Question

A network transport protocol uses Selective-Reject ARQ. The network data rate is 100 Mbps and the distance between the sender and the receiver is 36,000 km. Assuming an average packet size of 100,000 bits and a media propagation velocity of 3 x 108 m/s, find the window size required for maximum network utilization and the number of bits needed for the sequence numbers.

Answer 1

Round Trip Time= Transmission delay+(2*Propagation delay)

=100000/10⁸ +(2*36000*10³)/3*10⁸

=1ms+240ms

=241 ms

Number of packets=(Bandwidth * RoundTripTime ) / Packet Size

(10⁸ * 1ms)/100000=1000packets

Because of using SR protocol we need 1000 + 1000 =2000 sequence numbers at most.

Sender Window Size Depends on Link Capacity and Receiver Window Size. Receiver Window Size depends on its processing speed and all other factors.

So Sender Window Size=min(Link Capacity,Receiver Window Size)

But no other information have been given so we can say sender and receiver window size should be 1000 at most.