Pollaczek-Khinchine formula

For a stable first-come first-served M/G/1 queue, write $\rho =\lambda \mathbb{E}S<1$ and $\tilde{S}(s)=\mathbb{E}{e}^{-sS}$. The stationary waiting-time Laplace transform isWith finite second service moment, differentiating at zero gives the Pollaczek-Khinchine mean formula. These are the transform and mean forms of the same queueing identity.

For a stationary first come first served M/G/1 queue, take independent service times $S$ independent of the Poisson process of arrivals. Let $m=\mathbb{E}S$, $\rho =\lambda m<1$ and $\mathbb{E}{S}^2<\infty$. Then the mean queueing delay $w$ and mean queue population $L$ areTo see why the second moment appears, a completed service of length $S$ contributes triangular area $S^2/2$ under the residual service time curve. The renewal-reward theorem gives mean residual duration $r=\lambda \mathbb{E}{S}^2/2$, counting zero while idle. Poisson arrivals see time averages and Little's law then give $w=r+m\lambda w$. Solving this identity proves the formula.^[ref]

For the same mean $m$, deterministic service has $\mathbb{E}{S}^2=m^2$, while exponential service has $2m^2$, doubling the mean wait at fixed arrival rate. With $m=1$ minute and $\lambda =1/2$ per minute, these waits are respectively $1/2$ and one minute. Stability alone does not guarantee finite mean delay: for positive $\lambda$ and infinite second moment, the mean FCFS wait is infinite even when $\rho <1$.