Broken cell phones arrive to a phone repair store according to a Poisson process

Question

Broken cell phones arrive to a phone repair store according to a Poisson process, with rate of 3 phones per hour. Assume that only one phone can be repaired at any given time, and that the time it takes to repair a cell phone is exponentially distributed with mean 15 minutes.

(a) How long on average does it take newly arrived phone to be repaired?

(b) What is the average number of broken phones currently in the store (including the one being repaired at the moment)?

(c) What is the fraction of time the worker in the store is actually fixing phones?

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Explanation / Answer

The given problem can be modeled into classic problem of queuing theory with one server.

Here, arrival rate = 3 per hour

and service rate = 4 per hour

utilization p = 3/4 = 0.75

Number of customers in queue, Lq = p^2/(1-p) = 2.25

Wait time in queue Wq = Lq/lambda = 2.25/3 = 0.75 hour

Wait time in system = W = Wq + 1/4 = 0.75 + 0.25 = 1 hour

(A) Hence it takes 1 hour to newly arrived phone to be repaired.

(B)

Number of broken phones in the system are L = W(3) = 1 * 3 = 3

(C)

Fraction of time the worker in the store is actually fixing phones is p (utilization) = 0.75

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