# ~Poisson process

• Sep 27th 2009, 05:31 PM
bamby
~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?

• Sep 27th 2009, 10:05 PM
bling23
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
• Sep 27th 2009, 10:19 PM
bamby
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!
• Sep 27th 2009, 10:22 PM
bling23
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)
• Sep 27th 2009, 10:23 PM
bling23
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.