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With empty initial state, Poisson arrivals of rate $\lambda$ and exponential lifetimes of rate $\mu$ produce a Poisson occupancy of mean $\lambda (1-e^{-\mu t})/\mu$ at time $t$.

In Kendall notation, the first two letters $M$ mean Poisson arrivals and exponential service times. An M-M-1 queue has arrival rate $\lambda$, one server of rate $\mu$, and queue-length transitionsIt has a stationary distribution of an M-M-1 queue exactly whenand then

An M-M-infinity queue has infinitely many servers, so every customer begins service immediately. Its occupancy transitions areThe detailed balance for a birth-death process equations giveThus for every $\lambda ,\mu >0$ the stationary distribution of an M-M-infinity queue exists and isthe Poisson distribution of mean $\rho$. The Burke theorem for an M-M-infinity queue states that in this stationary regime the departure process is a Poisson process of rate $\lambda$; equivalently, time reversal turns departures into arrivals. The past departure process is also independent of the number in the system at the observation time.

Let $X_t$ be the number of telephone lines in use. The memorylessness of the exponential distribution makes $(X_t{)}_{t\ge 0}$ a continuous-time Markov chain, specifically the M-M-infinity queue with generator rates

Suppose $X_0=n$ and put $p_t=e^{-\mu t}$. Each initial call is still active at time $t$ with the exponential survival probability $p_t$, independently of the others. Their surviving number is thereforeA call arriving at time $u\in [0,t]$ survives until $t$ with probability $e^{-\mu (t-u)}$. By Poisson thinning, the number of surviving new calls is independent of $Y_t$ and has lawHence $X_t=Y_t+Z_t$ in distribution. Multiplying the probability generating functions of the two independent random variables gives, for $-1\le s\le 1$,Equivalently, the transient distribution of an M-M-infinity queue iswith independent summands. As $t\to \infty$, the binomial term converges to zero and the Poisson mean tends to $\lambda /\mu$. Thereforewhich proves the stated large-time approximation.

For an M-M-1 queue, the first service lasts on average $1/\mu$. During it, the mean number of arrivals is $\lambda /\mu$, and each arrival initiates another statistically identical busy-period family. Thus, when $\lambda <\mu$,so the mean busy period of an M-M-1 queue isFor $\lambda =\mu$ the busy period is almost surely finite but has infinite mean, while for $\lambda >\mu$ it is infinite with positive probability; in either case its expectation is infinite.

For an M-M-infinity queue, stationary occupancy is Poisson with mean $\rho =\lambda /\mu$, so the long-run probability of an empty system isThe process alternates between idle periods and busy periods. An idle period waits for the next Poisson arrival, so its mean is $1/\lambda$. The renewal-reward theorem, with reward equal to time spent empty, givesSolving yields the mean busy period of an M-M-infinity queue: