Question

Speedy Oil provides a single-server automobile oil change and lubrication service. Customers provide an arrival rate of 2.5 cars per hour. The service rate is 5 cars per hour. Assume that arrivals follow a Poisson probability distribution and that service times follow an exponential probability distribution.
If required, round your answer to the nearest whole number.
(a) What is the average number of cars in the system?
(b) What is the average time that a car waits for the oil and lubrication service to begin?
(c) What is the average time a car spends in the system?
(d) What is the probability that an arrival has to wait for service?

Answer 1

Answer:

(a) Average number of cars in the system is 1

(b) Average time a car waits is 12 minutes

(c) Average time a car spends in the system is 2 minutes

(d) Probability that an arrival has to wait for service is 0.08.

Step-by-step explanation:

We are given the following

Arrival Rate, A = 2.5

Service Rate B = 5

(a) Average Number of Cars in the System is determined by dividing the Arrival Rate A by the difference between the Service Rate B, and Arrival Rate A.

Average number of cars = A/(B - A)

= 2.5/(5 - 2.5)

= 2.5/2.5 = 1

There is an average of 1 car.

(b) Average time a car waits = A/B(B - A)

= 2.5/5(5 - 2.5)

= 2.5/(5 × 2.5)

= 2.5/12.5

= 1/5

= 0.20 hours

Which is 12 minutes

(c) Average time a car spends in the system is the ratio of the average time a car waits to the service rate.

Average time = 0.2/5

= 0.04 hours

= 2.4 minutes

Which is approximately 2 minutes.

(d) Probability that an arrival has to wait for service is the ratio of the average time a car waits to rate of arrivals.

Probability = 0.2/2.5

= 0.08