VIPSolutionsGive step-by-step solution with explanation and final answer:ia 2 > = 5 =. . . ~E — o 0 EE 5828 [ofof->[H] — 8g 3 His g Fo £ m5 = E @ 9% 3 E: Ei = Ea g Bee ie), = 2%2e 2 |o 3 $83 = 5 3825 g Bc g g E> 2 3 = : ED 3 2 dy 8 = $383 g — = 0-0 = = g§3988§ 3, - gs 3 = 22 3 = goa — S £35 - = 35 EE ao 0 = ° Eo E20 | =o slo 3]
Question:
Give step-by-step solution with explanation and final answer:
ia 2 >
= 5
=. . . ~E
— o 0 EE
5828 [ofof->[H]
— 8g 3
His g
Fo £ m5 =
E @ 9% 3 E: Ei
= Ea g
Bee ie), =
2%2e 2 |o 3
$83 = 5
3825 g
Bc g g
E> 2 3 =
: ED 3
2 dy 8
= $383 g
— = 0-0 =
= g§3988§ 3,
- gs 3
= 22 3
= goa
— S £35 -
= 35
EE ao 0
= °
Eo
E20 | =o slo 3]
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Asked by: Test User 100170
Created at: 2025-05-08 19:45:54
Asked by: Test User 100170
Created at: 2025-05-08 19:45:54
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Let's go step-by-step as required by the problem.
---
### **Step 1: Calculate Expected Time (TE) and Variance (σ²) for Each Activity**
For each activity, use the PERT formulas:
- **Expected Time (TE):** \(\displaystyle TE = \frac{O + 4M + P}{6}\)
- **Variance (σ²):** \(\displaystyle \sigma^2 = \left(\frac{P - O}{6}\right)^2\)
Let's calculate for each activity:
#### **Activity A**
- O = 2, M = 4, P = 6
\[
TE_A = \frac{2 + 4 \times 4 + 6}{6} = \frac{2 + 16 + 6}{6} = \frac{24}{6} = 4
\]
\[
\sigma_A^2 = \left(\frac{6 - 2}{6}\right)^2 = \left(\frac{4}{6}\right)^2 = \left(\frac{2}{3}\right)^2 = 0.444
\]
#### **Activity B**
- O = 5, M = 5, P = 7
\[
TE_B = \frac{5 + 4 \times 5 + 7}{6} = \frac{5 + 20 + 7}{6} = \frac{32}{6} = 5.33
\]
\[
\sigma_B^2 = \left(\frac{7 - 5}{6}\right)^2 = \left(\frac{2}{6}\right)^2 = \left(\frac{1}{3}\right)^2 = 0.111
\]
#### **Activity C**
- O = 8, M = 10, P = 9
\[
TE_C = \frac{8 + 4 \times 10 + 9}{6} = \frac{8 + 40 + 9}{6} = \frac{57}{6} = 9.5
\]
\[
\sigma_C^2 = \left(\frac{9 - 8}{6}\right)^2 = \left(\frac{1}{6}\right)^2 = 0.0278
\]
#### **Activity D**
- O = 5, M = 7, P = 9
\[
TE_D = \frac{5 + 4 \times 7 + 9}{6} = \frac{5 + 28 + 9}{6} = \frac{42}{6} = 7
\]
\[
\sigma_D^2 = \left(\frac{9 - 5}{6}\right)^2 = \left(\frac{4}{6}\right)^2 = \left(\frac{2}{3}\right)^2 = 0.444
\]
---
### **Step 2: Draw the PERT Network Diagram**
**Dependencies:**
- A has no predecessor.
- B follows A.
- C follows A.
- D follows B and C.
**Network:**
- Start → A
- A → B
- A → C
- B, C → D
- D → End
---
### **Step 3: Find the Critical Path and Project Completion Time**
Let's calculate the duration for each path:
#### **Paths:**
- Path 1: A → B → D: \(TE_A + TE_B + TE_D = 4 + 5.33 + 7 = 16.33\)
- Path 2: A → C → D: \(TE_A + TE_C + TE_D = 4 + 9.5 + 7 = 20.5\)
**Critical Path:** A → C → D (Longest Path)
**Expected Project Duration:** **20.5 days**
---
### **Step 4: Calculate Variance of the Critical Path**
- Variance on critical path = σ²_A + σ²_C + σ²_D
\[
0.444 + 0.0278 + 0.444 = 0.9158
\]
- **Standard deviation (σ):** \(\sqrt{0.9158} = 0.957\)
---
### **Step 5: Probability of Completion Within 20 Days**
- \(TE_{\text{critical}} = 20.5\)
- \(σ = 0.957\)
- \(T = 20\) days
- \(Z = \frac{T - TE_{\text{critical}}}{σ} = \frac{20 - 20.5}{0.957} = \frac{-0.5}{0.957} = -0.522\)
Look up Z = -0.52 in standard normal tables:
- Probability (P) ≈ **0.3015**
---
## **Final Answers**
**a) PERT Diagram:** (as above, A → B → D and A → C → D, with D depending on both B and C)
**b) Expected Project Completion Time:** **20.5 days**
**c) Variance of Critical Path:** **0.9158**
**d) Probability of Completion Within 20 Days:** **0.3015 (or 30.15%)**
---
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