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Suppose customers arrive in a one-server queue according to a Poisson distribution with rate lambda=1 (in hours). Suppose that the service times equal 1/4 hour, 1/2 hour, or one hour each with probability 1/3.

(a) Assume that the queue is empty and a customer arrives. What is the expected amount of time until that customer leaves?

(b) Assume that the queue is empty and a customer arrives. What is the expected amount of time until the queue is empty again?

(c) At a large time t what is the probability that there are no customers in the queue?

I'm trying to do couple of practice problems involving queuing before my exam and I am really confused, I would really appreciate it if someone can show me how to do this problem.

Thanks

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  • $\begingroup$ Part (a) is easy: $$\left( \frac 1 4 \times \frac 1 3 \right) + \left( \frac 1 2 \times \frac 1 3 \right) + \left( 1 \times \frac 1 3\right).$$ Part (c) is about the equilibrium distribution of a Markov chain. $\qquad$ $\endgroup$ Commented May 2, 2020 at 5:18

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I actually figured it out myself and decided to post the answer in case someone needed it.

(a) It is the mean service time: μ = (1/4 + 1/2 + 1)/3 = 7/12.

(b) This is μ/(1 − λμ) = μ/(1 − μ) = 7/5 hours.

(c) This is 1/(1 + 7/5) = 5/12.

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    $\begingroup$ We usually use $\mu$ to refer to the service rate, not the mean service duration. $\endgroup$
    – Brian Tung
    Commented Jul 17, 2018 at 20:18
  • $\begingroup$ Your result in $(b)$ can't be right. It tends to $0$ as $\lambda\to\infty$, whereas the expected time until the queue is empty is in fact infinite if $\lambda$ is greater than the inverse mean service time. $\endgroup$
    – joriki
    Commented May 2, 2020 at 9:40

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