Go Back N

Question

Consider a network which has : Bandwidth (B) = 10 Mbps Distance between two hosts (d) 500 kilometers Average packet size(L) = 103 bits Propagation speed = 2 × 108 m/s It is designed a Go – Back N sliding window protocol for this network. The network is to be used to its full capacity. Assume that processing delays at nodes are negligible. Then the minimum window size has to be ___________.

Correct Answer: 51    Status: unattempted

Answer 1

Transmission Time($T_{t}$)=$\frac{Average\; packet \:size}{Bandwidth}=\frac{10^{3}}{10*10^{6}}=10^{-4}sec$

Propagation delay($T_{p}$)=$\frac{Distance\; between \:two \: hosts}{Propagation Speed}=\frac{500*10^{3}}{2*10^{8}}=0.0025sec$

a=$\frac{T_{p}}{T_{t}}=\frac{0.0025\:sec}{0.0001\:sec}=25$

now,

Window Size(W)=1+2*a=1+2*(25)=1+50=51.....:-)

Please correct me if I'm wrong....!!!

Comments

right bro :)

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agree with your solution just a little bit of confusion.... the window size u got as an answer is  (sws+rws)??

and what if question asked for maximum window size? should we go through sequence number approach then!!!

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Window size 51 here is only for the sender window.

If seq bits are given in the question then we would have to select min(2^k-1,1+2a),k is no of seq. bits, for the sender window size.