I have two part question involving an M/M/1 queuing process. The first part I did but the second I'm not sure about.
The question involves a gas-station pump where customers arrive at a rate of 20 cars/hour. They pull into the station if there is 2 or less cars present. So, $N=3$ (for the total locations where you can service your car). And they leave after the required 5 mins to get serviced elapses. So the first part asked:
1) What proportion of the time is the attendant busy?
$ \mu = \frac{1}{5/60} = 12$ cars/hr are serviced.
$\rho = \frac{\lambda}{\mu} = \frac{5}{3}$
steady-state probabilities are $p_n=\rho^n p_0, n=0,1,…,N$
$p_0=(1+\sum_{n=1}^N \rho^n)^{-1} =(\frac{1-\rho^{N+1}}{1-\rho})^{-1}$
and I get $p_0=0.0993, p_1=0.1655, p_2=0.2758, p_3=0.4597$
Thus, the proportion the attendant is busy is $p_1+p_2+p_3=0.901$
Then it says…
If the service time is exponential with mean 3 minutes (instead of 5),
how many more customers per hour would enter the station on average?
My initial idea was to use a similar strategy to the one above but I run into an issue finding my new probabilities, since now
$ \mu = \frac{1}{3/60} = 20$ cars/hr are serviced.
So, $\rho = \frac{\lambda}{\mu} = \frac{20}{20} = 1$, which causes my steady-state probabilities $p_n$ to be 0.
So, I guess my question is: can I approach the second part the same way as I did the first or it there another method?
Thank you kindly!
Best Answer
This is an $M/M/1/N$ queuing system. For such a system, the formulas have a special case if $\lambda=\mu$, i.e., if $\rho=1$. For example, instead of $$\begin{align*} p_0 &= \frac{1-\rho}{1-\rho^{N+1}} \\ p_n & = \rho^n p_0, \end{align*}$$ we have $$p_0 = p_n = \frac{1}{N+1}$$ for $n=1,\ldots,N$.
In the second part of your problem, $\rho=1$, so you should be using the alternate formulas.