MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION

(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
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WINTER – 2023 EXAMINATION
Model Answer – Only for the Use of RAC Assessors

Subject Name: Advanced Computer Network Subject Code: 22520

Important Instructions to examiners: XXXXX

1) The answers should be examined by key words and not as word-to-word as given in the model answer scheme.
2) The model answer and the answer written by candidate may vary but the examiner may try to assess the
understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given more Importance (Not applicable
for subject English and Communication Skills.
4) While assessing figures, examiner may give credit for principal components indicated in the figure. The figures
drawn by candidate and model answer may vary. The examiner may give credit for any equivalent figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed constant values may vary
and there may be some difference in the candidate’s answers and model answer.
6) In case of some questions credit may be given by judgement on part of examiner of relevant answer based on
candidate’s understanding.
7) For programming language papers, credit may be given to any other program based on equivalent concept.
8) As per the policy decision of Maharashtra State Government, teaching in English/Marathi and Bilingual (English
+ Marathi) medium is introduced at first year of AICTE diploma Programme from academic year 2021-2022.
Hence if the students write answers in Marathi or bilingual language (English +Marathi), the Examiner shall
consider the same and assess the answer based on matching of concepts with model answer.

Sub
Q. Marking
Q. Answer
No. Scheme
N.
1 Attempt any FIVE of the following: 10 M
a Differentiate between IPv4 and IPv6 on the basis of length and security. 2M
Ans IPv4 IPv6 Any 2
points 1
Addresses are 32 bit Addresses are 128bits (16 M each
(4bytes) in length bytes) in length.

Deployed in 1981 Deployed in 1999

Header includes checksum Header does not include

checksum

Header includes options Optional data is supported

as extension header

Configuration is either Does not require manual

manually or through configuration or DHCP
DHCP

Address format in dotted Address format in

decimal notation hexadecimal notation

Both routers and the Routers do not support

sending host fragment packet fragmentation
packets sending host fragment
packets.

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b State the need of IPv6. 2M

Ans The need of IPv6 are as follows: Any 2
correct
1. Address Space: IPv6 provides a significantly larger address space compared to IPv4, allowing need 1M
for the accommodation of the growing number of devices connected to the internet. each

2. Addressing Efficiency: IPv6 simplifies address assignment and management, eliminating the
need for Network Address Translation (NAT) and making subnetting more efficient.

3. Auto-Configuration: IPv6 supports stateless address auto-configuration, enabling devices to

automatically configure their IPv6 addresses without the need for DHCP (Dynamic Host
Configuration Protocol).

4. Security Improvements: IPv6 includes features like IPsec (Internet Protocol Security) as a
standard, enhancing the security of communications between devices.

5. Multicast and Any cast: IPv6 incorporates improved support for multicast communication,
enabling efficient one-to-many communication. Anycast is also more easily implemented in IPv6.

6. Mobility Support: IPv6 is designed to better support mobile devices, ensuring seamless
connectivity as devices move between networks.

7. Simplified Header Structure: IPv6 has a simpler and more efficient header structure,
reducing processing overhead on networking devices.

8. Future-Proofing: As the successor to IPv4, IPv6 is essential for the continued growth of the
internet and the proliferation of connected devices, ensuring there are enough unique addresses
for future expansion.

c Elaborate need of domain name system. 2M

Ans 1. DNS ensures the internet is not only user-friendly but also works smoothly, loading Correct
whatever content we ask for quickly and efficiently. explanati
2. It allows the user to access remote system by entering human readable device hostnames on 2M
instead of IP address. It translates domain name into IP addresses so browser can load
internet resources.
3. It translates human readable domain names into the numerical identifiers associated with
networking equipment, enabling devices to be located and connected worldwide.
Analogous to a network “phone book,” DNS is how a browser can translate a domain
name (e.g., “facebook.com”) to the actual IP address of the server, which stores the
information requested by the browser.

d List any 2 features of TCP. 2M

Ans 1) Connection Oriented Protocol Any 2
2) Reliable correct
3) Congestion Control Feature
1 M each
4) Full Duplex
5) Error Control and Recovery
6) Flow Control
e List all 4 routing algorithms. 2M
Ans 1) Adaptive Routing Algorithm. 1/2 M
2) Link State Routing Algorithm. each
3) Distance Vector Routing Algorithm. algorithm

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4) Bellmen Ford Algorithm.
f Enlist any two services offered by UDP. 2M
Ans 1) Process to Process communication Any 2
2) Connectionless Service correct
3) Flow control services
4) Error control 1M each
5) Checksum
g State any three phases of mobile IP. 2M
Ans The mobile IP process works in 3 main phases: Correct 3
1) Agent discovery Phase phases
2) Agent Registration Phase 2M
3) Tunneling
2 Attempt any THREE of the following: 12 M
a Describe packet format of IPv6. 4M
Ans 2M
Explanati
on
&
2M
Diagram

Version (4-bits): It represents the version of Internet Protocol, i.e. 0110.

Traffic Class (8-bits): These 8 bits are divided into two parts. The most significant 6 bits
are used for Type of Service to let the Router Known what services should be provided to
this packet.
Flow Label (20-bits): This label is used to maintain the sequential flow of the packets
belonging to a communication. The source labels the sequence to help the router identify
that a particular packet belongs to a specific flow of information. This field helps avoid
re-ordering of data packets. It is designed for streaming/real-time media.
Payload Length (16-bits): This field is used to tell the routers how much information a
particular packet contains in its payload. Payload is composed of Extension Headers and
Upper Layer data.
Next Header (8-bits): This field is used to indicate either the type of Extension Header,
or if the Extension Header is not present then it indicates the Upper Layer PDU. The
values for the type of Upper Layer PDU are same as IPv4’s.
Hop Limit (8-bits): This field is used to stop packet to loop in the network infinitely.
This is same as TTL in IPv4. The value of Hop Limit field is decremented by 1 as it
passes a link. When the field reaches 0 the packet is discarded.
Source Address (128-bits): This field indicates the address of originator of the packet.

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Destination Address (128-bits): This field provides the address of intended recipient of
the packet.

Extension Headers When Extension Headers are used, IPv6 Fixed Header’s Next
Header field points to the first Extension Header. If there is one more Extension Header,
then the first Extension Header’s ‘Next-Header’ field points to the second one, and so on.
The last Extension Header’s ‘Next-Header’ field points to the Upper Layer Header. Thus,
all the headers points to the next one in a linked list manner.
b Explain Bellman Ford algorithm with suitable example. 4M
Ans Dynamic Programming is used in the Bellman-Ford algorithm. It begins with a starting 2M
vertex and calculates the distances between other vertices that a single edge can reach. It Algorith
then searches for a path with two edges, and so on. The Bellman-Ford algorithm uses the m,
2M
bottom-up approach.
Example
Bellman Ford's algorithm

The Bellman-Ford algorithm works by grossly underestimating the length of the path
from the starting vertex to all other vertices.

Step 1: Make a list of all the graph's edges. This is simple if an adjacency list represents
the graph.

Step 2: "V - 1" is used to calculate the number of iterations. Because the shortest distance
to an edge can be adjusted V - 1 time at most, the number of iterations will increase the
same number of vertices.

Step 3: Begin with an arbitrary vertex and a minimum distance of zero. Because you are
exaggerating the actual distances, all other nodes should be assigned infinity.

For each edge u-v, relax the path lengths for the vertices:

If distance[v] is greater than distance[u] + edge weight uv, then

distance[v] = distance[u] + edge weight uv

Step 4: If the new distance is less than the previous one, update the distance for each
Edge in each iteration. The distance to each node is the total distance from the starting
node to this specific node.

Step 5: To ensure that all possible paths are considered, you must consider alliterations.
You will end up with the shortest distance if you do this.

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c Explain working of world wide web. 4M

Ans 2M
Diagram
and
2M
Working

The World Wide Web (WWW), also known as the Web, is an interconnected network of
web pages and documents accessible through the Internet. Tim Berners-Lee created it in
1989 as a way for researchers to share information through linked documents.

Working of WWW:

1. A web browser is a software application that allows users to access and view web
pages on the Internet.

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2. It acts as an interface between the user and the World Wide Web by displaying
web pages Web browsers communicate with web servers using the HTTP or
HTTPS protocol, which allows users to access websites hosted on remote servers.
3. A web server is a computer program that serves web pages to clients, such as web
browsers, upon request. It is responsible for hosting websites, processing HTTP
requests, and delivering web content to users online.
4. Hyperlinks one of the key features of the Web is hyperlinks, which allow you to
navigate between web pages by clicking on links.
5. Uniform Resource Locators (URLs) Web pages are identified by URLs, which are
unique addresses that point to the location of the web page on the Internet.

d If an address in a block given in CIDR classless notation as 64.32.16.8/27 then find 4M

the following:
i. Number of addresses given in block(N)
ii. The first address
iii. The last address
iv. Find Prefix bit(n)
Ans 1) Number of addresses given in block(N) 1 M for
The CIDR notation is /27 i.e. 27 bits out of 32 bits are network & remaining No. of
5 bits are host address addresses,
1M- first
i.e. address,
1M- Last
2^5=32 address
and
N=32 1M-
Prefix
2) The first address bit(n)
To obtain the first address in block for this we have AND the given
address with the network mask

n (32-n)

network mask=27ones 5 zeros

network mask=255.255.255.224

First address =64.32.16.0

3) The last address

To obtained last address in the block we have to keep the ledt most 27
bits in the given address as it is and set the remaining 5 bits to 1s

Last address = 64.32.16.31

4) Find Prefix bit(n)

The CIDR notation “/27” implies the 27 bits are n/w bits so the prefix
bit(n)=27

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3 Attempt any THREE of the following: 12 M

a Differentiate between distance vector routing and link state routing. 4M
Ans Sr. Any Four
Distance Vector Routing Link State Routing points 1M
No.
Routing tables are updated by each
Complete topology is Distributed to
1 exchanging information with the
every router to update a routing table.
neighbors.
2 It updates full routing table. It updates only link states.
3 It uses Bellman-Ford algorithm It uses Dijkstra algorithm.
Distance Vector routing doesn’t have Link state routing works best
4
any hierarchical structure. for hierarchical routing design.
Higher utilization of CPU and
CPU and memory utilization are
5 memory than distance vector
lower than Link state routing.
routing.
6 Slow convergence. Fast convergence.
Example protocols are RIP Example protocols are OSPF
7
and IGRP. and IS-IS.
8 Count to infinity problem No count to infinity problem
b From below list, explain any two different transition method from IPv4 to IPv6.
i) Dual Stack
4M
ii) Tunneling
iii) Header translation
Ans 1. Dual Stack 2M for
In this kind of strategy, a station has a dual stack of protocols run IPv4 and IPv6 any 2
simultaneously. transition
To determine which version to use when sending a packet to a destination, the source host methods
with
queries the DNS. If the DNS returns an IPv4 address, the source host sends an IPv4 packet.
diagram
If the DNS returns an IPv6 address, the source host sends an IPv6 packet.

Fig. Dual Stack

2. Tunneling
Tunneling is a strategy used when two computers using IPv6 want to communicate with
each other and the packet must pass through a region that uses IPv4.
To pass through this region, the packet must have an IPv4 address. So, the IPv6 packet is
encapsulated in an IPv4 packet when it enters the region.
To make it clear that the IPv4 packet is carrying an IPv6 packet as data the protocol value
is set to 41.

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Fig Tunneling

3. Header Translation
In this case, the header format must be totally changed through header translation. The
header of the IPv6 packet is converted to an Ipv4 header see figure.

Fig. Header Translation

c Explain the working of TELNET. 4M
Ans TELNET: TELNET is an abbreviation for TErminaLNETwork. It is the standard TCP/IP Working
protocol for virtual terminal service. of Telnet
TELNET Working: 2M
• TELNET is a client-server application that allows a user to log on to a remote And
machine, giving the user access to the remote system. diagram
• The user sends the keystrokes to the terminal driver, where the local operating 2M
system accepts the characters but does not interpret them.
• A terminal driver correctly interprets the keystrokes on the local terminal or
terminal emulator.
The characters are sent to the TELNET client, which transforms the characters to a
universal character set called network virtual terminal (NVT) characters and delivers them
to the local TCP/IP protocol stack.
• The commands or text, in NVT form, travel through the Internet and arrive at the
TCP/IP stack at the remote machine.
• Here the characters are delivered to the operating system and passed to the TELNET
server, which changes the characters to the corresponding characters understandable by the
remote computer.
• However, the characters cannot be passed directly to the operating system because
the remote operating system is not designed to receive characters from a TELNET server:
It is designed to receive characters from a terminal driver.
• A piece of software called a pseudo terminal driver is added which pretends that
the characters are coming from a terminal.

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• The operating system then passes the characters to the appropriate application
program.

Fig. Working of TELNET

d The dump of a UDP header in hexadecimal format is as follows: 4M
BC82D00D002B001D
Obtain the following:
(i) Source port number
(ii) Destination port number
(iii) Total length
(iv) Packet direction
Ans Considering hexadecimal format as: 1 M for
BC82D00D002B001D Each
The UDP header has four parts, each of two bytes. correct
answer
That means we get the following interpretation of the header.
i) Source port number = BC8216 = 48258
ii) Destination port number = D00D16= 53261
iii) Total length = 002B16 = 43 bytes
iv) Packet direction: The provided dump does not contain information about the packet
direction. The UDP header alone does not specify the direction.

OR
Considering hexadecimal format as:
BC82000D002B001D
The UDP header has four parts, each of two bytes.
That means we get the following interpretation of the header.
i. Source port number = BC8216 = 48258
ii. Destination port number = 000D16= 13
iii. Total length = 002B16 = 43 bytes
iv. Packet direction: Since the destination port number is 13 (well-known port), the
packet is from the client to the server.

4 Attempt any THREE of the following: 12 M

Construct a suitable diagram for each below commands of FTP to show its use 4M
a i) get
ii) mget
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iii) put
iv) mput

Ans Command Purpose Syntax Each

command
get Download a single get<filename> and its
file use with
diagram
mget Download multiple mget<filename1 filename2 filename3> 1M
files

put Upload a single file put<filename>

mput Upload multiple files mput< filename1 filename2 filename3 >

i) get

get (Single File Download))

Local System

Local System Remote FTP Server

ii) mget

mget (Download Multiple files)

Local System Remote FTP Server

ii) put

put (upload single file)

Local System Remote FTP Server

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iv)mput

MPUT (Upload multiple Files)

Local System Remote FTP Server

b Describe RIP message format in detail. 4M

Ans RIP (Routing Information Protocol) message format 2M
•RIP is routing protocol based on Distance Vector Routing algorithm which is an intra Descripti
domain (interior) routing protocol used inside an autonomous system. on
•The metric used by RIP is the distance which is defined as the number of links (networks) &
2M
that have to be used to reach the destination. For this reason, the metric in RIP is called a
Diagram
hop count.
•Infinity is defined as 16, which means that any route in an autonomous system using RIP
cannot have more than 15 hops.
The next node column defines the address of the router to which the packet is to be sent to
reach its destination.

• Command: 8-bit
The type of message: request (1) or response (2)
• Version: 8-bit
Define the RIP version
• All 0s
This field is not actually used by RFC 1058 RIP; it was added solely to provide
backward compatibility with pre-standard varieties of RIP. Its name comes from its
defaulted value, zero.
• Family:
16-bit field defines the family of the protocol used. For TCP/IP, value is 2
• IP Address Network Address:
14 bytes n Defines the address of the destination network and
14 bytes for this field to be applicable to any protocol. However, IP currently uses only 4
bytes, the rest are all 0s.

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• Distance:
32-bit field defines the hop count from the advertising router to the destination
network.
c Describe the header fields in message format of e-mail system. 4M
Ans Electronic Mail (e-mail) is one of the most widely used services of the Internet. This service 2M
allows an Internet user to send a message in a formatted manner (mail) to other Internet Descripti
users in any part of the world. Message in the mail not only contain text, but it also contains on
&
images, audio and videos data. The person who is sending mail is called sender and person
2M for
who receives mail is called the recipient. Format of E-mail: Message
An e-mail consists of three parts that are as follows: format
1. Envelope diagram
2. Header
3. Body

Header:
The header consists of a series of lines. Each header field consists of a single line of ASCII
text specifying field name, colon and value. The main header fields related to message
transport are:
1. To: It specifies the DNS address of the primary recipient(s).
2. Cc: It refers to carbon copy. It specifies address of secondary recipient(s).
3. BCC: It refers to blind carbon copy. It is very similar to Cc. The only difference between
Cc and Bcc is that it allows user to send copy to the third party without primary and
secondary recipient knowing about this.
4. From: It specifies name of person who wrote message.
5. Sender: It specifies e-mail address of person who has sent message.
6. Received: It refers to identity of sender’s, data and also time message was received. It
also contains the information which is used to find bugs in routing system.
7. Return-Path: It is added by the message transfer agent. This part is used to specify how
to get back to the sender.

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d Compare TCP with UDD on any four points. 4M
Ans Any
Note: Consider the term UDD as a UDP.
Correct
Characteristics TCP UDP four
TCP is connection UDP is connection less points 1M
Connection
oriented Protocol Protocol each
It provides reliable It provides unreliable delivery
Reliability
Delivery of messages of messages

TCP makes checks UDP does error checking but no

Error Handling
For errors and reporting reporting.

Flow controlling TCP has flow control UDP has no flow control
No guarantee of the data
TCP gives guarantee
transmission order
that the order of the
Data transmission
data at the receiving
order
end is the same as the
sending end
Header Size 20 bytes 8 bytes
TCP
UDP has no acknowledgment
Acknowledgment Acknowledges the data
Section
reception
Used where reliability is Used where time Sensitivity is
Use
important more important.
Stream-based: No Message based data: Data sent in
Data Interface to
particular structure for discrete
application
data packages by application
Overhead Low Very low
Speed High Very high
FTP, Telnet, SMTP, DNS, BOOTP, DHCP, TFTP,
Application
DNS, HTTP, POP RIP

e Compare POP3 with IMAD on below Points.

i) TCP Port used
ii) E-mail Stored at 4M
iii) Time required to connect
iv) Multiple mail boxes.
Ans Note: Consider the term IMAD as an IMAP
Points POP3 IMAP Each
correct
TCP Uses port 110 (unencrypted) or Uses port 143 (unencrypted) or port Point 1 M
Port port 995 (encrypted/SSL) 993 (encrypted/SSL)
used Note:
Consider
E-mail Emails are typically downloaded Emails are stored on the mail server.
the term
Stored at from the server to the client The client accesses and manages
IMAD as
device. The emails are then stored emails directly on the server, an IMAP
locally on the device, and the allowing for synchronization across
server copy is usually deleted. multiple devices.

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Time Generally faster to connect May take more time to connect as it
required because it involves downloading involves syncing with the server and
to emails to the client device. fetching email headers. However,
connect Connection time is minimal since subsequent access to emails is faster
it retrieves emails and disconnects as only headers are initially
from the server. downloaded.

Multiple Usually does not support Supports multiple mailboxes and

mail multiple mailboxes. Emails are folders Changes made on one device
boxes. typically downloaded to a single (e.g., marking an email as read) are
device, and managing emails on reflected on all devices since the
multiple devices can be emails are stored centrally on the
challenging. server.

5 Attempt any TWO of the following: 12 M

a Explain how TCP connections are established using 3 way handshake. 6M
Ans TCP Connection: 2M for
When you establish a new TCP connection (3-way handshake) then the initial sequence diagram,
number is a random 32-bit value. The receiver will use this sequence number and sends 1 M for
back an acknowledgment. Protocol analyzers like wireshark will often use a relative TCP
sequence number of 0 since it’s easier to read than some high random number. Connectio
n
and
TCP uses a three-way handshake to establish a reliable connection. The connection is full 3M for
duplex, and both sides synchronize (SYN) and acknowledge (ACK) each other. The 3-way
exchange of these four flags is performed in three steps: SYN, SYN-ACK, ACK, as shown handshak
in figure below e
explanati
on

The client chooses an initial sequence number, set in the first SYN packet. The server also
chooses its own initial sequence number, set in the SYN/ACK packet. Each side
acknowledges each other’s sequence number by incrementing it: this is the
acknowledgement number. The use of sequence and acknowledgement numbers allows
both sides to detect missing or out-of-order segments.
Once a connection is established, ACKs typically follow for each segment. The connection
will eventually end with a RST (reset or tear down the connection) or FIN (gracefully end
the connection).
Three-Way Handshake:
The algorithm used by TCP to establish and terminate a connection is called a three-way
handshake. We first describe the basic algorithm and then show how it is used by TCP.

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The three-way handshake involves the exchange of three messages between the client and
the server, as we see in below figure

b Demonstrate with suitable example of call collision in TCP connection. 6M

Ans The connection is established in TCP using the three-way handshake as discussed earlier 1 M for
to create a connection. One side, say the server, passively stays for an incoming link by diagram,
implementing the LISTEN and ACCEPT primitives, either determining a particular other 2 M for
TCP
side or nobody in particular. The other side performs a connect primitive specifying the I/O
connectio
port to which it wants to join. The maximum TCP segment size available, other options are n
optionally like some private data (example password). The CONNECT primitive transmits And
a TCP segment with the SYN bit on and the ACK bit off and waits for a response. 3M for
The sequence of TCP segments sent in the typical case, as shown in the figure below – call
collision

When the segment sent by Host-1 reaches the destination, i.e., host -2, the receiving server
checks to see if there is a process that has done a LISTEN on the port given in the
destination port field. If not, it sends a response with the RST bit on to refuse the
connection. Otherwise, it governs the TCP segment to the listing process, which can accept
or decline (for example, if it does not look similar to the client) the connection.
Call Collision: If two hosts try to establish a connection simultaneously between the same
two sockets, then the events sequence is demonstrated in the figure under such
circumstances. Only one connection is established. It cannot select both the links because
their endpoints identify connections.

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Suppose the first set up results in a connection identified by (x, y) and the second
connection are also released up. In that case, only tail enter will be made, i.e., for (x, y) for
the initial sequence number, a clock-based scheme is used, with a clock pulse coming after
every 4 microseconds. For ensuring additional safety when a host crashes, it may not reboot
for sec, which is the maximum packet lifetime. This is to make sure that no packets from
previous connections are roaming around.
c Explain following address types of IPv6: 6M
a) Unicast address
b) Multicast address
c) Anycast address

Ans The three types of IPv6 addresses are: unicast, anycast, and multicast. 2 M for
a) Unicast address: This type of information transfer is useful when there is a each
participation of single sender and single recipient. So, in short you can term it as a address
types
one-to-one transmission. For example, a device having IP address 10.1.2.0 in a
network wants to send the traffic stream (data packets) to the device with IP address
20.12.4.2 in the other network, then unicast comes into picture. This is the most
common form of data transfer over the networks.

b) Multicast address: In multicasting, one/more senders and one/more recipients

participate in data transfer traffic. In this method traffic recline between the
boundaries of unicast (one-to-one) and broadcast (one-to-all). Multicast lets server’s
direct single copies of data streams that are then simulated and routed to hosts that
request it. IP multicast requires support of some other protocols like IGMP (Internet
Group Management Protocol), Multicast routing for its working. Also, in Classful IP
addressing Class D is reserved for multicast groups.

c) Anycast address: An IPv6 anycast address is an address that is assigned to more than
one interface (typically belonging to different nodes), where a packet sent to an
anycast address is routed to the nearest interface having that address, according to the
routing protocol's measure of distance. Anycast addresses, when used as part of a
route sequence, permit a node to select which of several Internet service providers it
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 MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
wants to carry its traffic. This capability is sometimes called source selected policies.
You implement this by configuring anycast addresses to identify the set of routers
belonging to Internet service providers (for example, one anycast address per Internet
service provider). You can use these anycast addresses as intermediate addresses in
an IPv6 routing header, to cause a packet to be delivered by a particular provider or
sequence of providers. You can also use anycast addresses to identify the set of routers
attached to a particular subnet or the set of routers providing entry into a particular
routing domain. You can locate anycast addresses from the unicast address space by
using any of the defined unicast address formats. Thus, anycast addresses are
syntactically indistinguishable from unicast addresses. When you assign a unicast
address to more than one interface, that is, turning it into an anycast address, you must
explicitly configure the nodes to which the address is assigned in order to know that
it is an anycast address.
6 Attempt any TWO of the following: 12 M
Explain Distance vector routing and open shortest path first routing protocol in
a 6M
detail.
Ans 1. Distance Vector Routing Protocol: A distance-vector routing protocol in data networks 3 M for
determines the best route for data packets based on distance. each
Distance-vector routing protocols measure the distance by the number of routers a packet Protocol
has to pass, one router counts as one hop.
Some distance-vector protocols also take into account network latency and other factors
that influence traffic on a given route.
To determine the best route across a network, routers, on which a distance-vector protocol
is implemented, exchange information with one another, usually routing tables plus hop
counts for destination networks and possibly other traffic information.
Distance-vector routing protocols also require that a router informs its neighbours of
network topology changes periodically.
Distance Vector Algorithm –
• A router transmits its distance vector to each of its neighbours in a routing packet.
• Each router receives and saves the most recently received distance vector from
each of its neighbours.
• A router recalculates its distance vector when:
• It receives a distance vector from a neighbour containing different information than
before.
• It discovers that a link to a neighbour has gone down.
2. Open Shortest Path First (OSPF) Protocol:
The OSPF (Open Shortest Path First) protocol is one of a family of IP Routing protocols,
and is an Interior Gateway Protocol (IGP) for the Internet, used to distribute IP routing
information throughout a single Autonomous System (AS) in an IP network.
The OSPF protocol is a link-state routing protocol, which means that the routers exchange
topology information with their nearest neighbours. The topology information is flooded
throughout the AS, so that every router within the AS has a complete picture of the topology
of the AS. This picture is then used to calculate end-to-end paths through the AS, normally
using a variant of the Dijkstra algorithm. Therefore, in a link-state routing protocol, the
next hop address to which data is forwarded is determined by choosing the best end-to-end
path to the eventual destination.

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 MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
Each OSPF router distributes information about its local state (usable interfaces and
reachable neighbors, and the cost of using each interface) to other routers using a Link State
Advertisement (LSA) message. Each router uses the received messages to build up an
identical database that describes the topology of the AS.
From this database, each router calculates its own routing table using a Shortest Path First
(SPF) or Dijkstra algorithm. This routing table contains all the destinations the routing
protocol knows about, associated with a next hop IP address and outgoing interface.
The protocol recalculates routes when network topology changes, using the Dijkstra
algorithm, and minimizes the routing protocol traffic that it generates.
It provides support for multiple paths of equal cost.
It provides a multi-level hierarchy (two-level for OSPF) called "area routing," so that
information about the topology within a defined area of the AS is hidden from routers
outside this area. This enables an additional level of routing protection and a reduction in
routing protocol traffic.
All protocol exchanges can be authenticated so that only trusted routers can join in the
routing exchanges for the AS.
b For the IP address given below: 6M
1) Identify the classes to which IP address belongs to
2) Identify Network address section
3) Identify Host address section
4) Calculate number of hosts can be assigned with each network
i)122.34.45.133
ii)12.12.12.12
iii)192.10.233.26
Ans 1.Identify the classes to which IP address belongs to ½ M for
i)122.34.45.133 = Class A identifyin
ii)12.12.12.12 = Class A g each
correct
iii)192.10.233.26 = Class C
class,

2. Identify Network address section ½ M for

i)122.34.45.133 = 122.0.0.0 identifyin
ii)12.12.12.12 = 12.0.0.0 g each
correct
iii)192.10.233.26 = 192.10.233.0
network
address
3. Identify Host address section section,
i)122.34.45.133 = 0.34.45.133
ii)12.12.12.12 = 0.12.12.12 ½ M for
identifyin
iii)192.10.233.26 = 0.0.0.26
g each
correct
4. Calculate number of hosts can be assigned with each network host
i)122.34.45.133 = 232-28=224 address
ii)12.12.12.12 = 232-28=224 section
iii)192.10.233.26 = 232-224=28
and
½ M for
calculatin
g number
of hosts

Page No: 19/20

 MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
c Describe DHCP operations, when DHCP client and server on same network. 6M
Ans DHCP is based on a client-server model and based on discovery, offer, request, and 2 M for
ACK.DHCP client and server can either be on the same network or on different networks. diagram
DHCP (Dynamic Host Configuration Protocol) is a network management protocol used to and
4 M for
dynamically assign an IP address to any device, or node, on a network so it can
Explanati
communicate using IP. DHCP automates and centrally manages these configurations rather on
than requiring network administrators to manually assign IP addresses to all network
devices. DHCP can be implemented on small local networks, as well as large enterprise
networks. DHCP assigns new IP addresses in each location when devices are moved from
place to place, which means network administrators do not have to manually configure
each device with a valid IP address or reconfigure the device with a new IP address if it
moves to a new location on the network.

In this case, the operation can be described as follows:

1. The DHCP server issues a passive open command on UDP port number 67 and
waits for a client.
2. A booted client issues an active open command on port number 68. The message is
encapsulated in a UDP user datagram, using the destination port number 67 and the source
port number 68.
3. The server responds with either a broadcast or a unicast message using UDP source
port number 67 and destination port number 68.

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