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.

A) The average number of customers awaiting repairs is; 0.06

B) The system utilization is; 21.98%

C) The amount of time during an eight-hour day that the repairman is not out on a call is; 6.24 hours

D) The probability of two or more customers in the system is; P₂ = 0.0483

A) We are given;

Arrival rate; a = 1.6 calls per 8 hours = 1.6/8 = 0.2 calls per hour

Repair time; s = 1.1 hours per call = 1/1.1 = 0.91 call per hour

Formula for average number of customers waiting repairs is;

a²/(s( s - a))

⇒ (0.2²)/(0.91 × (0.91 – 0.2)

= 0.04/(0.91 × 0.71)

= 0.06

B) Formula for System utilization is; a/s × 100

⇒ (0.2/0.91) × 100 = 21.98%

C) Percentage of time the repairman is not out on a call = 100 – 21.98% = 78.02%

∴ amount of time in a 8 hour day repairman is not on a call = 78.02% of 8 hours = 6.24 hours

D) Probability that zero customers are in the system; P₀ = 1 – (a/s)

P₀ = 1 – (0.2/0.91)

P₀ = 1 – 0.2198

P₀ = 0.7802

Probability that 1 customer is in the system; P₁= ( a/s ) × P₀

P₁ = (0.2/0.91) × P₀

P₁ =0.2198  × 0.7802

P₁ = 0.1715

Probability that 2 or more customers are in the system;

P₂ = 1 - (P₀ + P₁)

P₂ = 1 – (0.7802 + 0.1715)

P₂ = 0.0483

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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

= = 5 per day

Based on the above information

a. The average number of customers is

= 151

b. The system utilization is

=

= 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

=

= 5.44 hours

d. Now the probability of two or more customers is

= 0.1024

Therefore we simply applied the above formulas