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Consider a network connecting two systems located 4000 kilometers apart. The bandwidth of the network is 64 Mbps. The propagation speed of the media is 2/3 of the speed of light in vacuum. It is needed to design selective repeat sliding window protocol for this network. The average packet size is of 8 Kb. The network is to be used to its full capacity.
Assume that processing delays at nodes are negligible. Then, the minimum size in bits of the sequence number field has to be _________.

2 Answers

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2 votes

maximum 9 bits required...

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1 votes

bandwidth of the network is 64 Mbps

average packet size is of 8 Kb

distance between  two system =4000 kilometers

 transmission time Tt=8*1024/64*106=128*10-6 s

 propagation time Tp=4000*1000 meter/1.5*3*108 m/s=20 ms

maximum window size=1+2a

=1+2*20 ms/128*10-6 s = 1+2* 20*10-3 s /128*10-6 s =313.5

in SRA protocol sender window size=receiver window size

sequence no=2* window size =2*313.5=627

so sequence bit =log(627)=10 bits

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