The estimated times and immediate predecessors for the activities in a project at Caesar Douglas’s screening company are given in the following table. Assume that the activity times are independent. All times are in weeks.
| Activity | Predecessors | Optimistic Time (a) | Most Likely Time (m) | Pessimistic Time (b) |
| A | — | 9 | 10 | 11 |
| B | — | 4 | 10 | 16 |
| C | A | 9 | 10 | 11 |
| D | B | 5 | 8 | 11 |
1. What is the critical path based on expected times? You may use any method to determine this?
2. What is the probability of completing the project in exactly 20 weeks?
3. What is the probability of completing the project in 21 weeks or less?
Project management involves the meticulous planning, scheduling, and execution of tasks to achieve project objectives efficiently. In this analysis, we will delve into Caesar Douglas’s screening company project, considering estimated activity times and their predecessors, to determine the critical path, the probability of project completion in 20 weeks, and the probability of finishing in 21 weeks or less.
The critical path in a project is the sequence of activities that, when summed, results in the longest duration. It represents the path that, if delayed, will cause the overall project to be delayed. To determine the critical path based on expected times, we can use the PERT (Program Evaluation and Review Technique) analysis, which considers optimistic (a), most likely (m), and pessimistic (b) time estimates for each activity.
The expected duration (TE) for each activity can be calculated using the formula:
TE = (a + 4m + b) / 6
Let’s calculate TE for each activity:
Activity A: TE = (9 + 4(10) + 11) / 6 = 8.83 weeks
Activity B: TE = (4 + 4(10) + 16) / 6 = 10.67 weeks
Activity C (dependent on A): TE = (9 + 4(10) + 11) / 6 = 8.83 weeks
Activity D (dependent on B): TE = (5 + 4(8) + 11) / 6 = 7.83 weeks
Now, we identify the critical path, which consists of activities with the longest expected durations. In this case, activities B and D both have an expected duration of 10.67 weeks, making them the critical path activities.
Therefore, the critical path consists of activities B and D.
To calculate the probability of completing the project in a specific time frame, we can use the normal distribution. We’ll calculate the project’s mean completion time (μ) and standard deviation (σ) using the expected durations and variances of critical path activities.
μ = Sum of expected durations of critical path activities σ = Square root of the sum of variances of critical path activities
μ = 10.67 weeks (critical path)
σ = sqrt((10.67^2)) = 10.67 weeks
To find the probability of completing the project in exactly 20 weeks, we can use the Z-score formula:
Z = (X – μ) / σ
Where X is the desired completion time (20 weeks in this case).
Z = (20 – 10.67) / 10.67 ≈ 0.869
We can use a standard normal distribution table or calculator to find the probability associated with Z = 0.869, which is approximately 0.6451 or 64.51%.
To find the probability of completing the project in 21 weeks or less, we can calculate the Z-score for 21 weeks:
Z = (21 – 10.67) / 10.67 ≈ 1
Using the standard normal distribution table or calculator, the probability associated with Z = 1 is approximately 0.8413 or 84.13%.
In conclusion, for Caesar Douglas’s screening company project, the critical path comprises activities B and D. The probability of completing the project in exactly 20 weeks is approximately 64.51%, while the probability of completing it in 21 weeks or less is approximately 84.13%. This information is crucial for project managers to assess and manage project timelines effectively, considering both expected completion times and associated probabilities.
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