Virginia's Ron McPherson Electronics Corporation retains a service crew to repair machine breakdowns that occur on an average of λ = 4 per 8-hour workday (approximately Poisson in nature). The crew can service an average of µ = 10 machines per workday, with a repair time distribution that resembles the negative exponential distribution.

a. The utilization rate of this service system =_.
b. The average downtime for a broken machine =
c. The average number of machines waiting to receive service at any given time =
d. The probability that there is more than one machine that is in the system = ____

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27-04-2020
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Virginia's Ron McPherson Electronics Corporation retains a service crew to repair machine breakdowns that occur on an average of λ = 4 per 8-hour workday (approximately Poisson in nature). The crew can service an average of µ = 10 machines per workday, with a repair time distribution that resembles the negative exponential distribution.

a. The utilization rate of this service system =_.
b. The average downtime for a broken machine =
c. The average number of machines waiting to receive service at any given time =
d. The probability that there is more than one machine that is in the system = ____

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gbenewaeternity gbenewaeternity · Answer 1 · 2020-05-01T07:36:04+02:00

Answer:

(a) The utilization rate = 50%

(b) The average down time = 2 hours

(c) Average number of machines waiting = 0.5 Machines

(d) Pn>1 = 0.25

Explanation:

Given that:

Arrival Rate= λ = 4 per day

Service Rate = μ = 8 hour per day

Now,

a)

The utilization rate of this service system is given as;

The utilization rate = λ / μ

= 4/8

= 0.5 = 50%

b)

The average downtime for a broken machine is calculated as;

The average down time = 1/( μ- λ)

= 1/(8-4)

= 1/4

= 0.25 days

It is given that work day = 8 hours

Therefore, 0.25*8 = 2 hours

c)

Average number of machines waiting to be serviced at any given time is calculated as;

Average number of machines waiting = λ²/(μ*(μ-λ))

= 4²/(8*(8-4))

= 16/32

= 0.5 Machines

d) The probability that more than 1 machine is in the system is calculated as;

Pn>k = (λ/μ)^k+1

where k number of machine = 1 machine

Pn>1 = (4/8)¹⁺¹

=(4/8)²

= 0.5²

= 0.25