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GATE CSE 2026 Session 1 Q45: Minimum sliding window size for full link utilisation

Question

Consider the implementation of sliding window protocol over a lossless link, with a window size of WW frames, where each frame is of size 1000 bits (including header). The bandwidth of the link is 100 kbps (1k=1031\text{k} = 10^3) and the one-way propagation delay is 100 milliseconds. Assume that processing times at the sender and receiver are zero and the transmission time of acknowledgements is also zero. Which one of the following options gives the minimum size of WW (in number of frames) required to achieve 100% link utilization?

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Answer: B (official GATE 2026 final answer key)

Concept used: Full utilisation needs W⋅Tt≥Tt+2TpW \cdot T_t \ge T_t + 2T_p, i.e. W≥1+2aW \ge 1 + 2a.

First step: Compute Tt=1000/105=10T_t = 1000/10^5 = 10 ms and 2Tp=2002T_p = 200 ms.

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Source: official GATE 2026 Session 1 CS question paper and final answer key, organised by IIT Guwahati. Question paper · Answer key

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