Bandwidth R bps , Frame size = K bits , PT = t sec,/ kilometer, distance L kilometer. So total PT ( not RTT) = L*t sec .Processing delay =0.Assuming W is window size
Now, it says that We need to maximize utilization for sliding window protocol.
Maximum utilization if=> useful time/ total time =1 .
it is possible when ,useful time = Total time.
Useful time= W * {K/R } sec
Total time = K/R + 2* L*t sec
Now => (W * {K/R } ) / ( K/R + 2* L*t ) = 1
W * {K/R } = K/R + 2* L*t
W = K + 2LTR / K
Suppose W= 2n, where n = number of bits in sequence number
2n = K + 2LTR / K
n= log((K+2LTR)/K)