Question

Willow Brook National Bank operates a drive-up teller window that allows customers to complete bank transactions without getting out of their cars. On weekday mornings, arrivals to the drive-up teller window occur at random, with an arrival rate of 30 customers per hour or 0.5 customers per minute. In the same bank waiting line system, assume that the service times for the drive-up teller follow an exponential probability distribution with a service rate of 36 customers per hour, or 0.6 customers per minute. Determine the following operating characteristics for the system:

- The probability that no customers are in the system. If
required, round your answer to four decimal places.

*P*=_{0} - The average number of customers waiting. If required, round
your answer to four decimal places.

*L*=_{q} - The average number of customers in the system. If required,
round your answer to nearest whole number.

*L*= - The average time a customer spends waiting. If required, round
your answer to four decimal places.

*W*= _________ min_{q} - The average time a customer spends in the system. If required,
round your answer to nearest whole number.

*W*= __________ min - The probability that arriving customers will have to wait for
service. If required, round your answer to four decimal
places.

*P*=_{w}

Answer #1

λ = 0.5

µ =0.6

ρ = λ/µ = 0.5/0.6 = 0.83

a) Probability that no customers are in the system: P0 = 1-ρ = 1-0.83 = 0.1667

b) Average number of customers waiting

Lq = ρ2/(1-ρ) = (0.83*0.83)/(1-0.83) = 4.0524 ~ 4

c) Average number of customers in the system

L = λW

W = Wq + 1/µ

Wq = Lq/λ = 4.0524/0.5 = 8.1047

W = Wq + 1/µ = 8.1047 + 1/0.6 = 9.7714

L = λW = 0.5*9.7714 = 4.8857 ~ 5

d) Average time a customer spends waiting:

Wq = Lq/λ = 4.0524/0.5 = 8.1047 minutes

e) Average time a customer spends in the system:

W = Wq + 1/µ = 8.1047 + 1/0.6 = 9.7714 ~ 10 minutes

f) Probability that arriving customers will have to wait for service:

Pw = ρ = λ/µ = 0.5/0.6 = 0.83

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