1. Which of the following is a flow control technique used in the data link layer?
a) ARQ
b) TCP
c) Sliding Window
d) Error Control
Answer: c) Sliding Window
2. What is the main purpose of flow control in data communication?
a) To correct errors
b) To manage the rate of data transmission between sender and receiver
c) To ensure reliable delivery of data
d) To compress the data
Answer: b) To manage the rate of data transmission between sender and receiver
3. In Stop-and-Wait flow control, how many frames can the sender send before receiving an acknowledgment?
a) 1
b) 2
c) N
d) Infinite
Answer: a) 1
4. Which of the following is NOT a flow control technique?
a) Stop-and-Wait
b) Selective Repeat
c) Error detection
d) Go-Back-N
Answer: c) Error detection
5. In Go-Back-N ARQ, the receiver’s window size is:
a) 1
b) 2
c) N
d) Infinite
Answer: a) 1
6. The sliding window technique helps manage the flow of data by:
a) Ensuring the receiver processes frames in order
b) Allowing the sender to send multiple frames at once
c) Requiring acknowledgments after every frame
d) Sending one frame at a time
Answer: b) Allowing the sender to send multiple frames at once
7. Which of the following ARQ protocols uses a window size greater than 1 for both the sender and receiver?
a) Stop-and-Wait ARQ
b) Go-Back-N ARQ
c) Selective Repeat ARQ
d) Both b and c
Answer: d) Both b and c
8. Which of the following flow control techniques has the highest efficiency in terms of retransmissions?
a) Stop-and-Wait
b) Go-Back-N ARQ
c) Selective Repeat ARQ
d) None of the above
Answer: c) Selective Repeat ARQ
9. In which scenario is Stop-and-Wait ARQ most suitable?
a) High-bandwidth, low-latency networks
b) Low-bandwidth, high-latency networks
c) Networks with packet loss
d) Large networks with many connections
Answer: b) Low-bandwidth, high-latency networks
10. The sliding window mechanism used in Go-Back-N ARQ can lead to which of the following?
a) Low throughput if frames are lost
b) Efficient use of bandwidth
c) The receiver accepting out-of-order frames
d) Increased protocol complexity
Answer: a) Low throughput if frames are lost
11. In flow control, what does the term “window size” refer to?
a) The number of frames that can be sent without receiving an acknowledgment
b) The number of bits allowed per frame
c) The number of errors detected per second
d) The maximum distance between sender and receiver
Answer: a) The number of frames that can be sent without receiving an acknowledgment
12. In which of the following is the sender required to stop and wait for an acknowledgment before sending the next frame?
a) Go-Back-N ARQ
b) Stop-and-Wait ARQ
c) Selective Repeat ARQ
d) Sliding Window ARQ
Answer: b) Stop-and-Wait ARQ
13. What happens in the Stop-and-Wait flow control protocol if an acknowledgment is not received in a certain time period?
a) The sender discards the frame
b) The sender sends a negative acknowledgment (NAK)
c) The sender retransmits the frame
d) The receiver requests retransmission
Answer: c) The sender retransmits the frame
14. Which of the following flow control protocols allows the receiver to accept out-of-order frames?
a) Stop-and-Wait ARQ
b) Go-Back-N ARQ
c) Selective Repeat ARQ
d) Sliding Window ARQ
Answer: c) Selective Repeat ARQ
15. The “sliding” nature of a window in Go-Back-N ARQ helps to:
a) Increase the size of the window over time
b) Allow the sender to send new frames as old frames are acknowledged
c) Track the sequence of lost frames
d) Allow the receiver to process frames in any order
Answer: b) Allow the sender to send new frames as old frames are acknowledged
16. In Go-Back-N ARQ, if a single frame is lost, what happens to the subsequent frames?
a) Only the lost frame is retransmitted
b) All subsequent frames are retransmitted
c) No retransmission is necessary
d) The receiver discards the entire window
Answer: b) All subsequent frames are retransmitted
17. Flow control ensures that:
a) The receiver can handle the incoming data rate without overflowing its buffer
b) The sender can transmit without waiting for acknowledgments
c) The receiver can accept any out-of-order data
d) The sender does not retransmit lost packets
Answer: a) The receiver can handle the incoming data rate without overflowing its buffer
18. Which of the following flow control techniques is typically used in TCP?
a) Go-Back-N ARQ
b) Selective Repeat ARQ
c) Stop-and-Wait ARQ
d) Sliding Window
Answer: d) Sliding Window
19. Which of the following is a disadvantage of Stop-and-Wait flow control?
a) High throughput
b) Simple implementation
c) Inefficiency in high-latency networks
d) Inability to handle errors
Answer: c) Inefficiency in high-latency networks
20. In Selective Repeat ARQ, if a frame is lost, the sender:
a) Retransmits all frames in the window
b) Retransmits only the lost frame
c) Does not retransmit, as the receiver will accept the next frame
d) Waits for the receiver to request retransmission
Answer: b) Retransmits only the lost frame
11. In a Stop-and-Wait protocol, if the round-trip propagation time is 200 ms and the transmission time for each frame is 50 ms, what is the maximum throughput of the channel?
a) 4 frames/sec
b) 2 frames/sec
c) 1 frame/sec
d) 10 frames/sec
Answer: b) 2 frames/sec
2. If a sender’s window size in a Go-Back-N protocol is 4, and the round-trip time is 100 ms, what is the maximum throughput if the transmission rate is 1 Mbps?
a) 0.25 Mbps
b) 0.5 Mbps
c) 0.75 Mbps
d) 1 Mbps
Answer: b) 0.5 Mbps
3. In Selective Repeat ARQ, if the sender window size is 5 and the receiver window size is also 5, how many frames can be sent before waiting for an acknowledgment?
a) 5
b) 10
c) 15
d) 1
Answer: b) 10
Explanation:
In Selective Repeat ARQ, both sender and receiver have a window size of 5. Hence, up to 10 frames can be sent before requiring acknowledgment, 5 from the sender’s side and 5 from the receiver’s side.
4. If the sliding window size in a Go-Back-N ARQ is 8, and the transmission time per frame is 100 ms, what is the total time taken to transmit all frames in the window, assuming no losses and acknowledgments are instantaneous?
a) 100 ms
b) 800 ms
c) 400 ms
d) 1000 ms
Answer: b) 800 ms
Explanation:
For Go-Back-N ARQ, the sender can send multiple frames, but the total time to transmit all frames is simply the transmission time multiplied by the window size. Thus:
5. In a Go-Back-N ARQ system, if the window size is 3, and the round-trip propagation time is 150 ms, how much time does it take to send 6 frames assuming each frame takes 50 ms to transmit?
a) 250 ms
b) 450 ms
c) 600 ms
d) 700 ms
Answer: b) 450 ms
Explanation:
With a window size of 3, the sender can send 3 frames before waiting for acknowledgment. For the first 3 frames, the total transmission time will be:
6. In Stop-and-Wait ARQ, if the propagation time is 50 ms and the transmission time per frame is 10 ms, what is the throughput?
a) 0.02 frames/sec
b) 0.1 frames/sec
c) 0.5 frames/sec
d) 2 frames/sec
Answer: b) 0.1 frames/sec
Explanation:
The throughput TTT in Stop-and-Wait ARQ is given by:
7. If the window size in a Sliding Window protocol is 4, and the transmission time is 20 ms per frame, how long will it take to send 8 frames?
a) 80 ms
b) 100 ms
c) 160 ms
d) 200 ms
Answer: c) 160 ms
Explanation:
In Sliding Window protocol, as long as the window size is greater than 1, the sender can continue sending frames. However, for each window of 4 frames, the transmission time will be:
8. If the round-trip time (RTT) is 200 ms, and the transmission time per frame is 25 ms in Go-Back-N ARQ with a window size of 4, what is the throughput?
a) 0.25 frames/sec
b) 0.5 frames/sec
c) 1 frames/sec
d) 4 frames/sec
Answer: b) 0.5 frames/sec
Explanation:
Throughput TTT for Go-Back-N ARQ is given by:
9. In Selective Repeat ARQ, if the sender window size is 6, and the receiver window size is also 6, how many frames can be sent before waiting for acknowledgment?
a) 6
b) 12
c) 24
d) 1
Answer: b) 12
Explanation:
In Selective Repeat ARQ, both sender and receiver can hold 6 frames in their window. Therefore, a total of 12 frames can be sent before waiting for acknowledgment.
10. In Stop-and-Wait, if the transmission time for a frame is 15 ms and the propagation time is 10 ms, what is the total time to transmit 2 frames?
a) 30 ms
b) 50 ms
c) 60 ms
d) 40 ms
Answer: d) 40 ms
Explanation:
For Stop-and-Wait, the total time is the sum of the transmission time and round-trip propagation time for each frame. For 2 frames, the total time will be:
