Consider an instance of TCP’s Additive Increase Multiplicative Decrease (AIMD)…
2012
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 timeout occurs during the fifth transmission. Find the congestion window size at the end of the tenth transmission.
- A.
8 MSS
- B.
14 MSS
- C.
7 MSS
- D.
12 MSS
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Correct answer: C
Start:
cwnd = 2 MSS,ssthresh = 8 MSS.Transmission 1 (end): slow start doubles →
cwnd = 4.Transmission 2 (end): slow start doubles →
cwnd = 8(equalsssthresh, so next RTT will be congestion avoidance).Transmission 3 (end): congestion avoidance (+1) →
cwnd = 9.Transmission 4 (end): congestion avoidance (+1) →
cwnd = 10.Timeout occurs during 5th transmission: immediately set
ssthresh = cwnd/2 = 10/2 = 5 MSS, andcwnd = 1 MSS.Transmission 5 (end, after timeout):
cwnd = 1.Transmission 6 (end): slow start doubles →
cwnd = 2.Transmission 7 (end): slow start doubles →
cwnd = 4.Transmission 8 (end): doubling would give 8 which exceeds
ssthresh = 5, socwndis set tossthresh = 5and we switch to congestion avoidance.Transmission 9 (end): congestion avoidance →
cwnd = 6.Transmission 10 (end): congestion avoidance →
cwnd = 7.
So the congestion window at the end of the tenth transmission = 7 MSS (option C).
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