---
title: "CPM network, 10 activities (A=5 to J=5)"
subject: "Project Engineering"
chapter: 3
key: "semester_7/project_engineering:cpm-network-ten-act-k68"
source: https://www.bctengineer.com/q/semester_7/project_engineering/cpm-network-ten-act-k68
site: BCT Engineering
---

# CPM network, 10 activities (A=5 to J=5)

Project Engineering · Chapter 3: Project Planning and Scheduling

- Asked in: 2068 Baisakh (old course), 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-k68
- Whole chapter: https://www.bctengineer.com/md/semester_7/project_engineering/chapter-3

## Question

For a project, the following durations are given. Find EST, EFT, LST, LFT, TF, FF. Show the critical path and find the duration.

| SN | Activity | Duration | Precedence | Successor |
|---|---|---|---|---|
| 1 | A | 5 | - | E |
| 2 | B | 6 | - | F |
| 3 | C | 7 | - | G |
| 4 | D | 8 | - | H |
| 5 | E | 9 | A | I, J |
| 6 | F | 7 | B | I, J |
| 7 | G | 5 | C | I, J |
| 8 | H | 3 | D | I, J |
| 9 | I | 4 | E, F, G, H | - |
| 10 | J | 5 | E, F, G, H | - |

## Short answer

Duration 19 time units; critical path A-E-J. EST/EFT/LST/LFT/TF/FF: A 0,5,0,5,0,0; B 0,6,1,7,1,0; C 0,7,2,9,2,0; D 0,8,3,11,3,0; E 5,14,5,14,0,0; F 6,13,7,14,1,1; G 7,12,9,14,2,2; H 8,11,11,14,3,3; I 14,18,15,19,1,1; J 14,19,14,19,0,0.

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

```text
(1) --A(5)--> (2)
(1) --B(6)--> (3)
(1) --C(7)--> (4)
(1) --D(8)--> (5)
(2) --E(9)--> (6)
(3) --F(7)--> (6)
(4) --G(5)--> (6)
(5) --H(3)--> (6)
(6) --J(5)--> (7)
(6) --I(4)--> (8)
(7) - - dummy - -> (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 + 5 = 5
- B: EST = 0; EFT = 0 + 6 = 6
- C: EST = 0; EFT = 0 + 7 = 7
- D: EST = 0; EFT = 0 + 8 = 8
- E: EST = EFT of A = 5; EFT = 5 + 9 = 14
- F: EST = EFT of B = 6; EFT = 6 + 7 = 13
- G: EST = EFT of C = 7; EFT = 7 + 5 = 12
- H: EST = EFT of D = 8; EFT = 8 + 3 = 11
- I: EST = max(E:14, F:13, G:12, H:11) = 14; EFT = 14 + 4 = 18
- J: EST = max(E:14, F:13, G:12, H:11) = 14; EFT = 14 + 5 = 19

Project duration = largest EFT = **19 time units**.

#### 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 = 19; LST = 19 - 5 = 14
- I: LFT = project duration = 19; LST = 19 - 4 = 15
- H: LFT = min(I:15, J:14) = 14; LST = 14 - 3 = 11
- G: LFT = min(I:15, J:14) = 14; LST = 14 - 5 = 9
- F: LFT = min(I:15, J:14) = 14; LST = 14 - 7 = 7
- E: LFT = min(I:15, J:14) = 14; LST = 14 - 9 = 5
- D: LFT = LST of H = 11; LST = 11 - 8 = 3
- C: LFT = LST of G = 9; LST = 9 - 7 = 2
- B: LFT = LST of F = 7; LST = 7 - 6 = 1
- A: LFT = LST of E = 5; LST = 5 - 5 = 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 | 5 | 0 | 5 | 0 | 5 | 0 | 0 | 0 | 0 |
| B | 6 | 0 | 6 | 1 | 7 | 1 | 0 | 1 | 0 |
| C | 7 | 0 | 7 | 2 | 9 | 2 | 0 | 2 | 0 |
| D | 8 | 0 | 8 | 3 | 11 | 3 | 0 | 3 | 0 |
| E | 9 | 5 | 14 | 5 | 14 | 0 | 0 | 0 | 0 |
| F | 7 | 6 | 13 | 7 | 14 | 1 | 1 | 0 | 0 |
| G | 5 | 7 | 12 | 9 | 14 | 2 | 2 | 0 | 0 |
| H | 3 | 8 | 11 | 11 | 14 | 3 | 3 | 0 | 0 |
| I | 4 | 14 | 18 | 15 | 19 | 1 | 1 | 0 | 1 |
| J | 5 | 14 | 19 | 14 | 19 | 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 | 5 | 5 | 0 |
| 3 | 6 | 7 | 1 |
| 4 | 7 | 9 | 2 |
| 5 | 8 | 11 | 3 |
| 6 | 14 | 14 | 0 |
| 7 | 19 | 19 | 0 |
| 8 | 19 | 19 | 0 |

#### Critical path and project duration

- Critical activities (TF = 0): A, E, J
- Critical path: A - E - J (= 5+9+5 = 19 time units)

**Answer: project duration = 19 time units; critical path = A - E - J**

## Check yourself

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

1. What are the project duration and critical path?
   A. 21 units, B-F-J
   B. 18 units, A-E-I
   C. 19 units, A-E-J
   D. 19 units, D-H-J
   Answer: C. J ends at 19 (EST 14 + 5), and only A, E, J have TF = 0: 5+9+5 = 19. (Key point: Duration and critical path)

2. Why do I and J both have EST = 14?
   A. Both depend on E, F, G, H and the largest EFT among them is 14 (E)
   B. Each starts after the other
   C. They depend only on A
   D. They depend on D
   Answer: A. I and J need all of E, F, G, H complete; EFTs are 14, 13, 12, 11, so the largest, 14, controls. (Key point: EST of I and J)

3. LFT of E is 14. Where does it come from?
   A. EFT of A
   B. min(LST of I = 15, LST of J = 14)
   C. Project duration of 19
   D. LST of H
   Answer: B. LFT of a predecessor is the smallest LST of the successors: min(15, 14) = 14. (Key point: Backward pass for E)

4. Activity I has TF = 1, FF = 1. What does this mean?
   A. I is critical
   B. I can be delayed by 1 day without delaying the project
   C. I must start at day 15
   D. I has no float
   Answer: B. I ends at 18 against LFT 19, so it can slip 1 day without affecting the project; it is not critical. (Key point: Float on I)

5. Why is a dummy drawn from event 7 to event 8?
   A. To link D and H
   B. To show A is critical
   C. To add one day
   D. To make I and J end at a single end event
   Answer: D. I and J are both last activities, so a dummy brings them to one common end event (8). (Key point: Dummy at the end)

## 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, 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, 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, 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 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, 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×

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