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GATE CSE 2026 Session 2 Q58: Slow-start rounds before TCP reaches congestion avoidance

Question

Consider a new TCP connection between a sender and a receiver. The receiver advertised window is constant at 48 KB, the maximum segment size (MSS) is 2 KB, and the slow start threshold for TCP congestion control is 16 KB. Assume that there are no timeouts or duplicate acknowledgements. The number of rounds of transmission required for the congestion control algorithm of the TCP connection to reach the congestion avoidance phase is ___________. (answer in integer)

Note: 1K=2101\text{K} = 2^{10}

Try it: numerical answer: type the value

Numerical answer. No negative marking.
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Answer: 4 (official GATE 2026 final answer key)

Concept used: In slow start the congestion window doubles every round until it reaches the threshold.

First step: Start with cwnd = 1 MSS = 2 KB and double each round until it reaches 16 KB = 8 MSS.

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Source: official GATE 2026 Session 2 CS question paper and final answer key, organised by IIT Guwahati. Question paper · Answer key

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