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Sagot :
Sure, let's walk through each part of the solution step-by-step:
### a) Finding the Expected (Estimated) Time for Activity C
The expected time (TE) for an activity is calculated using the formula:
[tex]\[ TE = \frac{a + 4m + b}{6} \][/tex]
For activity C, the optimistic time \( a \), most likely time \( m \), and pessimistic time \( b \) are given as 9, 12, and 18 weeks respectively. Plugging these values into the formula:
[tex]\[ TE_C = \frac{9 + 4(12) + 18}{6} = \frac{9 + 48 + 18}{6} = \frac{75}{6} = 12.5 \text{ weeks} \][/tex]
Thus, the expected time for activity C is 12.5 weeks.
### b) Finding the Variance for Activity C
The variance (V) for an activity is calculated using the formula:
[tex]\[ V = \left(\frac{b - a}{6}\right)^2 \][/tex]
For activity C:
[tex]\[ V_C = \left(\frac{18 - 9}{6}\right)^2 = \left(\frac{9}{6}\right)^2 = \left(1.5\right)^2 = 2.25 \][/tex]
Thus, the variance for activity C is 2.25 weeks.
### c) Identifying the Critical Path
The critical path is determined as the longest path through the project network. The critical path given in the question is:
[tex]\[ A \to B \to E \to F \to H \to J \to K \][/tex]
### d) Estimating the Time for the Critical Path
The estimated time for the critical path is the sum of the expected times for all activities along the critical path. Given values for these activities are as follows:
- A: 11 weeks (calculated earlier)
- B: 13 weeks (calculated earlier)
- E: 2.5 weeks (calculated earlier)
- F: 10.5 weeks (calculated earlier)
- H: 2 weeks (calculated earlier)
- J: 8 weeks (calculated earlier)
- K: 1.5 weeks (calculated earlier)
Summing these values:
[tex]\[ TE_{\text{Critical Path}} = 11 + 13 + 2.5 + 10.5 + 2 + 8 + 1.5 = 42.5 \text{ weeks} \][/tex]
Thus, the estimated time for the critical path is 42.5 weeks.
### e) Calculating the Activity Variance Along the Critical Path
The total variance for the critical path is the sum of variances of the activities along that path. Given variances for these activities are as follows:
- A: 1
- B: 19.36
- E: 0.25
- F: 6.25
- H: 0
- J: 1.7778
- K: 0.25
Summing these values:
[tex]\[ V_{\text{Critical Path}} = 1 + 19.36 + 0.25 + 6.25 + 0 + 1.7778 + 0.25 = 28.8878 \text{ weeks} \][/tex]
Thus, the activity variance along the critical path is 28.8878 weeks.
Summarizing:
- a) The expected (estimated) time for activity C is 12.5 weeks.
- b) The variance for activity C is 2.25 weeks.
- c) The critical path is A-B-E-F-H-J-K.
- d) The estimated time for the critical path is 42.5 weeks.
- e) The activity variance along the critical path is 28.8878 weeks.
### a) Finding the Expected (Estimated) Time for Activity C
The expected time (TE) for an activity is calculated using the formula:
[tex]\[ TE = \frac{a + 4m + b}{6} \][/tex]
For activity C, the optimistic time \( a \), most likely time \( m \), and pessimistic time \( b \) are given as 9, 12, and 18 weeks respectively. Plugging these values into the formula:
[tex]\[ TE_C = \frac{9 + 4(12) + 18}{6} = \frac{9 + 48 + 18}{6} = \frac{75}{6} = 12.5 \text{ weeks} \][/tex]
Thus, the expected time for activity C is 12.5 weeks.
### b) Finding the Variance for Activity C
The variance (V) for an activity is calculated using the formula:
[tex]\[ V = \left(\frac{b - a}{6}\right)^2 \][/tex]
For activity C:
[tex]\[ V_C = \left(\frac{18 - 9}{6}\right)^2 = \left(\frac{9}{6}\right)^2 = \left(1.5\right)^2 = 2.25 \][/tex]
Thus, the variance for activity C is 2.25 weeks.
### c) Identifying the Critical Path
The critical path is determined as the longest path through the project network. The critical path given in the question is:
[tex]\[ A \to B \to E \to F \to H \to J \to K \][/tex]
### d) Estimating the Time for the Critical Path
The estimated time for the critical path is the sum of the expected times for all activities along the critical path. Given values for these activities are as follows:
- A: 11 weeks (calculated earlier)
- B: 13 weeks (calculated earlier)
- E: 2.5 weeks (calculated earlier)
- F: 10.5 weeks (calculated earlier)
- H: 2 weeks (calculated earlier)
- J: 8 weeks (calculated earlier)
- K: 1.5 weeks (calculated earlier)
Summing these values:
[tex]\[ TE_{\text{Critical Path}} = 11 + 13 + 2.5 + 10.5 + 2 + 8 + 1.5 = 42.5 \text{ weeks} \][/tex]
Thus, the estimated time for the critical path is 42.5 weeks.
### e) Calculating the Activity Variance Along the Critical Path
The total variance for the critical path is the sum of variances of the activities along that path. Given variances for these activities are as follows:
- A: 1
- B: 19.36
- E: 0.25
- F: 6.25
- H: 0
- J: 1.7778
- K: 0.25
Summing these values:
[tex]\[ V_{\text{Critical Path}} = 1 + 19.36 + 0.25 + 6.25 + 0 + 1.7778 + 0.25 = 28.8878 \text{ weeks} \][/tex]
Thus, the activity variance along the critical path is 28.8878 weeks.
Summarizing:
- a) The expected (estimated) time for activity C is 12.5 weeks.
- b) The variance for activity C is 2.25 weeks.
- c) The critical path is A-B-E-F-H-J-K.
- d) The estimated time for the critical path is 42.5 weeks.
- e) The activity variance along the critical path is 28.8878 weeks.
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