M/M/1 queue

Example

Let $\lambda > 0$ and $\mu > 0$. In the M/M/1 queue of queueing theory, customers arrive as a Poisson process of rate $\lambda$ and there is one server. Service times are independent exponential random variables of rate $\mu$. The number $Q(t)$ of customers in the system is a birth–death process on $\set{0, 1, 2, \ldots}$. Its birth rates are $\lambda_n = \lambda$ for $n \ge 0$ and its death rates are $\mu_n = \mu$ for $n \ge 1$. This holds because the exponential law has no memory. At each moment, the time to the next arrival and the remaining service time are again exponential, of rates $\lambda$ and $\mu$.

Let $\rho = \lambda / \mu$. The balance equations $\pi_n \lambda = \pi_{n+1} \mu$ give $\pi_n = \rho^n \pi_0$. So a stationary distribution exists if and only if $\rho < 1$, and then $$\pi_n = (1 - \rho) \rho^n, \qquad n \ge 0,$$ a geometric law. The mean number in the system is $L = \sum_n n \pi_n = \rho / (1 - \rho)$. By Little's law, the mean time in the system is $W = L / \lambda = 1/(\mu - \lambda)$. If $\rho \ge 1$ the queue has no stationary distribution: it is null recurrent for $\rho = 1$ and transient for $\rho > 1$.