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514 CHAPTER 12 SIMULATION

Customer 1
l An interarrival time of IAT ¼ 1.4 minutes is generated.
l Because the simulation run begins at time 0, the arrival time for customer 1 is
0 þ 1.4 ¼ 1.4 minutes.
l Customer 1 may begin service immediately with a start time of 1.4 minutes.
l The waiting time for customer 1 is the start time minus the arrival time:
1.4 1.4 ¼ 0 minutes.
l A service time of ST ¼ 2.3 minutes is generated for customer 1.
l The completion time for customer 1 is the start time plus the service time:
1.4 þ 2.3 ¼ 3.7 minutes.
l The time in the system for customer 1 is the completion time minus the arrival
time: 3.7 1.4 ¼ 2.3 minutes.

Customer 2
l An interarrival time of IAT ¼ 1.3 minutes is generated.
l Because the arrival time of customer 1 is 1.4, the arrival time for customer 2 is
1.4 þ 1.3 ¼ 2.7 minutes.
l Because the completion time of customer 1 is 3.7 minutes, the arrival time of
customer 2 is not greater than the completion time of customer 1; so, the ATM
is busy when customer 2 arrives.
l Customer 2 must wait for customer 1 to complete service before beginning
service. Customer 1 completes service at 3.7 minutes, which becomes the start
time for customer 2.
l The waiting time for customer 2 is the start time minus the arrival time:
3.7 2.7 ¼ 1minute.
l A service time of ST ¼ 1.5 minutes is generated for customer 2.
l The completion time for customer 2 is the start time plus the service time:
3.7 þ 1.5 ¼ 5.2 minutes.
l The time in the system for customer 2 is the completion time minus the arrival
time: 5.2 2.7 ¼ 2.5 minutes.

Customer 3
l An interarrival time of IAT ¼ 4.9 minutes is generated.
l Because the arrival time of customer 2 was 2.7 minutes, the arrival time for
customer 3 is 2.7 þ 4.9 ¼ 7.6 minutes.
l The completion time of customer 2 is 5.2 minutes, so the arrival time for
customer 3 is greater than the completion time of customer 2. Thus, the ATM
is idle when customer 3 arrives.
l Customer 3 begins service immediately with a start time of 7.6 minutes.
l The waiting time for customer 3 is the start time minus the arrival time:
7.6 7.6 ¼ 0 minutes.
l A service time of ST¼2.2 minutes is generated for customer 3.
l The completion time for customer 3 is the start time plus the service time:
7.6 þ 2.2 ¼ 9.8 minutes.
l The time in the system for customer 3 is the completion time minus the arrival
time: 9.8 7.6 ¼ 2.2 minutes.

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