PROBLEM SET-1

1. Consequences of Connection Failures in Different Topologies

A. Mesh Topology (Five Devices)

 Failure Impact: Highly resilient; failure of one connection doesn't affect the overall network. Each device
has multiple paths to others.

B. Star Topology (Five Devices, Not Counting the Hub)

 Failure Impact: If a device fails, only that device is affected. However, if the central hub fails, the entire
network becomes inoperable.

C. Bus Topology (Five Devices)

 Failure Impact: A break in the main cable (bus) can disrupt communication for all devices. However,
individual device failures typically don't affect the network.

D. Ring Topology (Five Devices)

 Failure Impact: A failure in any single device or connection can disrupt the entire network, as data cannot
complete its loop.

2. Star Topology Explained

Definition: In a star topology, each device is connected to a central hub or switch.

Advantages:

 Easy to install and manage.

 Failure of one device doesn't affect others.
 Simplifies fault isolation.
 Scalable; adding devices is straightforward.

Disadvantages:

 Central hub failure leads to network-wide disruption.

 Requires more cabling than some other topologies.
 Performance depends on the capacity of the central hub.

3. Repeater

Definition: A repeater is an electronic device that receives a signal and retransmits it to extend the transmission
distance. Purpose:

 Amplifies or regenerates signals to overcome distance limitations.

  Used in various communication systems, including optical and radio frequencies.

4. Hub

Definition: A hub is a network device that connects multiple Ethernet devices, making them act as a single network
segment.

Purpose:

 Broadcasts incoming data packets to all ports.

 Operates at the physical layer of the OSI model.
 Simple and cost-effective, but less efficient than switches.

5. Data Communication Modes

Simplex:

 Definition: Data flows in one direction only.

 Advantages: Simple implementation, no need for complex protocols.
 Disadvantages: No feedback or error correction.
 Examples: Broadcast TV, radio transmissions.

Half-Duplex:

 Definition: Data flows in both directions, but not simultaneously.

 Advantages: Allows two-way communication with less complexity.
 Disadvantages: Cannot send and receive data at the same time.
 Examples: Walkie-talkies, CB radios.

Full-Duplex:

 Definition: Data flows in both directions simultaneously.

 Advantages: Efficient communication, reduces wait times.
 Disadvantages: Requires more complex hardware and protocols.
 Examples: Telephone calls, video conferencing.

6. OSI Model Layers and Functions

1. Physical Layer: Transmits raw bitstream over the physical medium.

2. Data Link Layer: Provides node-to-node data transfer and error correction.
3. Network Layer: Determines the best path for data transmission.
4. Transport Layer: Ensures reliable data transfer with error correction.
5. Session Layer: Manages sessions between applications.
6. Presentation Layer: Translates data formats and encryption.
7. Application Layer: Provides network services to end-user applications.
7. TCP/IP Model Layers and Functions

1. Link Layer: Handles physical addressing and data transfer.

2. Internet Layer: Manages logical addressing and routing (e.g., IP).
3. Transport Layer: Ensures reliable data transfer (e.g., TCP, UDP).
4. Application Layer: Provides network services to applications (e.g., HTTP, FTP).

8. Comparison: TCP/IP vs. OSI Model

Feature OSI Model TCP/IP Model

Layers 7 4
Developed By ISO DARPA
Protocol Examples HTTP, FTP, TCP, IP HTTP, FTP, TCP, UDP, IP
Layered Approach Strictly layered More flexible, less strict
Usage Theoretical model Practical implementation

9. Data Transmission Time Calculations

A. Sending 10 bits at 1000 bps:

 Time = 10 bits ÷ 1000 bps = 0.01 seconds.

B. Sending 1 byte (8 bits) at 1000 bps:

 Time = 8 bits ÷ 1000 bps = 0.008 seconds.

C. Sending 100,000 characters (800,000 bits) at 1000 bps:

 Time = 800,000 bits ÷ 1000 bps = 800 seconds = 13 minutes and 20 seconds.

10. If the bandwidth of the channel is 5 Kbps, how long does it take to send a
frame of 100,000 bits?

Given:

 Bandwidth = 5 Kbps = 5000 bits per second

 Frame size = 100,000 bits

Calculation:

Time=Frame SizeBandwidth=100,000 bits5000 bps=20 seconds\text{Time} = \frac{\text{Frame

Size}}{\text{Bandwidth}} = \frac{100,000 \text{ bits}}{5000 \text{ bps}} = 20 \
text{ seconds}Time=BandwidthFrame Size=5000 bps100,000 bits=20 seconds
Answer:
20 seconds

11. A file contains 28 million bytes. How long does it take to download this file
using a 56 Kbps channel?

Given:

 File size = 28 million bytes = 28 × 10⁶ bytes = 28 × 8 × 10⁶ bits = 224 × 10⁶ bits
 Bandwidth = 56 Kbps = 56,000 bps

Calculation:

Time=224×10656,000=4000 seconds\text{Time} = \frac{224 \times 10^6}{56,000} = 4000 \

text{ seconds}Time=56,000224×106=4000 seconds

Convert to minutes:

4000÷60=66.67 minutes4000 \div 60 = 66.67 \text{ minutes}4000÷60=66.67 minutes

Answer:
4000 seconds or approximately 66.67 minutes

12. A file contains 56 million bytes. How long does it take to download this file
using a 280 Mbps channel?

Given:

 File size = 56 million bytes = 56 × 8 = 448 million bits

 Bandwidth = 280 Mbps = 280 × 10⁶ bps

Calculation:

Time=448×106280×106=1.6 seconds\text{Time} = \frac{448 \times 10^6}{280 \times 10^6} =

1.6 \text{ seconds}Time=280×106448×106=1.6 seconds

Answer:
1.6 seconds
13. A file contains 2 million bytes. How long does it take to download this file
using:

 a) 56 Kbps channel
 b) 1 Mbps channel

File size = 2 million bytes = 2 × 8 = 16 million bits

a) 56 Kbps = 56,000 bps

Time=16×10656,000=285.71 seconds\text{Time} = \frac{16 \times 10^6}{56,000} = 285.71 \

text{ seconds}Time=56,00016×106=285.71 seconds

b) 1 Mbps = 1,000,000 bps

Time=16×1061,000,000=16 seconds\text{Time} = \frac{16 \times 10^6}{1,000,000} = 16 \text{

seconds}Time=1,000,00016×106=16 seconds

Answer:

 56 Kbps: 285.71 seconds (~4.76 minutes)

 1 Mbps: 16 seconds

14. A light signal is travelling through a fiber. What is the delay if the fiber-optic
cable is 10 m, 100 m, and 1 Km?

Given:

 Propagation speed = 2×1082 \times 10^82×108 m/s

Calculations:

 10 m:

Delay=102×108=5×10−8 seconds=50 nanoseconds\text{Delay} = \frac{10}{2 \times 10^8} = 5 \

times 10^{-8} \text{ seconds} = 50 \text{ nanoseconds}Delay=2×10810
=5×10−8 seconds=50 nanoseconds

 100 m:

Delay=1002×108=5×10−7 seconds=500 nanoseconds\text{Delay} = \frac{100}{2 \times 10^8}

= 5 \times 10^{-7} \text{ seconds} = 500 \text{ nanoseconds}Delay=2×108100
=5×10−7 seconds=500 nanoseconds
  1 km = 1000 m:

Delay=10002×108=5×10−6 seconds=5 microseconds\text{Delay} = \frac{1000}{2 \times 10^8}

= 5 \times 10^{-6} \text{ seconds} = 5 \text{ microseconds}Delay=2×1081000
=5×10−6 seconds=5 microseconds

Answer:

 10 m: 50 ns
 100 m: 500 ns
 1 km: 5 µs

15. What is topology? Discuss various topologies with their advantages and
disadvantages.

Definition:
Network topology refers to the layout or organizational hierarchy of interconnected
nodes/devices in a computer network.

Types of Network Topologies:

1. Bus Topology:
o Advantages: Easy to install, requires less cable.
o Disadvantages: Difficult to troubleshoot; failure of main cable disrupts entire
network.
2. Star Topology:
o Advantages: Easy to manage, failure of one device doesn't affect others.
o Disadvantages: If the central hub fails, the entire network fails.
3. Ring Topology:
o Advantages: Data flows in one direction, reducing packet collisions.
o Disadvantages: One failure can bring down the whole network.
4. Mesh Topology:
o Advantages: High fault tolerance, reliable communication.
o Disadvantages: Expensive, complex installation.
5. Tree Topology:
o Advantages: Scalable, hierarchical management.
o Disadvantages: Depends heavily on the main bus/hub.
6. Hybrid Topology:
o Advantages: Combines benefits of multiple topologies.
o Disadvantages: Complex and costly to implement.
16. A packet of 10,000 bits is transmitted over a link with a bandwidth of 1
Mbps. What is the transmission delay?

Given:

 Packet size = 10,000 bits

 Bandwidth = 1 Mbps = 1,000,000 bps

Calculation:

Transmission delay=10,0001,000,000=0.01 seconds=10 milliseconds\text{Transmission delay}

= \frac{10,000}{1,000,000} = 0.01 \text{ seconds} = 10 \
text{ milliseconds}Transmission delay=1,000,00010,000=0.01 seconds=10 milliseconds

Answer:
10 milliseconds

17. What is the propagation delay if a signal travels through a 400 km cable at 8
× 10⁸ m/s?

Given:

 Distance = 400 km = 400,000 meters

 Propagation speed = 8×1088 \times 10^88×108 m/s

Calculation:

Propagation Delay=DistancePropagation Speed=400,0008×108=5×10−4 seconds=0.5 millisecon

ds\text{Propagation Delay} = \frac{\text{Distance}}{\text{Propagation Speed}} = \
frac{400,000}{8 \times 10^8} = 5 \times 10^{-4} \text{ seconds} = 0.5 \
text{ milliseconds}Propagation Delay=Propagation SpeedDistance=8×108400,000
=5×10−4 seconds=0.5 milliseconds

Answer:
0.5 milliseconds

18. What is World Wide Web (WWW)? How is it different from the Internet?

World Wide Web (WWW):

 A system of interlinked hypertext documents accessed via the Internet.

 Uses web browsers (e.g., Chrome, Firefox) to access web pages written in HTML.
 Invented by Tim Berners-Lee in 1989.
Difference Between WWW and Internet:

Feature World Wide Web (WWW) Internet

Definition A collection of web pages A global network of networks
Access Accessed via browsers Provides underlying infrastructure
Protocols Uses HTTP/HTTPS Uses TCP/IP, FTP, SMTP, etc.
Nature A service on the Internet The physical and logical network

19. A packet of 20,000 bits is transmitted over a link with a bandwidth of 4

Mbps. What is the transmission delay? Show your calculation.

To calculate the transmission delay, we use the following formula:

Transmission Delay=Packet SizeBandwidth\text{Transmission Delay} = \frac{\text{Packet

Size}}{\text{Bandwidth}}

Where:

 Packet Size = 20,000 bits

 Bandwidth = 4 Mbps (which is 4,000,000 bits per second)

Now, calculate the transmission delay:

Transmission Delay=20,000 bits4,000,000 bps=0.005 seconds\text{Transmission Delay} = \

frac{20,000 \text{ bits}}{4,000,000 \text{ bps}} = 0.005 \text{ seconds}

So, the transmission delay is 0.005 seconds or 5 milliseconds.

20. For n devices in a network, what is the number of cable links required for a
mesh, ring, bus, and star topology?

 Mesh Topology: In a mesh topology, each device is connected to every other device. The
number of links required is:

Number of links=n(n−1)2\text{Number of links} = \frac{n(n-1)}{2}

where n is the number of devices in the network. This is because each device is
connected to every other device, and there are n(n−1)2\frac{n(n-1)}{2} unique pairs.

 Ring Topology: In a ring topology, each device is connected to exactly two other
devices, forming a circular connection. The number of links required is:
 Number of links=n\text{Number of links} = n

where n is the number of devices, as each device needs to be connected to the next one in
a ring.

 Bus Topology: In a bus topology, all devices are connected to a single central bus or
backbone. The number of links required is:

Number of links=1\text{Number of links} = 1

because all devices share the same backbone.

 Star Topology: In a star topology, each device is connected to a central hub or switch.
The number of links required is:

Number of links=n\text{Number of links} = n

where n is the number of devices, as each device connects to the central hub.

21. Performance is inversely related to delay. When you use the Internet, which
of the following applications are more sensitive to delay?

A. Sending an e-mail
B. Copying a file
C. Surfing the Internet

 C. Surfing the Internet is more sensitive to delay because websites and interactive
applications (like video streaming or gaming) require low latency for real-time content
delivery. A high delay can cause a poor user experience with slower page loads, buffering
videos, or lag in online games.
 A. Sending an email and B. Copying a file are less sensitive to delay since these tasks
can tolerate some delay without significantly affecting their functionality.

22. You have two computers connected by an Ethernet hub at home. Is this a
LAN, a MAN, or a WAN? Explain your reason.

 This is a LAN (Local Area Network). A LAN is a network of devices connected within
a limited area, such as a home, office, or campus. Since the two computers are connected
within your home using an Ethernet hub, they are part of a local area network. A MAN
(Metropolitan Area Network) covers a larger area like a city, and a WAN (Wide Area
Network) spans even larger areas, like countries or continents.