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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