Question

Repair calls are handled by one repairman at a photocopy shop. Repair time, including travel time, is exponentially distributed, with a mean of 1.1 hours per call. Requests for copier repairs come in at a mean rate of 1.6 per eight-hour day (assume Poisson). Determine the following:
Required:
a. Determine the average number of customers awaiting repairs.
b. Determine system utilization.
c. Determine the amount of time during an eight-hour day that the repairman is not out on a call.
d. Determine the probability of two or more customers in the system.

Answer 1

Answer and Step-by-step explanation:

Data provided in the question

Mean = 1.1 hours per call =

R = Mean rate = 1.6 per eight hour day

$\mu$ = $\frac{8}{1.6}$ = 5 per day

Based on the above information

a. The average number of customers is

$= \frac{R^2}{\mu(\mu- R)}$

$= \frac{1.6^2}{5(5- 1.6)}$

= 151

b. The system utilization is

$= \frac{R}{\mu}$

= $\frac{1.6}{5}$

= 0.32

c. The amount of time required is

= 1 - system utilization

= 1 - 0.32

= 0.68

And, there is 8 hours per day

So, it would be

= $0.68 \times 8$

= 5.44 hours

d. Now the probability of two or more customers is

$= 1 - (0.68 + 0.68 \times 0.32)$

= 0.1024

Therefore we simply applied the above formulas