I am trying to understand Poisson processes and queues. I have this exercise:
Consider a fuel station with two fuel pumps and one park. Each car that comes to the fuel station when the pumps and the park are all busy, leaves immediately. We have Poisson arrivals with intensity $\lambda=1$ car/15min and the service time follows an exponential distribution with mean 15 mins.
Find:
- the probability that a car will be rejected because the system is full
- The mean time of a customer in the system
I should not use Continuous time Markov Chains to solve this problem. I understood that we are talking about a M/M/2/1 queue
Could please someone give me some hints? Thank you
You are dealing with a $M/M/2/3$ system, where the last $3$ describes the total number of spots that can be occupied. Let $X(t)$ be the number of cars in the system at time $t$ and let the equilibrium probabilities be defined as
\begin{align} p(i) := \lim_{t \to \infty} \mathbb{P}(X(t) = i), \quad i = 0,1,2,3. \end{align}
Because you have Poisson arrivals, the system has the Poisson arrivals see time averages (PASTA) property. Therefore, for question $1$, you are indeed interested in $p(3)$, which is the probability that an arriving car sees $3$ cars in the system. I do not see how to solve this without setting up some balance equations related to the continuous-time Markov chain. If you want, I can solve it using the balance equations.
For question 2, you could use a mean value approach or just compute the mean number of cars in the system as $\mathbb{E}[L] := \sum_{i = 0}^3 i p(i)$ and use Little's law to obtain the mean time spent in the system as $\mathbb{E}[S] = \mathbb{E}[L] / \lambda$.
Exercises 12 and 16 of these lecture notes are very related to your question. The lecture notes also gives some insight in the mean value approach.