computer netwk

Question

Compute approximate optimal window size when packet size is 53 bytes, the RTT is 60 micro seconds and bottleneck bandwidth is 155 Mbps?

Comments

window size = $\frac{B*RTT}{packet size}$ = $\frac{155*10^{6}*60*10^{-6}}{53*8}$ = 21.93

that means answer should be 21

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why we not take 22 ??

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becoz we can accomodate only 21.93, that means 22th packet is partial. If we consider it whole then it will overflow the window size.

Consider this using this example,

Their is a vessel of capacity 21.93lt now u cant put 22lt in it. But u can put 21lt

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k thanks

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why not 22 ? we know window size =1+2a =22.93 =22 .please clarify me

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read digvijay comment :)

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Yes, bibhas kumar is right.

AS per Sliding window is concern, window size is calc. by:

Window size= 1+2a   = floor(1+ 21.93)  = floor(22.93) = 22

Correct me if i am wrong.

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refer https://gateoverflow.in/1820/gate2006-44

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@stanchion its not mentioned that its sliding window so dont assume what digvijay singh said is right

Answer 1

BD product= RTT*Bandwidth

=155∗10⁶∗60∗10⁻⁶=155*60

packet size 53*8

so optimal window size=(155*60)/(53*8)=21.93

=22

Comments

Why you are taking 22?

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Window size= floor ( BW*RTT / packet size)

                           =floor (155∗10^6∗60∗10^−6/53*8) = floor(21.93)=21