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=2^10
ANSWER
The connection reaches congestion avoidance in round 4, so the answer is 4.
Official answer: 4Source: Official answer key, GATE 2026 CS2 (IIT Guwahati)
Work in units of MSS. , so the slow start threshold is MSS and the receiver window is MSS. The receiver window is larger than the threshold, so it never limits the sender before the threshold is reached.
A new connection starts in slow start with . With no timeouts or duplicate ACKs, every ACK adds one MSS, so the window doubles every round.
Take the window used in each round (the congestion window at the start of that round, which is also the amount sent in it):
In rounds 1 to 3 the window is , all below the threshold of , so the sender is still in slow start.
In round 4 the window is , which equals the threshold. Slow start ends here (the window is not allowed to grow past the threshold in slow start), and from this point the window grows by one MSS per round, which is congestion avoidance.
The connection therefore reaches the congestion avoidance phase in round 4, so the number of rounds of transmission needed is .
KEY IDEA
In slow start the window doubles every round from 1 MSS. Count the rounds until the window used in a round equals the slow start threshold: 2, 4, 8, 16 KB, which is round 4.
WATCH OUT FOR
Mixing two conventions for the window, such as listing the window after the ACKs in one row and before them in another, or letting slow start run on to 32 KB. Slow start stops at the threshold. Another slip is forgetting that the starting window is 1 MSS = 2 KB, not 1 KB, which gives the wrong number of doublings.
From the GATE 2026 CS2 paper. Every question of this paper · Download the full paper · All 12 TCP questions, with notes
