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