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}).

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s