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+2 votes
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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 _________.
asked in Computer Networks by Junior (613 points) | 194 views

2 Answers

+2 votes

maximum 9 bits required...

answered by Boss (10.5k points)
0
In answers, it is given10 bits.
+2

It's a question from madeeasy test series.

0
bro they are 100% wrong ....how can they take packet size in powers of 10 ??? they had done it wrong !!..

packet size should be in power of 2
+3
Shubham.. Your complete solution is right.. In last you have done a little bit of mistake.. Since whatever you have got in answer is the window size for one side.. But in SRA protocol we don't have only 1 window size on other side... Instead we have rwnd =swnd therefore multiply the sequence number >= (swnd+rwnd) =625 approx

Now you take log of it will lead you to correct number of bits required..  i.e 10
+1
oh!!..i got it bro thanks...
+1 vote

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

answered by Active (5k points)

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