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 8 boats per hour, and that service times follow an exponential probability distribution, with a service rate of 10 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.
A. What is the probability that the boat dock will be idle? Round your answer to four decimal places.
P0=
B. What is the average number of boats that will be waiting for service? Round your answer to four decimal places.
Lq=
C. What is the average time a boat will spend waiting for service? Round your answer to four decimal places.
Wq=
D. What is the average time a boat will spend at the dock? Round your answer to four decimal places.
W=
E. If you were the manager of Fore and Aft Marina, would you be satisfied with the service level your system will be providing? Round your answers to whole numbers.
Because the average wait time is ____ seconds. Each channel is idle ____ % of the time.

Answer 1

Answer:

A.) 0.2

B.) 3.2

C.) 0.4

D.) 0.5

Step-by-step explanation:

Arrival rate, λ = 8 per hour

Service rate, μ = 10 boats per hour

1.)

Probability that boat dock will be idle, P0:

P0 = 1 - λ/μ

P0 = 1 - 8/10

P0 = 1 - 0.8

P0 = 0.2

Average number of boats waiting for service :

Lq = λ² ÷ μ(μ - λ)

Lq = 8² ÷ 10(10- 8)

Lq = 64 ÷ 10(2)

Lq = 64 ÷ 20

Lq = 3.2

Average time spent waiting for service, Wq :

Wq = λ ÷ μ(μ - λ)

Wq = 8 ÷ 10(10- 8)

Wq = 8 ÷ 10(2)

Wq = 8 / 20

Wq = 0.4 hour

Wq = 0.4 * 60 = 24 minutes

Average time spent at the dock :

1 ÷ (μ - λ)

1 ÷ (10 - 8)

1 ÷ 2

= 0.5 hour

= 0.5 * 60 = 30 minutes