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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 2.1 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 the nearest whole percent. 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. (Use your rounded answer from Part b. 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

The problem of repair calls at photocopy shop will be solved using formula for queuing theory M/M/1 type as per Kendall’s notation
.Let, Arrival rate ( 2.1 calls per 8 hour ) = a = 2.1/8 = 0.2625 calls per hour
Repair time( 2 hours per call) = s = 0.5 call per hour
Average number of customers waiting repairs
= a^2/ s x ( s -a)
= ( 0.2625 x 0.2625) / 0.5 x ( 0.5 – 0.2625 )
= ( 0.2625 x 0.2625)/ ( 0.5 x 0.2375)
= 0.58
AVERAGE NUMBER OF CUSTOMERS AWITING REPAIRS = 0.58
System utilization = a/s x 100 = 0.2625/0.5 x 100 = 52.5%
SYSTEM UTILIZATION = 52.5%
Percentage of time the repairman is not out on a call = 100 – 52,5 = 47.5%
Thus, amount of time in a 8 hour day repairman is not on a call = 47.5% of 8 hours = 3.8 hours
AMOUNT OF TIME = 3.80 HOURS
Probability that ZERO customers are in the system = Po = 1 – a/s = 1 – 0.2625/0.5 = 1 – 0.525 = 0.475
Probability that 1 customer is in the system = P1 = ( a/s ) x Po = ( 0.2625/0.5) x Po = 0.525 x 0.475 = 0.2493
Probability that 2 or more customers are in the system
= 1 – Probability that ZERO customers are in the system – Probability of 1 customer in the system
= 1 – 0.475 – 0.2493
= 0.2757
PROBABILITY THAT 2 OR MORE CUSTOMERS ARE IN THE SYSTEM = 0.2757
 
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