0
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.)

$\endgroup$
1
  • $\begingroup$ Ah thank you, that clears things up quite a bit! I was pretty much stuck on the ``wrong entry" in the solution.. $\endgroup$
    – Stoner
    Nov 19 '17 at 8:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.