# Transient and limiting distribution of the M/M/∞ queue

Suppose customers arrive to a queue according to a Poisson process with intensity ${\lambda}$, have independent service times with distribution ${G(t)}$, and there is an infinite number of servers available, so that service begins immediately as a customer enters the system. Assume further that the system is empty at time ${0}$. Let ${X(t)}$ be the number of customers in the system at time ${t}$; what is the distribution of ${X(t)}$?

A customer who enters the system at time ${s\leqslant t}$ will still be in the system at time ${t}$ with probability ${1-G(t-s)}$. Let ${\{N(t):t\in\mathbb R_+\}}$ be a Poisson process with intensity ${\lambda}$ and ${N_1(t) = (1-G(t-s))N(t)}$, ${0\leqslant s\leqslant t}$. Then ${N_1(t)=X(t)}$, and so for each nonnegative integer ${k}$

\displaystyle \begin{aligned} \mathbb P_0(X(t) = k) &= \mathbb P_0(N_1(t) = k)\\ &= \exp\left( -\lambda \int_0^t (1-(G(t-s)))\ \mathsf ds\right) \frac{\left(\lambda\int_0^t (1-(G(t-s)))\ \mathsf ds \right)^k}{k!}\\ &= \exp\left( -\lambda \int_0^t (1-(G(s)))\ \mathsf ds\right) \frac{\left(\lambda\int_0^t (1-(G(s)))\ \mathsf ds \right)^k}{k!}.\\ \end{aligned}

In the case where ${G(t) = 1-e^{-\mu(t)}}$, that is, the service times are exponentially distributed with rate ${\mu}$, we have

$\displaystyle \int_0^t (1-G(s))\ \mathsf ds = \int_0^t e^{-\mu t} = \frac1\mu \left(1 - e^{-\mu t}\right),$

and hence

$\displaystyle \mathbb P_0(X(t) = k) = \exp\left(-\frac\lambda\mu\left(1-e^{-\mu t}\right)\right) \frac{\left(\frac\lambda\mu\left(1-e^{-\mu t}\right)\right)^k}{k!}.$

The expected number of customers at time ${t}$ is then

\displaystyle \begin{aligned} \mathbb E_0[X(t)] &= \sum_{k=1}^\infty k\cdot \mathbb P_0(X(t)=k)\\ &= \sum_{k=1}^\infty k\cdot \exp\left(-\frac\lambda\mu\left(1-e^{-\mu t}\right)\right) \frac{\left(\frac\lambda\mu\left(1-e^{-\mu t}\right)\right)^k}{k!}\\ &= \frac\lambda\mu\left(1-e^{-\mu t}\right)\exp\left(-\frac\lambda\mu\left(1-e^{-\mu t}\right)\right)\sum_{k=0}^\infty \frac{\left(\frac\lambda\mu\left(1-e^{-\mu t}\right)\right)^k}{k!}\\ &= \frac\lambda\mu\left(1-e^{-\mu t}\right)\exp\left(-\frac\lambda\mu\left(1-e^{-\mu t}\right)\right)\exp\left(\frac\lambda\mu\left(1-e^{-\mu t}\right)\right)\\ &= \frac\lambda\mu\left(1-e^{-\mu t}\right). \end{aligned}

Moreover, we observe that

$\displaystyle \lim_{t\rightarrow\infty} \mathbb P_0(X(t)=k) = e^{-\frac\lambda\mu}\frac{\left(\frac\lambda\mu\right)}{k!},$

which is the same as the stationary distribution ${\pi}$ of ${\{X(t)\}}$ derived from the generalized global balance equations

$\displaystyle \lambda \pi_k = (k+1)\mu \pi_{k+1},\quad n=0,1,2,\ldots$

(normalized by the condition ${\sum_{n=0}^\infty \pi_k=1}$).