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Compute the fraction of the bandwidth that is wasted on overhead(headers and retransmissions) for selective repeat on a heavily loaded 50-kbps satellite channel with data frames consisting of 40 header and 3960 data bits. Assume that the signal propagation time from the earth to the satellite is 270-msec. ACK frames never occur. NAK frames are 40 bits. The error rate for data frames is 1% and the error rate for NAK frames is negligible. The sequence numbers are 8 bits.
asked in Computer Networks by (35 points) | 25 views

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selective repeat protocol,and 8 bits sequence number

we know that ,sender window size+receiver window size<=available sequence number

(Ws+Wr<=ASN)

in case of selective repeat ,Ws=Wr =N so N+N<=2^8 =N=128

data =3960 bits(useful bits in one frame) and header =40 bits ,NAK frame =40 bits ,error rate =1%

propagation time =270 msec ,bandwidth = 50 kbps

efficiency=useful bits/total bits=total useful bits/(total bits +error bits+NAK bits) =>(3960*128)/((3960+40)*128+(3960+40)*128*(1/100)+(40))=>.98

bandwidth used =efficiency*Bandwidth =.98*50 kbps =49 kbps

so bandwidth wasted =50-49=1 kbps
answered by Active (2.9k points)
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@ Anil Ji So according to the calculation , 2% of the total bandwidth is wasted . But in answer it is given as 1.9%. Is that wrong?
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have they provided solution? did you get that? 2% is right
0
No they did not provide the solution.
0
i think 2% is right ..did u understand my solution ?


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