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At a single-phase, multiple-channel service facility, customers arrive randomly. Statistical analysis of past data shows that the interarrival time has a mean of 20 minutes and a standard deviation of 4 minutes. The service time per customer has a mean of 15 minutes and a standard deviation of 5 minutes. The waiting cost is $200 per customer per hour. The server cost is $25 per server per hour. Assume general probability distribution and no buffer capacity restriction

Required:
a. Find the optimal number of servers to be employed to minimize the total of waiting and server costs.
b. Find the average waiting time and the average total time through the system for the optimal case.
c. Find the cost per hour, average waiting time, and average flow time for one server if the probability distributions for the interarrival time and service time are assumed to be exponential and the mean values remain the same. .

Sagot :

Talanat
Cost per hour with one server = $ 59.00

Cost with 2 servers = $ 52.19

Cost with 2 servers = $ 75.40

Total cost with 2 servers is the lowest ($ 52.19). Therefore, two servers are optimal.

b) With 2 servers,

Average waiting time, Tq = 0.2188 minutes

Total time = Tq+p = 0.2188+15 = 15.2188 minutes

c) Arrival rate, \lambda = 60/20 = 3 per hour

Service rate, \mu = 60/15 = 4 per hour

Lq = \lambda 2/(\mu*(\mu-\lambda)) = 32/(4*(4-3)) = 2.25

Cost per hour = Lq*Cw+Cs = 2.25*200 + 25 = $ 475

Waiting time, Wq = Lq/\lambda = 2.25/3 = 0.75 hour = 45 min

Flow time = Wq+1/\mu = 0.75+1/4 = 1 hour = 60 min
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