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Consider the effect of using slow start on a line with propagation delay of 10ms.The receive window is 24KB and the maximum segment size is 2KB.How long does it take to send full window size segment in ms initial threshold of 20KB?

My ans is

2 | 4 | 8 | 16 | 18 | 20 | 22 | 24 |    Thus 8 RTT's= 160ms

Given ans is 

2 | 4 | 8 | 16 | 20 |  24 |       Thus 6 RTT's =120ms 

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Best answer
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given answer is right.

we have threshold as 20KB..means 10 MSS..go exponentially till threshold then linearly from there...

1 2 4 8 10(threshold) 11 12
so after 6 RTTs u reached maximum of 12MSS...

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2 | 4 | 8 | 16 | 20 |  22 | 24       Thus 6 RTT's =120ms 

Because Its slow start therfore start exponentially from starting MSS ie.,given 2 KB till threshold (given 20KB)

After theshold go linearly ie increment by MSS given 2KB till Max receiver window size (given 24 KB)

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Geet asked Dec 23, 2016
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How is the initial slow-start threshold value (ssthresh) determined for the very first time of data transfer?