How to apply queuing theory to find the long run proportion of customers who leave the system?

Question

I am trying to apply queuing theory / birth and death process to the following.

Suppose customers arrive in a restaurant according to a Poisson process with rate $\lambda = 1$.

Suppose there are $2$ counters in the restaurant. Customers are served in counter $1$ in Exponential $(1)$ amount of time, and customers are served in counter $2$ in Exponential $(2)$ amount of time.

Usage of counters by each customer is independent of the usage of counters by other customers. When a customer comes, s/he chooses counter $2$ if it's free/available, otherwise the customer goes to counter $1$.

If both counters are occupied, the customer leaves.

Here, I am trying to calculate the long run proportion of customers that leave because both counters are occupied.

I have been trying to set up the balance equations but I am not sure that's the correct approach. I'm preparing for the upcoming finals and I'd be grateful for any help on how to solve this. Thank you so much.

$EDIT:$

I tried to model this as an $M/M/2$ queue, and I get the following results:

I take the state space to be $S = (0, 1, 2),$ where each state refers to the number of customers in the system.

The balance equations are

$\lambda P_0 = (\mu_1 + \mu_2)P_1$

$\lambda P_1 = (\mu_1 + \mu_2)P_2$

And the normalizing equation is: $P_0 + P_1 + P_2 = 1$

Solving this gives me $P_0 = \frac{9}{13}, P_1 = \frac{3}{13},$ and $P_2 = \frac{1}{13}$

Is this correct? I am doubtful about the logic I have used to arrive at the balance equations, especially because of the fact that customers prefer counter $2$ over counter $1$.

If this is correct, then, I have $P_2$, which is the long run probability that a customer leaves because both counters are occupied. With this, how can I find the long run proportion of customers that leave because both counters are occupied? Thank you for any help.

Answer 1

It sounds like you are modeling an $M/M/2/2$ queue. It may be helpful to view the transition diagram, as follows.

Suppose that "deaths" in the process are given by a constant rate, $\mu_{i}$, corresponding to the service times at counters $i=1,2$. The long-run proportion (steady-state distribution) you seek can be found by first considering the rate in and out of each state. In particular, by viewing the graph, the "flow" out/in of state $0$ is \begin{equation*} \lambda\pi_{0} = \mu_{2}\pi_{1}; \end{equation*} while the flow out/in of states $1$ and $2$, respectively, are \begin{alignat*}{1} \pi_{1}(\lambda+\mu_{2}) &= \lambda\pi_{0}+(\mu_{2}+\mu_{1})\pi_{2},\\ (\mu_{2}+\mu_{1})\pi_{2} &= \lambda\pi_{1}. \end{alignat*} Solving the above system of equations (while also accounting for the normalizing equation $\sum \pi_{i}=1$) will yield the long-run proportion. There may be some additional interpretation needed to get the exact long-run proportion you are looking for, though.