= General-service infinite-server queue
{title2=$M/G/\infty$}
= M-G-infinity queue
{c}
{synonym}
Poisson arrivals of rate $\lambda$ receive independent service times distributed as $S$, with infinitely many servers so service starts on arrival. Starting empty, the occupancy at time $t$ is Poisson with mean $\lambda\int_0^t\mathbb P(S>u)\,du$, by independent thinning of the marked arrivals. If $\mathbb ES<\infty$, the equilibrium count is Poisson with mean $\lambda\mathbb ES$. Occupancy need not be a Markov process for a general service law; remaining service times carry additional state information.
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