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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).

ActivityDuration (week)PredecessorSuccessor
A1-C, E
B6-C, D
C2A, BF
D2BH
E4AG
F3CG, H
G4E, FI
H2D, FI
I5G, HJ
J3I-

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 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
A101565050
B606060000
C268680000
D26811135323
E4157116601
F38118110000
G4111511150000
H2111313152200
I5152015200000
J3202320230000

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

EventTE (earliest)TL (latest)Slack
1000
2165
3660
4660
5880
611110
711132
811110
915150
1020200
1123230

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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