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Let the size of congestion window of a TCP connection be $32$ KB when a timeout occurs. The round trip time of the connection is $100$ msec and the maximum segment size used is $2$ KB. The time taken (in msec) by the TCP connection to get back to $32$ KB congestion window is _________.

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1 MSS = 2KB so 16 MSS= 32KB

Note: The initial size of cwnd is 1MSS so

cwnd : 1 2 4 8(ssthresh) 9 10 11 12 13 14 15 16

RTT    : 0 1 2 3                  4    5   6  7   8  9   10   11

So at the 11th RTT we got to cwnd of size 16.

11*100 = 1100ms
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WSsize =32KB

threshhold value = 32/2=16 KB

as MSS = 2KB

START FROM 2 KB

2, 4 , 8, 16(THRESH HOLD VALUE),17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32

NOW CALCULATE RTT BETWEEN ABOVE = 19 + 1= 20

THESE +1 IS BECAUSE GET BACK TO 32 AS MENTION INTO THE QUESTION
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is it maximum segment size or minimum segment size is give as 2kb?.
Answer:

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