Suppose the bandwidth on a pointtopoint link is 1 Mbps and t

Suppose the bandwidth on a point-to-point link is 1 Mbps and the round-trip delay is 9 milliseconds. Assume that the size of each packet is 1000 bits. Also, assume that the size of ACK packet is negligible.

Answer the following.

(a) What would be utilization with stop-and-wait?

(b) Assuming that the link is error-free, what should be the minimum window size to achieve at least 50% utilization.

Solution

link utilization or efficiency of stop and wait protocol is ,

efficiency = Tx / ( Tx + 2Tp ) = 1 / ( 1 + 2(Tp / Tx)) = 1 / (1 +2a) ,

where , Transmission time = Tx = packet size / bandwidth = L / B

Propagation time = Tp = distance / speed = d / v ,

and ,

     a = Propagation time / Transmission time = ( Tp / Tx ) ,

now for 50% efficiency ,

efficiency = 1 / (1 +2a)

50% = 1 / (1 +2a)

1 / 2 = 1 / (1 +2a) ,

2 = (1 +2a)

2 - 1 = 2a

                                        1 = 2.(Tp / Tx)

                                      Tx = 2.Tp

                                   L / B = 2* 9 ms

                                        L = 2 * 9 ms * B = 2* 9 * 10-3 * 64 k bits = 2* 9*10-3 * 177 bits

                                        L = 40 * 64 bits = 40 * 64/8 bytes = 40* 8 bytes = 320 bytes