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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.NActivityDurationPredecessorSuccessor
1A3-D
2B6-E, G, I
3C2-F
4D2AG
5E1BH
6F3CI
7G7B, D-
8H3E-
9I4B, 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 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
A303141010
B606060000
C202464040
D235461100
E1679103030
F325694130
G76136130000
H371010133300
I46109133300

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

EventTE (earliest)TL (latest)Slack
1000
2341
3660
4264
57103
6660
7693
813130

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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