edited by
917 views
1 votes
1 votes
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.
edited by

1 Answer

1 votes
1 votes
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

Related questions