Thread: ~Poisson process

Hi,
I have this very complicated problem. Can you help me, please!

In the bank you can be served by server 1 or servwer 2.
The service time from the server i is exponent with rate Mu_i
When you arrive at the bank you wait for the server 1 even if he is busy.
After you finish with server 1 you are going to server 2 if that server is free.
If server 2 is not free you are waiting at server 1 until server 2 is free/
Suppose when you arrive there is 1 customer in the bank and he is served by server 1.
What is the expected total time you spend in the system?

Thank you for your help!

basically,
you know that you need to wait for server 1 to be done with the guy, before you can go to server 1.
once you are at server 1, if you finish earlier, u have to wait for server 2.
if you finish later, u go immediately to server 2
lastly, you will have to wait to be done with server 2.

therefore,

E(time of you being done) = E(server1 serving guy)+ E(Max(server2 serving guy, server1 serving you)) + E(Server2 serving you)

= 1/mu_1 + (1/mu_1 + 1/mu_2 - 1/(mu_1+mu_2)) + 1/mu_2

Thanks, that is really helpful.
Just to be sure miu_1 is 'server1 serving guy' or 'server1 serving you'?
Because if it is 'server1 serving guy' where is the time of serving you?

Thanks!

and just so you know.
the reason why Max(server 1, server 2) = 1/mu_1 + 1/mu_2 - 1/(mu_1+mu_2)

is that:
min(X,Y) = V
P(V>x) = P(X>x,Y>x) = P(X>x)P(Y>x) independance property
= exp(-mu_1x)exp(-mu_2x) = exp(-(mu_1+mu_2))x
=> V ~ exponential (mu_1 + mu_2)

W = max(X,Y)

and therefore, W+V = X+Y

E(W) = E(X) + E(Y) - E(V) = 1/mu_1 + 1/mu_2 - 1/(mu_1 + mu_2)

it doesn't matter because it is the same server
and exponential R.V has the same mean everytime regardless of who the person serves.