Question

Speedy Oil provides a single-server automobile oil change and lubrication service. Customers provide an arrival rate of 4 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.

A) What is the average number of cars in the system? If required, round your answer to two decimal places

L =

B) What is the average time that a car waits for the oil and lubrication service to begin? If required, round your answer to two decimal places.

W_q = ____hours

C) What is the average time a car spends in the system? If required, round your answer to two decimal places.

W = ____hours

D) What is the probability that an arrival has to wait for service? If required, round your answer to two decimal places.

P_w = _____

Answer 1

Given:

Mean arrival rate = 4 cars per hour (poisson dist)

Service rate = 5 cars per hour

Service time follows exponential distribution

i.e. average service time is  

Thus, A) Average numbers of cars in system (L) = = = 4 cars

Thus at an average there are 4 cars in the system. i.e. L = 4

B) Average time car waits before service begins (W_q) = p*W

Here p = Average utilization = = = 0.8

and W =average time spent waiting in system = = = 1 hour

Thus W_q = 0.8*1 = 0.8 hours

C) Average time car spends in system W = 1 hour ( as calculated above in part B)

D) Probablity that an arrival has to wait (P_w)

If an arrival has to wait then there must be more than 5 cars in the system as the service rate is 5 cars per hour

Thus probability of more than 5 cars in the system = 1 - probability of 5 or less cars in system

Thus P_w = Also,

P_w =

Here p is the utilization and n is the number of cars in system

Thus P_w = 1 - (1-0.8)(0.8⁰+0.8¹+0.8²+0.8³+0.8⁴+0.8⁵)

Thus P_w = 1- 0.2 (1+0.8+0.64+0.512+0.4096+0.327)

Thus P_w = 1 - 0.2*3.69 = 1 - 0.737

Thus P_w = 0.2621 i.e. 26.21%

So 26.21% chances are there that an arrival has to wait for the service