Question

Assuming that you are designing the sliding window protocol for a 1 mbps point to point the link to the moon. Which has a one-way latency(delay) of 1.25 sec. Assume that each frame carries 2 kb data, what is the minimum number of bits needed for the sequences number?

Answer 1

window size(w)=(bandwidth*two way propagation delay)/frame size

so here w=(10^6*2*1.25)/(1*1024*8)

=305

by default sliding window means go_back n ARQ

so here 2^n-1=305

n=9 bits