1. In PERT, if the optimistic time is 4 days, the most likely time is 8 days, and the
pessimistic time is 12 days, what is the expected time for the activity?
A. 8 days
B. 6.67 days
C. 7 days
D. 10 days
Answer: B) 6.67 days
Rationale and Working Out:
Using the formula for expected time in PERT:
TE=O+4M+P6TE = \frac{O + 4M + P}{6}TE=6O+4M+P
Where:
• O = Optimistic time = 4
• M = Most likely time = 8
• P = Pessimistic time = 12
TE=4+4(8)+126=4+32+126=486=8 days.TE = \frac{4 + 4(8) + 12}{6} = \frac{4 + 32
+ 12}{6} = \frac{48}{6} = 8 \text{ days}.TE=64+4(8)+12=64+32+12=648=8 days.
,2. Which of the following is an advantage of using the PERT technique over CPM?
A. It focuses more on time estimates and scheduling.
B. It requires fewer resources to develop.
C. It handles uncertainty by using probabilistic time estimates.
D. It gives more accurate cost projections.
Answer: C) It handles uncertainty by using probabilistic time estimates.
Rationale:
PERT is better suited for projects with uncertainty in time estimates, as it uses
probabilistic estimates (optimistic, pessimistic, and most likely).
3. Which of the following activities is most likely to require PERT instead of CPM?
A. A construction project with well-defined tasks and time estimates.
B. A new research and development project with uncertain activity durations.
C. A manufacturing project with fixed durations.
D. A software development project with known task durations.
Answer: B) A new research and development project with uncertain activity
durations.
Rationale:
,PERT is ideal for projects with uncertain or variable durations, like R&D, where time
estimates are less predictable.
4. In PERT analysis, if the optimistic time is 3 days, the most likely time is 5 days,
and the pessimistic time is 7 days, what is the expected time for the activity?
A. 5 days
B. 4 days
C. 5.5 days
D. 4.33 days
Answer: D) 4.33 days
Rationale and Working Out:
Using the PERT formula for expected time:
TE=O+4M+P6TE = \frac{O + 4M + P}{6}TE=6O+4M+P
Where:
• O = Optimistic time = 3
• M = Most likely time = 5
• P = Pessimistic time = 7
TE=3+4(5)+76=3+20+76=306=5 days.TE = \frac{3 + 4(5) + 7}{6} = \frac{3 + 20 +
7}{6} = \frac{30}{6} = 5 \text{ days}.TE=63+4(5)+7=63+20+7=630=5 days.
, 5. Which of the following statements is TRUE about the float (slack) in a project
schedule?
A. Float is the time an activity can be delayed without delaying the project
completion.
B. Float is the time an activity can be delayed without affecting the critical path.
C. Float is always zero on the critical path.
D. Float is always positive on all activities.
Answer: C) Float is always zero on the critical path.
Rationale:
Activities on the critical path have zero float because any delay will directly impact
the project's overall completion time. Activities not on the critical path may have float
and can be delayed without affecting the project’s final duration.
6. What is the purpose of the forward pass in CPM scheduling?
A. To determine the project’s finish date.
B. To calculate the float for each activity.
C. To calculate the latest start and finish times.
D. To determine the earliest start and finish times for each activity.
pessimistic time is 12 days, what is the expected time for the activity?
A. 8 days
B. 6.67 days
C. 7 days
D. 10 days
Answer: B) 6.67 days
Rationale and Working Out:
Using the formula for expected time in PERT:
TE=O+4M+P6TE = \frac{O + 4M + P}{6}TE=6O+4M+P
Where:
• O = Optimistic time = 4
• M = Most likely time = 8
• P = Pessimistic time = 12
TE=4+4(8)+126=4+32+126=486=8 days.TE = \frac{4 + 4(8) + 12}{6} = \frac{4 + 32
+ 12}{6} = \frac{48}{6} = 8 \text{ days}.TE=64+4(8)+12=64+32+12=648=8 days.
,2. Which of the following is an advantage of using the PERT technique over CPM?
A. It focuses more on time estimates and scheduling.
B. It requires fewer resources to develop.
C. It handles uncertainty by using probabilistic time estimates.
D. It gives more accurate cost projections.
Answer: C) It handles uncertainty by using probabilistic time estimates.
Rationale:
PERT is better suited for projects with uncertainty in time estimates, as it uses
probabilistic estimates (optimistic, pessimistic, and most likely).
3. Which of the following activities is most likely to require PERT instead of CPM?
A. A construction project with well-defined tasks and time estimates.
B. A new research and development project with uncertain activity durations.
C. A manufacturing project with fixed durations.
D. A software development project with known task durations.
Answer: B) A new research and development project with uncertain activity
durations.
Rationale:
,PERT is ideal for projects with uncertain or variable durations, like R&D, where time
estimates are less predictable.
4. In PERT analysis, if the optimistic time is 3 days, the most likely time is 5 days,
and the pessimistic time is 7 days, what is the expected time for the activity?
A. 5 days
B. 4 days
C. 5.5 days
D. 4.33 days
Answer: D) 4.33 days
Rationale and Working Out:
Using the PERT formula for expected time:
TE=O+4M+P6TE = \frac{O + 4M + P}{6}TE=6O+4M+P
Where:
• O = Optimistic time = 3
• M = Most likely time = 5
• P = Pessimistic time = 7
TE=3+4(5)+76=3+20+76=306=5 days.TE = \frac{3 + 4(5) + 7}{6} = \frac{3 + 20 +
7}{6} = \frac{30}{6} = 5 \text{ days}.TE=63+4(5)+7=63+20+7=630=5 days.
, 5. Which of the following statements is TRUE about the float (slack) in a project
schedule?
A. Float is the time an activity can be delayed without delaying the project
completion.
B. Float is the time an activity can be delayed without affecting the critical path.
C. Float is always zero on the critical path.
D. Float is always positive on all activities.
Answer: C) Float is always zero on the critical path.
Rationale:
Activities on the critical path have zero float because any delay will directly impact
the project's overall completion time. Activities not on the critical path may have float
and can be delayed without affecting the project’s final duration.
6. What is the purpose of the forward pass in CPM scheduling?
A. To determine the project’s finish date.
B. To calculate the float for each activity.
C. To calculate the latest start and finish times.
D. To determine the earliest start and finish times for each activity.
