Repair calls are handled by one repairman at a photocopy shop. Repair time, incl
ID: 335908 • Letter: R
Question
Repair calls are handled by one repairman at a photocopy shop. Repair time, including travel time, is exponentially distributed, with a mean of 2.7 hours per call. Requests for copier repairs come in at a mean rate of 1.8 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 %
c. 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 places.)
Amount of time 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
L = average arrival rate = 1.8 per day = 0.225 per hour
M = average service rate = 1 in 2.7 hours = (1/2.7) per hour
(a)
Lq = average number of customers waiting = L2 / (M*(M - L)) = (0.225^2)/((1/2.7)*((1/2.7) - 0.225)) = 0.94
(b)
Utilization = L/M = 0.225*2.7 = 0.6075 = 60.75%
(c)
P0 = probability of idle repairman = 1 - Utilization = 0.3925
% of time free = 8 hrs x 0.3925 = 3.14 hours
(d)
Pn>1 = Utilization1+1 = 0.6075^2 = 0.3691