---
title: "CPM network, road project (days)"
subject: "Project Engineering"
chapter: 3
key: "semester_7/project_engineering:cpm-network-road-project-b79"
source: https://www.bctengineer.com/q/semester_7/project_engineering/cpm-network-road-project-b79
site: BCT Engineering
---

# CPM network, road project (days)

Project Engineering · Chapter 3: Project Planning and Scheduling

- Asked in: 2079 Bhadra, 12 marks
- Video: [Finding the Critical Path, duration and Project Duration, Critical Path Method, float, EST, EFT, LST](https://www.youtube.com/watch?v=OKKcjeJt4cI) — 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-road-project-b79
- Whole chapter: https://www.bctengineer.com/md/semester_7/project_engineering/chapter-3

## Question

Draw a network diagram for a road project having information as in the table. Find out ES, EF, LS, LF, TF, FF, Int F, Ind F and then analyse the situation of the project stating the critical path, project completion time and critical activities.

| Activity | Duration (days) | Predecessor |
|---|---|---|
| A | 6 | - |
| B | 4 | - |
| C | 5 | - |
| D | 2 | A, B |
| E | 4 | C, D |
| F | 2 | D |
| G | 5 | D |
| H | 8 | D |
| I | 5 | G |
| J | 6 | E, G |

## Short answer

Using AOA network with dummies, forward/backward pass gives project duration 19 days. Critical path A-D-G-J with critical activities A, D, G, J; all floats (TF, FF, Int.F, Ind.F) are tabulated, with zero total float on critical activities.

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

The network has 9 events and 4 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 + 6 = 6
- B: EST = 0; EFT = 0 + 4 = 4
- C: EST = 0; EFT = 0 + 5 = 5
- D: EST = max(A:6, B:4) = 6; EFT = 6 + 2 = 8
- E: EST = max(C:5, D:8) = 8; EFT = 8 + 4 = 12
- F: EST = EFT of D = 8; EFT = 8 + 2 = 10
- G: EST = EFT of D = 8; EFT = 8 + 5 = 13
- H: EST = EFT of D = 8; EFT = 8 + 8 = 16
- I: EST = EFT of G = 13; EFT = 13 + 5 = 18
- J: EST = max(E:12, G:13) = 13; EFT = 13 + 6 = 19

Project duration = largest EFT = **19 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 = 19; LST = 19 - 6 = 13
- I: LFT = project duration = 19; LST = 19 - 5 = 14
- H: LFT = project duration = 19; LST = 19 - 8 = 11
- G: LFT = min(I:14, J:13) = 13; LST = 13 - 5 = 8
- F: LFT = project duration = 19; LST = 19 - 2 = 17
- E: LFT = LST of J = 13; LST = 13 - 4 = 9
- D: LFT = min(E:9, F:17, G:8, H:11) = 8; LST = 8 - 2 = 6
- C: LFT = LST of E = 9; LST = 9 - 5 = 4
- B: LFT = LST of D = 6; LST = 6 - 4 = 2
- A: LFT = LST of D = 6; LST = 6 - 6 = 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 | 6 | 0 | 6 | 0 | 6 | 0 | 0 | 0 | 0 |
| B | 4 | 0 | 4 | 2 | 6 | 2 | 2 | 0 | 2 |
| C | 5 | 0 | 5 | 4 | 9 | 4 | 3 | 1 | 3 |
| D | 2 | 6 | 8 | 6 | 8 | 0 | 0 | 0 | 0 |
| E | 4 | 8 | 12 | 9 | 13 | 1 | 1 | 0 | 0 |
| F | 2 | 8 | 10 | 17 | 19 | 9 | 9 | 0 | 9 |
| G | 5 | 8 | 13 | 8 | 13 | 0 | 0 | 0 | 0 |
| H | 8 | 8 | 16 | 11 | 19 | 3 | 3 | 0 | 3 |
| I | 5 | 13 | 18 | 14 | 19 | 1 | 1 | 0 | 1 |
| J | 6 | 13 | 19 | 13 | 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 | 4 | 6 | 2 |
| 3 | 6 | 6 | 0 |
| 4 | 8 | 8 | 0 |
| 5 | 13 | 13 | 0 |
| 6 | 16 | 19 | 3 |
| 7 | 8 | 9 | 1 |
| 8 | 13 | 13 | 0 |
| 9 | 19 | 19 | 0 |

#### Critical path and project duration

- Critical activities (TF = 0): A, D, G, J
- Critical path: A - D - G - J (= 6+2+5+6 = 19 days)

**Answer: project duration = 19 days; critical path = A - D - G - J**

#### Analysis of the project situation

- Project completion time = **19 days**.
- Critical path = **A - D - G - J** (6 + 2 + 5 + 6 = 19 days); critical activities = A, D, G and J (zero float). A delay in any of them delays the road project by the same number of days.
- Non-critical activities: B, C, E, F, H and I. E and I have only 1 day of float, so they are near-critical and must be watched; F (9 days) and H (3 days) have the most float.
- Management should give priority in resources and supervision to the critical activities, and use the float of F, H, B and C to level labour and equipment.

## Check yourself

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

1. Activity D has predecessors A (EFT 6) and B (EFT 4). What is the EST of D?
   A. 4
   B. 2
   C. 8
   D. 6
   Answer: D. EST = max(6, 4) = 6, so EFT = 8. (Key point: Forward pass rule)

2. What are the project duration and critical path?
   A. 19 days; A - D - G - J
   B. 19 days; B - D - H
   C. 18 days; A - D - G - I
   D. 16 days; A - D - H
   Answer: A. A 6 + D 2 + G 5 + J 6 = 19 days, with zero total float. (Key point: Project duration and critical path)

3. Activity G has successors I (LST 14) and J (LST 13). What is its LFT?
   A. 14
   B. 13
   C. 19
   D. 8
   Answer: B. LFT = min(14, 13) = 13, so LST = 13 - 5 = 8, equal to its EST, hence zero float. (Key point: Backward pass rule)

4. Activity E has EST 8 and LST 9. What is its total float?
   A. 0
   B. 9
   C. 1
   D. 4
   Answer: C. TF = LST - EST = 9 - 8 = 1 day. (Key point: Total float)

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