A Company Wants to Develop a Project Using PERT to Modify

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Examlex · Accepted Answer

The Answer is To answer these questions, we will use the Program Evaluation and Review Technique (PERT) calculations. PERT is a project management tool used to schedule, organize, and coordinate tasks within a project.a. Expected duration for activity a:
The expected duration (TE) for an activity in PERT is calculated using the formula:

TE = \frac{t_o + 4t_m + t_p}{6}

For activity a:

TE_a = \frac{4 + 4 imes 6 + 7}{6} = \frac{4 + 24 + 7}{6} = \frac{35}{6} \approx 5.83 ext{ days}

b. Variance for activity d:
The variance of an activity in PERT is calculated using the formula:

ext{Variance} = \left( \frac{t_p - t_o}{6} ight)^2

For activity d:

ext{Variance}_d = \left( \frac{12 - 7}{6} ight)^2 = \left( \frac{5}{6} ight)^2 \approx 0.69 ext{ days}^2

c. Number of paths in the network:
To determine the number of paths, we need to trace the paths from the start to the end of the project. The paths are:
1. a -> b -> d -> f
2. a -> c -> f
3. a -> b -> e (Note: This is not a complete path to the end of the project, so it is not counted)
So, there are 2 paths in the network.
d. Earliest start (ES) time for activity e:
The earliest start time for an activity is the earliest finish time of its immediate predecessor(s). Activity e can only start after activity b is finished.

ES_e = EF_b

To find EF_b, we need to calculate the expected duration for activity b:

TE_b = \frac{1 + 4 imes 2 + 3}{6} = \frac{1 + 8 + 3}{6} = \frac{12}{6} = 2 ext{ days}

Since activity a is the predecessor of b and its expected duration is 5.83 days, the earliest finish for b is:

EF_b = ES_b + TE_b = 5.83 + 2 = 7.83 ext{ days}

Therefore, the earliest start for e is 7.83 days.
e. Slack in days for activity b:
Slack or float is the amount of time that an activity can be delayed without delaying the project. Since activity b is on the critical path (as we will see in part f), its slack is 0 days.
f. Critical path:
The critical path is the longest path through the network, which determines the shortest time to complete the project. To find the critical path, we calculate the expected duration for all activities and paths:
Path 1: a -> b -> d -> f

TE_{Path 1} = TE_a + TE_b + TE_d + TE_f

TE_{Path 1} = 5.83 + 2 + \frac{7 + 4 imes 9 + 12}{6} + \frac{6 + 4 imes 9 + 10}{6}

TE_{Path 1} = 5.83 + 2 + 9 + 9 = 25.83 ext{ days}

Path 2: a -> c -> f

TE_{Path 2} = TE_a + TE_c + TE_f

TE_{Path 2} = 5.83 + 5 + 9 = 19.83 ext{ days}

The critical path is Path 1 (a -> b -> d -> f) since it has the longest duration.
g. Duration of the critical path in days:
From part f, the duration of the critical path (Path 1) is 25.83 days.
h. Standard deviation of the critical path:
The standard deviation of the critical path is the square root of the sum of the variances of the activities on the critical path. We only calculated the variance for activity d, but we would need the variances for activities a, b, and f as well to calculate the standard deviation of the critical path. Assuming we have those variances, the standard deviation would be:

\sigma_{Path 1} = \sqrt{ ext{Variance}_a + ext{Variance}_b + ext{Variance}_d + ext{Variance}_f}

Without the variances for a, b, and f, we cannot calculate the standard deviation.
i. Probability that the project will take longer than 28 weeks to complete:
To calculate this probability, we would need the standard deviation of the critical path, which we do not have. Additionally, 28 weeks is equivalent to 196 days, which is significantly longer than the duration of the critical path calculated in part g. Therefore, without further information, we can assume the probability of the project taking longer than 28 weeks is extremely low or effectively zero, given the current estimates.

A Company Wants to Develop a Project Using PERT to Modify

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