Suppose there are N equal-bandwidth links on the path between A and B, as in the

Question

Suppose there are N equal-bandwidth links on the path between A and B, as in the diagram below, and we wish to send M consecutive packets.

A S1 ... SN-1 B

Let BD be the bandwidth delay of a single packet on a single link, and let PD be the propagation delay on a single link. Show that the total bandwidth delay is (M+N-1)×BD, and the total propagation delay is N×PD. Hint: the total time is the sum of the time A takes to begin transmitting the last packet, and the time that last packet (or any other packet) takes to travel from A to B. Note that no packets ever have to wait at any Si because the ith packet takes exactly as long to arrive as the (i-1)th packet takes to depart.

Explanation / Answer

The time to transmit the first of M packets to the destination(B) = N x BD . Every BD seconds, a new packet from M-1 packets arrives at the destination. Thus, the total Bandwidth delay = N x BD + (M-1) x BD = (M+N-1) x BD

Since, PD is the propagation delay on a single link and we have N equal-bandwidth links, then the total propagation delay = N x PD [ Propagation delay is the time taken by a bit to travel from one router/host to the next. It is purely a function of distance between routers or distance from source to destination hosts.]

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