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Consider a 2 server system in which a customer is served first by server 1, then by server 2, then departs. The service times at server $i$ are exponentially distributed with rate $u_i$ ($i=1,2$). When you arrive, you find server 1 free and two customers at server 2: Lee in service and Yang waiting in line.

a) Find $P_L$, the probability Lee is still in service when you move over to server 2.

b) Find $P_Y$, the probability that Yang is still in the system when you move over to server 2.

c) Your total time $T$ in the system can be written as $T=S_1+S_2+W_L+W_Y$, where $S_i$ is your service time at server $i$, $W_L$ is the amount of time you wait in queue while Lee is being served, and $W_Y$ is the amount of time you wait in queue while yang is being served. Find $E[T]$, by determining each of $ES_1, ES_2,EW_L, EW_Y$.


Attempt:

Let L=Lee's service time; X=my service time, Y= Yang's service time.

A) (Sure). Since they are all exponential, then $P(X<L)=\mu_1/(\mu_1+\mu_2)$.

B) (Unsure). I think this situation has two possibilities:

P(finish service 1, and Yang is still waiting)+P(finish service 1, and Yang is being service)$=P(X<L)+P(X>L)P(X<Y)$.

In the second term, I have used the non-memory property of exponential. First, my service time will be greater than Lee, and given that happen, the exponential variable will refresh, and my service time would have to be less than Young's.

C) (Very Unsure). $E[S_1]+E[S_2]=1/\mu_1 + 1/\mu_2$.

Let $W=W_L+W_Y$

$E[W]=E[W|X<L]P(X<L)+E[W|X>L]P(x>L)=2/\mu_2*(\mu_1/(\mu_1+\mu_2)+1/\mu_2*\mu_2/(\mu_2+\mu_1)$.

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  • $\begingroup$ Sorry, I should have included my solution before posting this question. I am unsure about the memory-less property of the exponential distribution. Usually, we could utilize such property to solve the problem in a very clean and nice way. In this problem, I am not sure how to apply this property in part b and part c. Anyhow, I am 80% confident about part b) and very unconfident about part c. $\endgroup$
    – kuku
    Commented Mar 19, 2016 at 23:34
  • $\begingroup$ see my comment below, is this correct? $\endgroup$
    – kuku
    Commented Mar 21, 2016 at 2:27

1 Answer 1

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$(B)$ $$P_Y = \frac{\mu_1}{\mu_1 + \mu_2} + \frac{\mu_2}{\mu_1 + \mu_2}\frac{\mu_1}{\mu_1 + \mu_2}$$

$(C)$

$$\begin{align} E[T] & = E[S_1] + E[S_2] + E[W_L] + E[W_Y] \\ & \\ & = \frac{1}{\mu_1} + \frac{1}{\mu_2} + \frac{1}{\mu_2}\frac{\mu_1}{\mu_1 + \mu_2} + \frac{1}{\mu_2}\left(\frac{\mu_1}{\mu_1 + \mu_2} + \frac{\mu_2}{\mu_1 + \mu_2}\frac{\mu_1}{\mu_1 + \mu_2}\right) \\ \end{align}$$

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  • $\begingroup$ $E[W_L]=P(X<L)*(2*E[S_2])$ , this is so because once I finish server 1, then I find Lee is still being served (i.e, $P(X<L)$). Then I will have to wait Lee to finish, and Young to finish, this gives $2*E[S_2]$. $\endgroup$
    – kuku
    Commented Mar 21, 2016 at 2:20
  • $\begingroup$ $E[W_Y]=P(X>L)P(X<Y)*(E[S_2])$, this is so because once I finish server 1, then I find Lee is finish (i.e., $P(X>L)$, and Young is being served (i.e., $P(X<Y)$. Then I will have to wait Young to finish, this gives $E[S_2]$. $\endgroup$
    – kuku
    Commented Mar 21, 2016 at 2:22
  • $\begingroup$ and off course, $P(X>L)=\frac{\mu_2}{\mu_1+\mu_2}$, and similiarly, $P(X<L)=\frac{\mu_1}{\mu_1+\mu_2}$ $\endgroup$
    – kuku
    Commented Mar 21, 2016 at 2:25
  • $\begingroup$ In your comment above, is it inconsistent with your answer? (Your third term, you have $2P_L/\mu_2$, whereas in your answer, you have $P_L/\mu_2$ $\endgroup$
    – randy
    Commented Mar 21, 2016 at 15:46
  • $\begingroup$ @randy Yes, that was wrong. I am fairly confident the answer should be as in my answer. $\endgroup$
    – JKnecht
    Commented Mar 21, 2016 at 16:52

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