Neve Commercial Bank is the only bank in the town of York, Pennsylvania. On a typical Friday, an average of 12 customers per hour arrive at the bank to transact business. There is one teller at the bank, and the average time required to transact business is 4 minutes. It is assumed that service times may be described by the negative exponential distribution. A single line would be used, and the customer at the front of the line would go to the first available bank teller. If a single teller is used: a) The average time in the line16 minutes (round your response to two decimal places). b) The average number in the line 3.20 customers (round your response to two decimal places). c) The average time in the system = 20 minutes (round your response to two decimal places). d) The average number in the system4 customers (round your response to two decimal places). e) The probability that the bank is empty 0.20 (round your response to two decimal places). f CEO Benjamin Neve is considering adding a second teller (who would work at the same rate as the first) to reduce the waiting time for customers. He assumes that this will cut the waiting time in half. If a second teller is added, the average time a customer spends in the queue0.80 minutes (round your response to two decimal places). If a second teller is added, the average number of customers in the queuecustomers (round your response to four decimal places).

The above senario, where a second teller is added, is the single line, multi server and sigle phased model since both the tellers work at the same rate. Let,

s = number of tellers = 2

lambda = the number of customers arriving/hour = 12

mu = the number of customers serviced by a teller in one hour = frac{60}{4} = 15

p = the utilization = frac{lambda }{smu } = frac{12}{2times 15} = frac{2}{5}

The average number of cutomers in the queue is given by

No. of customers = frac{Qtimes (frac{lambda }{mu })^{s}times p}{s!times (1-p)^{2}} —— (1)

where,

Q = [sum_{n=0}^{s-1}frac{(frac{lambda }{mu })^{n}}{n!}+frac{(frac{lambda }{mu })^{2}}{s!}times (frac{1}{1-p})]^{-1}

Q = [sum_{n=0}^{1}frac{(frac{12 }{15 })^{n}}{n!}+frac{(frac{12}{15 })^{2}}{2!}times (frac{1}{1-frac{2}{5}})]^{-1}

Q = [sum_{n=0}^{1}frac{(frac{12 }{15 })^{n}}{n!}+frac{(frac{12}{15 })^{2}}{2}times frac{5}{3}]^{-1}

Q = [frac{(frac{12}{15})^{0}}{0!}+frac{12}{15}+frac{(frac{12}{15 })^{2}}{2}times frac{5}{3}]^{-1}

Q = [1+frac{12}{15}+frac{8}{15}]^{-1}

Q = [1+frac{20}{15}]^{-1}

Q = [1+frac{4}{3}]^{-1} = [frac{7}{3}]^{-1}=frac{3}{7}

By subtituing the values in equation (1), we get the number of customers =

frac{frac{3}{7}times (frac{12}{15})^{2}times frac{2}{5}}{2times (1-frac{2}{5})^{2}}

frac{frac{3}{7}times frac{4}{5}times frac{4}{5}times frac{2}{5}}{2times frac{3}{5}times frac{3}{5}}

frac{frac{16}{35}}{3} = frac{16}{35}times frac{1}{3} = frac{16}{35times 3} = 0.1523

Thus, if a second teller is added, the average number of customers in the queue = 0.1523 customers.

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