---
title: "CPM components (AOA), 9 activities"
subject: "Project Engineering"
chapter: 3
key: "semester_7/project_engineering:cpm-aoa-nine-act-b78"
source: https://www.bctengineer.com/q/semester_7/project_engineering/cpm-aoa-nine-act-b78
site: BCT Engineering
---

# CPM components (AOA), 9 activities

Project Engineering · Chapter 3: Project Planning and Scheduling

- Asked in: 2078 Bhadra, 14 marks
- Video: [Use forward and backward pass to determine project duration and critical path](https://www.youtube.com/watch?v=4oDLMs11Exs) — Engineer4Free
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/project_engineering/cpm-aoa-nine-act-b78
- Whole chapter: https://www.bctengineer.com/md/semester_7/project_engineering/chapter-3

## Question

Find all components of CPM from the following information using the AOA method.

| Activity | Duration (week) | Predecessor |
|---|---|---|
| A | 3 | - |
| B | 2 | - |
| C | 4 | A |
| D | 3 | B |
| E | 3 | B |
| F | 3 | C |
| G | 2 | C, D |
| H | 5 | E |
| I | 3 | F, G, H |

## Short answer

AOA network with one dummy; forward/backward pass gives 13 weeks. Two critical paths A-C-F-I and B-E-H-I; critical activities A, B, C, E, F, H, I; all floats computed, with zero total float on critical activities.

## Answer

The components of CPM are: the network, event times (TE, TL), activity times (EST, EFT, LST, LFT), floats (TF, FF, Int.F, Ind.F), the critical path and the project duration. All are found below with the activity-on-arrow (AOA) method.

Notation: EST/EFT = earliest start/finish, LST/LFT = latest start/finish, TF = total float, FF = free float, Int.F = interfering float, Ind.F = independent float.

#### Network diagram (activity-on-arrow)

Each activity is an arrow with its duration in brackets (weeks). Events are circled numbers; dashed arrows are dummy activities (zero duration) that show dependencies only.

```text
(1) --A(3)--> (2)
(1) --B(2)--> (3)
(2) --C(4)--> (4)
(3) --E(3)--> (5)
(3) --D(3)--> (6)
(4) - - dummy - -> (6)
(4) --F(3)--> (7)
(5) --H(5)--> (7)
(6) --G(2)--> (7)
(7) --I(3)--> (8)
```

The network has 8 events and 1 dummy activity. Start event is (1) and the end event is (8).

#### Forward pass (EST and EFT)

Rule: EST = largest EFT of the preceding activities (0 for start activities); EFT = EST + D.

- A: EST = 0; EFT = 0 + 3 = 3
- B: EST = 0; EFT = 0 + 2 = 2
- D: EST = EFT of B = 2; EFT = 2 + 3 = 5
- E: EST = EFT of B = 2; EFT = 2 + 3 = 5
- C: EST = EFT of A = 3; EFT = 3 + 4 = 7
- H: EST = EFT of E = 5; EFT = 5 + 5 = 10
- F: EST = EFT of C = 7; EFT = 7 + 3 = 10
- G: EST = max(C:7, D:5) = 7; EFT = 7 + 2 = 9
- I: EST = max(F:10, G:9, H:10) = 10; EFT = 10 + 3 = 13

Project duration = largest EFT = **13 weeks**.

#### Backward pass (LFT and LST)

Rule: LFT = smallest LST of the succeeding activities (project duration for end activities); LST = LFT - D.

- I: LFT = project duration = 13; LST = 13 - 3 = 10
- G: LFT = LST of I = 10; LST = 10 - 2 = 8
- F: LFT = LST of I = 10; LST = 10 - 3 = 7
- H: LFT = LST of I = 10; LST = 10 - 5 = 5
- C: LFT = min(F:7, G:8) = 7; LST = 7 - 4 = 3
- E: LFT = LST of H = 5; LST = 5 - 3 = 2
- D: LFT = LST of G = 8; LST = 8 - 3 = 5
- B: LFT = min(D:5, E:2) = 2; LST = 2 - 2 = 0
- A: LFT = LST of C = 3; LST = 3 - 3 = 0

#### Floats

- **Total float** $TF = LST - EST = LFT - EFT$
- **Free float** $FF = \min(\text{EST of successors}) - EFT$
- **Interfering float** $IntF = TF - FF$
- **Independent float** $IndF = \max\left(0,\ \min(\text{EST of successors}) - \max(\text{LFT of predecessors}) - D\right)$

#### Result table

| Activity | D | EST | EFT | LST | LFT | TF | FF | Int.F | Ind.F |
|---|---|---|---|---|---|---|---|---|---|
| A | 3 | 0 | 3 | 0 | 3 | 0 | 0 | 0 | 0 |
| B | 2 | 0 | 2 | 0 | 2 | 0 | 0 | 0 | 0 |
| C | 4 | 3 | 7 | 3 | 7 | 0 | 0 | 0 | 0 |
| D | 3 | 2 | 5 | 5 | 8 | 3 | 2 | 1 | 2 |
| E | 3 | 2 | 5 | 2 | 5 | 0 | 0 | 0 | 0 |
| F | 3 | 7 | 10 | 7 | 10 | 0 | 0 | 0 | 0 |
| G | 2 | 7 | 9 | 8 | 10 | 1 | 1 | 0 | 0 |
| H | 5 | 5 | 10 | 5 | 10 | 0 | 0 | 0 | 0 |
| I | 3 | 10 | 13 | 10 | 13 | 0 | 0 | 0 | 0 |

Event times (TE from the forward pass, TL from the backward pass):

| Event | TE (earliest) | TL (latest) | Slack |
|---|---|---|---|
| 1 | 0 | 0 | 0 |
| 2 | 3 | 3 | 0 |
| 3 | 2 | 2 | 0 |
| 4 | 7 | 7 | 0 |
| 5 | 5 | 5 | 0 |
| 6 | 7 | 8 | 1 |
| 7 | 10 | 10 | 0 |
| 8 | 13 | 13 | 0 |

#### Critical path and project duration

- Critical activities (TF = 0): A, B, C, E, F, H, I
- Critical path: A - C - F - I (= 3+4+3+3 = 13 weeks)
- Critical path: B - E - H - I (= 2+3+5+3 = 13 weeks)
- There are two critical paths of the same length, so a delay on either one delays the project.

**Answer: project duration = 13 weeks; critical paths = A - C - F - I and B - E - H - I**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. G depends on C and D, but F depends only on C. Why does the AOA network need a dummy from event 4 to event 6?
   A. To add extra time to activity F
   B. Because every network must contain at least one dummy
   C. To show that G follows C as well as D without making F depend on D
   D. To make the project duration longer
   Answer: C. A dummy of zero duration carries only the dependency, so C's end event feeds G while F stays dependent on C alone. (Key point: Need for a dummy activity)

2. G has predecessors C (EFT 7) and D (EFT 5). What are EST and EFT of G (duration 2)?
   A. EST 5, EFT 7
   B. EST 7, EFT 9
   C. EST 12, EFT 14
   D. EST 7, EFT 7
   Answer: B. EST is the largest predecessor EFT = 7, so EFT = 7 + 2 = 9. (Key point: Forward pass with two predecessors)

3. B is followed by D (LST 5) and E (LST 2). What are LFT and LST of B (duration 2)?
   A. LFT 5, LST 3
   B. LFT 2, LST 0
   C. LFT 7, LST 5
   D. LFT 2, LST 2
   Answer: B. LFT = smallest successor LST = 2, so LST = 2 - 2 = 0. B is therefore critical. (Key point: Backward pass for B)

4. The project has duration 13 weeks. What are the critical paths and what does it mean?
   A. Only A-C-F-I; the other path has float
   B. Only B-E-H-I; the other path has float
   C. A-C-F-I and B-E-H-I, both 13 weeks, so a delay on either delays the project
   D. A-C-G-I and B-D-G-I, because G has float
   Answer: C. Both A+C+F+I = 3+4+3+3 and B+E+H+I = 2+3+5+3 equal 13 with all TF = 0, so both are critical. (Key point: Multiple critical paths)

## Close variants set in other exams

- [CPM components (AOA), activities A to I](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-aoa-nine-act-c76) — Project Engineering, chapter 3, asked 4×
- [CPM components (AOA), 10 activities](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-aoa-ten-act-b80) — Project Engineering, chapter 3, asked 5×
- [CPM components (AOA), 13 activities (months)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-aoa-thirteen-act-c75) — Project Engineering, chapter 3, asked 5×

## Related in Project Engineering

- [CPM components, 9 activities with successors](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-components-nine-act-c70) — Project Engineering, chapter 3, asked 3×
- [CPM network, 9 activities (A=3 to I=1)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-nine-act-b82) — Project Engineering, chapter 3, asked 5×
- [CPM network, 9 activities (A=5 to I=2)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-nine-act-a75) — Project Engineering, chapter 3, asked 5×
- [CPM precedence diagram, 9 activities](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-precedence-k72) — Project Engineering, chapter 3
- [CPM network, 13 activities (A to M)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-thirteen-act-k82) — Project Engineering, chapter 3, asked 5×
- [CPM network, 13 activities (A=2 to M=7)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-thirteen-act-b81) — Project Engineering, chapter 3, asked 5×

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