Question: Repair calls are handled by one repairman at a photocopy shop. Repair

Question

Question:

Repair calls are handled by one repairman at a photocopy shop. Repair time, including travel
time, is exponentially distributed, with a mean of 1 hour and 20 minutes per call. Requests for copier repairs
come in at a mean rate of five per eight-hour day (assume Poisson distribution). Determine:

a. The average number of customers awaiting repairs.

b. System utilization.

c. The amount of time during an eight-hour day that the repairman is not out on a call.

d. The probability of two or more customers in the system.

Please show show how you did it.

Explanation / Answer

Mean rate of five per eight-hour day implies arrival rate is 5 per day (per 8hours).

All inclusive repair time has mean 1 hour and 20 minutes implies service rate is 6 (480/80) per day (per 8 hours).

As arrivals follows Poisson distribution and service time follows exponential distribution so formulas for M/M/1 queue system are applicable.

The average number of customers awaiting repairs =(Arrival rate)2 / (service rate*(service rate-arrival rate))=25/6

say 4.17approx. and average number in the queue system is arrival rate/(service rate-arrival rate) = 5/1=5 per day

b. System utilization = arrival rate/service rate = 5/6=83.33%

c. The amount of time repairman idle = 1 - system utilization = 1 - (5/6) = 1/6 of day(8 hours)

That is =80 minutes ( 1hour and 20minutes)

d. The probability of two or more customers in the system is using formula for Prob.( n>= 2) = (arrival rate/service rate)2 where n represents number of customers in the queue system, 25/36 = .69444

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