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Consider a queueing system where customers arrive according to a Poisson process with rate $\lambda$, but the service facility consists of two parallel servers. A customer upon entry into the service facility will proceed to either server 1 with probability 0.25 or to server 2 with probability 0.75. While the customer in the service facility is receiving his/her service, no other customer is allowed into the service facility. If the service rates of these two servers are exponentially distributed with rate $\mu_i$ (i = 1, 2), calculate the mean waiting time of a customer in this queueing system.

My approach is to look at this system as an M/G/1 with variable service times, since, from the point of view of a general customer in queue, there are Poisson arrival with rate $\lambda$ in the system, and the departure rates are of rate $\mu_1$ with probability 0.25 and $\mu_2$ with probability 0.75, so I used the Polaczek-Khinchin formula:

$W = \frac{\lambda \bar {x^2}}{2(1-\rho)}$

where we have $\mu = \mu_1/4 + 3\mu_2/4$, $\rho = \lambda\mu$, but I have some doubts on calculate $\bar x$ and $\bar {x^2}$, since my new departure rate $\mu$ now is linear combination of two dependent random variables with exponential distribution, and so I don't know how to treat it. Is this can be considered solved, or am I wrong to use this approach? Can someone help me, please? Thanks

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The question's wording "While the customer in the service facility is receiving his/her service, no other customer is allowed into the service facility" is a bit imprecise, as it doesn't specify whether a queue of waiting customers is considered part of the "service facility." The more natural interpretation to me is that it is, in which case:

Sojourn times are equal to their service times - which in this case has hyperexponential distribution with density $f_S(t)=\frac14 \mu_1 e^{-\mu_1 t}+\frac34 \mu_2 e^{-\mu_2 t}$. The mean would of course be $\frac14\mu_1^{-1} + \frac34\mu_2^{-1}$.

If the queue is not considered part of the "service facility," then this is a $M/G/1$ queue with the service distribution again having hyperexponential distribution with density $f_S(t)=\frac14 \mu_1 e^{-\mu_1 t}+\frac34 \mu_2 e^{-\mu_2 t}$. We may use the Pollaczek–Khinchine formula to compute the mean waiting time: $$ W' = \frac{\lambda\mathbb E[S^2]}{2(1-\rho)} = \frac{\lambda \left(3 \mu _1+\mu _2\right) \left(3 \mu _1^2+\mu _2^2\right)}{4 \mu _1^2 \mu _2^2 \left(4 \lambda \mu _2 \mu _1-3 \mu _1-\mu _2\right)}, $$ and the mean sojourn time by adding to this the mean service time: $$ W = W' + \mathbb E[S]^{-1} =\small \frac{64 \lambda \mu _2^4 \mu _1^4-27 \lambda \mu _1^4-18 \lambda \mu _2 \mu _1^3-12 \lambda \mu _2^2 \mu _1^2-6 \lambda \mu _2^3 \mu _1-\lambda \mu _2^4-48 \mu _2^3 \mu _1^4-16 \mu _2^4 \mu _1^3}{4 \mu _1^2 \mu _2^2 \left(3 \mu _1+\mu _2\right) \left(4 \lambda \mu _2 \mu _1-3 \mu _1-\mu _2\right)}. $$

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  • $\begingroup$ "Customers can only arrive to an empty system" is one way to interpret "no other customer is allowed into the service facility", but is it possible for that to simply mean that they line up outside the facility? $\endgroup$ Commented Aug 22, 2023 at 17:58
  • $\begingroup$ @MishaLavrov Indeed, the wording in the question is a bit imprecise. I'll revise my answer to include both interpretations. $\endgroup$
    – Math1000
    Commented Aug 22, 2023 at 21:31

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