---
title: "CPM network, 10 activities (A=4 to J=9)"
subject: "Project Engineering"
chapter: 3
key: "semester_7/project_engineering:cpm-network-ten-act-as70"
source: https://www.bctengineer.com/q/semester_7/project_engineering/cpm-network-ten-act-as70
site: BCT Engineering
---

# CPM network, 10 activities (A=4 to J=9)

Project Engineering · Chapter 3: Project Planning and Scheduling

- Asked in: 2070 Asar, 16 marks
- Video: [How to draw a CPM network diagram](https://www.youtube.com/watch?v=gEBMPP0SOnU) — 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-network-ten-act-as70
- Whole chapter: https://www.bctengineer.com/md/semester_7/project_engineering/chapter-3

## Question

Construct the CPM network for a project with the following activities. Find: (i) critical path, (ii) project completion time, (iii) EST, EFT, LST, LFT, total float, free float (FF), independent float (IdF) and interfering float (If).

| Activity | Days | Predecessor |
|---|---|---|
| A | 4 | - |
| B | 7 | - |
| C | 4 | A, B |
| D | 3 | B |
| E | 2 | A |
| F | 1 | C |
| G | 6 | E, F |
| H | 5 | D, F |
| I | 8 | G, H |
| J | 9 | I |

## Short answer

Duration 35 days; critical path B-C-F-G-I-J. EST/EFT/LST/LFT/TF/FF: A 0,4,3,7,3,0; B 0,7,0,7,0,0; C 7,11,7,11,0,0; D 7,10,10,13,3,2; E 4,6,10,12,6,6; F 11,12,11,12,0,0; G 12,18,12,18,0,0; H 12,17,13,18,1,1; I 18,26,18,26,0,0; J 26,35,26,35,0,0.

## Answer

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

```text
(1) --A(4)--> (2)
(1) --B(7)--> (3)
(2) - - dummy - -> (4)
(2) --E(2)--> (8)
(3) - - dummy - -> (4)
(3) --D(3)--> (7)
(4) --C(4)--> (5)
(5) --F(1)--> (6)
(6) - - dummy - -> (7)
(6) - - dummy - -> (8)
(7) --H(5)--> (9)
(8) --G(6)--> (9)
(9) --I(8)--> (10)
(10) --J(9)--> (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 + 4 = 4
- B: EST = 0; EFT = 0 + 7 = 7
- E: EST = EFT of A = 4; EFT = 4 + 2 = 6
- C: EST = max(A:4, B:7) = 7; EFT = 7 + 4 = 11
- D: EST = EFT of B = 7; EFT = 7 + 3 = 10
- F: EST = EFT of C = 11; EFT = 11 + 1 = 12
- G: EST = max(E:6, F:12) = 12; EFT = 12 + 6 = 18
- H: EST = max(D:10, F:12) = 12; EFT = 12 + 5 = 17
- I: EST = max(G:18, H:17) = 18; EFT = 18 + 8 = 26
- J: EST = EFT of I = 26; EFT = 26 + 9 = 35

Project duration = largest EFT = **35 days**.

#### 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 = 35; LST = 35 - 9 = 26
- I: LFT = LST of J = 26; LST = 26 - 8 = 18
- H: LFT = LST of I = 18; LST = 18 - 5 = 13
- G: LFT = LST of I = 18; LST = 18 - 6 = 12
- F: LFT = min(G:12, H:13) = 12; LST = 12 - 1 = 11
- D: LFT = LST of H = 13; LST = 13 - 3 = 10
- C: LFT = LST of F = 11; LST = 11 - 4 = 7
- E: LFT = LST of G = 12; LST = 12 - 2 = 10
- B: LFT = min(C:7, D:10) = 7; LST = 7 - 7 = 0
- A: LFT = min(C:7, E:10) = 7; LST = 7 - 4 = 3

#### 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 | 4 | 0 | 4 | 3 | 7 | 3 | 0 | 3 | 0 |
| B | 7 | 0 | 7 | 0 | 7 | 0 | 0 | 0 | 0 |
| C | 4 | 7 | 11 | 7 | 11 | 0 | 0 | 0 | 0 |
| D | 3 | 7 | 10 | 10 | 13 | 3 | 2 | 1 | 2 |
| E | 2 | 4 | 6 | 10 | 12 | 6 | 6 | 0 | 3 |
| F | 1 | 11 | 12 | 11 | 12 | 0 | 0 | 0 | 0 |
| G | 6 | 12 | 18 | 12 | 18 | 0 | 0 | 0 | 0 |
| H | 5 | 12 | 17 | 13 | 18 | 1 | 1 | 0 | 0 |
| I | 8 | 18 | 26 | 18 | 26 | 0 | 0 | 0 | 0 |
| J | 9 | 26 | 35 | 26 | 35 | 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 | 4 | 7 | 3 |
| 3 | 7 | 7 | 0 |
| 4 | 7 | 7 | 0 |
| 5 | 11 | 11 | 0 |
| 6 | 12 | 12 | 0 |
| 7 | 12 | 13 | 1 |
| 8 | 12 | 12 | 0 |
| 9 | 18 | 18 | 0 |
| 10 | 26 | 26 | 0 |
| 11 | 35 | 35 | 0 |

#### Critical path and project duration

- Critical activities (TF = 0): B, C, F, G, I, J
- Critical path: B - C - F - G - I - J (= 7+4+1+6+8+9 = 35 days)

**Answer: project duration = 35 days; 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. What are the project completion time and critical path?
   A. 26 days, B-C-F-G-I
   B. 35 days, B-C-F-G-I-J
   C. 32 days, B-D-H-I-J
   D. 35 days, A-C-F-G-I-J
   Answer: B. J ends at 35, and B, C, F, G, I, J all have zero total float: 7+4+1+6+8+9 = 35. (Key point: Project duration)

2. Why is EST of G equal to 12?
   A. It is the EFT of H
   B. It is the EFT of E (6)
   C. It is the larger of EFT of E (6) and EFT of F (12)
   D. It is the project duration minus D
   Answer: C. G follows E and F, and EST is the largest predecessor EFT: max(6, 12) = 12. (Key point: EST of G)

3. What is LST of B?
   A. 7
   B. 0
   C. 2
   D. 3
   Answer: B. LFT of B = min(LST of C = 7, LST of D = 10) = 7, so LST = 7 - 7 = 0. (Key point: Backward pass for B)

4. For D (TF = 3, FF = 2), what is the interfering float?
   A. 2
   B. 0
   C. 5
   D. 1
   Answer: D. Interfering float = TF - FF = 3 - 2 = 1. (Key point: Floats of D)

5. Which formula gives independent float?
   A. TF - FF
   B. max(0, min EST of successors - max LFT of predecessors - D)
   C. LST - EST
   D. min EST of successors - EFT
   Answer: B. IndF is the float left when predecessors finish as late as possible and successors start as early as possible; negative values are taken as 0. (Key point: Independent float)

## Related in Project Engineering

- [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, building project (A to K)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-eleven-act-s73) — Project Engineering, chapter 3
- [CPM network, 12 activities (A=10 to L=8)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-twelve-act-as67) — Project Engineering, chapter 3
- [CPM network, 12 activities (A=8 to L=3)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-twelve-act-b66) — Project Engineering, chapter 3
- [CPM network, 10 activities (A=10 to J=4)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-ten-act-k79) — Project Engineering, chapter 3
- [CPM network, 9 activities with successors](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-nine-act-a76) — Project Engineering, chapter 3

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