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 [tex]I_i[/tex] :
[tex]I_i = \dfrac{R^2}{\mu (\mu-R)}[/tex]
[tex]I_i = \dfrac{2.1^2}{5 (5-2.1)}[/tex]
[tex]I_i = \dfrac{4.41}{5 (2.9)}[/tex]
[tex]I_i = \dfrac{4.41}{14.5}[/tex]
[tex]\mathbf{I_i = 0.30}[/tex]
the average number of customers awaiting repairs = 0.30
(b) Determine system utilization.
The system utilization is determined as follows:
[tex]\delta = \dfrac{R}{\mu}[/tex]
[tex]\delta = \dfrac{2.1}{5}[/tex]
[tex]{\delta = 0.42}[/tex]
[tex]\mathbf{\delta = 42}[/tex]
(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 - [tex]\delta[/tex]
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
