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Assume the scenario where size of the congestion window of a TCP connection be 40KB when timeout occurs. The MSS is 2KB. Propagation delay be 200msec. Time taken by TCP connection to get back to 40KB congestion window is …...
asked in Computer Networks by Loyal (7.4k points) | 47 views
0
Mine is coming 5600, i.e. at t = 14 it reaches 20KB window size.

But answer is 6000 :/
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2 4 8 16 20 22 24 26 28 30 32 34 36 38 40

14*2*200 => 5600.

Got the same :)
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Don't we take the transition from 0 to 2KB? I am confused.Please help
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Actually we can start with 1MSS instead of 2MSS that would give the answer as 6000

But not sure when to start.

Check this here we started with 2 , i feel it should be mentioned in the question.

https://gateoverflow.in/101484/cn-tcp-congestion-control

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Different questions are taking it in different ways
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No brother we should start with 0 i have solved various GATE Question and got the exact answer by starting from 0, at it make sense as well
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how 0!! either we should start from 1 or 2 mss which makes sense.
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@Hemanth_13 I mean the transition starting from 1 upto 2.do we have to consider it?

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Actually speaking we have to start from 1MSS but see few problems were solved with 2MSS if they want from 2MSS it should be specified in the question.
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Ok thank you

1 Answer

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thresh hold =40/2= 20 kb
                                                        time out then it will incremented by 2
the seq goes like this 2 | 4 | 8 | 16 | 20 |  22 | 24 | 26 | 28 | 30 | 32 | 34 | 36 | 38 | 40|

so total will be number of times | occurs * rtt

so total will be 15*400=6000 msec
answered ago by (379 points)
edited ago by

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