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 4 boats per hour, and that service times follow an exponential probability distribution, with a service rate of 8 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.

Answer 1

Number of servers of boat facilities, M = 2.

Arrival rate = 4 boats per hour.

Service rate of the boats = 8 boats per hour.

Probability of zero boats in the system,\mathrm{PO}=1 /\{[(1 / 0 !) \time(4 / 8) 0+(1 / 1 !) \time(4 / 8) 1]+[time(4 / 8) 2 /(2 ! \(1-(4 /(2 \8)))]\}PO=1/{[(1/0!)×time(4/8)0+(1/1!)×time(4/8)1]+[time(4/8)2/(2!×(1−(4/(2×8)) = 0.5475

Standard number of boats awaiting in line for service:

Lq =[\lambda.\mu.( \lambda / \mu )M / {(M – 1)! (M. \mu – \lambda )2}] x P0[λ.μ.(λ/μ)M/(M–1)!(M.μ–λ)2]xP0

= [\{4 \ 8 \time(4 / 8) 2\} /\{(2-1) ! \((2/ 8)-4) 2\}] \ 0.547[{4/8×times(4/8)2}/{(2−1)!×((2×8)−4)2}]×0.547 = 0.0556 boats.

The standard time a boat will devote waiting for service, Wq = 0.0556 divide by 4= 0.0139 hours = 0.834minutes.

The standard time a boat will take at the dock is 4.2345 minutes.

So there is the possibility of enlarging the docking facility.