25 Problem 18-1 Repair calls are handled by one repairman at a photocopy shop. R

Question

25 Problem 18-1 Repair calls are handled by one repairman at a photocopy shop. Repair time, including travel time, is exponentially distributed, with a mean of 2.0 hours per call. Requests for copier repairs come in at a mean rate of 2.5 per eight-hour day (assume Poisson) a. Determine the average number of customers awaiting repairs. (Round your answer to 2 decimal places.) Number of customers b. Determine system utilization. (Round your answer to 2 decimal places. Omit the "%" sign in your response.) System utilization e. Determine the amount of time during an eight-hour day that the repairman is not out on a call. (Round your answer to 2 decimal polints eBook Hint Print --% Heterences aces. Amount of time 2 2.9 hours d. Determine the probability of two or more customers in the system. (Do not round intermediate calculations. Round your answer to 4 decimal places. Probability

Explanation / Answer

Mean rate () = 2.5 per day

= Total hours in a day/no of hours per call

= 8/2 = 4 per day

= / = 2.5/4 = 0.625

a) Average no. of customer awaiting= Ii= 2 / (1- )

=0.6252 / (1-0.625)

=0.390625/(0.375)

=1.041 customers

(b) Utilization: = / = 2.5/4 = 0.625=62.5%

(c) Percentage of time idle = 1- = 1- 62.5% = 37.5%

So, idle time = 37.5%*8 = 3 hours

(d) Probability (N2) = (1- P0– P1) = 1 – 0.3437 – 0.3437*0.6563 = 0.4307

Here P1= Utilization*percentage of idle time

=1-0.375-(0.375*0.625)

=1-0.375-0.234375

=1-0.609375

=0.390625

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