GATE 2026 CS (CS2) – Question 58
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 there are no timeouts or duplicate acknowledgements. The number of rounds of transmission required for the congestion control algorithm to reach the congestion avoidance phase is __________. (answer in integer)
Note: $1\text{ K}=2^{10}$.
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Correct answer: 3
Explanation
The slow-start threshold is 16 KB and the MSS is 2 KB, so the threshold is 8 MSS. Starting from 1 MSS, the congestion window in slow start evolves as 1, 2, 4, and then 8 MSS. Thus the connection reaches the congestion-avoidance threshold after 3 rounds of transmission. Hence the answer is 3. This answer is marked tentative in the source because some conventions count the round in which the threshold is reached and would report 4 instead.