Activity Times and Predecessors
Activity	Description	Predecessors	 (a)
(m)
(b)
A	Drain Oil
---	3	5	10
B	Replace Filter
A	2	4	9
C	Refill Oil
B, E	3	5	8
D	Check Tyres
---	1	5	9
E	Wash Windows
D	2	5	8
F	Fill Fluids
E	2	5	10
G	Final Test
C, F
2	6	9
NB: Time periods are in weeks

Draw the activity network for this problem (either AON or AOA).

Determine the expected time and variance for each activty.

Show the activity schedule (ES,EF, LS, LF) as well as the slack. Show the results in a table.

Determine and state the critical path for this project. this must be based on your work at C, not by inspection.

What is the expected time, variance and standard deviation of this project?

What is the probability that the project will be finished within 59 days?

This problem involves project management techniques such as drawing an Activity Network, calculating the expected time and variance for each activity, identifying the critical path, and determining the probability of completing the project within a certain time frame. Below, I'll guide you through each step.

### 1. Draw the Activity Network (AON)
I'll provide a general approach to drawing an Activity on Node (AON) network.

**Steps:**
- **Nodes:** Represent each activity.
- **Arrows:** Show dependencies (predecessors).

**Network:**
- **A** (Drain Oil) has no predecessors.
- **B** (Replace Filter) depends on A.
- **D** (Check Tyres) has no predecessors.
- **E** (Wash Windows) depends on D.
- **C** (Refill Oil) depends on B and E.
- **F** (Fill Fluids) depends on E.
- **G** (Final Test) depends on C and F.

### 2. Determine the Expected Time and Variance
Use the formula:
\[
\text{Expected Time (TE)} = \frac{a + 4m + b}{6}
\]
\[
\text{Variance} = \left(\frac{b - a}{6}\right)^2
\]

Let’s compute these for each activity:

| Activity | a  | m  | b  | Expected Time (TE) | Variance  |
|----------|----|----|----|--------------------|-----------|
| A        | 3  | 5  | 10 | \( \frac{3 + 4(5) + 10}{6} = 5.5 \)   | \( \left(\frac{10 - 3}{6}\right)^2 = 1.36 \) |
| B        | 2  | 4  | 9  | \( \frac{2 + 4(4) + 9}{6} = 4.5 \)   | \( \left(\frac{9 - 2}{6}\right)^2 = 1.36 \) |
| C        | 3  | 5  | 8  | \( \frac{3 + 4(5) + 8}{6} = 5.0 \)   | \( \left(\frac{8 - 3}{6}\right)^2 = 0.69 \) |
| D        | 1  | 5  | 9  | \( \frac{1 + 4(5) + 9}{6} = 5.0 \)   | \( \left(\frac{9 - 1}{6}\right)^2 = 1.78 \) |
| E        | 2  | 5  | 8  | \( \frac{2 + 4(5) + 8}{6} = 5.0 \)   | \( \left(\frac{8 - 2}{6}\right)^2 = 1.00 \) |
| F        | 2  | 5  | 10 | \( \frac{2 + 4(5) + 10}{6} = 5.5 \)  | \( \left(\frac{10 - 2}{6}\right)^2 = 1.78 \) |
| G        | 2  | 6  | 9  | \( \frac{2 + 4(6) + 9}{6} = 5.83 \) | \( \left(\frac{9 - 2}{6}\right)^2 = 1.36 \) |

### 3. Activity Schedule (ES, EF, LS, LF) and Slack
This step involves determining the earliest start (ES), earliest finish (EF), latest start (LS), latest finish (LF), and Slack (LS-ES) for each activity.

First, calculate forward pass for ES and EF:
- **A:** ES = 0, EF = ES + TE = 5.5
- **B:** ES = EF(A) = 5.5, EF = 5.5 + 4.5 = 10
- **D:** ES = 0, EF = 5
- **E:** ES = EF(D) = 5, EF = 10
- **C:** ES = max(EF(B), EF(E)) = 10, EF = 15
- **F:** ES = EF(E) = 10, EF = 15.5
- **G:** ES = max(EF(C), EF(F)) = 15.5, EF = 21.33

Then, calculate backward pass for LS and LF:
- **G:** LF = 21.33, LS = 15.5
- **C:** LF = min(LS(G)) = 15.5, LS = 10
- **F:** LF = min(LS(G)) = 15.5, LS = 10
- **E:** LF = min(LS(C), LS(F)) = 10, LS = 5
- **D:** LF = LS(E) = 5, LS = 0
- **B:** LF = LS(C) = 10, LS = 5.5
- **A:** LF = LS(B) = 5.5, LS = 0

Slack Calculation:
- **A:** 0 (Critical)
- **B:** 0 (Critical)
- **C:** 0 (Critical)
- **D:** 0
- **E:** 0
- **F:** 0
- **G:** 0

### 4. Determine the Critical Path
The Critical Path has zero slack. The path is A -> B -> C -> G.

### 5. Expected Time, Variance, and Standard Deviation of the Project
Expected time for the project (TE project) = Sum of expected times on the critical path = 5.5 + 4.5 + 5.0 + 5.83 = 20.83 weeks

Variance for the project = Sum of variances on the critical path = 1.36 + 1.36 + 0.69 + 1.36 = 4.77 weeks^2

Standard Deviation for the project = \( \sqrt{4.77} \approx 2.18 \) weeks

### 6. Probability of Finishing the Project within 59 Days
Convert days to weeks: \( \text{59 days} \approx \frac{59}{7} \approx 8.43 \) weeks

To find the probability, calculate the Z-score:
\[
Z = \frac{\text{Target Time} - \text{Expected Time}}{\text{Standard Deviation}} = \frac{8.43 - 20.83}{2.18} \approx -5.68
\]
The Z-score of -5.68 is extremely low, meaning the probability of completing the project in 59 days is almost 0.

---

Would you like more detailed explanations or further questions about any specific part of the process? 

### Further Exploration Questions
1. How would changes in activity duration affect the critical path?
2. What would happen if the variance for a critical activity increased significantly?
3. How would the project schedule change if a new dependency is added?
4. Can you create a Gantt chart based on the given data?
5. What methods could be used to reduce the project duration?
6. How do you handle situations with multiple critical paths?
7. How would adding resources to reduce time affect the project cost?
8. What are the implications of having zero slack in the critical path?

### Tip:
To better understand the critical path, try plotting the entire network diagram and highlighting activities with zero slack. This visual representation can make it easier to comprehend the sequence and dependencies.

Learn how to manage projects using activity networks and calculate expected time, variance, and standard deviation. Understand the critical path method and determine project completion probabilities based on statistical analysis. Suitable for advanced learners and project management professionals.

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