Fall’06 Computer Networks TelCom 2310 http://www.cs.pitt.edu/∼znati/tel2310.html

Homework 2 Solution

Problem 1

Assume that packets arrive at a node to be transmitted. The packets arrive at random times T_{1}, T_{2}, ⋯, T_{n}, ⋯, and are transmitted in the order of their arrivals (FIFO).

1. The diagram showing how many bits are stored in the node buffer as a function of time is depicted in Figure ??.

Figure 1: Number of bits, N(t), in the buffer at time t

2. Let D_{n} denote the delay faced by the n-th packet. We have D_{n} = \frac{N(T_{n})}{R}. For packet 1, D_{1} = \frac{P}{R}, since N(T_{1}) = P. To identify D_{2}, we observe that the queing time of the second packet is equal to the residual transmission time of the first packet, i.e., max (D_{1} − (T_{2} − T_{1}), 0). Consequently, we have:

D_{2} = max (D_{1} − (T_{2} − T_{1}), 0) + \frac{P}{R}.  (1)

In general, we find that:

D_{n+1} = max (D_{n} − (T_{n+1} − T_{n}), 0) + \frac{P}{R}.  (2)

3. To show that large average queueing delays can oocur, even with low average arrival rate, consider batches of k packets that arrive every T seconds. We choose T larger than \frac{kP}{R} so that the successive batches find the buffer empty. The first packet of a batch has a delay D_{1} = \frac{P}{R}, the second packet has a delay D_{2} = \frac{2P}{R}, and so on. The k-th packet in a batch has a delay D_{k} = \frac{kP}{R}. Thus the average delay per packet is \frac{(D_{1}+D_{2}+···+D_{k})}{k} = \frac{(k+1)P}{2R}. By choosing k large, we can make the average delay large, even when T is so large that the average packet arrival rate \frac{k}{T} is very small.