In a supermarket customers arrive at the cash desk with a Poisson process with an average of 30 customers per hour. There is one cash desk and the service time is exponential with an average of 2 minutes.
If there are 3 or more customers in the queue, then:
Model 1: The checker is assisted by a clerk. The service time is still exponential, but with an average of 1 minute.
Model 2: There will be opened another cash desk, with also a service time which is negative exponential with an average of 2 minutes.
I am interested in the balance equations for both models. In my opinion we have a $M/M/2$ queue for model 2.
For model 1 ($\lambda = 30, \mu = 30$) I got:
$\lambda P_o = \mu P_1$
$(\lambda + \mu) P_1 = \lambda P_0 + \mu P_2$
$(\lambda + \mu) P_2 = \lambda P_1 + 2 \mu P_3$
$(\lambda + 2 \mu) P_n = \lambda P_{n-1} + 2 \mu P_{n+1}$ , with $n\geq 3$
But this is exactly what I got for the second model. I doubt wheter this is correct. Is something wrong?
Model 1 is a $M/M/1$ queue with the service rate increased from $\mu$ to $a\mu$ (where $a>1$) when the queue length exceeds a threshold $Q_L$, while model 2 is a $M/M/1$ queue which adds a server when the queue length exceeds $Q_L$. However, the analysis of these systems is the same, because $a=2$ and $Q_L=3$, so for $n\geqslant Q_L$ the transition rate $q_{n+1,n}=1$ (as the minimum of two $\operatorname{Exp}\left(\frac12\right)$ random variables has $\operatorname{Exp}(1)$ distribution).