---
title: "CPM components (AOA), 10 activities"
subject: "Project Engineering"
chapter: 3
key: "semester_7/project_engineering:cpm-aoa-ten-act-b80"
source: https://www.bctengineer.com/q/semester_7/project_engineering/cpm-aoa-ten-act-b80
site: BCT Engineering
---

# CPM components (AOA), 10 activities

Project Engineering · Chapter 3: Project Planning and Scheduling

- Asked in: 2080 Bhadra, 12 marks
- Video: [Finding the Critical Path, duration and Project Duration, Critical Path Method, float, EST, EFT, LST](https://www.youtube.com/watch?v=OKKcjeJt4cI) — Civil Engineering Exam
- 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-ten-act-b80
- Whole chapter: https://www.bctengineer.com/md/semester_7/project_engineering/chapter-3

## Question

Find all the components of CPM from the following information (use AOA).

| Activity | Duration (week) | Predecessor | Successor |
|---|---|---|---|
| A | 1 | - | C, E |
| B | 6 | - | C, D |
| C | 2 | A, B | F |
| D | 2 | B | H |
| E | 4 | A | G |
| F | 3 | C | G, H |
| G | 4 | E, F | I |
| H | 2 | D, F | I |
| I | 5 | G, H | J |
| J | 3 | I | - |

## Short answer

AOA network with dummies; forward/backward pass yields 23 weeks. Critical path B-C-F-G-I-J; critical activities B, C, F, G, I, J; all activity times and 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(1)--> (2)
(1) --B(6)--> (3)
(2) - - dummy - -> (4)
(2) --E(4)--> (8)
(3) - - dummy - -> (4)
(3) --D(2)--> (7)
(4) --C(2)--> (5)
(5) --F(3)--> (6)
(6) - - dummy - -> (7)
(6) - - dummy - -> (8)
(7) --H(2)--> (9)
(8) --G(4)--> (9)
(9) --I(5)--> (10)
(10) --J(3)--> (11)
```

The network has 11 events and 4 dummy activities. Start event is (1) and the end event is (11).

#### 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 + 1 = 1
- B: EST = 0; EFT = 0 + 6 = 6
- E: EST = EFT of A = 1; EFT = 1 + 4 = 5
- C: EST = max(A:1, B:6) = 6; EFT = 6 + 2 = 8
- D: EST = EFT of B = 6; EFT = 6 + 2 = 8
- F: EST = EFT of C = 8; EFT = 8 + 3 = 11
- G: EST = max(E:5, F:11) = 11; EFT = 11 + 4 = 15
- H: EST = max(D:8, F:11) = 11; EFT = 11 + 2 = 13
- I: EST = max(G:15, H:13) = 15; EFT = 15 + 5 = 20
- J: EST = EFT of I = 20; EFT = 20 + 3 = 23

Project duration = largest EFT = **23 weeks**.

#### Backward pass (LFT and LST)

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

- J: LFT = project duration = 23; LST = 23 - 3 = 20
- I: LFT = LST of J = 20; LST = 20 - 5 = 15
- H: LFT = LST of I = 15; LST = 15 - 2 = 13
- G: LFT = LST of I = 15; LST = 15 - 4 = 11
- F: LFT = min(G:11, H:13) = 11; LST = 11 - 3 = 8
- D: LFT = LST of H = 13; LST = 13 - 2 = 11
- C: LFT = LST of F = 8; LST = 8 - 2 = 6
- E: LFT = LST of G = 11; LST = 11 - 4 = 7
- B: LFT = min(C:6, D:11) = 6; LST = 6 - 6 = 0
- A: LFT = min(C:6, E:7) = 6; LST = 6 - 1 = 5

#### 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 | 1 | 0 | 1 | 5 | 6 | 5 | 0 | 5 | 0 |
| B | 6 | 0 | 6 | 0 | 6 | 0 | 0 | 0 | 0 |
| C | 2 | 6 | 8 | 6 | 8 | 0 | 0 | 0 | 0 |
| D | 2 | 6 | 8 | 11 | 13 | 5 | 3 | 2 | 3 |
| E | 4 | 1 | 5 | 7 | 11 | 6 | 6 | 0 | 1 |
| F | 3 | 8 | 11 | 8 | 11 | 0 | 0 | 0 | 0 |
| G | 4 | 11 | 15 | 11 | 15 | 0 | 0 | 0 | 0 |
| H | 2 | 11 | 13 | 13 | 15 | 2 | 2 | 0 | 0 |
| I | 5 | 15 | 20 | 15 | 20 | 0 | 0 | 0 | 0 |
| J | 3 | 20 | 23 | 20 | 23 | 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 | 1 | 6 | 5 |
| 3 | 6 | 6 | 0 |
| 4 | 6 | 6 | 0 |
| 5 | 8 | 8 | 0 |
| 6 | 11 | 11 | 0 |
| 7 | 11 | 13 | 2 |
| 8 | 11 | 11 | 0 |
| 9 | 15 | 15 | 0 |
| 10 | 20 | 20 | 0 |
| 11 | 23 | 23 | 0 |

#### Critical path and project duration

- Critical activities (TF = 0): B, C, F, G, I, J
- Critical path: B - C - F - G - I - J (= 6+2+3+4+5+3 = 23 weeks)

**Answer: project duration = 23 weeks; critical path = B - C - F - G - I - J**

## Check yourself

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

1. C has predecessors A (EFT 1) and B (EFT 6). What is its EST and EFT (duration 2)?
   A. EST 1, EFT 3
   B. EST 7, EFT 9
   C. EST 6, EFT 8
   D. EST 0, EFT 2
   Answer: C. EST is the largest predecessor EFT = 6; EFT = 6 + 2 = 8. (Key point: Forward pass for C)

2. F is followed by G (LST 11) and H (LST 13). What is the LFT and LST of F (duration 3)?
   A. LFT 13, LST 10
   B. LFT 11, LST 8
   C. LFT 24, LST 21
   D. LFT 11, LST 11
   Answer: B. LFT is the smallest successor LST = 11, so LST = 11 - 3 = 8. (Key point: Backward pass for F)

3. E has EFT = 5 and LST = 7, EST = 1. Its successor G has EST = 11. What are TF and FF?
   A. TF 6, FF 6
   B. TF 6, FF 0
   C. TF 2, FF 6
   D. TF 10, FF 6
   Answer: A. TF = LST - EST = 7 - 1 = 6. FF = EST of G - EFT of E = 11 - 5 = 6, so interfering float is 0. (Key point: Free float of E)

4. What are the project duration and critical path?
   A. 20 weeks, B-C-F-G-I
   B. 23 weeks, A-E-G-I-J
   C. 23 weeks, B-C-F-H-I-J
   D. 23 weeks, B-C-F-G-I-J
   Answer: D. B+C+F+G+I+J = 6+2+3+4+5+3 = 23 weeks, and all these have TF = 0. (Key point: Critical path and duration)

## 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), 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×
- [CPM components (AOA), 9 activities](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-aoa-nine-act-b78) — Project Engineering, chapter 3, asked 4×
- [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×

## Related in Project Engineering

- [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, 10 activities with successors](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-ten-act-a74) — Project Engineering, chapter 3
- [CPM network, 10 activities (A=4 to J=9)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-ten-act-as70) — Project Engineering, chapter 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 network, 10 activities (A=5 to J=5)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-ten-act-k68) — Project Engineering, chapter 3

## Related in other subjects

- [TCTO numerical: least cost schedule, Rs 2000/week](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Atcto-2068-bhadra) — Construction Management, chapter 2