1.The critical path of a network is the
A) shortest time path through the network.
B) path with the fewest activities.
C) path with the most activities.
D) longest time path through the network.
E) None of the above
2.The first step in planning and scheduling a project is to develop the
A) employee scheduling plan.
B) PERT/CPM network diagram.
C) critical path.
D) work breakdown structure.
E) variance calculations for each activity.
Table 134 


The following represents a project with known activity times. All times are in weeks. 


Activity 
Immediate Predecessor 
Time 
A 
 
4 
B 
 
3 
C 
A 
2 
D 
B 
7 
E 
C, D 
4 
F 
B 
5 
G 
E, F 
4 
3. Using the data in Table 134, what is the minimum possible time required for
completing the project?
(a) 8
(b) 12
(c) 18
(d) 10
(e) none of the above
4. Using the data in Table 134, what is the latest possible time that C may be started
without delaying completion of the project?
(a) 0
(b) 4
(c) 8
(d) 10
(e) none of the above
5. Using the data in Table 134, compute the slack time for activity D.
(a) 0
(b) 5
(c) 3
(d) 6
(e) none of the above
6. Consider a project that has an expected completion time of 50 weeks and a standard deviation of 9 weeks. What is the probability that the project is finished in 57 weeks or fewer? (Round to two decimals.)
(a) 0.68
(b) 0.78
(c) 0.22
(d) 0.32
(e) none of the above
The following table provides information for the next two questions.
Table 136 


Activity 
Immediate Predecessor 
Optimistic 
Most Likely 
Pessimistic 
Expected t 
s 
s ^{2} 
A 
 
2 
3 
4 
3 
0.333 
0.111 
B 
 
2 
5 
8 
5 
1.000 
1.000 
C 
A 
1 
2 
9 
3 
1.330 
1.780 
D 
A 
5 
5 
5 
5 
0.000 
0.000 
E 
B, C 
6 
7 
8 
7 
0.333 
0.111 
F 
B 
12 
12 
12 
12 
0.000 
0.000 
G 
D, E 
1 
5 
9 
5 
1.333 
1.780 
H 
G, F 
1 
4 
8 
4.167 
1.167 
1.362 
7. Which activities are part of the critical path?
(a) A, B, E, G, H
(b) A, C, E, G, H
(c) A, D, G, H
(d) B, F, H
(e) none of the above
8. What is the variance of the critical path?
(a) 5.222
(b) 4.364
(c) 1.362
(d) 5.144
(e) none of the above
Table 141 


M/M/2 


Mean Arrival Rate: 
9 occurrences per minute 


Mean Service Rate: 
7 occurrences per minute 


Number of Servers: 
2 




Queue Statistics: 


Mean Number of Units in the System: 
2.191 


Mean Number of Units in the Queue: 
0.905 


Mean Time in the System: 
14.609 minutes 

Mean Time in the Queue: 
6.037 minutes 

Service Facility Utilization Factor: 
0.643 

Probability of No Units in System: 
0.217 

9. According to the information provided in Table 141, on average, how many units are in the line?
(a) 0.643
(b) 2.191
(c) 2.307
(d) 0.217
(e) 0.905
10. According to the information provided in Table 141, what proportion of time is at least one server busy?
(a) 0.643
(b) 0.905
(c) 0.783
(d) 0.091
(e) none of the above
11. Using the information provided in Table 141 and counting each person being served and the people in line, on average, how many people would be in this system?
(a) 0.905
(b) 2.191
(c) 6.037
(d) 14.609
(e) none of the above
12. According to the information provided in Table 141, what is the average time spent by a person in this system?
(a) 0.905 minutes
(b) 2.191 minutes
(c) 6.037 minutes
(d) 14.609 minutes
(e) none of the above
13. According to the information provided in Table 141, what percentage of the total available service time is being used?
(a) 90.5%
(b) 21.7%
(c) 64.3%
(d) It could be any of the above, depending on other factors.
(e) none of the above
Table 145 

M/D/1 

Mean Arrival Rate: 
5 occurrences per minute 
Constant Service Rate: 
7 occurrences per minute 
Queue Statistics: 

Mean Number of Units in the System: 
1.607 
Mean Number of Units in the Queue: 
0.893 
Mean Time in the System: 
0.321 minutes 
Mean Time in the Queue: 
0.179 minutes 
Service Facility Utilization Factor: 
0.714 
14. According to the information provided in Table 145, which presents the solution for a queuing problem with a constant service rate, on average, how much time is spent waiting in line?
(a) 1.607 minutes
(b) 0.714 minutes
(c) 0.179 minutes
(d) 0.893 minutes
(e) none of the above
15. According to the information provided in Table 145, which presents the solution for a queuing problem with a constant service rate, on average, how many customers are in the system?
(a) 0.893
(b) 0.714
(c) 1.607
(d) 0.375
(e) none of the above
16. According to the information provided in Table 145, which presents a queuing problem solution for a queuing problem with a constant service rate, on average, how many customers arrive per time period?
(a) 5
(b) 7
(c) 1.607
(d) 0.893
(e) none of the above
Table 152 


A pharmacy is considering hiring another pharmacist to better serve customers. To help analyze this situation, records are kept to determine how many customers will arrive in any 10minute interval. Based on 100 tenminute intervals, the following probability distribution has been developed and random numbers assigned to each event. 


Number of Arrivals 
Probability 
Interval of Random Numbers 
6 
0.2 
0120 
7 
0.3 
2150 
8 
0.3 
5180 
9 
0.1 
8190 
10 
0.1 
9100 
17. According to Table 152, the number of arrivals in any 10minute period is between 6 and 10, inclusive. Suppose the next three random numbers were 18, 89, and 67, and these were used to simulate arrivals in the next three 10minute intervals. How many customers would have arrived during this 30minute time period?
(a) 22
(b) 23
(c) 24
(d)25
(e) none of the above
18. According to Table 152, the number of arrivals in any 10minute period is between 6 and 10, inclusive. Suppose the next three random numbers were 20, 50, and 79, and these were used to simulate arrivals in the next three 10minute intervals. How many customers would have arrived during this 30minute time period?
(a) 18
(b) 19
(c) 20
(d)21
(e) none of the above
19. According to Table 152, the number of arrivals in any 10minute period is between 6 and 10 inclusive. Suppose the next 3 random numbers were 02, 81, and 18. These numbers are used to simulate arrivals into the pharmacy. What would the average number of arrivals per 10minute period be based on this set of occurrences?
(a) 6
(b) 7
(c) 8
(d)9
(e) none of the above
Table 15.3 


A pawn shop in Arlington, Texas, has a drivethrough window to better serve customers. The following tables provide information about the time between arrivals and the service times required at the window on a particularly busy day of the week. All times are in minutes. 


Time Between Arrivals 
Probability 
Interval of Random Numbers 
1 
0.1 
0110 
2 
0.3 
1140 
3 
0.4 
4180 
4 
0.2 
8100 
Service Time 
Probability 
Interval of Random Numbers 
1 
0.2 
0120 
2 
0.4 
2160 
3 
0.3 
6190 
4 
0.1 
9100 
The first random number generated for arrivals is used to tell when the first customer arrives after opening. 
20. According to Table 153, the time between successive arrivals is 1, 2, 3, or 4 minutes. If the store opens at 8:00a.m. and random numbers are used to generate arrivals, what time would the first customer arrive if the first random number were 02?
(a) 8:01
(b) 8:02
(c) 8:03
(d)8:04
(e) none of the above
21. According to Table 153, the time between successive arrivals is 1, 2, 3, or 4 minutes. The store opens at 8:00a.m. and random numbers are used to generate arrivals and service times. The first random number to generate an arrival is 39, while the first service time is generated by the random number 94. What time would the first customer finish transacting business?
(a) 8:03
(b) 8:04
(c) 8:05
(d)8:06
(e) none of the above
22. According to Table 153, the time between successive arrivals is 1, 2, 3, or 4 minutes. The store opens at 8:00a.m. and random numbers are used to generate arrivals and service times. The first 3 random numbers to generate arrivals are 09, 89, and 26. What time does the third customer arrive?
(a) 8:07
(b) 8:08
(c) 8:09
(d)8:10
(e) none of the above
23. According to Table 153, the time between successive arrivals is 1, 2, 3, or 4 minutes. The store opens at 8:00a.m. and random numbers are used to generate arrivals and service times. The first two random numbers for arrivals are 95 and 08. The first two random numbers for service times are 92 and 18. At what time does the second customer finish transacting business?
(a) 8:07
(b) 8:08
(c) 8:09
(d)8:10
(e) none of the above
Table 154 

Variable Value 
Probability 
Cumulative Probability 
0 
0.08 
0.08 
1 
0.23 
0.31 
2 
0.32 
0.63 
3 
0.28 
0.91 
4 
0.09 
1.00 
Number of Runs 
200 

Average Value 
2.10 
24. According to Table 154, which presents a summary of the Monte Carlo output from a simulation of 200 runs, there are 5 possible values for the variable of concern. If this variable represents the number of machine breakdowns during a day, what is the probability that the number of breakdowns is 2 or fewer?
(a) 0.23
(b) 0.31
(c) 0.32
(d)0.63
(e) none of the above
25. According to Table 154, which presents a summary of the Monte Carlo output from a simulation of 200 runs, there are 5 possible values for the variable of concern. If this variable represents the number of machine breakdowns during a day, what is the probability that the number of breakdowns is more than 4?
(a) 0
(b) 0.08
(c) 0.09
(d)1.00
(e) none of the above
26. According to Table 154, which presents a summary of the Monte Carlo output from a simulation of 200 runs, there are 5 possible values for the variable of concern. If this variable represents the number of machine breakdowns during a day, based on this simulation run, what is the average number of breakdowns per day?
(a) 2.00
(b) 2.10
(c) 2.50
(d)200
(e) none of the above
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