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GATE CSE 2025 Session 2 Q36: Channel utilisation of stop-and-wait on a slow link

Question

Suppose we are transmitting frames between two nodes using Stop-and-Wait protocol. The frame size is 3000 bits. The transmission rate of the channel is 2000 bps (bits/second) and the propagation delay between the two nodes is 100 milliseconds. Assume that the processing times at the source and destination are negligible. Also, assume that the size of the acknowledgement packet is negligible.

Which ONE of the following most accurately gives the channel utilization for the above scenario in percentage?

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Answer: A (official GATE 2025 final answer key)

Concept used: Stop-and-wait utilisation =Tt/(Tt+2Tp)= T_t / (T_t + 2T_p).

First step: Tt=3000/2000=1.5T_t = 3000/2000 = 1.5 s and 2Tp=0.22T_p = 0.2 s.

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Source: official GATE 2025 Session 2 CS question paper and final answer key, organised by IIT Roorkee. Question paper · Answer key

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