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Let the size of congestion window of TCP connection be 38 KB when a timeout occurs. The propagation time of the connection is 100 msec and the maximum segment size used is 2 KB. The time taken (in sec) by the TCP connection to get back to 37 KB congestion window is ______

What is the correct answer for above question?
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I'm getting 14 RTTs - so 1400ms, using the discussion and method established by https://gateoverflow.in/1794/gate2014-1-27. What is your answer ?

There was a lot of discussion between 1100 and 1200 in that 2014 case. The official answer key was then given as a range of 1100-1300.

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Even i am getting 14 RTT.

1 RTT = 200ms

so answer should be according to me 14*200 = 2800 ms = 2.8 ms.

But the answer provided by made easy is different and their solution is quite different.

See the solution below provided by them,

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I'd love to know how they went from 16 to 19. Only way for that is 16KB + 2KB + 1KB, i.e., they've reduced the MSS itself from 2KB to 1KB.

Based on all the examples, literature and references I've seen - the MSS is kept constant, what is changed is the window size.
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Yes u are right shashin.
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but they are not reducing MSS they are just sending less data in a segment which must be allowed .

correct me if i am wrong