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Module 4: Transport Layer & Congestion Control

20212m

Consider an instance of TCP's Additive Increase Multiplicative Decrease (AIMD) algorithm where the window size at the start of the slow start phase is 2 MSS and the threshold at the start of the first transmission is 8 MSS. Assume that a time-out occurs during the fifth transmission. Find the congestion window size at the end of the tenth transmission.

(i) 8 MSS
(ii) 14 MSS
(iii) 7 MSS
(iv) 12 MSS

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Answer: (iii) 7 MSS

Working: Slow start doubles the window each transmission until the threshold is reached; after that congestion avoidance adds 1 MSS per transmission. On a time-out, ssthresh = cwnd / 2 and slow start resumes from the initial window.

Transmission Window at start Phase Result
1st 2 MSS Slow start doubles → 4 MSS
2nd 4 MSS Slow start doubles → 8 MSS
3rd 8 MSS Threshold (8 MSS) reached → congestion avoidance +1 → 9 MSS
4th 9 MSS Congestion avoidance +1 → 10 MSS
5th 10 MSS Congestion avoidance time-out occurs
6th 2 MSS Slow start restarts (new threshold = 10/2 = 5 MSS) doubles → 4 MSS
7th 4 MSS Slow start would double to 8, but not permitted (8 > 5) → capped at 5 MSS
8th 5 MSS Congestion avoidance +1 → 6 MSS
9th 6 MSS Congestion avoidance +1 → 7 MSS
10th 7 MSS Congestion avoidance +1 → 8 MSS

So the congestion window at the tenth transmission is 7 MSS.

Why not the others: 14 MSS assumes the time-out was ignored and growth stayed linear (10 → 11 → … → 14); 12 MSS likewise ignores the multiplicative decrease; 8 MSS would require doubling past the new 5 MSS threshold, which congestion avoidance forbids.

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