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 $W=W_L+W_Y$
$E[W]=E[W|X<L]P(X<L)+E[W|X>L]P(x>L)=2/\mu_2*(\mu_1/(\mu_1+\mu_2)+1/\mu_2*\mu_2/(\mu_2+\mu_1)$.
Best Answer
$(B)$ $$P_Y = \frac{\mu_1}{\mu_1 + \mu_2} + \frac{\mu_2}{\mu_1 + \mu_2}\frac{\mu_1}{\mu_1 + \mu_2}$$
$(C)$
$$\begin{align} E[T] & = E[S_1] + E[S_2] + E[W_L] + E[W_Y] \\ & \\ & = \frac{1}{\mu_1} + \frac{1}{\mu_2} + \frac{1}{\mu_2}\frac{\mu_1}{\mu_1 + \mu_2} + \frac{1}{\mu_2}\left(\frac{\mu_1}{\mu_1 + \mu_2} + \frac{\mu_2}{\mu_1 + \mu_2}\frac{\mu_1}{\mu_1 + \mu_2}\right) \\ \end{align}$$