---
title: "CPM components (AOA), 13 activities (months)"
subject: "Project Engineering"
chapter: 3
key: "semester_7/project_engineering:cpm-aoa-thirteen-act-c75"
source: https://www.bctengineer.com/q/semester_7/project_engineering/cpm-aoa-thirteen-act-c75
site: BCT Engineering
---

# CPM components (AOA), 13 activities (months)

Project Engineering · Chapter 3: Project Planning and Scheduling

- Asked in: 2075 Chaitra, 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-thirteen-act-c75
- 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 the AOA method.

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

## Short answer

Using AOA forward/backward pass, all CPM components (network, event/activity times, floats) are determined. Project duration is 14 months. Critical paths are B-E-I-L and B-E-K-M, each with total duration 14 months.

## 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 (months). Events are circled numbers; dashed arrows are dummy activities (zero duration) that show dependencies only.

```text
(1) --A(1)--> (2)
(1) --B(3)--> (3)
(2) --C(2)--> (4)
(2) --D(2)--> (5)
(3) --E(5)--> (6)
(4) --G(3)--> (6)
(4) --F(1)--> (8)
(5) --H(3)--> (6)
(6) --K(4)--> (7)
(6) - - dummy - -> (8)
(6) --J(1)--> (9)
(7) --M(2)--> (10)
(8) --I(5)--> (9)
(9) --L(1)--> (10)
```

The network has 10 events and 1 dummy activity. Start event is (1) and the end event is (10).

#### 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 + 3 = 3
- C: EST = EFT of A = 1; EFT = 1 + 2 = 3
- D: EST = EFT of A = 1; EFT = 1 + 2 = 3
- E: EST = EFT of B = 3; EFT = 3 + 5 = 8
- F: EST = EFT of C = 3; EFT = 3 + 1 = 4
- G: EST = EFT of C = 3; EFT = 3 + 3 = 6
- H: EST = EFT of D = 3; EFT = 3 + 3 = 6
- I: EST = max(E:8, F:4, G:6, H:6) = 8; EFT = 8 + 5 = 13
- J: EST = max(E:8, G:6, H:6) = 8; EFT = 8 + 1 = 9
- K: EST = max(E:8, G:6, H:6) = 8; EFT = 8 + 4 = 12
- M: EST = EFT of K = 12; EFT = 12 + 2 = 14
- L: EST = max(I:13, J:9) = 13; EFT = 13 + 1 = 14

Project duration = largest EFT = **14 months**.

#### Backward pass (LFT and LST)

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

- L: LFT = project duration = 14; LST = 14 - 1 = 13
- M: LFT = project duration = 14; LST = 14 - 2 = 12
- K: LFT = LST of M = 12; LST = 12 - 4 = 8
- J: LFT = LST of L = 13; LST = 13 - 1 = 12
- I: LFT = LST of L = 13; LST = 13 - 5 = 8
- H: LFT = min(I:8, J:12, K:8) = 8; LST = 8 - 3 = 5
- G: LFT = min(I:8, J:12, K:8) = 8; LST = 8 - 3 = 5
- F: LFT = LST of I = 8; LST = 8 - 1 = 7
- E: LFT = min(I:8, J:12, K:8) = 8; LST = 8 - 5 = 3
- D: LFT = LST of H = 5; LST = 5 - 2 = 3
- C: LFT = min(F:7, G:5) = 5; LST = 5 - 2 = 3
- B: LFT = LST of E = 3; LST = 3 - 3 = 0
- A: LFT = min(C:3, D:3) = 3; LST = 3 - 1 = 2

#### 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 | 2 | 3 | 2 | 0 | 2 | 0 |
| B | 3 | 0 | 3 | 0 | 3 | 0 | 0 | 0 | 0 |
| C | 2 | 1 | 3 | 3 | 5 | 2 | 0 | 2 | 0 |
| D | 2 | 1 | 3 | 3 | 5 | 2 | 0 | 2 | 0 |
| E | 5 | 3 | 8 | 3 | 8 | 0 | 0 | 0 | 0 |
| F | 1 | 3 | 4 | 7 | 8 | 4 | 4 | 0 | 2 |
| G | 3 | 3 | 6 | 5 | 8 | 2 | 2 | 0 | 0 |
| H | 3 | 3 | 6 | 5 | 8 | 2 | 2 | 0 | 0 |
| I | 5 | 8 | 13 | 8 | 13 | 0 | 0 | 0 | 0 |
| J | 1 | 8 | 9 | 12 | 13 | 4 | 4 | 0 | 4 |
| K | 4 | 8 | 12 | 8 | 12 | 0 | 0 | 0 | 0 |
| L | 1 | 13 | 14 | 13 | 14 | 0 | 0 | 0 | 0 |
| M | 2 | 12 | 14 | 12 | 14 | 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 | 3 | 2 |
| 3 | 3 | 3 | 0 |
| 4 | 3 | 5 | 2 |
| 5 | 3 | 5 | 2 |
| 6 | 8 | 8 | 0 |
| 7 | 12 | 12 | 0 |
| 8 | 8 | 8 | 0 |
| 9 | 13 | 13 | 0 |
| 10 | 14 | 14 | 0 |

#### Critical path and project duration

- Critical activities (TF = 0): B, E, I, K, M, L
- Critical path: B - E - I - L (= 3+5+5+1 = 14 months)
- Critical path: B - E - K - M (= 3+5+4+2 = 14 months)
- There are two critical paths of the same length, so delay on either one delays the project.

**Answer: project duration = 14 months; critical path = B - E - I - L ; B - E - K - M**

## Check yourself

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

1. I has predecessors E (EFT 8), F (EFT 4), G (EFT 6) and H (EFT 6). What is the EST of I?
   A. 4
   B. 6
   C. 8
   D. 24
   Answer: C. EST is the largest predecessor EFT, which is E at 8. EFT of I = 8 + 5 = 13. (Key point: Forward pass for I)

2. H is followed by I (LST 8), J (LST 12) and K (LST 8). What is LFT and LST of H (duration 3)?
   A. LFT 12, LST 9
   B. LFT 8, LST 5
   C. LFT 8, LST 8
   D. LFT 13, LST 10
   Answer: B. LFT is the smallest successor LST = 8, so LST = 8 - 3 = 5. (Key point: Backward pass for H)

3. Project duration is 14 months. Which are the critical paths?
   A. B-E-I-L and B-E-K-M
   B. A-C-G-I-L only
   C. B-E-J-L and B-E-K-M
   D. B-E-I-L only
   Answer: A. B+E+I+L = 3+5+5+1 = 14 and B+E+K+M = 3+5+4+2 = 14, both with zero float. (Key point: Two critical paths)

4. F has EST 3, EFT 4, LST 7 and LFT 8. What is its total float?
   A. 2 months
   B. 5 months
   C. 4 months
   D. 0 months
   Answer: C. TF = LST - EST = 7 - 3 = 4 (or LFT - EFT = 8 - 4 = 4). (Key point: Total float of F)

## Close variants set in other exams

- [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), 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), 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×

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