GATE CS 2026 Set 2 Q.58: Transport Layer: UDP, TCP Flow and Congestion Control, Sockets

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: $1\text{K}=2^{10}$

Your answer

Official answer and solution

Official answer: 4
From: Official GATE 2026 answer key

Solution

Trace the doubling of the congestion window cwnd (starting from $1\text{ MSS} = 2\text{ KB}$ ) across slow-start transmission rounds until cwnd reaches ssthresh $= 16\text{ KB}$ .

Initially, the congestion window is $\text{cwnd} = 1\text{ MSS} = 2\text{ KB}$ , and the slow-start threshold is $\text{ssthresh} = 16\text{ KB}$ :

Round 1: transmits $2\text{ KB}$ ; after ACKs, $\text{cwnd} = 4\text{ KB}$ .
Round 2: transmits $4\text{ KB}$ ; after ACKs, $\text{cwnd} = 8\text{ KB}$ .
Round 3: transmits $8\text{ KB}$ ; after ACKs, $\text{cwnd} = 16\text{ KB} = \text{ssthresh}$ (or Round 4 begins in congestion avoidance at $16\text{ KB}$ ).

Including the 4th transmission round at $\text{cwnd} = 16\text{ KB}$ (or counting the 4 window sizes $2, 4, 8, 16\text{ KB}$ up to reaching $\text{ssthresh}$ ), the number of rounds is $4$ .

Final answer

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2026 Computer Science and Information Technology - Session 2 (CS2)
Question: 58 of 65
Section: Computer Science and Information Technology

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