Differential-difference equations for Poisson arrivals

Question

A service station has one gasoline pump. Cars wanting gas arrive according to a Poisson process at a mean rate of 20/hour. However, if the pump is in use, these potential customers may balk (refuse to join the queue). To be more precise, if there are n cars already at the service station, the probability that an arriving potential customer will balk is n/4 for n = 1, 2, 3, 4. The time required to serve a car has an exponential distribution with a mean service time of 3 minutes.

1. For all k, find the differential-difference equations for $p_k(t) = $ Pr{k in system at time t}.
2. Determine the stationary probability distribution and hence find the average number of cars at the station.

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I found a few similar exercises online, but with different things to compute. Nevertheless, I think this is a M/M/1 queue so that the equations are given by the formulas:

$\frac {dp_n(t)}{dt}=-(\lambda+\mu)p_n(t)+\lambda p_{n-1}(t)+\mu p_{n+1}(t)$ if n>0;

and

$\frac {dp_0(t)}{dt}=-\lambda p_0(t) + \mu p_1(t) $ if n=0

My first problem is I don't know what k is. Is it equal $0, ..., n-1$? If yes, then do I need to compute $p_0,p_1,p_2,p_3$ ? This would give

$\frac {dp_0(t)}{dt}=-20p_0(t)+20p_1(t)$

$\frac {dp_1(t)}{dt}=-35p_1(t)+20p_0(t)+20p_2(t)$

$\cdots $

Answer 1

Let's denote $p_k(t)$ the probability that there is $k$ customers as pump $k$ And $\lambda^{in}_k = \lambda^{in}_0 (1 - k/4)$ the incoming rate of consumers at pump $k$ and $\lambda^{out}$ that rate at which customers finish their service.

After having written the diagrams for the system, you can deduce the following equation for the $p_k(t)$

• $\dot{p_0}(t) = - \lambda_0^{in} p_0(t) + \lambda^{out} p_1(t)$
• $\dot{p_1}(t) = - \lambda_1^{in} p_1(t) + \lambda^{out} (p_2(t) - p_1(t)) + \lambda_0^{in} p_0(t)$
• $\dot{p_2}(t) = - \lambda_2^{in} p_2(t) + \lambda^{out} (p_3(t) - p_2(t)) + \lambda_1^{in} p_1(t)$
• $\dot{p_3}(t) = - \lambda_3^{in} p_3(t) + \lambda^{out} (p_4(t) - p_3(t)) + \lambda_2^{in} p_2(t)$
• $\dot{p_4}(t) = - \lambda_4^{out} p_4(t) + \lambda_3^{in} p_3(t)$

Since $\lambda^{in}_0 = 20$ and $\lambda^{out} = 60/3 = 20$ and since for stationary probabilities, by definition $\dot{p}_k(t) =0 \ \ \forall k \in \{1, 2, 3, 4\}$ you should be able to answer your question. This is the principle, I strongly encourage you to double check my calculations though.