Question

Repair calls are handled by one repairman at a photocopy shop. Repair time, including travel time, is exponentially distributed, with a mean of 1.6 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. (Round your answer to 2 decimal places.)
Amount of time hoursd. Determine the probability of two or more customers in the system. (Do not round intermediate calculations. Round your answer to 4 decimal places.)
Probability

Answer 1

Answer:

the average number of customers awaiting repairs = 0.30

the system utilization = 42

the amount of time that the repairman is not out on a call is  = 4.64 hours

the probability of two or more customers in the system = 0.1764

Explanation:

Given that :

Repair time, including travel time =  mean of 1.6 hours per call.

Requests for copier repairs = mean rate of 2.1 per eight-hour day

i.e mean rate R = 2.1/day

Time = 8 hours

thus; mean rate μ = 8 hours/ 1.6 hours = 5

(a)

Let the average number of customers awaiting repairs be $I_i$ :

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

$I_i = \dfrac{2.1^2}{5 (5-2.1)}$

$I_i = \dfrac{4.41}{5 (2.9)}$

$I_i = \dfrac{4.41}{14.5}$

$\mathbf{I_i = 0.30}$

the average number of customers awaiting repairs = 0.30

(b) Determine system utilization.

The system utilization is determined as follows:

$\delta = \dfrac{R}{\mu}$

$\delta = \dfrac{2.1}{5}$

${\delta = 0.42}$

$\mathbf{\delta = 42}$

(c) The amount of time during an eight-hour day that the repairman is not out on a call is calculated as :

Percentage of Idle time = 1 - $\delta$

Percentage of Idle time = 1 - 0.42

Percentage of Idle time = 0.58

However during an 8 hour day; The amount of time that the repairman is not out on a call is = 0.58 × 8 = 4.64 hours

(d)

the probability of two or more customers in the system by assuming Poisson Distribution is:

P(N ≥ 2) = 1 - (P₀+ P₁)

where;

P₀ = 0.58

P₁ = 0.58  × 0.42 = 0.2436

P(N ≥ 2) = 1 - ( 0.58 + 0.2436)

P(N ≥ 2) = 1 - 0.8236

P(N ≥ 2) = 0.1764

Thus; the probability of two or more customers in the system is 0.1764