Question

​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​ = ____________

Answer 1

The student's question regarding queueing theory is solved by finding the utilization rate, average downtime, average number of waiting machines, and the probability of having more than one machine in the service system.

Step-by-step explanation:

The student is asking about a queueing theory problem, specifically related to the Poisson distribution and the negative exponential distribution within an electronics corporation's service system.

a. The utilization rate (ρ) of this service system is the ratio of the arrival rate (λ) to the service rate (μ). Here, ρ = λ / μ = 4/10 = 0.4 or 40%.

b. The average downtime for a broken machine is the inverse of the service rate, μ. It's given by 1/μ = 1/10 workday per machine = 0.8 hours (or 48 minutes) per machine.

c. The average number of machines waiting to receive service (Lq) can be calculated using the formula Lq = ρ^2 / (1 - ρ) when ρ < 1. Here, Lq = 0.4^2 / (1-0.4) = 0.16 / 0.6 ≈ 0.2667 machines.

d. The probability that there is more than one machine in the system (P(X > 1)) can be found by 1 - P(X ≤ 1), where P(X ≤ k) for the Poisson distribution with parameter λt can be found using the cumulative distribution function. For k = 1, this is 1 - [P(X = 0) + P(X = 1)] = 1 - e^(-λt)(1 + λt), where t is the length of the workday. Because we consider the entire workday, t = 1, so we get 1 - e^(-4)(1 + 4) ≈ 0.9084 or 90.84%.

Answer 2

(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

Step-by-step 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