computer networks

Question

Assume we need to design a Go- Back- N sliding- window protocol for a network in which the bandwidth is 100 Mbps and the average distance between the sender and receiver is 10,000 km. Assume the average packet size is 100,000 bits and the propagation speed in the media is 2 × 108 m/ s. Find the maximum size of the send and receive windows, the number of bits in the sequence number field ( m), and an appropriate time- out value for the timer.

Answer 1

window size = (TT+2PT)/TT=1+(2PT/TT)

here TT=data size/bandwidth

 100000/100Mbps =10⁵/10⁸ = 1 ms

and PT = distance/velocity

=10,000 km/2*10⁸ ms=10⁷/2*10⁸ = 50 ms

so window size = 1+(2*50)/1=101

so sender window size = 100

receiver window size =1

s.w.s.+ r.w.s.<= 2^m(where m is no. of sequence bits)

101<=2^m

m=7 bits