# common queue and two servers

in a checkout system, customers arrive according to Poisson rate λ. The system consists of two parallel boxes, in Box 1 time is exponential of rate μ1 and box 2 exponential of rate μ2.

There is only one common queue. If both boxes are empty, clients prefer the box 1. You want to study the number of customers in the system at any steady-state.

Under normal conditions, customers choose the shortest line.

Formulate a model for these effects and obtain the average wait in the common queue.

this is not homework, im studying for a test and I'm resolving lots of problem, but this seems to beat me, im not sure if it Jackson or M|D|S, what complicates me is the two different μ

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Thanks to the memoryless property of the exponential distribution, the analysis of your system is quite easy. It is basically a modification of the M/M/2: indeed you have two servers, working in parallel with different speed if they are not empty.

Consider the state space $S=\{0,1_a,1_b,2,3,\dots\}$, where state $s$ denotes that there are $s$ customers in the system. We need a double state $1$ because we need to distinguish the two possible situations with only one customer, that could be in box 1 (a) or 2 (b). From your assumptions we know that if the system is empty, a new customer will go directly to box 1, but when there are two customers in the systems and one of them finishes the service, we may be in the situation where there is a customer in box 2 and none in box 1.

The transition rates for your system are:

• $q_{0,1a}= \lambda$,
• $q_{1a,2}=\lambda$ and $q_{1_a,0}=\mu_1$,
• $q_{1b,2}=\lambda$ and $q_{1_b,0}=\mu_2$,
• $q_{2,3}=\lambda$, $q_{2,1_a}=\mu_2$ and $q_{2,1_b}=\mu_1$
• $q_{n,n+1}=\lambda$ and $q_{n,n-1}=\mu_1+\mu_2$ for $n\geq 3$.

Notice that we need to specify clearly all the possible transitions when there are two or less customers in the system, while when they are more than three in not necessary anymore: this is because when both servers are working, one of the two will finish first and the residual service time of the other one will be again exponential with the same mean, thanks to the memoryless property of the exponential distribution.

Once one has set the mathematical problem in this way, it should not be hard to derive the number of people in the system and the waiting time distribution.

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