Each entering customer must be served first by server $1$, then by server $2$, and finally by server $3$. The amount of time it takes to be served by server i is an exponential random variable with rate $\mu_i$, $i = 1, 2, 3$. Suppose you enter the system when it contains a single customer who is being served by server $3$.

Find the probability that server 3 will still be busy when you move over to server 2.

If we let $T_i\sim exp(\mu_i)$ the service time of server $i$, then we just want to find $$P(T_3>T_1)=P(T_1=min(T_1,T_3))=\frac{\mu_1}{\mu_1+\mu_3}$$

but the answer says $\frac{\mu_3}{\mu_1+\mu_3}$.

Where did that come from?

  • $\begingroup$ Unfortunately there are two different conventions for exponential variables: one has the pdf $\lambda e^{-\lambda x}$ with mean $1/\lambda$ and the other has the pdf $\frac{1}{\lambda} e^{-x/\lambda}$ with mean $\lambda$. The former can be said to have "rate $\lambda$" when the exponential variable itself is viewed as a time. Which convention are you using? $\endgroup$ – Ian Sep 30 '15 at 14:33
  • $\begingroup$ @Ian I'm using $\lambda e^{-\lambda x}$, but the exercise don't say nothing about it. $\endgroup$ – Roland Sep 30 '15 at 14:40
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    $\begingroup$ See, interchanging the two interchanges your answers, because $\frac{\frac{1}{\mu_1}}{\frac{1}{\mu_1}+\frac{1}{\mu_3}}=\frac{1}{1+\frac{\mu_1}{\mu_3}}=\frac{\mu_3}{\mu_3+\mu_1}$. $\endgroup$ – Ian Sep 30 '15 at 14:41
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    $\begingroup$ That's right (except for the obvious typo). The point is that when you take the minimum, you're waiting for the first event, and so the rates of (independent) exponentials will add together. You can make an electrical analogy to parallel resistors (resistance is to mean as conductance is to reciprocal of mean). $\endgroup$ – Ian Sep 30 '15 at 14:58
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    $\begingroup$ Let us assume that the given answer is the one they intended to give (not a good assumption!). Then the answer makes sense if the $\mu_i$ are the means. That interpretation is reinforced by the choice of the letter $\mu$. I would not call that the rate, it is certainly not the rate of the associated Poisson. Sadly, usage is not uniform. $\endgroup$ – André Nicolas Sep 30 '15 at 15:01

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