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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 (8.6k points) | 112 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
0

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

2 Answers

+1 vote
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 by Junior (553 points)
edited by
+1 vote
We send data in terms of MSS AND NOT KB SO WHENEVER DATA IS GIVEN IN TERMS OF KB,CONVERT IT TO MSS.
Here, it has been said that MSS is 2KB ==> It means the size of 1 segment is 2KB (Max.)
congestion window of a TCP connection be 40KB ===> 40KB/2KB = 20MSS
Threshold = 10MSS

You can either take the window Size in KB OR CONVERT IT TO MSS.....
2KB = 1MSS

1MSS = 2KB
2MSS = 4KB
4MSS = 8KB
8MSS = 16KB
10MSS = 20KB
11MSS = 22KB
12MSS = 24KB
13MSS = 26KB
14MSS = 28KB
15MSS = 30KB ...............

After RTT = 2*200 * 14 = 5600ms, we are able to achieve the Windows Size of 40KB and after RTT = 6000ms, we were able to send
20MSS OR 40KB DATA IN ONE WINDOW SUCCESSFULLY.

But Question is asking
Time taken by TCP connection to get back to 40KB congestion window is …...

it should be 5600ms.

I hope I am clear.
answered by (425 points)
edited by

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