---
title: "CPM network, 9 activities with successors"
subject: "Project Engineering"
chapter: 3
key: "semester_7/project_engineering:cpm-network-nine-act-a76"
source: https://www.bctengineer.com/q/semester_7/project_engineering/cpm-network-nine-act-a76
site: BCT Engineering
---

# CPM network, 9 activities with successors

Project Engineering · Chapter 3: Project Planning and Scheduling

- Asked in: 2076 Asoj, 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-nine-act-a76
- Whole chapter: https://www.bctengineer.com/md/semester_7/project_engineering/chapter-3

## Question

Draw network diagram. Compute EST, EFT, LST, LFT, TF, FF, interfering float and independent float. Write down the significance of calculating total float in network analysis.

| Activity | Duration (day) | Predecessor | Successor |
|---|---|---|---|
| A | 5 | - | B, C, D |
| B | 4 | A | E |
| C | 2 | A | F, H |
| D | 3 | A | G |
| E | 2 | B | H |
| F | 1 | C | I |
| G | 3 | D | I |
| H | 1 | C, E | - |
| I | 2 | F, G | - |

## Short answer

With the AOA network drawn, forward and backward passes give all event/activity times and floats. Project duration is 13 days. Critical path is A-D-G-I (5+3+3+2 days). Total float shows which activities are critical and how long non-critical ones may slip without delaying the project.

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

Project duration = largest EFT = **13 days**.

#### Backward pass (LFT and LST)

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

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

#### Critical path and project duration

- Critical activities (TF = 0): A, D, G, I
- Critical path: A - D - G - I (= 5+3+3+2 = 13 days)

**Answer: project duration = 13 days; critical path = A - D - G - I**

#### Significance of total float

- Activities with TF = 0 are critical; they decide the project duration, so they get the closest control and priority in resources.
- TF tells how long a non-critical activity may be delayed without delaying the project, which helps to decide its start date.
- It is used to level and smooth resources, shifting activities inside their float to cut peaks.
- It shows which activities can release men and machines to the critical activities when the project must be speeded up.
- It is the cushion against delay; when TF becomes negative the project is behind schedule and corrective action is needed.
- Float is shared along a path, so using it on one activity reduces the float of the others on the same path.

## Check yourself

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

1. H has predecessors C (EFT 7) and E (EFT 11). What are EST and EFT of H (duration 1)?
   A. EST 7, EFT 8
   B. EST 11, EFT 12
   C. EST 18, EFT 19
   D. EST 9, EFT 10
   Answer: B. EST is the larger predecessor EFT = 11, so EFT = 12. (Key point: Forward pass for H)

2. A is followed by B (LST 6), C (LST 8) and D (LST 5). What is LFT and LST of A (duration 5)?
   A. LFT 5, LST 0
   B. LFT 6, LST 1
   C. LFT 8, LST 3
   D. LFT 5, LST 5
   Answer: A. LFT = smallest successor LST = 5, so LST = 5 - 5 = 0. (Key point: Backward pass for A)

3. What is the project duration and critical path?
   A. 13 days, A-B-E-H
   B. 13 days, A-D-G-I
   C. 12 days, A-D-G-I
   D. 13 days, A-C-F-I
   Answer: B. A+D+G+I = 5+3+3+2 = 13 days, and they have TF = 0. A-B-E-H is 12 days. (Key point: Critical path)

4. Which statement best gives the significance of total float?
   A. It is the time an activity can be delayed without delaying project completion, so TF = 0 marks critical activities that need priority
   B. It is the time an activity can be delayed without affecting any successor
   C. It is the duration of the longest path
   D. It is always the same for every activity on a path, so it cannot be used for rescheduling
   Answer: A. TF is the delay allowed without delaying the project. Zero float marks critical activities, and positive float lets non-critical ones be shifted to level resources. (Key point: Significance of total 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, 9 activities with successors](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-nine-act-s65) — Project Engineering, chapter 3, asked 2×
- [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=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, 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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