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Math Help - Queuing Theory problem, M/M/1/K queue with twist

  1. #1
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    Queuing Theory problem, M/M/1/K queue with twist

    The problem statement
    Customers arrive to a register as a poisson process with arrival rate λ and are serviced with an exponential distribution with service rate μ. When a customer arrives he'll decide to join the line or not depending on how many people are currently in the system (not including himself). In other words, if at a given time n people are in the system then a new customer will join the line with probability βn, and not join the line and not take the service with probability 1-βn. However, 0 ≤ βn ≤ 1 (0 ≤ n ≤ N), βn = 0 (n > N).

    (1) Find the balance equation expressed with the balance probability state pn (where n is the number of customers in the system).

    (2) Solve the equation in (1).

    My attempt at a solution

    (1) So, my problem here is where the β will enter the problem. I'm thinking that the events "Customer stays" and "Customer arrives" are independent, resulting in the following:
    With this I would get:
    λ β0 p0 = μ p1
    λ βn pn + μ pn = μ pn+1 + λ β[sub]n-1/SUB] pn-1 for 1 ≤ n ≤ N-1
    λ βN-1 pN-1 = μ pN

    Is this correct? If not, any hint?

    (2) If the above is correct, how to solve it? Any hint is appreciated.
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  2. #2
    Member Abu-Khalil's Avatar
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    Re: Queuing Theory problem, M/M/1/K queue with twist

    You are right. Those events are independent.

    To solve that, just use the recursion to express any p_n in terms of p_0 and then remember \sum_{n=0}^N p_n=1.
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  3. #3
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    Re: Queuing Theory problem, M/M/1/K queue with twist

    Thanks! I calculated it and got a pretty messy, but hopefully correct, expression.
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