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randy patton wrote at 2013-07-14 08:52:18
This question needs to be answered using the results of queueing theory. The question uses the terminology of queueing theory so it is assumed that the questioner has access to these results. In particular, Little's Law is key,

mean number of units in system = (arrival rate of units)x(mean time of a unit in system)

E(n) = λE(Ts), where Ts = waiting time (in queue) plus service time = sojourn time.

The system described has exponential arrival and processing times (Poisson processes) with one processor and so has a Kendall representation of M/M/1.

From the given information, λ = 3/hr and μ = (1/(10 minutes/unit) = 6/hr.

i) The utilization factor is ρ = λ/λ = 3/6 = 1/2. Note that we need ρ < 1 or else the system overloads.

ii) The probability of finding n people in the system is given by P(n) = (1-ρ)ρ^n, so for 2 people, we have P(2) = 1/8.

iii) Expected number of units in the queue is E(Pq) = ρ^2/(1-ρ) = 1/2.

iv) For the mean waiting time in the system, I assume you mean the mean waiting time for a customer in the queue (as opposed to the sojourn time = waiting time in queue + processing time); this is given from iii) by Little's Law

expected number of units in the queue = (arrival rate)x(mean number of people in queue)

E(Pq) = λE(Wq)  or  E(Wq) = (1/μ)/(1-ρ) = (1/6)hr = 10 minutes.

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#### Scott A Wilson

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