M/G/1 queue

An $M/G/1$ queue has Poisson arrivals, independent identically distributed service times, and one server. If service has mean $m$ and $\lambda m<1$, its mean busy period isThis follows by decomposing the busy period into the first service and the busy-period descendants of arrivals during that service.

An M/D/1 queue has a Poisson process of arrivals, a deterministic service time $\tau$, and one work-conserving server. With arrival rate $\lambda$ and $\lambda \tau <1$, its equilibrium idle probability is $1-\lambda \tau$: the stationary completed work per unit time is $\lambda \tau$. Its output generally need not be a Poisson process.

Let $Q_n$ be the number left immediately after departure $n$, and let $A_n$ be the arrivals during the next service. ThenConditional on service duration $S$, $A_n$ is Poisson with parameter $\lambda S$, so

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 stable $M/G/1$ queue with traffic intensity $\rho =\lambda \mathbb{E}S<1$, the mean waiting time before service isEquivalently, the mean number in the system is