Pollaczek-Khinchine mean formula Created 2026-10-03 Updated 2026-11-05
For a stationary first come first served M/G/1 queue, take independent service times independent of the Poisson process of arrivals. Let , and . Then the mean queueing delay and mean queue population are
To see why the second moment appears, a completed service of length contributes triangular area under the residual service time curve. The renewal-reward theorem gives mean residual duration , counting zero while idle. Poisson arrivals see time averages and Little's law then give . Solving this identity proves the formula.[ref]
For the same mean , deterministic service has , while exponential service has , doubling the mean wait at fixed arrival rate. With minute and per minute, these waits are respectively and one minute. Stability alone does not guarantee finite mean delay: for positive and infinite second moment, the mean FCFS wait is infinite even when .
Residual service time 2026-11-05
A residual service time is the unserved part of a job's full-capacity service time. If the original service time is and units of service have already been delivered, its remaining amount is conditioned on the event . For ,
provided the denominator is positive.
Observation during a stationary sequence of independent and identically distributed services produces a different calculation. Conditional on sampling a busy instant uniformly in service time, with , the equilibrium residual satisfies
The second identity assumes a finite second moment. A service of deterministic duration gives a uniform distribution on and mean . Long services are encountered more often than short services, explaining why the observed residual need not resemble the original probability distribution.[ref]
Robert G. Gallager Created 2026-10-24 Updated 2026-11-05
Robert G. Gallager wrote teaching material on discrete stochastic processes. Probability models for evolving states connect repeated random events with long-term behavior and performance of systems.[ref]