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Problem 8: 10 points Consider a queuing system M/M/1 with one server. Customer arrivals form a Poisson process with the intensity A 15 per hour. Service times are exponentially distributed with the expectation3 minutes Assume that the number of customers at t-0, has the stationary distribution. 1. Find the average queue length, (L) 2. What is the expected waiting time, (W), for a customer? 3. Determine the expected number of customers that have completed their services within the 8-hour shift

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Answer #1

1)

M/M/1

L_q = \frac{\lambda ^2}{\mu (\mu -\lambda )} = \frac{15^2}{20*(20-15)} = 2.25

2)

Expected waiting time in queue = Lq/\lambda = 2.25/(15/hour) = 4*2.25 min = 9 min

Ws = 1/(20 - 15) = 1/5 hour = 12 min

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