A Finite Capacity Queueing System with $N=3$

The following is a question from Pinksy and Karlin's An Introduction to Stochastic Modelling:

The problem is to model a queueing system having finite capacity. We assume arrivals according to a Poisson process of rate λ, with independent exponentially distributed service times having mean 1/µ, a single server, and a finite system capacity N. By this we mean that if an arriving customer finds that there are already N customers in the system, then that customer does not enter the system and is lost. Let X(t) be the number of customers in the system at time t. Suppose that N = 3 (2 waiting, 1 being served).

(a) Specify the birth and death parameters for X(t).

The answer provided was in the form of a transition matrix $Q$:

$$\begin{equation} Q= \begin{bmatrix} -\lambda&\lambda & 0 & 0\\ \mu & -(\lambda +\mu)7\lambda &0 & 0\\ 0 & \mu & -(\lambda+\mu) & \lambda\\ 0 & 0 & \mu & \mu \end{bmatrix} \end{equation}$$

Is the solution correct, and how can I interpret the matrix $Q$ with regards to the birth and death parameters?

The entries $q_{ij}$ for $i\neq j$ are the rates of the exponential transitions from $i$ to $j$, while $-q_{ii}$ is the rate of the exponential holding time at state $i$. So e.g. $q_{01}=\lambda$ is the birth (arrival) rate when the system is empty. The matrix you provided is incorrect; it should be $$\begin{equation} Q= \begin{bmatrix} -\lambda&\lambda & 0 & 0\\ \mu & -(\lambda +\mu) &\lambda & 0\\ 0 & \mu & -(\lambda+\mu) & \lambda\\ 0 & 0 & \mu & -\mu \end{bmatrix} \end{equation}$$
(In the literature on continuous-time Markov chains, $Q$ is often called the rate matrix, or simply the $Q$-matrix.)
• Ah thank you, that clears things up quite a bit! I was pretty much stuck on the wrong entry" in the solution.. Nov 19 '17 at 8:42