- Asked 3/26
- 2070 Chaitra · 12 marks
CPM components, 9 activities with successors
Short answer
Project duration is 13 time units; critical path B–G. All CPM components (event times, activity times and floats) are computed; only B and G have zero total float and are critical.
Find all the components of CPM from the following information.
| S.N | Activity | Duration | Predecessor | Successor |
|---|---|---|---|---|
| 1 | A | 3 | - | D |
| 2 | B | 6 | - | E, G, I |
| 3 | C | 2 | - | F |
| 4 | D | 2 | A | G |
| 5 | E | 1 | B | H |
| 6 | F | 3 | C | I |
| 7 | G | 7 | B, D | - |
| 8 | H | 3 | E | - |
| 9 | I | 4 | B, F | - |
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 (time units). Events are circled numbers; dashed arrows are dummy activities (zero duration) that show dependencies only.
(1) --A(3)--> (2)
(1) --B(6)--> (3)
(1) --C(2)--> (4)
(2) --D(2)--> (6)
(3) --E(1)--> (5)
(3) - - dummy - -> (6)
(3) - - dummy - -> (7)
(4) --F(3)--> (7)
(5) --H(3)--> (8)
(6) --G(7)--> (8)
(7) --I(4)--> (8)
The network has 8 events and 2 dummy activities. 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 + 3 = 3
- B: EST = 0; EFT = 0 + 6 = 6
- C: EST = 0; EFT = 0 + 2 = 2
- F: EST = EFT of C = 2; EFT = 2 + 3 = 5
- D: EST = EFT of A = 3; EFT = 3 + 2 = 5
- E: EST = EFT of B = 6; EFT = 6 + 1 = 7
- G: EST = max(B:6, D:5) = 6; EFT = 6 + 7 = 13
- I: EST = max(B:6, F:5) = 6; EFT = 6 + 4 = 10
- H: EST = EFT of E = 7; EFT = 7 + 3 = 10
Project duration = largest EFT = 13 time units.
Backward pass (LFT and LST)
Rule: LFT = smallest LST of the succeeding activities (project duration for end activities); LST = LFT - D.
- H: LFT = project duration = 13; LST = 13 - 3 = 10
- I: LFT = project duration = 13; LST = 13 - 4 = 9
- G: LFT = project duration = 13; LST = 13 - 7 = 6
- E: LFT = LST of H = 10; LST = 10 - 1 = 9
- D: LFT = LST of G = 6; LST = 6 - 2 = 4
- F: LFT = LST of I = 9; LST = 9 - 3 = 6
- C: LFT = LST of F = 6; LST = 6 - 2 = 4
- B: LFT = min(E:9, G:6, I:9) = 6; LST = 6 - 6 = 0
- A: LFT = LST of D = 4; LST = 4 - 3 = 1
Floats
- Total float
- Free float
- Interfering float
- Independent float
Result table
| Activity | D | EST | EFT | LST | LFT | TF | FF | Int.F | Ind.F |
|---|---|---|---|---|---|---|---|---|---|
| A | 3 | 0 | 3 | 1 | 4 | 1 | 0 | 1 | 0 |
| B | 6 | 0 | 6 | 0 | 6 | 0 | 0 | 0 | 0 |
| C | 2 | 0 | 2 | 4 | 6 | 4 | 0 | 4 | 0 |
| D | 2 | 3 | 5 | 4 | 6 | 1 | 1 | 0 | 0 |
| E | 1 | 6 | 7 | 9 | 10 | 3 | 0 | 3 | 0 |
| F | 3 | 2 | 5 | 6 | 9 | 4 | 1 | 3 | 0 |
| G | 7 | 6 | 13 | 6 | 13 | 0 | 0 | 0 | 0 |
| H | 3 | 7 | 10 | 10 | 13 | 3 | 3 | 0 | 0 |
| I | 4 | 6 | 10 | 9 | 13 | 3 | 3 | 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 | 3 | 4 | 1 |
| 3 | 6 | 6 | 0 |
| 4 | 2 | 6 | 4 |
| 5 | 7 | 10 | 3 |
| 6 | 6 | 6 | 0 |
| 7 | 6 | 9 | 3 |
| 8 | 13 | 13 | 0 |
Critical path and project duration
- Critical activities (TF = 0): B, G
- Critical path: B - G (= 6+7 = 13 time units)
Answer: project duration = 13 time units; critical path = B - G
✓ Answer checked by Claude Opus 5.5 (high effort)
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