---
title: "CPM network, 12 activities (A=1 to L=4)"
subject: "Project Engineering"
chapter: 3
key: "semester_7/project_engineering:cpm-network-twelve-act-k81"
source: https://www.bctengineer.com/q/semester_7/project_engineering/cpm-network-twelve-act-k81
site: BCT Engineering
---

# CPM network, 12 activities (A=1 to L=4)

Project Engineering · Chapter 3: Project Planning and Scheduling

- Asked in: 2081 Baisakh, 10 marks; 2072 Chaitra, 2+5+6 marks (asked 2 times)
- Video: [Project Management: Finding the Critical Path, duration and Project Duration | Critical Path Method](https://www.youtube.com/watch?v=tOJ8IuIqBpw) — 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-network-twelve-act-k81
- Whole chapter: https://www.bctengineer.com/md/semester_7/project_engineering/chapter-3

## Question

Draw a network diagram from the following data. (i) Find critical path and critical activities. (ii) Find ES, EF, LS, LF, TF, FF, Ind.F and Int.F.

| Activity | Duration (weeks) | Predecessor |
|---|---|---|
| A | 1 | None |
| B | 3 | None |
| C | 2 | A |
| D | 4 | A, B |
| E | 3 | B |
| F | 5 | C, D |
| G | 1 | D |
| H | 2 | D |
| I | 6 | E, H |
| J | 3 | E, H |
| K | 2 | F, G, I |
| L | 4 | K, J |

## Short answer

Network analysis gives project duration 21 weeks. Critical activities: B, D, H, I, K, L; path B-D-H-I-K-L.

## Answer

#### Network diagram (activity on node)
Each node is: activity (duration). Arrows show precedence.

```text
A(1) --+--> C(2) -----------------> F(5) --------+
       |                             ^           |
       +--> D(4) --+-----------------+           |
            ^      |                             v
B(3) --+----+      +--> G(1) -----------------> K(2) ---> L(4)
       |           |                             ^         ^
       |           +--> H(2) --+--> I(6) --------+         |
       |                       |                           |
       +--> E(3) --------------+--> J(3) ------------------+
```

Links: A and B start; C after A; D after A and B; E after B; F after C and D; G and H after D; I and J after E and H; K after F, G and I; L after K and J.

#### Forward and backward passes
$EST$ = largest $EFT$ of predecessors; $LFT$ = smallest $LST$ of successors. Project duration $T = 21$ weeks. $TF = LST - EST$; $FF = \min(EST_{succ}) - EFT$; $Int.F = TF - FF$; $Ind.F = \max\{0, \min(EST_{succ}) - \max(LFT_{pred}) - D\}$.

| Act | D | ES | EF | LS | LF | TF | FF | Int.F | Ind.F | Critical |
|---|---|---|---|---|---|---|---|---|---|---|
| A | 1 | 0 | 1 | 2 | 3 | 2 | 0 | 2 | 0 | No |
| B | 3 | 0 | 3 | 0 | 3 | 0 | 0 | 0 | 0 | Yes |
| C | 2 | 1 | 3 | 8 | 10 | 7 | 4 | 3 | 2 | No |
| D | 4 | 3 | 7 | 3 | 7 | 0 | 0 | 0 | 0 | Yes |
| E | 3 | 3 | 6 | 6 | 9 | 3 | 3 | 0 | 3 | No |
| F | 5 | 7 | 12 | 10 | 15 | 3 | 3 | 0 | 0 | No |
| G | 1 | 7 | 8 | 14 | 15 | 7 | 7 | 0 | 7 | No |
| H | 2 | 7 | 9 | 7 | 9 | 0 | 0 | 0 | 0 | Yes |
| I | 6 | 9 | 15 | 9 | 15 | 0 | 0 | 0 | 0 | Yes |
| J | 3 | 9 | 12 | 14 | 17 | 5 | 5 | 0 | 5 | No |
| K | 2 | 15 | 17 | 15 | 17 | 0 | 0 | 0 | 0 | Yes |
| L | 4 | 17 | 21 | 17 | 21 | 0 | 0 | 0 | 0 | Yes |

Sample check, activity C: $ES = EF_A = 1$; $EF = 3$; $LF = LS_F = 10$; $LS = 8$; $TF = 8-1 = 7$; $FF = ES_F - EF_C = 7-3 = 4$; $Int.F = 3$; $Ind.F = 7 - 3 - 2 = 2$ (using $LF_A = 3$).

#### (i) Critical path
The activities with $TF = 0$ form the critical path **B - D - H - I - K - L**:
$3+4+2+6+2+4 = 21$ weeks.

**Answer:** Critical activities: B, D, H, I, K, L. Project duration = **21 weeks**.

## Check yourself

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

1. Project duration:
   A. 19 weeks
   B. 21 weeks
   C. 17 weeks
   D. 23 weeks
   Answer: B. B-D-H-I-K-L = 3+4+2+6+2+4 = 21. (Key point: Duration)

2. Critical activities are:
   A. A, C, F, K, L
   B. B, D, H, I, K, L
   C. B, E, J, L
   D. A, D, G, K
   Answer: B. They have TF = 0. (Key point: Critical activities)

3. Int.F is calculated as:
   A. TF - FF
   B. FF - TF
   C. LF - EF
   D. TF + FF
   Answer: A. Int.F = TF - FF; for C, 7 - 4 = 3. (Key point: Interfering float)

4. TF of J is:
   A. 0
   B. 3
   C. 5
   D. 7
   Answer: C. TF = LS - ES = 14 - 9 = 5. (Key point: Total float)

## 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, road project (days)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-road-project-b79) — 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, 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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- [TCTO numerical: crashing critical path, Rs 30/day](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Atcto-2069-bhadra) — Construction Management, chapter 2