Question

The new Fore and Aft Marina is to be located on the Ohio River near Madison, Indiana. Assume that Fore and Aft decides to build a docking facility where one boat at a time can stop for gas and servicing. Assume that arrivals follow a Poisson probability distribution, with an arrival rate of 5 boats per hour, and that service times follow an exponential probability distribution, with a service rate of 9 boats per hour. The manager of the Fore and Aft Marina wants to investigate the possibility of enlarging the docking facility so that two boats can stop for gas and servicing simultaneously.

Assume that the arrival rate is 5 boats per hour and that the service rate for each server is 9 boats per hour. What is the probability that the boat dock will be idle? If required, round your answer to two decimal places.

P0 =

What is the average number of boats that will be waiting for service? If required, round your answer to four decimal places

.Lq =

What is the average time a boat will spend waiting for service? If required, round your answer to four decimal places.

Wq =

What is the average time a boat will spend at the dock? If required, round your answer to four decimal places.

W =

If you were the manager of Fore and Aft Marina, would you be satisfied with the service level your system will be providing? why?

Answer 1

a. P0 = 0.5686

b Lq = 0.0467

c. Wq 0.0093 hours

d. W= 0.1205 hours.

e. Yes

Step-by-step explanation:

This is a case of a multiple-server queuing Model.

The average arrival rate, λ = 5 per hour .

The average service rate, µ = 9 per hour .

The number of servers (k) = 2 .

The ratio λ/µ = 5/9 = 0.556. The nearest value from the table (attached as image) is 0.55.

a. P0 = 0.5686

b. Lq can be obtained using the formula below.

Lq = λ.μ((λ/μ))^s)P0 /((s-1)!(s.μ) - λ)^2)

Lq = 5*9*((5/9)^2)*0.5686/((2-1)!*(2*9 - 5)^2) = 0.0467

c. Wq can be obtained using the formula below

Wq = Lq / λ = 0.0467/5 = 0.0093 hours

d. The average time a boat will spend at the dock, W

W= Wq + 1/µ = 0.0093 + 1/9 = 0.1205 hours.

e. The average service time is 6 minutes (60/10), and the average waiting time is also 6 minutes. Since the average time, a boat will spend waiting is just 6 minutes, which is acceptable. Hence, the level of service is satisfactory.