read or altered
by intruder as well as it enables receiver to recover the original message from disguised one. Various
components are involved in cryptography (see Figure 15.1), which are described as follows:
‰‰ Plaintext: It refers to the original unencrypted message that the sender wishes to send.
‰‰ Ciphertext: It refers to the encrypted message that is received by the receiver.
‰‰ Encryption: It is the process of encrypting the plaintext, so that ciphertext can be produced. The
plaintext is transformed to ciphertext using the encryption algorithm.
‰‰ Decryption: It is opposite of the encryption process. In this process, the ciphertext is converted
back to plaintext using a decryption algorithm.
‰‰ Ciphers: The encryption and decryption algorithms are together known as ciphers. Ciphers need
not necessarily be unique for each communicating pair; rather a single cipher can be used for
­communication between multiple pairs of sender and receiver.
‰‰ Key: A key is usually a number or a set of numbers on which the cipher operates. Encryption and
decryption algorithms make use of a key to encrypt or decrypt messages respectively. At the sender’s
end, the encryption algorithm and encryption key are required to convert the plaintext to ciphertext.
At the receiver’s end, the decryption algorithm uses the decryption key to convert the ciphertext
back to the plaintext. The longer the key is, the harder it is for an intruder to d­ ecrypt the message.
5. Defin the term cryptanalysis.
Ans: Cryptanalysis is the science and art of breaking the encrypted codes that are created by ­applying
some cryptography algorithm. The person who performs cryptanalysis is known as ­cryptanalyst. Crypta-
nalysis attack is done by cryptanalyst so as to obtain the plaintext or key that was used to encrypt the message.
6. Explain the categories of cryptography algorithms.
Ans: Cryptography relies upon two basic components: an algorithm (or cryptographic ­methodology)
and a key. Algorithms are the complex mathematical formulae and keys are strings of bits. For two ­parties
to communicate over a network (Internet), they must use the same algorithm (or algorithms that are
­designed to work together). In some cases, they must also use the same key. The cryptography algorithms
are broadly classified into two categories, namel , secret key cryptography and public key cryptography.

Secret Key Cryptography

The secret key cryptography also called symmetric key cryptography uses a single key (shared ­secret
key) for both encryption and decryption of data. Thus, it is obvious that the key must be known to
both the sender and the receiver. As shown in Figure 15.2, the sender uses this key and the encryption
algorithm to transform the plaintext into ciphertext. The ciphertext is then sent to the receiver via a
communication network. The receiver applies the same key and the decryption algorithm to decrypt the
ciphertext and recover the plaintext. Some examples of secret key cryptography algorithms include data
encryption standard (DES), triple DES and advanced encryption standard (AES).

CN_Chapter 15.indd 280 2/20/2012 6:41:11 PM

 Network Security 281

Shared
key

Sender Receiver

Encryption Decryption
Plaintext Ciphertext Plaintext

Figure 15.2 Message Exchange Using Secret Key

The main problem in secret key cryptography is getting the sender and receiver to agree on the
s­ ecret key without anyone else finding out. If the key is compromised, the security offered by secret key
cryptography is severely reduced or eliminated. Secret key cryptography assumes that the parties who
share a key rely upon each other not to disclose the key and protect it against modification. If they are in
separate physical locations, they must trust on a medium such as courier, or a phone system, to prevent
the disclosure of the secret key. Anyone who overhears or intercepts the key in transit can later read,
modify and forge all messages encrypted or authenticated using that key.

Public Key Cryptography

The public key cryptography also known as asymmetric key cryptography solves the problem found
in secret key cryptography by involving two different keys for encryption and decryption. These two
keys are referred to as the public key (used for encryption) and the private key (used for decryption).
Each authorized user has a pair of public key and private key. The public key of each user is known to
everyone, whereas, the private key is known to its owner only.
Now suppose that user A wants to transfer some information to user B securely. The user A encrypts
the data by using public key of user B and sends the encrypted message to user B. On receiving ­encrypted
message, user B decrypts it by using his or her private key. Since decryption process requires the private
key of user B, which is only known to user B, the information is transferred securely. Figure 15.3 illus-
trates the whole process. RSA is a well-known example of public key encryption algorithm.

B’s public B’s private

key key

A B

Encryption Decryption
Plaintext Ciphertext Plaintext

Figure 15.3 Message Exchange Using Public Key

The main advantage of public key cryptography is that the need for the sender and receiver to share
secret key is eliminated and all communications involve only public keys. Thus, no private key is ever
transmitted or shared. Anyone can send a confidential message using the public key, but the message can
only be decrypted with a private key, which is in the sole possession of the intended r­ ecipient.
7. Explain substitution ciphers and transposition ciphers.
Ans: All encryption and decryption methods have been divided into two categories, namely, substi-
tution ciphers and transposition ciphers. Both are the character-oriented ciphers.

CN_Chapter 15.indd 281 2/20/2012 6:41:12 PM

 282 Data Communications and Computer Networks

Substitution Cipher
This cipher replaces a symbol (a single letter or group of letters) of the plaintext with another symbol.
An example of substitution cipher is the Caesar cipher in which each alphabet of plaintext is replaced
by an alphabet obtained by shifting three letters from it. That is, A is replaced by D, B is replaced by E, Z
is replaced by C and so on. For example, cipher formed from the plaintext TACKLE will be WDFNOH.
A slight generalization of Caesar cipher is shift cipher in which the ciphertext alphabet can be obtained
by shifting n letters instead of 3; thus, n becomes the key. Substitution ciphers are further categorized
into two types, which are as follows.
‰‰ Monoalphabetic Cipher: In monoalphabetic cipher, the characters in the plaintext have a one-
to-one relationship with the characters in the ciphertext. It means that a character in the plaintext
will always be replaced by the same character in the ciphertext. For example, if it is decided that a
­ciphertext character will be obtained by shifting two positions from the character in the plaintext
and the given plaintext is HAPPY, then its ciphertext will be JCRRA.
‰‰ Polyalphabetic Cipher: In polyalphabetic cipher, the characters in the plaintext may have a
one-to-many relationship with the characters in the ciphertext. It means that the same character
appearing in plaintext can be replaced by a different character in the ciphertext. For example,
the plaintext HELLO can be encrypted to ARHIF using a polyalphabetic cipher. Due to one-to-
many relationship between the characters of plaintext and ciphertext, the key used must indi-
cate which of the possible characters can be used for replacing a character in the plaintext. For
this, the plaintext is divided into groups of characters and a set of keys is used for ­encrypting
the groups.

Transposition Cipher
This cipher changes the location of characters in plaintext to form the ciphertext. In this cipher, there
is no substitution of characters and thus, the order of characters in the plaintext is no longer preserved
in the ciphertext. Transposition cipher uses a key that maps the position of characters in the plaintext
to that of characters in the ciphertext. One of the commonly used transposition ciphers is columnar
transposition in which a word or phrase without containing any repeated letters is chosen as a key.
Each letter of the key is numbered to form the columns and the numbering is done in such a way that
column 1 is one under the key letter closest to the start of the 26-alphabet set. Then, the plaintext is
­arranged horizontally under the columns forming the rows. The rows are padded with extra characters
to fill the matrix, if required. The ciphertext is then read out column-wise starting from the first column
to the last column. For example, if the key is BACKIN and the plaintext is given as hellohowareyou,
then ciphertext will be formed as follows:
B   A   C   K   I   N
2   1   3   5   4   6
h   e   l   l   o   h
o   w   a   r   e   y
o   u   a   b   c   d
Thus, the ciphertext will be ewuhoolaaoeclrbhyd.
8. What is the difference between stream cipher and block cipher?
Ans: The stream cipher and block cipher are the categories of symmetric cipher—the ciphers that
use the same key for both encryption and decryption.

CN_Chapter 15.indd 282 2/20/2012 6:41:12 PM

 Network Security 283

The stream cipher operates on one symbol of plaintext at a time and using the key applied it ­produces
a symbol of ciphertext one at a time. The stream ciphers implement a feedback mechanism so that the key
is constantly changing. Thus, the same character in plaintext may be encrypted to different characters in
ciphertext. However, each character is encrypted and decrypted using the same key regardless of the fact
that multiple keys are being used. For example, consider the plaintext is user and three different keys (K1,
K2 and K3) are used to produce ciphertext, such that the characters u and r are encrypted ­using key K1, the
characters s is encrypted using key K2 and the character e is encrypted using K3. Then, ­during decryption
also, the same set of keys (K1, K2 and K3) are used, such that the characters u and r are decrypted using
key K1, the character s is decrypted using key K2 and the character e is decrypted using the key K3.
On the other hand, in block ciphers, an n-bit block of plaintext is encrypted together to produce an
n-bit block of ciphertext. Similarly, during decryption, n-bit block of ciphertext is converted back to n-bit
block of plaintext, one block at a time. Each block of bits is encrypted or decrypted using the same key.
Thus, the same block of plaintext will always be encrypted to same block of ciphertext.
9. Describe S-box and P-box.
Ans: S-box (substitution box) and P-box (permutation box) are used to perform substitution and
transposition function respectively. These are described as follows.
‰‰ S-box: This is a substitution box having same characteristics as that of substitution cipher except
that the substitution of several bits is performed in parallel. It takes n bits of plaintext at a time
as input and produces m bits of ciphertext as output where the value of n and m may be same or
­different. An S-box can be keyed or keyless. In a keyed S-box, the mapping of n inputs to m outputs
is decided with the help of a key, while in keyless S-box, the mapping from inputs to outputs is
predetermined.
‰‰ P-box: This is a permutation box having same characteristics as that of traditional transposition
­cipher except that it performs transposition at bit-level and transposition of several bits is per-
formed at the same time. The input bits are permutated to produce the output bits. For example, the
first input bit can be the second output bit, second input bit can be the third output bit and so on.
P-box is normally keyless and can be classified into the following three types based on the length
of input and output.
‰‰ Straight P-box: This P-box takes n bits as input, permutes them and produces n bits as output.
As the number of inputs and outputs is the same, there are total n! ways to map n inputs to n outputs.
‰‰ Compression P-box: This P-box takes n bits as input and permutes them in such a way that
an output of m bits is produced where m  <  n. This implies that two or more inputs are mapped to
the same output.
‰‰ Expansion P-box: This P-box takes n bits as input and permutes them in such a way that an output
of m bits is produced where m  >  n. This implies that a single input is mapped to more than one output.
10. Explain DES in detail.
Ans: DES is a symmetric-key cipher that was developed by IBM. This encryption standard was
adopted by the U.S. government for non-classified information and by various industries for the use in
security products. It is also called a block cipher, as it divides plaintext into blocks and same key is used
for encryption and decryption of blocks. DES involves multiple rounds to produce the ciphertext and
the key used in each round is the subset of the general key called round key produced by the round key
generator. That is, if there are P rounds in cipher, then P number of keys (K1, K2… KP) will be generated
where K1 will be used in first round, K2 in second round and so on.

CN_Chapter 15.indd 283 2/20/2012 6:41:12 PM

 284 Data Communications and Computer Networks

At the sender’s end, the DES takes 64-bit block of plaintext, encrypts it using the 56-bit round key
and produces 64-bit ciphertext. Originally, the round key is of 64 bits including eight parity bits, thus,
the usable bits in key are only 56. The whole process of producing ciphertext from plaintext comprises
19 stages (see Figure 15.4). The first stage is the keyless transposition on the 64-bit plaintext. Next, 16
stages are the rounds that are functionally similar and in each round, a different key Ki of 48 bits derived
from the original key of 56 bits is used. The second last stage performs a swap function in which leftmost
32 bits are exchanged with the rightmost 32 bits. The last stage is simply the opposite of the first stage,
that is, it performs inverse transposition on 64 bits. At the receiver’s end, the decryption is performed
using the same key as in encryption; however, now, the steps are performed in the reverse order.

64-bit plaintext

Initial transposition Stage 1

K1
48 bits Round 1 Stage 2
Round key generator

K2
48 bits Round 2 Stage 3
56-bit key

K16
48 bits Round 16 Stage 17

32-bit swap Stage 18

Final transposition Stage 19

64-bit ciphertext

Figure 15.4 Stages Involved in the DES

The structure of one of the 16 rounds (say, ith round) during the encryption in DES is shown in
Figure 15.5. It takes two inputs: the leftmost 32 bits as left input (Li) and the rightmost 32 bits as right
­input (Ri) and produces two outputs: left output (Li+1) and right output (Ri+1), each of 32 bits. The left

32 bits 32 bits
Left input Li Right input Ri

Ki
Li⊕f(Ri, Ki)
48 bits

Left output Li +1 Right output Ri +1

32 bits 32 bits
Figure 15.5 Structure of Encryption Round

CN_Chapter 15.indd 284 2/20/2012 6:41:12 PM

 Network Security 285

output (Li+1) is just the right input (Ri). The right output (Ri+1) is obtained by first applying the DES
f­ unction (f  ) on the right input (Ri) and the 48-bit key (Ki) being used in the ith round, denoted as
f(Ri, Ki), and then performing the bitwise XOR of the result of DES func-
tion and the left input (Li). The structure of decryption round in DES is
simply the opposite of the encryption round. Ri(32 bits)
The essence of DES is the DES function. The function f(Ri, Ki) comprises
  

four steps (see Figure 15.6), which need to be carried out sequentially. These Expansion P-box
steps are as follows: 48 bits
Ki(48 bits)
1. The right output (Ri) of 32 bits is fed into the expansion P-box XOR operation
which produces an output (say, E) of 48 bits. S-boxes 48 bits
2. A bitwise XOR is performed on 48-bit E and 48-bit key Ki gener-
ated for that round, resulting in 48 bits.
3. The 48-bit output of XOR operation is broken down into eight 32 bits
groups with each group consisting of six bits. Each group of six bits
is then fed to one of eight S-boxes. Each S-box maps six inputs to Straight P-box
four outputs and thus, total 32 bits are obtained from eight S-boxes.
4. The 32 bits obtained from S-boxes are input to a straight P-box, 32 bits
which permutes them and produces 32 bits as output.
Figure 15.6 DES Function
11. Write a short note on triple DES.
Ans: The length of the key used in DES was too short. Therefore,
triple DES (3DES) was developed to increase the key length, thereby making the DES more secure.
The encryption and decryption in 3DES are performed in three stages with the help of two keys, say K1
and K2 of 56 bits each. During encryption, the plaintext is encrypted using DES with key K1 in the first
stage, then the output of first stage is decrypted using DES with key K2 in the second stage and finall ,
the output of second stage is encrypted using DES with key K1 in the third stage thereby producing the
ciphertext. On the other hand, during decryption, the ciphertext is decrypted using DES with key K1
in the first stage, then the output of first stage is encrypted using DES with key K2 in the second stage
and finall , the output of second stage is decrypted using DES with key K1 in the third stage thereby
­producing the plaintext. The use of two keys and three stages in 3DES increased the key size to 112 bits
and provides more secured communication.
Another version of 3DES uses three keys of 56 bits each and a different key is used for encryption/
decryption in each stage. The use of three different keys further increases the key length to 168 bits;
however, it results in an increased overhead due to managing and transporting one more key.
12. Explain the RSA algorithm.
Ans: In 1978, a group at M.I.T. discovered a strong method for public key encryption. It is known as
RSA, the name derived from the initials of the three discoverers Ron Rivest, Adi Shamir and Len Adleman. It
is the most widely accepted public key scheme, in fact most of the practically implemented security is based
on RSA. The algorithm requires keys of at least 1024 bits for good security. This algorithm is based on some
principles from number theory, which states that determining the prime factors of a number is extremely dif-
ficult. The algorithm follows the following steps to determine the e­ ncryption and decryption keys.
1. Take two large distinct prime numbers, say m and n (about 1024 bits).
2. Calculate p  =  m*n and q  = (m – 1)*(n – 1).
3. Find a number which is relatively prime to q, say D. That number is the decryption key.
4. Find encryption key E such that E*D  =  1  mod q.

CN_Chapter 15.indd 285 2/20/2012 6:41:12 PM

 286 Data Communications and Computer Networks

Using these calculated keys, a block B of plaintext is encrypted as Te = BE mod  p. To recover the original
data, compute B =  Te)D  mod  p. Note that E and p are needed to perform encryption whereas D and p are
needed to perform decryption. Thus, the public key consists of (E , p) and the private key consists of (D, p).
An important property of RSA algorithm is that the roles of E and D can be interchanged. As the number
theory suggests that it is very hard to find prime factors of p, it is extremely difficult for an intruder to
determine decryption key D using just E and p, because it requires factoring p which is very hard.
13. What is digital signature? How it works?
Ans: The historical legal concept of “signature” is defined as any mark made with the intention of
authenticating the marked document. Digital signature refers to the digitized images of paper signature
used to verify the authenticity of an electronic document. In other words, digital signatures play the role
of physical handwritten signatures in verifying electronic documents. Digital signatures use public key
cryptography technique, which employs an algorithm using two different but mathematically related
keys: private and public keys. Both public and private keys have an important property that permits the
reversal of their roles; the encryption key (E) can be used for decryption and the decryption key (D) can
be used for encryption, that is, E(D(P)) = D(E(P)) where P denotes the plaintext. This property is used
for creating messages with digital signature.
The private key is known only to the signer who uses it for creating a digital signature or ­transforming
data into a seemingly unintelligible form and the signed document can be made public. The public key is
used for verifying the digital signature or returning the message to its original form. Any user can easily
verify the authenticity of the document by using the public key that means it can be easily verified that
the data is originated by the person who claims for it. However, no one can sign the document without
having the private key.
To have a clear understanding of how digital signature is used, refer Figure 15.7. Suppose A wants
to send his or her signed message (message with digital signature) to B through network. For this,
A encrypts the message (M) using his or her private key (EA) and this process results in an encrypted
message [EA(M)] bearing A’s signature on it. The signed message is then sent through the network to
B. Now, B attempts to decrypt the received message using A’s public key (DA) in order to verify that the
received message has really come from A. If the message gets decrypted {that is, DA[EA(M)] = M}, B can
believe that the message has come from A. However, if the message or the digital signature has been
modified during the transmission, it cannot be decrypted using A’s public key. From here, B can conclude
that either the message transmission has tampered or the message has not been generated by A.
Digital signatures also ensure non-repudiation. For example, on receiving the encrypted message, B
can keep a copy of that message, so that A cannot later deny of sending of message. Moreover, as B is
unaware of A’s private key (EA), he or she cannot alter the contents of the encrypted message. However,
the only problem with this mechanism is that the message can be tapped by anyone (other than the in-
tended user B) who knows the A’s public key (DA) thereby breaching confidentialit .
To ensure message confidentiali y, encryption and decryption are performed twice at A’s end and B’s
end respectively. At A’s end, first the message is encrypted using A’s private key (EA) and then a second

EA DA
A B
Signed message
Encryption Decryption
Message M EA(M) Message M

Figure 15.7 Digital Signature Using Private Key

CN_Chapter 15.indd 286 2/20/2012 6:41:13 PM

 Network Security 287

EA DB EB DA
A B
Signed
message
Encryption Encryption Decryption Decryption
Message M EA(M) DB(EA(M)) EA(M) Message M

Figure 15.8 Digital Signature Using Public and Private Keys

encryption is performed using the B’s public key (DB) as shown in Figure 15.8. Similarly, at B’s end,
first, the message is decrypted using B’s private key (EB) and then a second decryption is performed
­using A’s public key (DA). With this mechanism, only B can decrypt the encrypted message received
from A because only he or she knows his or her own private key.
14. Defin hash function. What are its properties?
Ans: A hash function is a cryptographic algorithm that transforms the given input (such as a message)
into a fixed-length string, referred to as the hash value. Formally, the hash value (h) can be expressed as:
h  =  H(M),
where M = message (string) of any length, H = hash function and H(M) = a fixed-length string
The hash value plays the role of a “signature” for the data being sent from the sender to receiver
through the network. Sometimes, the hash value is also referred to as message digest or simply digest,
or electronic form of fingerprint
An ideal hash function is characterized by the following properties:
‰‰ For any given message, the hash value can be computed very easily and efficientl .
‰‰ Given a hash value, it is difficult, nearly impossible, to dete mine the message having that hash value.
‰‰ No two messages, even being almost similar, are likely to have the same hash value.

15. What do you understand by the term fi ewall? Explain its use with the help of an example?
Ans: The ongoing occurrences of incidents pertaining to network security caused a great concern
to the people, using computers as their medium to exchange data across the country. A need was felt
for a method of controlling the traffic, which allows access of information to computers. Organizations
required an application that could protect and isolate their internal systems from the Internet. This applica-
tion is called fi ewall. Simply put, a firewall prevents certain outside connections from entering into the
network. It traps inbound or outbound packets, analyzes them and then permits access or discards them.
Generally, firewall system comprises software (embedded in a router), computer, host or a collection
of hosts set up specifically to shield a site or subnet from protocols and services that can be a threat from
hosts outside the subnet. It serves as the gatekeeper between an untrusted network (Internet) and the more
trusted internal networks. If a remote user tries to access the internal networks without going through the
firewall, its effectiveness is diluted. For example, if a travelling manager has an office computer that he
or she can dial into while travelling, and his or her computer is on the protected internal network, then an
attacker who can dial into that computer has circumvented the firewall. Similarly, if a user has a dial-up
Internet account, and sometimes connects to the Internet from his or her office computer, he or she opens
an unsecured connection to the Internet that circumvents the fi ewall.
To understand the use of firewall, consider an example where an organization is having hundreds of
computers on the network. In addition, the organization will have one or more connections to the In-
ternet lines. Now, without a firewall in place, all the computers are directly accessible to anyone on the
Internet. A person who knows what other people are doing can probe those computers; try to make FTP
(file transfer protocol) connections to them, or telnet connections and so on. If one employee makes a
mistake and leaves a security hole, hackers can get to the machine and exploit that hole.

CN_Chapter 15.indd 287 2/20/2012 6:41:13 PM

 288 Data Communications and Computer Networks

With a firewall in place, the network landscape becomes much different. An organization will place
a firewall at every connection to the Internet (for example, at every T1 line coming into the company).
The firewall can implement security rules. For example, one of the security rules may be: out of the 300
computers inside an organization, only one is permitted to receive public FTP traffic. A company can
set up rules like this for FTP servers, web servers, telnet servers and so on. In addition, an organization
can have control on how employees connect to websites, whether or not files can be sent as attachments
outside the company over the network and so on. Firewall provides incredible control over how people
use the network.
16. What is the role of packet filterin in the fi ewall?
Ans: A firewall intercepts the data between the Internet and the computer. All data traffic passes
through it and it allows only authorized data to pass into the corporate network. Firewalls are typically
implemented using packet filterin ,
Packet filtering is the most basic firewall protection technique used in an organization. It operates
at the network layer to examine incoming and outgoing packets and applies a fixed set of rules to the
packets to determine whether or not they will be allowed to pass. The packet filter firewall is typically
very fast because it does not examine any of the data in the packet. It simply examines the IP packet
header, the source and destination IP addresses and the port combinations and then it applies filtering
rules. For example, it is easy to filter out all packets destined for Port 80, which might be the port for a
web server. The administrator may decide that Port 80 is off limits except for specific IP subnets and a
packet filter would suffice for this. Packet filtering is fast, flexible, transparent (no changes are required
at the client) and cheap. This type of filter is commonly used in small to medium businesses that require
control over users to use the Internet.
17. Defin identificatio and authentication. Explain how users can be authenticated?
Ans: Often people confuse identification from authentication, as both have similar aspects. Identi-
ficatio is the means through which a user provides a claimed identity to the system. On the other hand,
authentication refers to establishing the validity of the claim. Computer systems make use of data authen-
tication for recognizing people, which the systems receive. Authentication presents several challenges
such as collecting authentication data, transmitting the data securely and identifying the same person who
was earlier authenticated and is still using the computer system.
Various methods can be used to authenticate a user, such as a secret password, some physical
­characteristics of the user, a smart card or a key given to the user.

Password Authentication
It is the simplest and most commonly used authentication scheme. In this scheme, the user is asked to
enter the user name and password to log in into the database. The DBMS then verifies the combination
of user name and password to authenticate the user and allows him or her to access the database if he
or she is the legitimate user, otherwise access is denied. Generally, password is asked once when a user
log in into the database; however, this process can be repeated for each operation when the user is trying
to access sensitive data.
Though the password scheme is widely used by database systems, this scheme has some limitations. In
this method, the security of database completely relies on the password. Thus, the password itself needs to
be secured from unauthorized access. One simple way to secure the password is to store it in an encrypted
form. Further, care should be taken to ensure that password would never be displayed on the screen in its
decrypted (non-encrypted) form.

CN_Chapter 15.indd 288 2/20/2012 6:41:13 PM

 Network Security 289

Physical Characteristics of User

In this method, the physical characteristics, such as fingerprints, voice, length of fingers of hand, and
face structure of the users are used for the identification purpose. These characteristics of users are
known to be unique and have a very low probability of duplication. Thus, the security of the database is
relatively high in this scheme as compared to the password scheme. However, this method requires the
use of some special hardware and software to identify physical characteristics of the user, which incurs
extra cost to the organization.

Smart Card
In this method, a database user is provided with a smart card that is used for identification. The smart
card has a key stored on an embedded chip and the operating system of smart card ensures that the key
can never be read. Instead, it allows data to be sent to the card for encryption or decryption using that
private key. The smart card is programmed in such a way that it is extremely difficult to extract the
­values from smart card; thus, it is considered as a secure device.
18. Write a short note on message authentication.
Ans: Message authentication is a means to verify that the message received by the receiver is from
the intended sender and not from the intruder. The sender needs to send some proof along the message,
so that the receiver can authenticate the message. To authenticate a message, the message authentication
code (MAC) is used. MAC uses a hash function (MAC algorithm) that generates a MAC (a tag-like)
with the help of a secret key shared between the sender and the receiver. Figure 15.9 depicts the use
of MAC to authenticate a message at the sender’s end and to verify the authenticity of message at the
receiver’s end.
At the sender’s end, the original message that is to be authenticated along with the secret key are given
as input to the MAC algorithm that produces a MAC as output. The MAC is attached with the original
message and both are sent to the receiver through the network. To verify the authenticity of message at
the receiver’s end, the message is distinguished from MAC and the MAC algorithm is again applied on
the message using the secret key to generate a new MAC. Then, the newly generated MAC is compared

Sender Receiver
Message
Message

Key Message Key MAC

MAC
algorithm algorithm

MAC
MAC MAC

Is Yes
MAC Accept
=?
message
No
Discard message

Figure 15.9 Message Authentication Using MAC

CN_Chapter 15.indd 289 2/20/2012 6:41:13 PM

 290 Data Communications and Computer Networks

with the received MAC to determine whether they are same or not. If so, the receiver knows that the
message has not been changed and is actually from the intended sender and thus, accepts the message.
Otherwise, the message is discarded.
19. Encrypt the plaintext 6 using RSA public key encryption algorithm. Use prime numbers
11 and 3 to compute the public and private keys. Moreover, decrypt the ciphertext using the
private key.
Ans: Here, m = 11 and n = 3
According to RSA algorithm (as explained in Q12)
p = m * n = 11 * 3 = 33
q = (m − 1) * (n − 1) = (11 − 1) * (3 − 1) = 10 * 2 = 20
We choose D = 3 (a number relatively prime to 20, that is, gcd (20, 3) = 1).
Now,
   E * D = 1 mod q
⇒ E * 3 = 1 mod 20
        ⇒ E = 7
As we know, the public key consists of (E, p) and the private key consists of (D, p). Therefore, the
public key is (7, 33) and the private key is (3, 33).
The plaintext 6 can be converted to ciphertext using the public key (7, 33) as follows.
T e= BE mod p
     ⇒ 67 mod 33
    ⇒ 30
On applying the private key to the ciphertext 30 to get original plaintext, we get
B = (Te)D mod p
   ⇒ (30)3 mod 33
⇒6
Multiple Choice Questions
1. Which of the following are necessary for (a) Ciphertext (b) Ciphers
secured communication? (c) Key (d) None of these
(a) Authentication (b) Confidentialit 5. In public key cryptography, ________ key is
(c) Integrity (d) All of these used for encryption.
2. In ______ attack, an opponent either alters (a) Public (b) Private
the original message or creates a fake mes- (c) Both (a) and (b) (d) Shared
sage. 6. _________ is the means through which a
(a) Passive (b) Inactive user provides a claimed identity to the sys-
(c) Active (d) Access tem.
3. __________ is a type of passive attack. (a) Authentication (b) Identificatio
(a) Replay (b) Traffic analysi (c) Encryption (d) Decryption
(c) Masquerade (d) Denial of service 7. In __________cipher, characters in the
4. Which of the following is not a component plaintext and ciphertext are related to each
of cryptography? other by one-to-many relationship.

CN_Chapter 15.indd 290 2/20/2012 6:41:13 PM

 Network Security 291

(a) Monoalphabetic (b) XOR (a) Digital signature

(c) Polyalphabetic (d) Rotation (b) Authentication
(c) Identificatio
8. DES takes _____-bit key as an input for
(d) Firewall
encrypting the text.
(a) 128 (b) 64 10. MDC stands for
(c) 56 (d) 168 (a) Message detection code
(b) Modification detection cod
9. Which of the following applications traps
(c) Masquerade detection code
inbound or outbound packets, analyze them
(d) None of these
and then permits access or discards them?

Answers
1. (d) 2. (c) 3. (b) 4. (d) 5. (a) 6. (b) 7. (c) 8. (c) 9. (d) 10. (b)

CN_Chapter 15.indd 291 2/20/2012 6:41:13 PM

 This page is intentionally left blank.

CN_Chapter 15.indd 292 2/20/2012 6:41:13 PM

 Index

A
Adaptive tree-walk algorithm, 155 Authentication protocols types, 140
Address mask, 198 challenge handshake authentication protocol, 141
Address resolution protocol, 29, 226, 228 password authentication protocol, 140
Advantage of token-passing protocol over CSMA/CD Authentication, 140–141, 171, 177,
protocol, 156 231, 233, 254, 272, 278–279, 288–291
Advantage and disadvantage of E-mail, 255 Authentication-request packet, 140
Advantage and disadvantage of fibre optic cables, 7 Autonegotiation, 165
Advantage of computer networking, 8
communication, 9 B
preserve information, 8
sharing hardware resources, 8 Bandpass channel, 45
sharing information, 8 Bandwidth in hertz, 46
sharing software resources, 8 Bandwidth utilization, 83
Advantages of FDDI over a basic Token Ring, 169 Basic types of guided media
Advantages of reference model, 23 coaxial cables, 71
ALOHA protocol limitation, 148 fibre-optic Cable, 7
American standard code for information, 3 stp cable, 71
hexadecimal values, 3 twisted-pair cable, 70
interchange (ASCII), 3 utp cable, 71
Analog and digital signals, 40–41 Bayone-Neill-Concelman, 74
Analog data, 40, 44, 59 Benefits of ATM, 190
Analog hierarchy, 90, 108 Berkeley sockets, 238
Analog modulation, 59 Binary countdown, 153–154
Analog transmission, 44 Binary exponential back-off algorithm, 146, 149, 153
Applications of a computer network, 9 Bit-map, 153–154
business application, 9 bit-oriented and byte-oriented protocol, 133
conferencing, 9 Block Coding, 51, 165
directory and Information services, 9 Bluetooth, 161, 174–178, 192
e-mail services, 9 architecture, 174–175
financial services, protocol stack, 175–178
manufacturing, 9 BNC terminator, 74
marketing and sales, 9 Bootstrap protocol, 228
mobile application, 9 Border Gateway Protocol, 212
Architecture of e-mail, 255 Broadcast mechanism, 96
ASCII-7 Coding Chart, 3–5
Asynchronous frame, 168
Asynchronous transfer mode, 184–185
C
architecture, 185 Cable TV network generation, 102–104
Authenticate-ack packet, 140 Call Setup and Call Clearing, 180–181

CN_Index.indd 293 2/20/2012 6:41:43 PM

 294 Index

Carrier Sense Multiple Access, 145, 149–150 Controlled access method, 155–156
Carrier Sense Multiple Access/ Conversion, 57
Collision Detection, 150 Cryptanalysis, 280
Carrier Signal, 56 Cryptography, 279–281, 286, 290
Challenge packet, 141 CSMA/CA protocol, 152
Channel allocation, 144
channel static allocation, 144
D
dynamic channel allocation, 144
Channelization protocol types, 156–157 Data Communication and Computer
Characteristics of line coding, 48 Network, 2 ,4, 6, 8, 10, 12, 14, 16, 18, 20
Check bit, 115 characteristics of an efficient data, 1–
CIDR, 199 differences between communication and
Classful and classless addressing, 195–196 transmission, 1
Coaxial cable, 73 Data link connection identifie , 182
Collision detection, 150 Data link layer design, 110
Collision-free protocol, 153–154 Data link layer functions in OSI model, 27
Communication channel, 84 access control, 28
Community antenna, 103 error control, 28
Companding, 53 flow control, 2
Comparison of Bus and Mesh Topology, 13–14 framing, 27
ring topology, 14 physical addressing, 28
advantages of ring topology, 14 Data link layer service, 111
disadvantages of ring topology, 14 Data representation, 2
star topology, 14 Data transfer mode, 134
concentrator, 14 Data transmission devices for cable network
advantages of star topology, 15 cable modem transmission system, 104
tree topology, 15 cable modem, 104
Comparison of Circuit Switching, Packet Switching Data Transmission Modes, 7
and Message Switching, 97 full duplex, 7
Components of data communication, 2 half duplex, 7
message, 2 Data transmission system design, 70
protocol, 2 Datagram network, 179
receiver, 2 Decibel (dB), 46, 61
sender, 2 Defination bit interval, bit rate and bit length, 4
transmission medium, 2 Defination of code rate, 14
Composite signal, 43 Defination of Codeword, 14
Compression, 274, 276–277 Defination of Data, 4
Computer network, 7 Defination of framing, 11
performance, 7 Defination of Signal, 40, 43, 5
reliability, 8 Definition of frequency-domain, 4
response time, 8 Definition of Hamming Weight of a Codeword, 114
security, 8 Definition of time-domain, 4
throughput., 8 Delta Modulation, 51, 53
transit time, 8 Demodulator, 54, 84, 98
Congestion control, 216 Demultiplexer, 83–87, 89, 108, 191
Congestion, 215 Diagram error, 113
Contentaion, 145, 149–150, 153–154, 171 Difference guided and unguided transmission media, 70
Control Field for I-frame, 135 Differences Between BRI and PRI, 106
Control Field for S-frame, 136 Differences Between P2P Architecture and
Control Field for U-Frame, 137 Client/Server Architecture, 10–11, 261

CN_Index.indd 294 2/20/2012 6:41:43 PM

 Index 295

Differences between serial and parallel Error correction, 113–114, 117, 187, 190, 238
transmissions, 54–55 Error-detection method, 113, 115–117
Different categories of a network, 16 Ethernet, 164
local Area Network, 16 Expanding, 53
metropolitan Area Network, 17 Extended Binary Coded Decimal Interchange
wide Area Network, 17 Code (EBCDIC), 5
Different categories of transmission media, 69 Extranet, 20
guided transmission media, 69
unguided transmission media, 70 F
Different propagation method for unguided signal, 78
ground wave propagation, 78 FDM differs from FDMA, 158
Ionospheric propagation, 78 Fibre distributed data interface, 165, 168
line-of-sight propagation, 78 Fibre-optic propagation modes, 74
Different types of unguided media, 79 multimode graded-index fibre, 7
Digital data, 40–41, 51, 53 multimode step-index fibre, 7
Digital signature, 278, 286–287, 291 File transfer protocol, 259, 287
Digital transmission, 44 Flooding, 201–202
Direct Sequence Spread Spectrum, 92 Flow control and error control, 124, 162
Dish antenna, 80 Flow-based Routing, 202
Distance vector routing, 200–201, 204–209, 213 Footprint, 81
Domain name system, 251 Formula of BFSK, 58
Downlink, 81 Fourier analysis, 44
DS hierarchy, 89 Frame Relay, 86, 104–105, 166, 182–184,
DSL technology, 99–100 186, 190, 193, 217
asymmetric digital subscriber line, 99 Framing bit, 89
high-bit-rate digital subscriber line, 100 Frequency Hopping Spread Spectrum, 91
spliterless ADSL, 100 Full-duplex synchronous transmission, 98–99
symmetric digital subscriber line, 100 Functionalities of network layer with transport layer, 28
very high-bit-rate digital subscriber line, 100
DSL, 71, 99–100, 104, 109 G
Duty of layer in OSI model, 25
Geostationary satellites, 81
data link layer, 26
Go-back-N ARQ, 130
datagram method, 26
dialog control, 26
error control, 26 H
network Layer, 26 Hamming code, 118, 121–123
physical Layer, 25 Hamming distance, 114
session Layer, 26 Hash function, 287
transport Layer, 26 HDLC protocol, 133–134
virtual circuit method, 26 Hexadecimal colon notation, 197
Dynamic domain name system, 254 Hop-by-hop choke packet, 218
Dynamic host control protocol, 229, 254 Horn antenna, 80
Hybrid topology, 20
E Hypertext transfer protocol, 178, 261, 263

EBCDIC Coding Chart, 5–6

I
simplex, half-duplex and full-duplex, 6
Electromagnetic spectrum, 25, 40, 46, 67, 69–70, 77 Identification, 22, 229–230, 233, 258, 288–289, 29
Electronic Industries Association, 73 Idle, 56, 135, 145, 149–153, 155, 158–159
E-mail, 254–257, 259, 264, 271–272, 279 IEEE 802 standards, 161

CN_Index.indd 295 2/20/2012 6:41:43 PM

 296 Index

Infrared data association, 80 Method of framing, 111–112

Inner product of two sequence, 157 Microwave relay station, 81
Integrated services digital network, 104 Modem, 98
Interdomain routing, 201 Moving picture experts group, 276
Interfaces and services terms, 27 Multilevel multiplexing, 87–88, 108
peer entities, 27 Multimedia, 105, 261, 267, 269–275, 277
Interframe space, 152 Multipath fading effect, 80
Internet control message protocol, 29, 234–235 Multiple-slot allocation, 87–88, 108
Internet, 18–19, 30–31, 97, 104, 182 Multiplexer, 83–87, 89, 107–108, 158, 186, 190–192
advanced research projects agency network, 18 Multiplexing, 83–91, 93, 95, 97–99, 176–177, 240
NSFNET, 18–19 Multipurpose Internet mail extension, 257
Internetworking, 35, 37, 39, 161, 195, 214–215, 229
Intradomain routing, 201 N
Intranet, 19–20
Inverse multiplexing, 91 Network address, 197
IP address, 195, 197–198, 229, 251 Network architecture, 10, 24
IP datagram frame format, 229 Network Control Protocol, 140
IP spoofi ng, 20 Network layer design issue, 194–195
ISDN architecture, 106–107 Network security attack, 279
ISDN channel, 105–106 Non-periodic signal, 41
JPEG process, 275–276 Novell Netware network model, 31
Nyquist bit rate, 46
Nyquist theorem, 52, 63
K
Kinds of light sources used with the fibre-optic O
cables, 76
light-emitting diodes, 76 One-bit sliding window protocol, 129
semiconductor laser diodes, 76 On-off keying, 57
Open Shortest Path First, 210–212
Open Systems Interconnection, 24, 161
L Optimality principle, 200
Layered architecture, 23 Orthogonal sequence, 157
Layering advantages in network, 23
Leaky bucket algorithm, 225 P
Limited-contention protocol, 154
Line Coding, 47 Packet filtering, 28
decoder, 47 Parameters that characterize the sine waves, 41–42
encoder, 47 Path vector routing, 201, 208–209, 212
Link Control Protocol, 12, 139 Performance of non-persistent and 1-persistent
Link state routing, 200–201, 206–208, 210 CSMA protocols, 150
Link Utilization, 125–126, 128–129, 132–133, 141, 143 Periodic signal, 40–41, 44, 67
Load shedding, 218 Permutation box, 283
Logical address, 30 Physical address, 30
Lost frame and damaged frame, 128 Piggybacking, 128, 143
Low-pass channel, 45 Plain old telephone system, 97
Port address, 30, 238
PPP frame format, 138
M PPP protocol, 133, 137–139
MACA for wireless, 174 Preamble bit, 157
Medium access control, 145 Presentation layer function, 28
Message authentication, 289 Principle of OSI layer, 24

CN_Index.indd 296 2/20/2012 6:41:43 PM

 Index 297

Propagation time formula, 47 SNR, 46, 52–53, 63, 66–67

Protocol data unit, 27, 162 Socket address, 238
logical link control (LLC) sublayer, 28 Socket primitives, 239
media access control (MAC) sublayer, 28 SONET regenerator, 191
Protocol, 2, 19–20, 22, 24–25 Specific add ess, 30
key elements of protocols, 19 spread spectrum technique principle, 91
Protocols used at the Link Level, 180 Spread spectrum, 91
Proxy server, 239, 265 Spreading code, 91
Pulse amplitude modulation, 51–52 Spreading process, 91
Pulse stuffing, 87–88, 10 Spreading, 83, 91–93
Pure ALOHA, 145–148, 159 Standard, 19
de facto, 19
Q de jure, 19
Statistical TDM Frame format, 85–87, 89
Quadrature amplitude modulation, 59 Stop-and-wait ARQ, 128
Quality of service, 223 Stop-and-wait flow control protocol, 12
characteristics of, 223 Store-and-forward mechanism, 96
Quantization error, 52 Structure of ATM adaptation layer, 188–190
Substitution Cipher, 281–283
R Supernetting, 199–200, 222
Switch and Hub differences, 38
Random access method, 145 Switch technology types, 101
Real-time transport protocol, 269 space-division switch, 101
Receive-not-ready, 127 time division switch, 102
Receive-ready, 127 Switched Virtual Circuit and Permanent
Redundancy concept, 113 Virtual Circui, 180
Reference models and network devices, 23–24 Symmetric key cryptography, 280–281
Regional cable head, 103 SYN flooding attack, 24
Remote procedure call, 241, 247–248 Synchronization bit, 51, 26, 86–87, 89
Resolver, 251–253 Synchronous Data Link Control, 137
Retransmission technique, 118 Synchronous frame, 168
Reverse address resolution protocol, 228 Synchronous TDM, 85–87, 190
Routing Information Protocol, 209–210, 241
Routing, 200, 202
RSA algorithm, 285
T
RTP packet header format, 270 TDM differs from TDMA, 158
TDM scheme, 84–87, 89, 102, 108, 144, 156–158,
S 160, 176, 190
Technique multiple frames, 127
S-box (substitution box), 283 Techniques for analog-to-analog
Selective-reject ARQ, 132 amplitude modulation, 59
Service access point, 27, 162, 239 frequency modulation, 60
Services of ISDN, 105 phase modulation, 61
Session initiation protocol, 272–273 Techniques for digital-to-analog conversion, 57
Session layer function, 28 amplitude shift Keying, 57
Shortest path routing, 200, 202–204 frequency shift Keying, 57
Simple mail transfer protocol, 242, 257–258 phase shift Keying, 57
Simple network management protocol, 265 Telecommunication, 250, 279
Simultaneous auditory masking, 275 Telephone network major components, 97–98
Sliding window flow control mechanism, 12 TELNET, 30, 238, 250, 265, 287–288
Slotted ALOHA, 145, 147–148, 158–159 Term channelization, 156

CN_Index.indd 297 2/20/2012 6:41:43 PM

 298 Index

Term firewall, 28 Types of multiplexing technique, 84

Term network topology, 12 frequency-division multiplexing, 84
bus topology, 13 time-division multiplexing, 85
mesh topology, 13 wavelength-division multiplexing, 85
Term switching, 93 Types of serial mode transmissions, 55–56
Three-way handshaking mechanism, 141, 244–246 asynchronous transmission, 55
Time slot interchange, 102 Isochronous transmission, 56
Token bucket algorithm, 226 synchronous transmission, 56
Token management, 26 Types of station, 133
application layer, 27 Types of switching technique, 94–97
presentation layer, 26 circuit switching, 94
Token Ring, 32, 36–37, 73, 159, 161–163, 166 message switching, 95
Transmission control protocol, 29, 176, 212, 229, packet switching, 95
238, 241, 250 Types of transmission impairments, 45–46
segment format of TCP 243–244 attenuation, 45
application layer, 30 delay distortion, 45
internet layer, 29 intermodulation noise, 46
network access layer, 29
transport layer, 30 U
Transmission media, 69–71, 73, 98, 105, 162
Transmission path, 25, 85, 185 Unicast and multicast routing, 213–214
Transmission process, 127 USp link, 81
Transmission technology, 16, 18 UTP cable, 73
broadcast network, 18
point-to-point network, 18 V
types of transmission technology, 18
Transmission time formula, 47 Variations of CSMA protocol, 149
Transmission, 150 Various network devices, 32–37
Transport layer duty, 237 bridge, 33
Transport service primitives, 238 gateway, 35
Transposition Cipher, 281–283 hub, 33
Trellis coding, 99 network interface card, 32
Twisted-pair cables, 72 remote bridge, 36
Types of line coding schemes, 49 repeater, 32
bipolar scheme, 50–51 router, 34
manchester scheme, 50 source routing bridge, 36
polar scheme, 49 switch, 34
unipolar scheme, 49 transparent bridge, 36
Types of computer network architecture, 10 Virtual circuit network, 178–179, 192, 216
client/server architecture, 10 Virtual tributary, 192
peer to peer (P2P)Architecture, 10 Vulnerable period, 148
Types of Connector Fibre-optic cable, 75
mt-rj connector, 75 W
sc connector, 75
st connector, 75 Wavelength, 42–43, 76, 84–85
Types of frames specified in IEEE 802. 1, 171–172 Wireless LAN, 80, 91, 161, 169–170, 172,
Types of ISDN, 105 174, 192
Types of Link Configurations, 13 World Wide Web, 30

CN_Index.indd 298 2/20/2012 6:41:43 PM