m/m/2 question in queueing theory

Question

Customers arrive at a serving-system according to a Poisson process with rate 1. In the system there are two serving stations, A and B, which only take care of one customer at a time. If a customer come to a full system, both A and B are busy, then he leaves. If a customer come to an empty system he chooses A. When a customer has been served he leaves. The serving time are independent of the arrivals and mutually independent following an exponential distribution with rate 1. Let Xt take the value (0; 0) if the system is empty, (1; 0) if A is being used but not B, (0; 1) if A is free and B is being used and (1; 1) if both A and B are busy. Then $(X_t)_{t \geq 0}$ is an irreducible Markov-chain with state-space {(0; 0); (1; 0); (0; 1); (1; 1)}. Determine the Q-matrix of the chain and the equilibrium distribution.

Answer 1

In state

• (0,0) arrivals happen at rate 1 and take you to the state (1,0)
• (1,0) arrivals happen at rate 1 and take you to the state (1,1) while services happen also at rate 1 and take you to the state (0,0)
• (0,1) arrivals happen at rate 1 and take you to the state (1,1) while services happen also at rate 1 and take you to the state (0,0)
• (1,1) no arrivals can happen (they are not admitted to the system) and two services are happening in parallel. A service at rate 1 which will take you to (0,1) and another at rate 1 which will take you to (1,0).

So $$ Q = \begin{pmatrix} -1 & 1 & 0 & 0 \\ 1 & -2 & 0 & 1 \\ 1 & 0 & -2 & 1 \\ 0 & 1 & 1 & -2 \end{pmatrix}.$$

We compute the stationary distribution of $Q$ by solving $\pi Q=0$ with the requirement that the elements of $\pi$ sum to 1. This gives $$\pi=\left(\frac{2}{5},\frac{3}{10},\frac{1}{10},\frac{1}{5}\right).$$