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I have one queue with two servers $S_1$ and $S_2$.The serving times are modeled $\sim exp(\mu_1)$ and $\sim exp(\mu_2)$ respectively.

The first server is free while the second has two clients, $A$ whose service is in progress and $B$ that is waiting. There is a path for every client so each of them must be served first by $S_1$ and then by $S_2$.
Then another client $X$ arrives. Now, help me developing this concept:

  1. $P_a$, the probability that customer A is still in service by $S_2$ when $S_1$ finishes to serve $X$.

For $P_a$, the only thing I know is that intuitively, I need to calculate the probability that the first service time is less than the second. But I stop here.

Thank you

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Once $S_1$ finishes serving the first customer A, it's a race whether $S_1$ can serve two more customers (B and X) before $S_2$ finishes serving A.

The probability is given by:

$$ \int_0^{\infty}\mu_2 e^{-\mu_2z}\int_0^z \mu_1e^{-\mu_1y}\int_0^{z-y}\mu_1e^{-\mu_1x} \:dx \:dy \:dz $$

Where $z$ represents the time for $S_2$ to serve A, $y$ the time for $S_1$ to serve B, and $x$ the time for $S_1$ to serve X.

Edit: sounds like you're saying that $S_1$ doesn't need to serve B before they can start serving X? In which case isn't the existence of B irrelevant? The integral would then just be:

$$ \int_0^{\infty}\mu_2 e^{-\mu_2z}\int_0^{y}\mu_1e^{-\mu_1x} \:dx \:dy $$

See https://math.stackexchange.com/questions/1332413/comparing-two-exponential-random-variables for the derivation you're looking for.

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  • $\begingroup$ In this case, given two service times, I need to have T1 < T2. But from the theory I know that the probability $P(X_1 < X_2)$ converges to $\frac{\mu_1}{\mu_1 + \mu_2}$. Thus is there therefore the possibility that those integrals converge to this value? Thank you $\endgroup$
    – docdev
    Commented Nov 28, 2020 at 21:36

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