---
title: "CPM network, 10 activities (A=10 to J=4)"
subject: "Project Engineering"
chapter: 3
key: "semester_7/project_engineering:cpm-network-ten-act-k79"
source: https://www.bctengineer.com/q/semester_7/project_engineering/cpm-network-ten-act-k79
site: BCT Engineering
---

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

Project Engineering · Chapter 3: Project Planning and Scheduling

- Asked in: 2079 Baisakh, 12 marks
- Video: [Project Management: Finding the Critical Path, duration and Project Duration | Critical Path Method](https://www.youtube.com/watch?v=vsl3PuFqRN4) — 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-ten-act-k79
- Whole chapter: https://www.bctengineer.com/md/semester_7/project_engineering/chapter-3

## Question

Draw the network diagram and compute EST, EFT, LST, LFT, TF, FF, IF and interfering floats of each activity of a project having the precedence relationship given below. The time durations are in days.

| Activity | Duration (days) | Predecessor |
|---|---|---|
| A | 10 | - |
| B | 12 | A |
| C | 8 | A |
| D | 6 | A |
| E | 10 | B |
| F | 8 | D |
| G | 4 | C |
| H | 10 | G |
| I | 6 | E |
| J | 4 | F, H, I |

## Short answer

AOA network without dummies; forward/backward pass yields 42 days. Critical path A-B-E-I-J with critical activities A, B, E, I, J; all activity times and floats computed, with zero total float on critical activities.

## Answer

Here IF is taken as the independent float; interfering float = TF - FF.

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(10)--> (2)
(2) --B(12)--> (3)
(2) --C(8)--> (4)
(2) --D(6)--> (5)
(3) --E(10)--> (6)
(4) --G(4)--> (7)
(5) --F(8)--> (8)
(6) --I(6)--> (8)
(7) --H(10)--> (8)
(8) --J(4)--> (9)
```

The network has 9 events and 0 dummy activities. Start event is (1) and the end event is (9).

#### 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 + 10 = 10
- B: EST = EFT of A = 10; EFT = 10 + 12 = 22
- C: EST = EFT of A = 10; EFT = 10 + 8 = 18
- D: EST = EFT of A = 10; EFT = 10 + 6 = 16
- F: EST = EFT of D = 16; EFT = 16 + 8 = 24
- G: EST = EFT of C = 18; EFT = 18 + 4 = 22
- E: EST = EFT of B = 22; EFT = 22 + 10 = 32
- H: EST = EFT of G = 22; EFT = 22 + 10 = 32
- I: EST = EFT of E = 32; EFT = 32 + 6 = 38
- J: EST = max(F:24, H:32, I:38) = 38; EFT = 38 + 4 = 42

Project duration = largest EFT = **42 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 = 42; LST = 42 - 4 = 38
- I: LFT = LST of J = 38; LST = 38 - 6 = 32
- H: LFT = LST of J = 38; LST = 38 - 10 = 28
- E: LFT = LST of I = 32; LST = 32 - 10 = 22
- G: LFT = LST of H = 28; LST = 28 - 4 = 24
- F: LFT = LST of J = 38; LST = 38 - 8 = 30
- D: LFT = LST of F = 30; LST = 30 - 6 = 24
- C: LFT = LST of G = 24; LST = 24 - 8 = 16
- B: LFT = LST of E = 22; LST = 22 - 12 = 10
- A: LFT = min(B:10, C:16, D:24) = 10; LST = 10 - 10 = 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 | 10 | 0 | 10 | 0 | 10 | 0 | 0 | 0 | 0 |
| B | 12 | 10 | 22 | 10 | 22 | 0 | 0 | 0 | 0 |
| C | 8 | 10 | 18 | 16 | 24 | 6 | 0 | 6 | 0 |
| D | 6 | 10 | 16 | 24 | 30 | 14 | 0 | 14 | 0 |
| E | 10 | 22 | 32 | 22 | 32 | 0 | 0 | 0 | 0 |
| F | 8 | 16 | 24 | 30 | 38 | 14 | 14 | 0 | 0 |
| G | 4 | 18 | 22 | 24 | 28 | 6 | 0 | 6 | 0 |
| H | 10 | 22 | 32 | 28 | 38 | 6 | 6 | 0 | 0 |
| I | 6 | 32 | 38 | 32 | 38 | 0 | 0 | 0 | 0 |
| J | 4 | 38 | 42 | 38 | 42 | 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 | 10 | 10 | 0 |
| 3 | 22 | 22 | 0 |
| 4 | 18 | 24 | 6 |
| 5 | 16 | 30 | 14 |
| 6 | 32 | 32 | 0 |
| 7 | 22 | 28 | 6 |
| 8 | 38 | 38 | 0 |
| 9 | 42 | 42 | 0 |

#### Critical path and project duration

- Critical activities (TF = 0): A, B, E, I, J
- Critical path: A - B - E - I - J (= 10+12+10+6+4 = 42 days)

**Answer: project duration = 42 days; critical path = A - B - E - I - J**

## Check yourself

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

1. Activity J has predecessors F (EFT 24), H (EFT 32) and I (EFT 38). What is the EST of J and why?
   A. 24, because F is the earliest to finish
   B. 38, because EST is the largest EFT of the predecessors
   C. 32, because H is the middle value
   D. 42, because durations are added to every predecessor
   Answer: B. EST of an activity is the largest EFT among its predecessors, because it cannot start until all of them are finished. Here that is I at 38. (Key point: Forward pass merge rule)

2. Activity A is followed by B (LST 10), C (LST 16) and D (LST 24). What is LFT of A?
   A. 24
   B. 16
   C. 10
   D. 50
   Answer: C. LFT is the smallest LST of the succeeding activities, so that none of them is delayed. The smallest is B at 10, and A has LST = 10 - 10 = 0. (Key point: Backward pass for a merge/burst node)

3. Activity C has EST = 10, EFT = 18, LST = 16 and LFT = 24. What is its total float?
   A. 6 days
   B. 8 days
   C. 14 days
   D. 0 days
   Answer: A. TF = LST - EST = 16 - 10 = 6 (also LFT - EFT = 24 - 18 = 6). It can be delayed 6 days without delaying the project. (Key point: Total float)

4. Activity F has EFT = 24, TF = 14, and its only successor J has EST = 38. What are its FF and interfering float?
   A. FF = 0, IntF = 14
   B. FF = 14, IntF = 14
   C. FF = 7, IntF = 7
   D. FF = 14, IntF = 0
   Answer: D. FF = EST of successor - EFT = 38 - 24 = 14. Interfering float = TF - FF = 14 - 14 = 0. (Key point: Free and interfering float)

## Related in Project Engineering

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