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1 votes
1 votes

Sender window size is 2^6 . So reciever window should be the same. i. e. 2^6. 

Sequence no = (sender window + rexiever window) =2^7 i.e. 7 bit 

2 Answers

Best answer
3 votes
3 votes

We know :

Efficiency  =   N / (1 + 2a) where  a  = tp  /  tf

Hence ,

           N / (1 + 2a)   =  1/2

 ==>    N                 =   ceil((1 + 2a) / 2)

 ==>    N                 =   ceil(( 1 + (1000/16) ) / 2)

 ==>    N                 =   32

 Hence senders window size  =  32

 Hence total sequence numbers needed  =  2 * 32  =  64

 Hence no of bits needed for sequence no  =  log2 64    

                                                               =  6

 Hence  6 is the correct answer..

selected by
1 votes
1 votes

Just one more approach for this kind of questions

calculate BD product first

Bw*RTT

=512*103*1000*10-3=512*1000 (considering bps )

now  packet size 8000 bit

so optimum window size= bd product/packet size

=512*1000/(8*1000)=64

now log264= 6 bit

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