Repair calls are handled by one repairman at a photocopy sho

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).

Determine the average number of customers awaiting repairs. (Round your answer to 2 decimal places.)

Determine system utilization. (Round your answer to the nearest whole percent. Omit the \"%\" sign in your response.)

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.)

Determine the probability of two or more customers in the system. (Do not round intermediate calculations. Round your answer to 4 decimal places.)

Solution

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