- Asked 5/26
- 2080 Bhadra · 12 marks
CPM components (AOA), 10 activities
Short answer
AOA network with dummies; forward/backward pass yields 23 weeks. Critical path B-C-F-G-I-J; critical activities B, C, F, G, I, J; all activity times and floats computed, with zero total float on critical activities.
Find all the components of CPM from the following information (use AOA).
| Activity | Duration (week) | Predecessor | Successor |
|---|---|---|---|
| A | 1 | - | C, E |
| B | 6 | - | C, D |
| C | 2 | A, B | F |
| D | 2 | B | H |
| E | 4 | A | G |
| F | 3 | C | G, H |
| G | 4 | E, F | I |
| H | 2 | D, F | I |
| I | 5 | G, H | J |
| J | 3 | I | - |
The components of CPM are: the network, event times (TE, TL), activity times (EST, EFT, LST, LFT), floats (TF, FF, Int.F, Ind.F), the critical path and the project duration. All are found below with the activity-on-arrow (AOA) method.
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 (weeks). Events are circled numbers; dashed arrows are dummy activities (zero duration) that show dependencies only.
(1) --A(1)--> (2)
(1) --B(6)--> (3)
(2) - - dummy - -> (4)
(2) --E(4)--> (8)
(3) - - dummy - -> (4)
(3) --D(2)--> (7)
(4) --C(2)--> (5)
(5) --F(3)--> (6)
(6) - - dummy - -> (7)
(6) - - dummy - -> (8)
(7) --H(2)--> (9)
(8) --G(4)--> (9)
(9) --I(5)--> (10)
(10) --J(3)--> (11)
The network has 11 events and 4 dummy activities. Start event is (1) and the end event is (11).
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 + 1 = 1
- B: EST = 0; EFT = 0 + 6 = 6
- E: EST = EFT of A = 1; EFT = 1 + 4 = 5
- C: EST = max(A:1, B:6) = 6; EFT = 6 + 2 = 8
- D: EST = EFT of B = 6; EFT = 6 + 2 = 8
- F: EST = EFT of C = 8; EFT = 8 + 3 = 11
- G: EST = max(E:5, F:11) = 11; EFT = 11 + 4 = 15
- H: EST = max(D:8, F:11) = 11; EFT = 11 + 2 = 13
- I: EST = max(G:15, H:13) = 15; EFT = 15 + 5 = 20
- J: EST = EFT of I = 20; EFT = 20 + 3 = 23
Project duration = largest EFT = 23 weeks.
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 = 23; LST = 23 - 3 = 20
- I: LFT = LST of J = 20; LST = 20 - 5 = 15
- H: LFT = LST of I = 15; LST = 15 - 2 = 13
- G: LFT = LST of I = 15; LST = 15 - 4 = 11
- F: LFT = min(G:11, H:13) = 11; LST = 11 - 3 = 8
- D: LFT = LST of H = 13; LST = 13 - 2 = 11
- C: LFT = LST of F = 8; LST = 8 - 2 = 6
- E: LFT = LST of G = 11; LST = 11 - 4 = 7
- B: LFT = min(C:6, D:11) = 6; LST = 6 - 6 = 0
- A: LFT = min(C:6, E:7) = 6; LST = 6 - 1 = 5
Floats
- Total float
- Free float
- Interfering float
- Independent float
Result table
| Activity | D | EST | EFT | LST | LFT | TF | FF | Int.F | Ind.F |
|---|---|---|---|---|---|---|---|---|---|
| A | 1 | 0 | 1 | 5 | 6 | 5 | 0 | 5 | 0 |
| B | 6 | 0 | 6 | 0 | 6 | 0 | 0 | 0 | 0 |
| C | 2 | 6 | 8 | 6 | 8 | 0 | 0 | 0 | 0 |
| D | 2 | 6 | 8 | 11 | 13 | 5 | 3 | 2 | 3 |
| E | 4 | 1 | 5 | 7 | 11 | 6 | 6 | 0 | 1 |
| F | 3 | 8 | 11 | 8 | 11 | 0 | 0 | 0 | 0 |
| G | 4 | 11 | 15 | 11 | 15 | 0 | 0 | 0 | 0 |
| H | 2 | 11 | 13 | 13 | 15 | 2 | 2 | 0 | 0 |
| I | 5 | 15 | 20 | 15 | 20 | 0 | 0 | 0 | 0 |
| J | 3 | 20 | 23 | 20 | 23 | 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 | 1 | 6 | 5 |
| 3 | 6 | 6 | 0 |
| 4 | 6 | 6 | 0 |
| 5 | 8 | 8 | 0 |
| 6 | 11 | 11 | 0 |
| 7 | 11 | 13 | 2 |
| 8 | 11 | 11 | 0 |
| 9 | 15 | 15 | 0 |
| 10 | 20 | 20 | 0 |
| 11 | 23 | 23 | 0 |
Critical path and project duration
- Critical activities (TF = 0): B, C, F, G, I, J
- Critical path: B - C - F - G - I - J (= 6+2+3+4+5+3 = 23 weeks)
Answer: project duration = 23 weeks; critical path = B - C - F - G - I - J
✓ Answer checked by Claude Opus 5.5 (high effort)
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