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I have been trying to wrap my head around an M/M/s queue problem but I can't seem to understand what's going on.

The problem is queuing at a gas station and I have done some research about the M/M/s queue model but can't seem to figure out how it works.

Here is what I know:

Expected vehicles in 1 hour is 70
Average service rate of 13.6 vehicles per hour per fueling position
10 fueling positions

So this tells me:

$\lambda$ = 70

$\mu$ = 13.6

$c$ = 10

I know how to get $P_0$ and $P_n$ but how do I solve for average customers in system, average customers in queue, average time spent in system, average time waiting in line?

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  • $\begingroup$ for stationary distribution in an MC you need to construct Kolmogorov forward equations for $\pi_k$ $\endgroup$
    – Alex
    Commented Feb 20, 2014 at 16:39
  • $\begingroup$ and by that what exactly do you mean? $\endgroup$
    – user130256
    Commented Feb 20, 2014 at 16:59

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Let the number of jobs in the system be $0,1, \ldots$. What you need is stationary distribution as $t \to \infty, \ \pi_k$. If the probability to observe $k$ jobs in the system at time $t$ is $P(X_t=k), \ \pi_k = \lim_{t \to \infty} P(X_t=k)$. This is found using Kolmogorov forward equations: $$ \pi_k = p \pi_{k-1}+q \pi_{k+1} $$ (keep in mind $p+q=1$). Once you have derived the expression for $\pi_k$, the definition of expectation of $X$: $$ \mathbf{E}X=\sum_{k=0}^{\infty}k \pi_k $$ and this will be the mean number of jobs in the system. For the mean waiting time you should use Little's formula: $L= \lambda W$. Can you handle from here?

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  • $\begingroup$ This makes sense, but when I was researching this somewhere I found that the average number of customers in the queue can be calculated by $$p/(1-p)*p_0*p^c/(c!(1-p))$$ $\endgroup$
    – user130256
    Commented Feb 20, 2014 at 17:39
  • $\begingroup$ given that arrivals follow Poisson distribution and service time is Exponential, you can derive this expression from the first principles I outlined $\endgroup$
    – Alex
    Commented Feb 20, 2014 at 19:15

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