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CPM components (AOA), activities A to I

Short answer

Using the AOA method, all CPM components are found: network with 8 events and 3 dummies, event times, EST/EFT/LST/LFT and all floats tabulated. Project duration is 17 weeks. Critical path is B-D-H-I (3+5+4+5 weeks).

Find all the components of CPM from the following information. Use the AOA method.

ActivityDuration (week)Predecessor
A1-
B3-
C2A, B
D5B
E3B
F1C, D
G3D
H4D, E
I5F, G, H

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

The network has 8 events and 3 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 + 1 = 1
  • B: EST = 0; EFT = 0 + 3 = 3
  • C: EST = max(A:1, B:3) = 3; EFT = 3 + 2 = 5
  • D: EST = EFT of B = 3; EFT = 3 + 5 = 8
  • E: EST = EFT of B = 3; EFT = 3 + 3 = 6
  • F: EST = max(C:5, D:8) = 8; EFT = 8 + 1 = 9
  • G: EST = EFT of D = 8; EFT = 8 + 3 = 11
  • H: EST = max(D:8, E:6) = 8; EFT = 8 + 4 = 12
  • I: EST = max(F:9, G:11, H:12) = 12; EFT = 12 + 5 = 17

Project duration = largest EFT = 17 weeks.

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 = 17; LST = 17 - 5 = 12
  • H: LFT = LST of I = 12; LST = 12 - 4 = 8
  • G: LFT = LST of I = 12; LST = 12 - 3 = 9
  • F: LFT = LST of I = 12; LST = 12 - 1 = 11
  • E: LFT = LST of H = 8; LST = 8 - 3 = 5
  • D: LFT = min(F:11, G:9, H:8) = 8; LST = 8 - 5 = 3
  • C: LFT = LST of F = 11; LST = 11 - 2 = 9
  • B: LFT = min(C:9, D:3, E:5) = 3; LST = 3 - 3 = 0
  • A: LFT = LST of C = 9; LST = 9 - 1 = 8

Floats

  • Total float TF=LST−EST=LFT−EFTTF = LST - EST = LFT - EFT
  • Free float FF=min⁡(EST of successors)−EFTFF = \min(\text{EST of successors}) - EFT
  • Interfering float IntF=TF−FFIntF = TF - FF
  • Independent float IndF=max⁡(0, min⁡(EST of successors)−max⁡(LFT of predecessors)−D)IndF = \max\left(0,\ \min(\text{EST of successors}) - \max(\text{LFT of predecessors}) - D\right)

Result table

ActivityDESTEFTLSTLFTTFFFInt.FInd.F
A101898262
B303030000
C2359116330
D538380000
E336582202
F18911123300
G38119121101
H48128120000
I5121712170000

Event times (TE from the forward pass, TL from the backward pass):

EventTE (earliest)TL (latest)Slack
1000
2330
3880
4396
5880
68113
712120
817170

Critical path and project duration

  • Critical activities (TF = 0): B, D, H, I
  • Critical path: B - D - H - I (= 3+5+4+5 = 17 weeks)

Answer: project duration = 17 weeks; critical path = B - D - H - I

✓ Answer checked by Claude Opus 5.5 (high effort)

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