## A little more on $$M(\lambda)/M(\mu)/1$$

• $$Q(t)$$ the number of items in the system at time $$t$$

• $$\rho = \lambda/\mu$$ is the traffic intensity

• $$Q$$ is a continuous-time Markov chain, with generator $$G$$ given by $G(n, n-1) = \mu \text{ for all n \geq 1}$ and $G(n, n+1) = \lambda \text{ for all n \geq 0 }$

## Theorem: stationary distribution

If $$\rho <1$$, then as $$t \to \infty$$ $\mathbb{P}(Q(t) = n) \to (1- \rho)\rho^n = \pi_n,$ were $$\pi$$ is the stationary distribution.

• Solving $$\pi G = \mathbf{0}$$, we obtain for $$n \geq 1$$

$\begin{eqnarray*} 0 &=&(\pi G)_n \\ &=& \pi_{n-1}G(n-1,n) + \pi_{n+1}G(n+1,n )+ \pi_nG(n,n) \end{eqnarray*}$

and

$0=(\pi G)_0 = \pi_1 G(1, 0) + \pi_0 G(0,0)$

## continued

• Thus

$\pi_{n+1} = \pi_n(\rho +1) - \pi_{n-1} \rho$ and

$\pi_1= \rho \pi_0$

• It is easy to see that

$\pi_n = \rho^n \pi_0.$

• Summing over all $$n$$, using the fact that $$\pi$$ is probability measure, we obtain $\pi_0 = 1- \rho$

## Burke’s theorem

• Theorem: The departures $$D(t)$$ of a stationary $$M(\lambda)/M(\mu)/1$$ queue are a Poisson process of rate $$\lambda$$. Furthermore, $$Q(t)$$ is independent of the $$(D(s), s \leq t)$$.

• We sketch a proof using reversibility. See Reich’s proof of Burke’s theorem

• It is easy to check that $$Q$$ is reversible.

• Given $$T>0$$, the process $$R(t)= Q(T-t)$$ has the same law as $$(Q(s), s\leq T)$$.

• The process $$R$$ adds an arrival at time $$t$$, when $$Q$$ has a departure at time $$T-t$$.

## continued

• The arrivals of $$R$$ and $$Q$$ are both Poisson

• The time-reversal of a Poisson is again a Poisson.

• Thus the departure process is also Poisson.

• $$Q(0)$$ is independent from the arrivals in $$[0, T]$$

• Reversing time, $$Q(T)$$ is independent from the departures in $$[0, T]$$.

## Two M/M/1 queues in series

• It is a general question, given a item has to pass through two queues in series, one after another, which order should the queues to placed in?

• This is in general a hard question

• However, using Burke’s theorem, we can show the order does not matter in the case of two M/M/1 queues in equilibrium.

Theorem Suppose items enter a $$M(\lambda)/M(\mu_1)/1$$ queue, then upon exiting report to another queues with exponential service rate $$\mu_2$$. If $$\lambda < \min(\mu_1, \mu_2)$$, then the continuous-time Markov chain $$(Q_1, Q_2)$$ has stationary distribution given by

$(m,n) \mapsto \rho_1^m(1-\rho_1) * \rho_2^n(1-\rho_2)$ where $$\rho_1 = \lambda/\mu_1$$ and $$\rho_2 = \lambda/\mu_2$$.

## Proof

• We already know that if the first queue is stationary, then by Burke’s theorem, the departures are again a Poisson process with rate $$\lambda$$

• Thus what feeds into the second queue is still a Poisson.

• We already know that the individual expressions correspond to the the stationary distributions of the individual queues

• Independence follows from the independence assertion is Burke’s theorem.

## Summary

• We found the stationary distribution for a M/M/1 queue.

• We used reversibility to prove Burke’s theorem.

• We applied Burke’s theorem to two M/M/1 queues in series.