Willow Brook National Bank operates a drive-up teller window that allows custome
ID: 449778 • Letter: W
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 18 customers per hour or 0.3 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 30 customers per hour, or 0.5 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.
Lq =
The average number of customers in the system. If required, round your answer to the nearest whole number.
L =
The average time a customer spends waiting. If required, round your answer to four decimal places.
Wq = min
The average time a customer spends in the system. If required, round your answer to the 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.
Pw =
Explanation / Answer
The probability that no customers are in the system. If required, round your answer to four decimal places.
P0 = 40%
Arrival rate
=
18
Service rate
=
30
Utilization rate of server
= /
60.0%
Probability of NO customers in system
P0
= 1-
40.0%
The average number of customers waiting. If required, round your answer to four decimal places.
Lq = 0.9
Average time in system
W
= L/
0.08333
hour
Average time waiting in line
Wq
= W-1/
0.05
hour
Average number of customers waiting in line
Lq
= *Wq
0.9000
customers
The average number of customers in the system. If required, round your answer to the nearest whole number.
L = 1.5
Average number of customers in system
L
=/(1-)
1.5
customers
The average time a customer spends waiting. If required, round your answer to four decimal places.
Wq = 3 min
Average time in system
W
= L/
0.08333
hour
Average time waiting in line
Wq
= W-1/
0.05
hour
The average time a customer spends in the system. If required, round your answer to the nearest whole number.
W = 5 min
Average time in system
W
= L/
0.08333
hour
The probability that arriving customers will have to wait for service. If required, round your answer to four decimal places.
Pw = 60%
Probability of waiting time > t
P(t)
=*e(-µ*(1-)*t)
60.0%
Distribution of time in queue
t =
0
Arrival rate
=
18
Service rate
=
30
Utilization rate of server
= /
60.0%
Probability of NO customers in system
P0
= 1-
40.0%
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