$M/M/2/3$ Queuing Theory Word-Problem

Question

A service center consists of two servers, each working at an exponential rate of two services per hour. If customers arrive at a Poisson rate of three per hour, then, assuming a system capacity of at most three customers.

1. What fraction of potential customers enter the system?
2. What would the value of part (1) be if there were only a single server and his rate were twice as fast ($\mu = 4$)?

I understand there are four states in the system : $0$, $1$, $2$ and $3$. The transition rates for the system are $\lambda _0=\lambda _1=\lambda _2=3$ and $\mu _1=2$, $\mu _2=\mu _3=4$. I am struggling to model this and find transition table to aid the questions.

Answer 1

This is popularly known as the Erlang-B model. You can use time reversibility of ctmcs(see corollary 1.5 of this) to get the transitional probabilities. As you pointed out the $\lambda$ and $\mu$ correctly, in general for any $i$ we have $$ \pi_i \lambda = \pi_{i+1}(i+1)\mu$$

You can easily solve this recursion to get $\pi_i$ from here.

Hint: It would come out to be something like $\pi_i \propto \frac{\lambda}{\mu}\frac{1}{i!}, i\geq1$, if done correctly.

The final step is a bit more subtle: let us find the long-run fraction of time a customer coming in sees the queue full(so they will not enter the system). Conveniently, by PASTA(which I hope you must have studied by now. If not, see this), it is just $\pi_3$. So the fraction of people who enter the system is $1-\pi_3$. The final answer should come out to be $$1-\frac{\frac{\rho^m}{m!}} { \sum_{i=0}^m \frac{\rho^i}{i!}}$$ where $\rho$ is $\frac{\lambda}{\mu}$. $m$ and $\rho$ you can define appropriately for parts a and b resp.

Answer 2

We can model this as a continuous-time Markov chain with generator matrix $$ Q = \begin{pmatrix} -3 & 3 & 0 & 0\\ 2 & -5 & 3 & 0\\ 0 & 4 & -7 & 3\\ 0 & 0 & 4 & -4 \end{pmatrix}. $$ The stationary distribution $\pi$ satisfies $\pi Q=0$ and $\sum_{i=0}^3 \pi_i=1$, from which we find $$ \pi = \begin{pmatrix}\frac{32}{159}&\frac{16}{53}&\frac{12}{53}&\frac 9{53}\end{pmatrix}. $$ It follows that the fraction of potential customers that enter the system is $1-\pi_3=\frac{44}{53}$.

If there were one server with a rate of $\mu=4$ then the generator matrix would be $$ Q= \begin{pmatrix}-3&3&0&0\\4&-7&3&0\\0&4&-7&3\\0&0&4&-4 \end{pmatrix}, $$ with $$ \pi = \begin{pmatrix}\frac{64}{175}&\frac{48}{175}&\frac{36}{175}&\frac{27}{175}\end{pmatrix} $$ and the fraction of potential customers that enter the system $\frac{148}{175}$.