The new Fore and Aft Marina is to be located on the Ohio River near Madison, Ind
ID: 339916 • Letter: T
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?
Explanation / Answer
Please find below answers to first 4 questions :
Following data are provided :
Arrival rate of boats = a = 5 / hour
Service rate of boats = s = 9 / hour
Probability that boat dock will be idle = Po = ( 1 – a/s) = 1 – 5/9 = 0.44
Average number of boats that will be waiting for service ,
Lq = a^2/ S x ( S – a ) = 5 x 5 / 9 x ( 9 -5 ) = 5x5/9x4 = 25/36 = 0.6944
Average time a boat will spend waiting for service , Wq
= a/ S x ( s – a ) hour
= 5/ 9 x ( 9 – 5)
= 5 / 9 x4
= 5/36 hours
= 0.1389 ( rounded to 4 decimal places )
Average time boat will spend at the dock
= Average time a boat will spend waiting for service + 1/S hour
= 0.1389 + 1/9 hour
= 0.1389 + 0.1111 hours
= 0.25 hours
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