---
title: "CPM network, 10 activities with successors"
subject: "Project Engineering"
chapter: 3
key: "semester_7/project_engineering:cpm-network-ten-act-a74"
source: https://www.bctengineer.com/q/semester_7/project_engineering/cpm-network-ten-act-a74
site: BCT Engineering
---

# CPM network, 10 activities with successors

Project Engineering · Chapter 3: Project Planning and Scheduling

- Asked in: 2074 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-ten-act-a74
- Whole chapter: https://www.bctengineer.com/md/semester_7/project_engineering/chapter-3

## Question

Define total float, free float and independent float. Draw a CPM network and find EST, EFT, LST, LFT, TF, FF, IntF and IndF. Show the critical path also.

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

## Short answer

Total float is maximum delay to an activity without delaying project completion; free float is delay without delaying earliest start of any successor; independent float is delay using latest predecessors and earliest successors without affecting others. CPM analysis gives project duration 14 days. Critical path is A-D-G-I-J.

## Answer

#### Definitions

- **Total float (TF):** the time by which an activity can be delayed without delaying the project completion. $TF = LST - EST = LFT - EFT$.
- **Free float (FF):** the time by which an activity can be delayed without delaying the earliest start of any following activity. $FF = \min(EST\ \text{of successors}) - EFT$.
- **Independent float (IndF):** the time available to an activity when its predecessors finish at their latest and its successors start at their earliest; using it affects no other activity. $IndF = \max(0,\ \min(EST_{succ}) - \max(LFT_{pred}) - D)$.
- Interfering float $= TF - FF$ (the part of the total float that is shared with the following activities).

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(2)--> (2)
(2) --B(3)--> (3)
(2) --C(4)--> (4)
(2) --D(5)--> (5)
(3) --E(4)--> (6)
(4) - - dummy - -> (6)
(4) - - dummy - -> (7)
(4) --F(3)--> (8)
(5) --G(2)--> (7)
(6) --H(1)--> (8)
(7) --I(2)--> (8)
(8) --J(3)--> (9)
```

The network has 9 events and 2 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 + 2 = 2
- B: EST = EFT of A = 2; EFT = 2 + 3 = 5
- C: EST = EFT of A = 2; EFT = 2 + 4 = 6
- D: EST = EFT of A = 2; EFT = 2 + 5 = 7
- E: EST = EFT of B = 5; EFT = 5 + 4 = 9
- F: EST = EFT of C = 6; EFT = 6 + 3 = 9
- G: EST = EFT of D = 7; EFT = 7 + 2 = 9
- H: EST = max(C:6, E:9) = 9; EFT = 9 + 1 = 10
- I: EST = max(C:6, G:9) = 9; EFT = 9 + 2 = 11
- J: EST = max(F:9, H:10, I:11) = 11; EFT = 11 + 3 = 14

Project duration = largest EFT = **14 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 = 14; LST = 14 - 3 = 11
- I: LFT = LST of J = 11; LST = 11 - 2 = 9
- H: LFT = LST of J = 11; LST = 11 - 1 = 10
- G: LFT = LST of I = 9; LST = 9 - 2 = 7
- F: LFT = LST of J = 11; LST = 11 - 3 = 8
- E: LFT = LST of H = 10; LST = 10 - 4 = 6
- D: LFT = LST of G = 7; LST = 7 - 5 = 2
- C: LFT = min(F:8, H:10, I:9) = 8; LST = 8 - 4 = 4
- B: LFT = LST of E = 6; LST = 6 - 3 = 3
- A: LFT = min(B:3, C:4, D:2) = 2; LST = 2 - 2 = 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 | 2 | 0 | 2 | 0 | 2 | 0 | 0 | 0 | 0 |
| B | 3 | 2 | 5 | 3 | 6 | 1 | 0 | 1 | 0 |
| C | 4 | 2 | 6 | 4 | 8 | 2 | 0 | 2 | 0 |
| D | 5 | 2 | 7 | 2 | 7 | 0 | 0 | 0 | 0 |
| E | 4 | 5 | 9 | 6 | 10 | 1 | 0 | 1 | 0 |
| F | 3 | 6 | 9 | 8 | 11 | 2 | 2 | 0 | 0 |
| G | 2 | 7 | 9 | 7 | 9 | 0 | 0 | 0 | 0 |
| H | 1 | 9 | 10 | 10 | 11 | 1 | 1 | 0 | 0 |
| I | 2 | 9 | 11 | 9 | 11 | 0 | 0 | 0 | 0 |
| J | 3 | 11 | 14 | 11 | 14 | 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 | 2 | 2 | 0 |
| 3 | 5 | 6 | 1 |
| 4 | 6 | 8 | 2 |
| 5 | 7 | 7 | 0 |
| 6 | 9 | 10 | 1 |
| 7 | 9 | 9 | 0 |
| 8 | 11 | 11 | 0 |
| 9 | 14 | 14 | 0 |

#### Critical path and project duration

- Critical activities (TF = 0): A, D, G, I, J
- Critical path: A - D - G - I - J (= 2+5+2+2+3 = 14 days)

**Answer: project duration = 14 days; critical path = A - D - G - I - J**

## Check yourself

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

1. Which definition describes free float?
   A. Delay allowed without delaying project completion
   B. Delay allowed without delaying the earliest start of any following activity
   C. Delay allowed when predecessors finish at their latest and successors start at their earliest
   D. TF - FF
   Answer: B. Free float is the delay that does not push the successors' earliest starts: FF = min(EST of successors) - EFT. (Key point: Definition of free float)

2. Independent float of an activity is based on which situation?
   A. Predecessors finish at earliest and successors start at latest
   B. All activities on the critical path are delayed
   C. Predecessors finish at their latest and successors start at their earliest
   D. The project ends at its latest time
   Answer: C. IndF = max(0, min EST of successors - max LFT of predecessors - D): the worst case for the activity, so using it affects no other activity. (Key point: Independent float)

3. I has predecessors C (EFT 6) and G (EFT 9). What are the EST and total float of I (duration 2, LST 9)?
   A. EST 6, TF 3
   B. EST 9, TF 0
   C. EST 9, TF 2
   D. EST 11, TF 0
   Answer: B. EST = largest predecessor EFT = 9, and TF = LST - EST = 9 - 9 = 0, so I is critical. (Key point: Forward pass for I)

4. What is the critical path and duration?
   A. A-B-E-H-J, 13 days
   B. A-C-F-J, 12 days
   C. A-D-G-I-J, 14 days
   D. A-D-G-I-J, 16 days
   Answer: C. A+D+G+I+J = 2+5+2+2+3 = 14 days, with TF = 0 on all of them. (Key point: Critical path)

## Related in Project Engineering

- [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
- [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, 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, 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=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 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×

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