There's in a banc two identical queues and totally separated : these are two queues of type $M/M/1$. For each of them, the arrivals are separated by exponential times of parameter $\nu$, the time of service are exponentials of parameter $\mu$ (we suppose $\nu < \mu$). All random variables of the system are independent. How many customers are on average is there in the bank in steady state (stationary distribution)?

Question : How could I find the stationary distribution in considering both queues?

I have a part solution, but I forgot that all random variables of the system are independent, and I blocked on how to solve it. Is anyone could solve this problem for me?

My solution :

Let $Z_i(t)$ be the number of customers in system $i$ at time $t$, $i=1,2$. Then $\{(Z_1(t), Z_2(t)):t\geqslant0\}$ is a continuous-time Markov chain on state space $\mathcal S=\{0,1,2,\ldots\}^2$. The transition rates are given by $$ q_{(i,j),(i',j')} = \begin{cases} \nu,& (i',j') = (i',j'+1), (i'+1,j'+1) \\ \mu,& (i',j') = (i',j'-1), (i'-1,j'). \end{cases} $$ This yields the global balance equations \begin{align} 2\nu\pi(0,0) &= \mu(\pi(0,1)+\pi(1,0))\\ (2\nu+\mu)\pi(i,0) &= \nu\pi(i-1,0) + \mu(\pi(i+1,0)+\pi(i,1)),\quad i\geqslant1\\ (2\nu+\mu)\pi(0,j) &= \nu\pi(0,j-1) + \mu(\pi(0,j+1),+\pi(1,j)),\quad j\geqslant1\\ 2(\nu+\mu)\pi(i,j) &= \nu(\pi(i-1,j)+\pi(i,j-1)) + \mu(\pi(i+1,j)+\pi(i,j+1)),\quad i,j\geqslant1. \end{align} The stationary distribution $\pi$ satisfies the above along with $$\sum_{i,j=0}^\infty \pi(i,j)=1. $$

  • $\begingroup$ In the definition of $q_{(i,j),(i',j')}$ there is a mistake with $(i' + 1,j' + 1)$; it should be $(i' + 1,j')$. $\endgroup$
    – Ritz
    Mar 31, 2016 at 13:44

1 Answer 1


The two systems are totally separated, as you put it. So the stationary probability

\begin{equation} p(i,j) := \lim_{t \to \infty} \mathbb{P}(Z_1(t) = i, ~ Z_2(t) = j) \end{equation}

can be written in product form as $p(i,j) = p_1(i) \, p_2(j)$ with

\begin{equation} p_n(i) = (1 - \nu/\mu) (\nu/\mu)^i, \quad n = 1,2. \end{equation}


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