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 6 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. find P0 Lq Wq W

Answer 1

Po = 0.5385, Lq = 0.0593 boats, Wq = 0.5930 minutes, W = 6.5930 minutes.

Explanation:

The problem is that of Multiple-server Queuing Model.

Number of servers, M = 2.

Arrival rate, $\lambda$= 6 boats per hour.

Service rate, $\mu$= 10 boats per hour.

Probability of zero boats in the system,$\mathrm{PO}=1 /\{[(1 / 0 !) \times(6 / 10) 0+(1 / 1 !) \times(6 / 10) 1]+[(6 / 10) 2 /(2 ! \times(1-(6 /(2 \times 10)))]\}$ = 0.5385

Average number of boats waiting in line for service:

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

= $[\{6 \times 10 \times(6 / 10) 2\} /\{(2-1) ! \times((2 \times 10)-6) 2\}] \times 0.5385$ = 0.0593 boats.

The average time a boat will spend waiting for service, Wq  =  0.0593 divide by 6 = 0.009883 hours = 0.5930 minutes.

The average time a boat will spend at the dock, W =  0.009883 plus (1 divide 10) = 0.109883 hours = 6.5930 minutes.