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GATE CS 2026 Set 2 · Computer Networks

Question 65

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Question

It is necessary to design a link-layer protocol between two hosts that are directly connected over a lossless link of length 3000 kilometers. Assume that the link bandwidth is 10810^8 bits per second and that the propagation delay in the link is 5 nanoseconds per meter. Every transmitted data byte is assigned a unique sequence number.

Let NN be the minimum number of bits needed for the sequence number field in the protocol header such that

i. the sequence numbers do not wrap around before 60 seconds, and
ii. the maximum utilization of the link is achieved.

The value of NN is ______. (answer in integer)

Official answer and solution
Official answer
30
From
Official GATE 2026 answer key

Solution

Compute both the number of bytes transmitted in 60 seconds at full link rate and the window size required for 100% link utilization.

  1. Bytes transmitted in 60 seconds at maximum (100%) link utilization:
    • Link bandwidth B=108 bits/s=1088 bytes/s=1.25×107 bytes/sB = 10^8\text{ bits/s} = \frac{10^8}{8}\text{ bytes/s} = 1.25 \times 10^7\text{ bytes/s}.
    • In T=60 secondsT = 60\text{ seconds}, the number of bytes transmitted is:
      1.25×107×60=7.5×108 bytes1.25 \times 10^7 \times 60 = 7.5 \times 10^8 \text{ bytes}
  2. Check condition (ii) (sliding window size for 100% utilization):
    • One-way propagation delay Tp=3000×103 m×5×10−9 s/m=0.015 sT_p = 3000 \times 10^3\text{ m} \times 5 \times 10^{-9}\text{ s/m} = 0.015\text{ s}, so RTT=0.03 s\text{RTT} = 0.03\text{ s} (much smaller than 60 s60\text{ s}).
  3. Minimum sequence number bits NN:
    • For sequence numbers not to wrap around within 60 s60\text{ s}, we need 2N≥7.5×1082^N \ge 7.5 \times 10^8.
    • Since 229=536,870,912<7.5×108≤230=1,073,741,8242^{29} = 536,870,912 < 7.5 \times 10^8 \le 2^{30} = 1,073,741,824, the minimum number of bits is N=30N = 30.
Final answer

30

Common trap

Forgetting to divide 10810^8 bits/s by 8 to convert bits into bytes (which would give N=33N = 33).

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2026 Computer Science and Information Technology - Session 2 (CS2)
Question
65 of 65
Section
Computer Science and Information Technology

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