Willow Brook National Bank operates a drive-up teller window that allows customers to complete bank transactions...

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 24 customers per hour or 0.4 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.

P₀ = ??
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_q = ?? min
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 = ??

Solutions

Expert Solution

Results		Formulas	Calculations
Average server utilization(r)	0.6667	=arrival rate/service rate	=24/36
Average number of customers in the queue(Lq)	1.3333	=arrival rate^2/(service rate*(service rate-arrival rate)	=24^2/(36*(36-24))
Average number of customers in the system(L)	2	=arrival rate/(service rate-arrival rate)	=24/(36-24)
Average waiting time in the queue(Wq)	0.0556	=arrival rate/(service rate*(service rate-arrival rate))	=24/(36*(36-24))
Average time in the system(W)	0.0833	=1/(service rate-arrival rate)	=1/(36-24)
Probability (% of time) system is empty (P0)	0.3333	=1-p	=1-0.6667

a. The probability that no customers are in the system. If required, round your answer to four decimal places.
P0	0.3333
b. The average number of customers waiting. If required, round your answer to four decimal places.
Lq	1.3333
c. The average number of customers in the system. If required, round your answer to nearest whole number.
L	2
d. The average time a customer spends waiting. If required, round your answer to four decimal places.
		=0.0556*60 minutes
Wq	0.0556	3.3333 minutes
e. The average time a customer spends in the system. If required, round your answer to nearest whole number.
		=0.0833*60 minutes
W	0.0833	5 minutes
f. The probability that arriving customers will have to wait for service. If required, round your answer to four decimal places.
Pw = p	0.6667

keosha answered 1 week ago

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