# Queue system with 2 parallel servers that works one at a time. Mean waiting time?

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

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: $$W = W' + \mathbb E[S]^{-1} =\small \frac{64 \lambda \mu _2^4 \mu _1^4-27 \lambda \mu _1^4-18 \lambda \mu _2 \mu _1^3-12 \lambda \mu _2^2 \mu _1^2-6 \lambda \mu _2^3 \mu _1-\lambda \mu _2^4-48 \mu _2^3 \mu _1^4-16 \mu _2^4 \mu _1^3}{4 \mu _1^2 \mu _2^2 \left(3 \mu _1+\mu _2\right) \left(4 \lambda \mu _2 \mu _1-3 \mu _1-\mu _2\right)}.$$