Window size

Question

Compute approx. optimal window size, when packet size is 53 B, RTT is 60 micro-sec and bottleneck bandwidth is 155Kbps.

Comments

@Ashwin,

yes, it is correct.

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The given answer is 21.

In 1 sec ------- 155 * 10⁶

In 60*10^-6---------155*60 bits.

So optimal window size should be floor(9300/(53*8)) = 21.

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Please check RTD is 60 micro seconds or 60 milliseconds ?

Answer 1

We use 1+2A only when it is given that there is full utilization of channel.

If nothing is mentioned then use this, Bw * RTT / Packet size

= ( 155 * 10^3 * 60 * 10^-3 ) / 53 * 8
= 21.93
= Take ceil(21.93) = 21

Comments

Any reference source  ?

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@Yogesh

Answer is  not correct by the above formula ,

Given RTD = 2*Tp = 60 micro seconds = 60*10^ ( -6) seconds , but in your answer you considered it as

milliseconds .

Ans 21 is only possible if in question RTD is 60 ms instead of 60 micro seconds

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@ Yogesh Mandge

It has been given in the question that we need to find the optimal window size, which I guess, corresponds to utilizing the condition for maximum efficiency...