p) (23.15) Since the probability that a customer has to wait Prob(wait) is just the server utiliza- tion p, we get Eq. (23.16) telling us the probability that a packet has to wait longer than a given time t: Prob(wait>t) = pe-t(1-p) (23.16) Assuming a packet switch delay requirement t, and assuming that we know the uti- lization p and the average packet duration 1/, Eq. (23.16) gives us the fraction of pack- ets that will not clear the packet switch within time t. As we said earlier, this is not the entire story of packet traffic engineering. The dis- tribution of packet sizes is not exponential, and the interarrival time between packets has a bursty characteristic that the M/M/1 model ignores. We can deal with some of this by treating the packet bursts as customers in the queue and choosing SO that 1/ is the average size of a packet burst. Also, packet routing often involves multiple packet switches, and several of these packet switches can be congested at the same time. make the statistics problem more difficult, when one packet switch is congested, the others are more likely to be congested as well. 6f the server is busy p of the time and the server has one customer when busy, then the aver- age number being served is p. Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. Traffic Engineering for Voice and Data Traffic Engineering for Voice and Data However, while we are looking at reasons why the M/M/1 queue model is not perfect, we might be missing the point. As long as p stays low enough, the packets will seldom have to wait very long. Increasing the capacity and speed of the packet pipe reduces both the average burst duration 1/ and the packet pipe utilization p. Equation (23.16) gives us a good indication of just how much extra capacity the pipe needs SO that few enough packets have excess delay. And this is what traffic engineering is al