Question

# Problem 11-17 (Algorithmic) The new Fore and Aft Marina is to be located on the Ohio...

Problem 11-17 (Algorithmic)

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 4 boats per hour, and that service times follow an exponential probability distribution, with a service rate of 5 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.

Note: Use P0 values from Table 11.4 to answer the questions below.

1. What is the probability that the boat dock will be idle? Round your answer to four decimal places.

P0 =
2. What is the average number of boats that will be waiting for service? Round your answer to four decimal places.

Lq =
3. What is the average time a boat will spend waiting for service? Round your answer to four decimal places.

Wq =  hours
4. What is the average time a boat will spend at the dock? Round your answer to four decimal places.

W =  hours
5. If you were the manager of Fore and Aft Marina, would you be satisfied with the service level your system will be providing?

Yes

Because the average wait time is  seconds. Each channel is idle  % of the time.

Arrival rate lambda = 4 per hour

Service rate Mu = 5 per hour.

Utilization rho = 4/5 = 0.8

P(0) with 2 channels

= 1/ [ 1+rho+ (rho)^2*mu /(2*mu-lambda)]

= 1/ [ 1+ 0.8+0.64*5/(10-4)]

= 1/2.33

= 0.4291

Lq = P(0) *lambda*mu*(rho)^2 / (2*mu-lambda)^2

= 0 4291 *20*0 .64 /36

= 0.1525

Wq. = Lq / lambda = 0.1525/4 = 0.0381 hours

= 2.286 min.

L = Lq +rho = 0.1525+0.8 = 0.9525

W= L / lambda = 0.9525/4 = 0.23812 hour

= 14.2875 min.

Note: As per policies, I can answer first 4 parts of a problem. Inconvenience is regretted

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