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Assume a scenario where the size of congestion window of a TCP connection be 40 KB when a timeout occurs. The maximum segment size (MSS) be 2 KB. Let the propagation delay be 200 msec. The time taken by the TCP connection to get back to 40 KB congestion window is _________ msec.

i am getting 5600msec but answer given is 6000msec.I am taking MSS as 2KB.ls someone explain 

NOTE:same question also asked here https://gateoverflow.in/1794/gate2014-1-27

in Computer Networks by Loyal (7.7k points) | 513 views
0
yes 5600
+1
2|4|8|16|20|22|24|26|28|30|32|34|36|38|40

14 RTT requires

14*400msec = 5600 msec.
0
yes that is correct
but, one thing tell me
when first 2KB is taking, why there is no RTT?
+2

hpe this will help you.

0
yes, that I understand
I want to know
at first window size is 0. Then as max MSS=2, we make window size 2.
here slow start phase also not started.
So, in 0 to 2, is there no work done?
Is there no RTT too?
0
there will be no RTT from 0 to 2, its just we are waiting at sender side till sender's buffer fills 2kb, as soon as it become 2kb we send it.

4 Answers

0 votes
Answer should be 13*400=5200
by (13 points)
0 votes
Take 1mss=2KB

          2mss=4KB

          4mss=8kb

          8mss=16kb

          9mss=18kb

threshold=40/2=>20kb..i.e10mss

        ...

        ....

Increase upto 20 miss=40kb

Hence total RTT is 15 so 15*400=6000
by Junior (653 points)
0 votes
Given cwnd = 40KB.

threshold =  20 KB (congestion window of a TCP connection be 40 KB when a timeout occurs)

Assuming 1 mss = 2KB.

cwnd =  40KB equivalent to 20 MSS (since 1MSS =  2KB)

threshold =  20 KB equivalent to 10 MSS

1 | 2 | 4 | 8 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20  ------->14 RTT's

RTT = 2 *  propagation delay

Time taken = 14 * 400 = 5600msec
by Active (1.8k points)
0
Why are you start from 1 it shoul be 2 ??
0
Because I've written in terms of MSS, not KB.  2KB = 1 MSS

To get in KB , you can multiply each term by 2KB. Like 1 MSS = (1*2KB) 2KB, 2 MSS = (2 * 2KB) 4KB, 4 MSS = (4*2KB) 8KB ,.......,20 MSS = (20*2KB) 40KB .
0 votes

At timeout time congestion window size is 40KB.

So,New Threshold=40KB/2=20KB.


Now,start from 0 KB exponentially increasing upto Threshold.

0kb-2kb,2kb-4kb,4kb-8kb,8kb-16kb,16kb-20kb(here 16kb-32kb is not possible because it exceed Thrshold),

And now increase only one MSS at a time

20kb-22kb,22kb-24kb,24kb-26kb,26kb-28kb,28kb-30kb,30kb-32kb,32kb-34kb,34kb-36kb,36kb-38kb,38kb-40kb.

Total 15 times.

so total time=RTT*15

RTT=2*PD=2*200msec=400msec.

total time=15*400=6000msec.

 

by (71 points)

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