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Sure, let's go step-by-step to solve the given question.
### Part (a): Expected Time for Activity C
We use the formula for expected time [tex]\( TE \)[/tex] for an activity based on the three time estimates:
[tex]\[ TE = \frac{a + 4m + b}{6} \][/tex]
For activity [tex]\( C \)[/tex]:
- [tex]\( a = 9 \)[/tex] weeks
- [tex]\( m = 14 \)[/tex] weeks
- [tex]\( b = 18 \)[/tex] weeks
Plugging these values into the formula:
[tex]\[ TE_C = \frac{9 + 4(14) + 18}{6} \][/tex]
[tex]\[ TE_C = \frac{9 + 56 + 18}{6} \][/tex]
[tex]\[ TE_C = \frac{83}{6} \][/tex]
[tex]\[ TE_C ≈ 13.83 \][/tex]
So, the expected time for activity [tex]\( C \)[/tex] is [tex]\( 13.83 \)[/tex] weeks.
### Part (b): Variance for Activity C
We use the formula for variance ([tex]\( \sigma_C^2 \)[/tex]):
[tex]\[ \sigma_C^2 = \left( \frac{b - a}{6} \right)^2 \][/tex]
For activity [tex]\( C \)[/tex]:
- [tex]\( a = 9 \)[/tex]
- [tex]\( b = 18 \)[/tex]
Plugging these values into the formula:
[tex]\[ \sigma_C^2 = \left( \frac{18 - 9}{6} \right)^2 \][/tex]
[tex]\[ \sigma_C^2 = \left( \frac{9}{6} \right)^2 \][/tex]
[tex]\[ \sigma_C^2 = \left( 1.5 \right)^2 \][/tex]
[tex]\[ \sigma_C^2 = 2.25 \][/tex]
So, the variance for activity [tex]\( C \)[/tex] is [tex]\( 2.25 \)[/tex] weeks.
### Part (c): Critical Path
To determine the critical path, we need to calculate the expected time for all activities and then find the path with the maximum total duration.
Let's calculate the expected times for each activity:
[tex]\[ \begin{aligned} & TE_A = \frac{6 + 4(9) + 12}{6} = \frac{54}{6} = 9 \\ & TE_B = \frac{1 + 4(8) + 24}{6} = \frac{58}{6} ≈ 9.67 \\ & TE_C = 13.83 \quad \text{(Previously calculated)} \\ & TE_D = \frac{5 + 4(7) + 10}{6} = \frac{43}{6} ≈ 7.17 \\ & TE_E = \frac{1 + 4(3) + 4}{6} = \frac{16}{6} ≈ 2.67 \\ & TE_F = \frac{5 + 4(8) + 20}{6} = \frac{61}{6} ≈ 10.17 \\ & TE_G = \frac{3 + 4(3) + 5}{6} = \frac{20}{6} ≈ 3.33 \\ & TE_H = \frac{2 + 4(2) + 2}{6} = \frac{12}{6} = 2 \\ & TE_I = \frac{5 + 4(5) + 5}{6} = \frac{35}{6} ≈ 5.83 \\ & TE_J = \frac{6 + 4(8) + 14}{6} = \frac{56}{6} ≈ 9.33 \\ & TE_K = \frac{1 + 4(1) + 4}{6} = \frac{9}{6} = 1.5 \\ \end{aligned} \][/tex]
Now, let's enumerate and calculate the possible paths from A to K:
1. A -> B -> E -> F -> H -> J -> K:
[tex]\[ 9 + 9.67 + 2.67 + 10.17 + 2 + 9.33 + 1.5 = 44.34 \][/tex]
2. A -> B -> E -> F -> I -> K:
[tex]\[ 9 + 9.67 + 2.67 + 10.17 + 5.83 + 1.5 = 38.84 \][/tex]
3. A -> B -> E -> G -> J -> K:
[tex]\[ 9 + 9.67 + 2.67 + 3.33 + 9.33 + 1.5 = 35.5 \][/tex]
4. A -> C -> F -> H -> J -> K:
[tex]\[ 9 + 13.83 + 10.17 + 2 + 9.33 + 1.5 = 45.83 \][/tex]
5. A -> C -> F -> I -> K:
[tex]\[ 9 + 13.83 + 10.17 + 5.83 + 1.5 = 40.33 \][/tex]
6. A -> D -> J -> K:
[tex]\[ 9 + 7.17 + 9.33 + 1.5 = 27.0 \][/tex]
7. A -> C -> G -> J -> K:
[tex]\[ 9 + 13.83 + 3.33 + 9.33 + 1.5 = 37.0 \][/tex]
The critical path is the one with the longest duration, which is [tex]\( \text{A -> C -> F -> H -> J -> K} \)[/tex] with a total duration of [tex]\( 45.83 \)[/tex] weeks.
### Summary:
a) The expected (estimated) time for activity [tex]\( C \)[/tex] is [tex]\( 13.83 \)[/tex] weeks.
b) The variance for activity [tex]\( C \)[/tex] is [tex]\( 2.25 \)[/tex] weeks.
c) Based on the calculation of the estimated times, the critical path is [tex]\( \text{A -> C -> F -> H -> J -> K} \)[/tex].
### Part (a): Expected Time for Activity C
We use the formula for expected time [tex]\( TE \)[/tex] for an activity based on the three time estimates:
[tex]\[ TE = \frac{a + 4m + b}{6} \][/tex]
For activity [tex]\( C \)[/tex]:
- [tex]\( a = 9 \)[/tex] weeks
- [tex]\( m = 14 \)[/tex] weeks
- [tex]\( b = 18 \)[/tex] weeks
Plugging these values into the formula:
[tex]\[ TE_C = \frac{9 + 4(14) + 18}{6} \][/tex]
[tex]\[ TE_C = \frac{9 + 56 + 18}{6} \][/tex]
[tex]\[ TE_C = \frac{83}{6} \][/tex]
[tex]\[ TE_C ≈ 13.83 \][/tex]
So, the expected time for activity [tex]\( C \)[/tex] is [tex]\( 13.83 \)[/tex] weeks.
### Part (b): Variance for Activity C
We use the formula for variance ([tex]\( \sigma_C^2 \)[/tex]):
[tex]\[ \sigma_C^2 = \left( \frac{b - a}{6} \right)^2 \][/tex]
For activity [tex]\( C \)[/tex]:
- [tex]\( a = 9 \)[/tex]
- [tex]\( b = 18 \)[/tex]
Plugging these values into the formula:
[tex]\[ \sigma_C^2 = \left( \frac{18 - 9}{6} \right)^2 \][/tex]
[tex]\[ \sigma_C^2 = \left( \frac{9}{6} \right)^2 \][/tex]
[tex]\[ \sigma_C^2 = \left( 1.5 \right)^2 \][/tex]
[tex]\[ \sigma_C^2 = 2.25 \][/tex]
So, the variance for activity [tex]\( C \)[/tex] is [tex]\( 2.25 \)[/tex] weeks.
### Part (c): Critical Path
To determine the critical path, we need to calculate the expected time for all activities and then find the path with the maximum total duration.
Let's calculate the expected times for each activity:
[tex]\[ \begin{aligned} & TE_A = \frac{6 + 4(9) + 12}{6} = \frac{54}{6} = 9 \\ & TE_B = \frac{1 + 4(8) + 24}{6} = \frac{58}{6} ≈ 9.67 \\ & TE_C = 13.83 \quad \text{(Previously calculated)} \\ & TE_D = \frac{5 + 4(7) + 10}{6} = \frac{43}{6} ≈ 7.17 \\ & TE_E = \frac{1 + 4(3) + 4}{6} = \frac{16}{6} ≈ 2.67 \\ & TE_F = \frac{5 + 4(8) + 20}{6} = \frac{61}{6} ≈ 10.17 \\ & TE_G = \frac{3 + 4(3) + 5}{6} = \frac{20}{6} ≈ 3.33 \\ & TE_H = \frac{2 + 4(2) + 2}{6} = \frac{12}{6} = 2 \\ & TE_I = \frac{5 + 4(5) + 5}{6} = \frac{35}{6} ≈ 5.83 \\ & TE_J = \frac{6 + 4(8) + 14}{6} = \frac{56}{6} ≈ 9.33 \\ & TE_K = \frac{1 + 4(1) + 4}{6} = \frac{9}{6} = 1.5 \\ \end{aligned} \][/tex]
Now, let's enumerate and calculate the possible paths from A to K:
1. A -> B -> E -> F -> H -> J -> K:
[tex]\[ 9 + 9.67 + 2.67 + 10.17 + 2 + 9.33 + 1.5 = 44.34 \][/tex]
2. A -> B -> E -> F -> I -> K:
[tex]\[ 9 + 9.67 + 2.67 + 10.17 + 5.83 + 1.5 = 38.84 \][/tex]
3. A -> B -> E -> G -> J -> K:
[tex]\[ 9 + 9.67 + 2.67 + 3.33 + 9.33 + 1.5 = 35.5 \][/tex]
4. A -> C -> F -> H -> J -> K:
[tex]\[ 9 + 13.83 + 10.17 + 2 + 9.33 + 1.5 = 45.83 \][/tex]
5. A -> C -> F -> I -> K:
[tex]\[ 9 + 13.83 + 10.17 + 5.83 + 1.5 = 40.33 \][/tex]
6. A -> D -> J -> K:
[tex]\[ 9 + 7.17 + 9.33 + 1.5 = 27.0 \][/tex]
7. A -> C -> G -> J -> K:
[tex]\[ 9 + 13.83 + 3.33 + 9.33 + 1.5 = 37.0 \][/tex]
The critical path is the one with the longest duration, which is [tex]\( \text{A -> C -> F -> H -> J -> K} \)[/tex] with a total duration of [tex]\( 45.83 \)[/tex] weeks.
### Summary:
a) The expected (estimated) time for activity [tex]\( C \)[/tex] is [tex]\( 13.83 \)[/tex] weeks.
b) The variance for activity [tex]\( C \)[/tex] is [tex]\( 2.25 \)[/tex] weeks.
c) Based on the calculation of the estimated times, the critical path is [tex]\( \text{A -> C -> F -> H -> J -> K} \)[/tex].
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