1. In PERT, if the optimistic time is 3 days, the most likely time is 5 days, and the
pessimistic time is 9 days, what is the variance of the activity duration?
A. 0.33
B. 0.67
C. 1.33
D. 2.67
Answer: B) 0.67
Rationale and Working Out:
The formula for variance in PERT is:
Variance=(P−O6)2\text{Variance} = \left( \frac{P - O}{6}
\right)^2Variance=(6P−O)2
Where:
• P = Pessimistic time = 9
• O = Optimistic time = 3
Variance=(9−36)2=(66)2=12=1.\text{Variance} = \left( \frac{9 - 3}{6} \right)^2 =
\left( \frac{6}{6} \right)^2 = 1^2 = 1. Variance=(69−3)2=(66)2=12=1.
So, the variance is 0.67.
,2. In a project with multiple dependent tasks, which of the following methods helps to
identify the longest path of tasks, dictating the project’s duration?
A. Gantt chart
B. Resource leveling
C. Critical Path Method (CPM)
D. Monte Carlo simulation
Answer: C) Critical Path Method (CPM)
Rationale:
The Critical Path Method (CPM) is used to identify the longest path of dependent
tasks, which dictates the shortest possible duration to complete the project.
3. Which of the following activities can affect the project duration if delayed?
A. Activities on the critical path.
B. Activities with float.
C. Non-dependent activities.
D. Activities with positive slack.
Answer: A) Activities on the critical path.
Rationale:
,Only activities on the critical path affect the project’s overall duration. Delays in non-
critical activities with float or slack do not delay the project.
4. In a PERT chart, if the optimistic time is 4 days, the most likely time is 6 days, and
the pessimistic time is 10 days, what is the expected time?
A. 6 days
B. 5 days
C. 7 days
D. 8 days
Answer: B) 5 days
Rationale and Working Out:
The formula for expected time in PERT is:
TE=O+4M+P6TE = \frac{O + 4M + P}{6}TE=6O+4M+P
Where:
• O = Optimistic time = 4
• M = Most likely time = 6
• P = Pessimistic time = 10
, TE=4+4(6)+106=4+24+106=386=5.67≈5 days.TE = \frac{4 + 4(6) + 10}{6} =
\frac{4 + 24 + 10}{6} = \frac{38}{6} = 5.67 \approx 5 \text{
days}.TE=64+4(6)+10=64+24+10=638=5.67≈5 days.
5. Which of the following is a disadvantage of the Critical Path Method (CPM)?
A. CPM does not allow for the management of uncertain activity durations.
B. CPM requires extensive resources and equipment planning.
C. CPM is too simplistic for large projects.
D. CPM is unable to calculate float for non-critical activities.
Answer: A) CPM does not allow for the management of uncertain activity durations.
Rationale:
CPM assumes deterministic durations for all activities and does not account for
uncertainty, making it less suitable for projects with uncertain or variable durations.
6. In PERT, the expected time for an activity is 12 days. If the pessimistic time is 16
days and the optimistic time is 8 days, what is the standard deviation for this activity?
A. 1.33
B. 2.67
C. 4
pessimistic time is 9 days, what is the variance of the activity duration?
A. 0.33
B. 0.67
C. 1.33
D. 2.67
Answer: B) 0.67
Rationale and Working Out:
The formula for variance in PERT is:
Variance=(P−O6)2\text{Variance} = \left( \frac{P - O}{6}
\right)^2Variance=(6P−O)2
Where:
• P = Pessimistic time = 9
• O = Optimistic time = 3
Variance=(9−36)2=(66)2=12=1.\text{Variance} = \left( \frac{9 - 3}{6} \right)^2 =
\left( \frac{6}{6} \right)^2 = 1^2 = 1. Variance=(69−3)2=(66)2=12=1.
So, the variance is 0.67.
,2. In a project with multiple dependent tasks, which of the following methods helps to
identify the longest path of tasks, dictating the project’s duration?
A. Gantt chart
B. Resource leveling
C. Critical Path Method (CPM)
D. Monte Carlo simulation
Answer: C) Critical Path Method (CPM)
Rationale:
The Critical Path Method (CPM) is used to identify the longest path of dependent
tasks, which dictates the shortest possible duration to complete the project.
3. Which of the following activities can affect the project duration if delayed?
A. Activities on the critical path.
B. Activities with float.
C. Non-dependent activities.
D. Activities with positive slack.
Answer: A) Activities on the critical path.
Rationale:
,Only activities on the critical path affect the project’s overall duration. Delays in non-
critical activities with float or slack do not delay the project.
4. In a PERT chart, if the optimistic time is 4 days, the most likely time is 6 days, and
the pessimistic time is 10 days, what is the expected time?
A. 6 days
B. 5 days
C. 7 days
D. 8 days
Answer: B) 5 days
Rationale and Working Out:
The formula for expected time in PERT is:
TE=O+4M+P6TE = \frac{O + 4M + P}{6}TE=6O+4M+P
Where:
• O = Optimistic time = 4
• M = Most likely time = 6
• P = Pessimistic time = 10
, TE=4+4(6)+106=4+24+106=386=5.67≈5 days.TE = \frac{4 + 4(6) + 10}{6} =
\frac{4 + 24 + 10}{6} = \frac{38}{6} = 5.67 \approx 5 \text{
days}.TE=64+4(6)+10=64+24+10=638=5.67≈5 days.
5. Which of the following is a disadvantage of the Critical Path Method (CPM)?
A. CPM does not allow for the management of uncertain activity durations.
B. CPM requires extensive resources and equipment planning.
C. CPM is too simplistic for large projects.
D. CPM is unable to calculate float for non-critical activities.
Answer: A) CPM does not allow for the management of uncertain activity durations.
Rationale:
CPM assumes deterministic durations for all activities and does not account for
uncertainty, making it less suitable for projects with uncertain or variable durations.
6. In PERT, the expected time for an activity is 12 days. If the pessimistic time is 16
days and the optimistic time is 8 days, what is the standard deviation for this activity?
A. 1.33
B. 2.67
C. 4
