Respuesta :
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
[tex]\mu[/tex] = [tex]\frac{8}{1.6}[/tex] = 5 per day
Based on the above information
a. The average number of customers is
[tex]= \frac{R^2}{\mu(\mu- R)}[/tex]
[tex]= \frac{1.6^2}{5(5- 1.6)}[/tex]
= 151
b. The system utilization is
[tex]= \frac{R}{\mu}[/tex]
= [tex]\frac{1.6}{5}[/tex]
= 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
= [tex]0.68 \times 8[/tex]
= 5.44 hours
d. Now the probability of two or more customers is
[tex]= 1 - (0.68 + 0.68 \times 0.32)[/tex]
= 0.1024
Therefore we simply applied the above formulas
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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