GATE2012_45 CN

+2 votes
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

After Timeout We should start from 1 MSS irrespectice of What size is at the start or After Timeout again we have to take from 2 MSS??    AND

The question asks After the end of 10th Transmission so My approach is

At t=1, =>2mss t=2, =>4mss t=3, =>8mss t=4, =>9mss (after threshold additive increase) t=5, =>10mss (fails) t=6, =>2mss t=7 =>4mss t=8, =>5mss t=9, =>6mss t=10, =>7mss.So at the end of 10th transmission 7 segments will be transmitted.   

So the Answer Should be 8MSS or 7MSS???
asked Dec 16, 2016 in Computer Networks by hi2saifgmail-com (3,200 points)

1 Answer

+1 vote
 
Best answer
It would depend on how we define end of the transmission. If it is the time when the ack hits receiver then it is 7 MSS, otherwise if it is after getting ack then after that incrementing window size then answer would be 8 MSS.

But first case is convincing right.
answered Jan 5 by getgatebook (38,090 points)
selected Jan 8 by hi2saifgmail-com
in this question MSS is 2 so why in congestion avoidance phase we are adding 1MSS? Should we not add 2MSS in Congestion Avoidance as we have fixed our segment size to 2MSS
incrementing cwnd doesn't depend on initialization.
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