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Computer Science Engineering (CSE) Exam  >  Computer Science Engineering (CSE) Test  >  GATE Computer Science Engineering(CSE) 2027 Mock Test Series  >  Test: Communication & Congestion Control - Computer Science Engineering (CSE) MCQ

GATE Computer Science Engineering(CSE) 2027 Test: Communication & Congestion


MCQ Practice Test & Solutions: Test: Communication & Congestion Control (10 Questions)

You can prepare effectively for Computer Science Engineering (CSE) GATE Computer Science Engineering(CSE) 2027 Mock Test Series with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Communication & Congestion Control". These 10 questions have been designed by the experts with the latest curriculum of Computer Science Engineering (CSE) 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 30 minutes
  • - Number of Questions: 10

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Test: Communication & Congestion Control - Question 1
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In a sliding window protocol, the sender's window size is N and the receiver's window size is 1. If the protocol is Go-Back-N, what is the relationship between the sequence number space (S) and the sender window size (N) to avoid the ambiguity of duplicate acknowledgments?

Detailed Solution: Question 1

In the Go-Back-N (GBN) ARQ protocol, the receiver window size is always 1. To distinguish between new frames and retransmitted frames (in case of ACK loss), the available sequence numbers (S) must be greater than the sender's window size ( N ).
The condition is: Available Sequence Numbers ≥ Sender Window Size + 1 .
Therefore, S ≥ N + 1 .

Test: Communication & Congestion Control - Question 2
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A network uses CSMA/CD protocol with a bandwidth of 10 Mbps. The maximum propagation delay on the channel is 25 μs. What is the minimum size of the frame required to ensure proper collision detection?

Detailed Solution: Question 2

For CSMA/CD to work correctly, the transmission time (Tt) must be at least twice the propagation delay (Tp) so the sender is still transmitting when the collision signal arrives back.Condition:
Tt ​≥ 2 × Tp​
Let Frame size be L . L/Bandwidth ≥ 2 × 25μs
L/(10 × 106) ≥ 50 × 10−6
L ≥ 50 × 10−6 × 10 × 106 L ≥ 500 bits.

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Test: Communication & Congestion Control - Question 3
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An organization is granted the block 16.0.0.0/8. The administrator wants to create 500 fixed-length subnets. What is the optimal subnet mask to accommodate this requirement while maximizing the number of hosts per subnet?

Detailed Solution: Question 3

We need 500 subnets.We must borrow bits from the host part such that 2x ≥ 500.
28=256 (not enough)
29=512 (enough)
So we need to borrow 9 bits for the subnet ID.The default mask for /8 is 255.0.0.0.
New mask adds 9 bits: 11111111.11111111.10000000.00000000
In decimal: 255.255.128.0.

Test: Communication & Congestion Control - Question 4
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Consider a TCP connection where the available bandwidth is 1 Gbps. The TCP sequence number field is 32 bits long. Approximately how long will it take for the sequence numbers to wrap around (be reused)?

Detailed Solution: Question 4

Total number of unique bytes that can be numbered with 32 bits = 232 bytes ≈ 4.29 GB.
Bandwidth = 1 Gbps = 109 bits per second = 109/8 bytes per second = 125 MBps.
Wrap Around Time = (Total Data) / (Data Rate)Time = 232/(125 x 106) Time ≈ (4.29 x 109)/(0.125 x 109) ≈ 34.35 seconds.

Test: Communication & Congestion Control - Question 5
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Which of the following HTTP methods is NOT idempotent?

Detailed Solution: Question 5

An idempotent method means that multiple identical requests have the same effect as a single request.

  • GET: Retrieving data multiple times doesn't change the server state (Idempotent).
  • PUT: Uploading the same file to the same location multiple times results in the same file being there (Idempotent).
  • DELETE: Deleting a resource; once deleted, subsequent deletes effectively do nothing (Idempotent).
  • POST: Submitting data (like a form) multiple times creates multiple entries or transactions (NOT Idempotent).

Test: Communication & Congestion Control - Question 6
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Let us consider a statistical time division multiplexing of packets. The number of sources is 10. In a time unit, a source transmits a packet of 1000 bits. The number of sources sending data for the first 20 time units is 6, 9, 3, 7, 2, 2, 2, 3, 4, 6,1, 10, 7, 5, 8, 3, 6, 2, 9, 5 respectively. The output capacity of multiplexer is 5000 bits per time unit. Then the average number of back logged of packets per time unit during the given period is

Detailed Solution: Question 6

Answer is B
Here we can spent at max 5 packets per Time unit 5000 /1000.
So Whatever which is not sent is backlog.
So First Time Unit => 6 ,
Backlog in First time unit => 6-5 => 1 This one gets added to next Time units load
Second time unit => 9 + 1 (One from Previous Time Unit)
Backlog in Second time Unit = 10-5 => 5 (This one gets added to next Time Units load.)
Total Backlog this way = 1+5+3+5+2+0+0+0+0+1+0+5+7+7+10+8+9+6+10+10=89
Avg Backlog=89/20=4.45
The average number of backlogged of packets per time unit during the given period is 4.45 , (Option B) .

Test: Communication & Congestion Control - Question 7
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On a TCP connection, current congestion window size is Congestion Window = 4 KB. The window size advertised by the receiver is Advertise Window = 6 KB. The last byte sent by the sender is Last Byte Sent = 10240 and the last byte acknowledged by the receiver is Last Byte Acked = 8192. The current window size at the sender is

Detailed Solution: Question 7

Ans should be (B)
Current Sender window = min (Congestion Window, Advertised Window)= min(4KB, 6KB)= 4KB

Test: Communication & Congestion Control - Question 8
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In the slow start phase of the TCP congestion algorithm, the size of the congestion window

Detailed Solution: Question 8

increase is exponential in the Slow Start Phase.
answer = option D

Test: Communication & Congestion Control - Question 9
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Consider an instance of TCP’s Additive Increase Multiplicative Decrease (AIMD) algorithm where the window size at the startof the slow start phase is 2 MSS and the threshold at the start of the first transmission is 8 MSS. Assume that a time out occurs during the fifth transmission. Find the congestion window size at the end of the tenth transmission.

Detailed Solution: Question 9

The Answer is correct , but method of solving is wrong .
At
t=1, =>2mss
t=2, =>4mss
t=3, =>8mss
t=4, =>9mss (after threshold additive increase)
t=5, =>10mss (fails)
Threshold will be reduced by n/2 i.e. 10/2 = 5.
t=6, =>2mss
t=7 =>4mss
t=8, =>5mss
t=9, =>6mss
t=10, =>7mss.
So at the end of 10th transmission congestion window size will be 8 mss.

Test: Communication & Congestion Control - Question 10
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Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of the connection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.


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About this Computer Science Engineering (CSE) Practice Test

This test contains 10 MCQs on Communication & Congestion Control with detailed solutions for each question. It is part of the GATE Computer Science Engineering(CSE) 2027 Mock Test Series course on EduRev, designed to align with the current Computer Science Engineering (CSE) syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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