GATE CS 2024 Set 2 — Question 54

NAT+2 / -0MediumTCP Congestion ControlTransport LayerComputer Networks

Computer Networks → Transport Layer → TCP Congestion Control

Last updated 20 August 2026

Question

Consider a TCP connection operating at a point of time with the congestion window of size 12 MSS (Maximum Segment Size), when a timeout occurs due to packet loss. Assuming that all the segments transmitted in the next two RTTs (Round Trip Time) are acknowledged correctly, the congestion window size (in MSS) during the third RTT will be _________

Correct answer

4 to 4

Solution

When a timeout occurs in TCP (standard implementations like Tahoe, Reno, or NewReno), the congestion control mechanism performs the following steps:

1.Update ssthresh: The slow start threshold (

ssthresh

) is set to half of the current congestion window (

CWND

ssthresh = \max(2, \lfloor \frac{CWND}{2} \rfloor) = \frac{12}{2} = 6 \text{ MSS}

2.Reset CWND: The congestion window is reset to 1 MSS.

3.Enter Slow Start: The algorithm restarts in the Slow Start phase.

Now, we trace the value of

CWND

over the subsequent Round Trip Times (RTTs):

During 1st RTT:
$CWND$ starts at $1$ MSS.
1 segment is transmitted.
Since it is acknowledged correctly, $CWND$ increases by 1 (exponential growth in Slow Start).
End of 1st RTT: $CWND = 1 + 1 = 2$ MSS.
During 2nd RTT:
$CWND$ starts at $2$ MSS.
2 segments are transmitted.
Both are acknowledged correctly. $CWND$ increases by 1 for each ACK (total +2).
End of 2nd RTT: $CWND = 2 + 2 = 4$ MSS.
During 3rd RTT:
$CWND$ starts at $4$ MSS.
Since $CWND < ssthresh$ ( $4 < 6$ ), we are still in the Slow Start phase.

The question asks for the congestion window size during the third RTT, which is the value at the start of that round.

Answer: 4

GATE CS 2024 Set 2 — Question 54

Question

Correct answer

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