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Math Help - ~Poisson process

  1. #1
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    Red face ~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!
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  2. #2
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    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
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  3. #3
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    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!
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  4. #4
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    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)
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  5. #5
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    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.
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