Networking Basics Questions and Answers

Q: What does the term 'bandwidth' refer to in networking ?

Bandwidth is the amount of data a network can send in a given amount of time.In networking, bandwidth is the maximum data transfer rate of a link, measured in bits per second (for example 100 Mbps). It is a capacity, not the number of devices or the cable length.The number of connected devices, the length of a cable and the number of connections are different properties of a network.

Q: In the OSI model, which layer is responsible for end-to-end error recovery and flow control ?

The layer is the transport layer (layer 4).The transport layer runs between the two end hosts. Protocols such as TCP number segments, acknowledge them, retransmit lost or damaged ones (end-to-end error recovery) and use a receive window to keep the sender from overwhelming the receiver (flow control).The data link layer also does error and flow control, but only hop by hop over a single link, not end to end. The network layer handles addressing and routing. The session layer manages dialogues and synchronisation between applications.

Q: What is the window size in the sliding window protocol if the sequence number field is 3 bits ?

The window size is 7.A 3-bit sequence number field gives \(2^{3} = 8\) numbers, 0 to 7. In a go-back-N sliding window protocol the sender's window can be at most one less than that:\[ W_{max} = 2^{m} - 1 = 2^{3} - 1 = 7 \]A window of 8 would let the receiver confuse a new frame 0 with a retransmitted old frame 0. 3 is the field width, not a window. 15 would need 4 bits. Selective repeat would allow only \(2^{m-1} = 4\), which is not offered, so the key follows the go-back-N figure.

Q: In Ethernet, what does CSMA/CD stand for ?

CSMA/CD stands for Carrier Sense Multiple Access with Collision Detection.It is the access method of classic shared Ethernet (IEEE 802.3). A station listens for a carrier before sending (carrier sense). Many stations share the medium (multiple access). If two transmit at once, they detect the collision, send a jam signal and retry after a random binary exponential backoff (collision detection).The other expansions mix in words such as circuit, switch, amplitude, domain and delay that have nothing to do with the protocol.

Q: The Protocol Data Unit (PDU) for the application layer in the Internet stack is

The application layer PDU is the Message.Each layer of the Internet stack has its own name for its data unit: message at the application layer, segment at the transport layer (TCP), datagram (or packet) at the network layer (IP), and frame at the link layer.So segment, datagram and frame are the PDUs of the lower layers.Encapsulation adds a header at each layer on the way down: the application's message becomes the payload of a TCP segment, which becomes the payload of an IP datagram, which is carried inside a link-layer frame.

Q: Consider a source computer (S) transmitting a file of size 106 bits to a destination computer (D) over a network of two routers (R1 and R2) and three links (L1, L2, and L3). L1 connects S to R1 ; L2 connects R1 to R2; and L3 connects R2 to D. Let each link be of length 100 km. Assume signals travel over each link at a speed of 10^8 meters per second. Assume that the link bandwidth on each link is 1Mbps. Let the file be broken down into 1000 packets, each of size 1000 bits. Find the total sum of transmission and propagation delays in transmitting the file from S to D.

The total delay is 1005 ms.The file is \(10^{6}\) bits (the booklet prints 106, but 1000 packets of 1000 bits make \(10^{6}\)). Each packet takes 1 ms to transmit at 1 Mbps, and each 100 km link adds \(10^{5} / 10^{8}\) s = 1 ms of propagation. Packets are pipelined: the source sends for 1000 ms, and the last packet then needs two more transmissions at the routers.\[ \begin{aligned} \text{transmission} &= 1000 + 2 = 1002\ \text{ms} \\ \text{propagation} &= 3 \times 1 = 3\ \text{ms} \end{aligned} \]So the total is 1002 + 3 = 1005 ms. 1010 ms overcounts the router delays, and 3000 ms and 3003 ms would assume each router waits for the whole file before forwarding, which is not how packet switching works.

Q: A bit-stuffing based framing protocol uses an 8-bit delimiter pattern of 01111110. If the output bit-string after stuffing is 01111100101, then the input bit-string is

The input bit-string is 0111110101.With the flag 01111110, the sender inserts a 0 after every run of five consecutive 1s in the data, so the flag can never appear inside it. The receiver removes a 0 that follows five 1s.In 01111100101 the bits 2 to 6 are five 1s and the next bit is a stuffed 0. Removing it: 011111 + 0101 = 0111110101. The other options either keep the stuffed bit or drop the wrong one.

Q: Which of the following does not represent a valid physical or logical network topology used in computer networks ?

The one that is not a network topology is Solar.A topology is the way the nodes of a network are arranged and connected. In a star, every node connects to a central hub or switch. In a ring, each node connects to two neighbours and data passes round the loop. In a bus, all nodes share one backbone cable. Mesh, tree and hybrid are the other standard topologies.There is no 'solar' topology; the word is a distractor that sounds like a technical term.

Q: Which device connects two different networks ?

The device that connects two different networks is a router.A router works at the network layer. It reads the destination IP address of each packet and forwards it from one network to another along the best path, which is how a home or office LAN is joined to the internet.A switch connects devices within one network using MAC addresses. A hub simply repeats every signal to all its ports, and a repeater regenerates a signal to extend the reach of a single network; neither joins separate networks.

Q: Which switching technique guarantees constant bandwidth and fixed delay but is inefficient for bursty data ?

The technique is circuit switching.Circuit switching sets up a dedicated path between the two ends before any data is sent, as in the traditional telephone network. The reserved path gives a constant bandwidth and a fixed delay, but it stays reserved even when nothing is being sent, so it wastes capacity on bursty data traffic.Packet switching, frame relay and message switching all share links on demand, so they use capacity efficiently for bursty traffic but cannot guarantee a fixed delay.

51.

What does the term 'bandwidth' refer to in networking ?

A.The amount of data a network can send in a given amount of time.
B.The maximum number of devices connected to a network.
C.The physical length of a network cable.
D.The number of connections between devices in a network.

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: A

Bandwidth is the amount of data a network can send in a given amount of time.

In networking, bandwidth is the maximum data transfer rate of a link, measured in bits per second (for example 100 Mbps). It is a capacity, not the number of devices or the cable length.

The number of connected devices, the length of a cable and the number of connections are different properties of a network.

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52.

In the OSI model, which layer is responsible for end-to-end error recovery and flow control ?

A.Network layer
B.Transport layer
C.Session layer
D.Data link layer

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: B

The layer is the transport layer (layer 4).

The transport layer runs between the two end hosts. Protocols such as TCP number segments, acknowledge them, retransmit lost or damaged ones (end-to-end error recovery) and use a receive window to keep the sender from overwhelming the receiver (flow control).

The data link layer also does error and flow control, but only hop by hop over a single link, not end to end. The network layer handles addressing and routing. The session layer manages dialogues and synchronisation between applications.

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53.

What is the window size in the sliding window protocol if the sequence number field is 3 bits ?

A.3
B.7
C.8
D.15

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: B

The window size is 7.

A 3-bit sequence number field gives \(2^{3} = 8\) numbers, 0 to 7. In a go-back-N sliding window protocol the sender's window can be at most one less than that:

\[ W_{max} = 2^{m} - 1 = 2^{3} - 1 = 7 \]

A window of 8 would let the receiver confuse a new frame 0 with a retransmitted old frame 0. 3 is the field width, not a window. 15 would need 4 bits. Selective repeat would allow only \(2^{m-1} = 4\), which is not offered, so the key follows the go-back-N figure.

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54.

In Ethernet, what does CSMA/CD stand for ?

A.Carrier Sense Multiple Access with Collision Detection
B.Circuit Switch Multiple Access with Collision Domain
C.Carrier Signal Multiple Amplitude with Collision Delay
D.Circuit Sense Multiple Amplitude with Collision Detection

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: A

CSMA/CD stands for Carrier Sense Multiple Access with Collision Detection.

It is the access method of classic shared Ethernet (IEEE 802.3). A station listens for a carrier before sending (carrier sense). Many stations share the medium (multiple access). If two transmit at once, they detect the collision, send a jam signal and retry after a random binary exponential backoff (collision detection).

The other expansions mix in words such as circuit, switch, amplitude, domain and delay that have nothing to do with the protocol.

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55.

The Protocol Data Unit (PDU) for the application layer in the Internet stack is

A.Segment
B.Datagram
C.Message
D.Frame

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: C

The application layer PDU is the Message.

Each layer of the Internet stack has its own name for its data unit: message at the application layer, segment at the transport layer (TCP), datagram (or packet) at the network layer (IP), and frame at the link layer.

So segment, datagram and frame are the PDUs of the lower layers.

Encapsulation adds a header at each layer on the way down: the application's message becomes the payload of a TCP segment, which becomes the payload of an IP datagram, which is carried inside a link-layer frame.

Discuss in Forum

56.

Consider a source computer (S) transmitting a file of size 106 bits to a destination computer (D) over a network of two routers (R1 and R2) and three links (L1, L2, and L3). L1 connects S to R1 ; L2 connects R1 to R2; and L3 connects R2 to D. Let each link be of length 100 km. Assume signals travel over each link at a speed of 10^8 meters per second. Assume that the link bandwidth on each link is 1Mbps. Let the file be broken down into 1000 packets, each of size 1000 bits. Find the total sum of transmission and propagation delays in transmitting the file from S to D.

A line from S through link L1 (100 km) to router R1, link L2 (100 km) to router R2, link L3 (100 km) to D

A.1005 ms
B.1010 ms
C.3000 ms
D.3003 ms

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: A

The total delay is 1005 ms.

The file is \(10^{6}\) bits (the booklet prints 106, but 1000 packets of 1000 bits make \(10^{6}\)). Each packet takes 1 ms to transmit at 1 Mbps, and each 100 km link adds \(10^{5} / 10^{8}\) s = 1 ms of propagation. Packets are pipelined: the source sends for 1000 ms, and the last packet then needs two more transmissions at the routers.

\[ \begin{aligned} \text{transmission} &= 1000 + 2 = 1002\ \text{ms} \\ \text{propagation} &= 3 \times 1 = 3\ \text{ms} \end{aligned} \]

So the total is 1002 + 3 = 1005 ms. 1010 ms overcounts the router delays, and 3000 ms and 3003 ms would assume each router waits for the whole file before forwarding, which is not how packet switching works.

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57.

A bit-stuffing based framing protocol uses an 8-bit delimiter pattern of 01111110. If the output bit-string after stuffing is 01111100101, then the input bit-string is

A.0111110100
B.0111110101
C.0111111101
D.0111111111

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: B

The input bit-string is 0111110101.

With the flag 01111110, the sender inserts a 0 after every run of five consecutive 1s in the data, so the flag can never appear inside it. The receiver removes a 0 that follows five 1s.

In 01111100101 the bits 2 to 6 are five 1s and the next bit is a stuffed 0. Removing it: 011111 + 0101 = 0111110101. The other options either keep the stuffed bit or drop the wrong one.

Discuss in Forum

58.

Which of the following does not represent a valid physical or logical network topology used in computer networks ?

A.Star
B.Ring
C.Bus
D.Solar

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: D

The one that is not a network topology is Solar.

A topology is the way the nodes of a network are arranged and connected. In a star, every node connects to a central hub or switch. In a ring, each node connects to two neighbours and data passes round the loop. In a bus, all nodes share one backbone cable. Mesh, tree and hybrid are the other standard topologies.

There is no 'solar' topology; the word is a distractor that sounds like a technical term.

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59.

Which device connects two different networks ?

A.Router
B.Switch
C.Hub
D.Repeater.

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: A

The device that connects two different networks is a router.

A router works at the network layer. It reads the destination IP address of each packet and forwards it from one network to another along the best path, which is how a home or office LAN is joined to the internet.

A switch connects devices within one network using MAC addresses. A hub simply repeats every signal to all its ports, and a repeater regenerates a signal to extend the reach of a single network; neither joins separate networks.

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60.

Which switching technique guarantees constant bandwidth and fixed delay but is inefficient for bursty data ?

A.Packet Switching
B.Circuit Switching
C.Frame Relay
D.Message Switching

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: B

The technique is circuit switching.

Circuit switching sets up a dedicated path between the two ends before any data is sent, as in the traditional telephone network. The reserved path gives a constant bandwidth and a fixed delay, but it stays reserved even when nothing is being sent, so it wastes capacity on bursty data traffic.

Packet switching, frame relay and message switching all share links on demand, so they use capacity efficiently for bursty traffic but cannot guarantee a fixed delay.

Discuss in Forum

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