CPM components (AOA), 9 activities

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

The network has 8 events and 1 dummy activity. 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 + 2 = 2
• D: EST = EFT of B = 2; EFT = 2 + 3 = 5
• E: EST = EFT of B = 2; EFT = 2 + 3 = 5
• C: EST = EFT of A = 3; EFT = 3 + 4 = 7
• H: EST = EFT of E = 5; EFT = 5 + 5 = 10
• F: EST = EFT of C = 7; EFT = 7 + 3 = 10
• G: EST = max(C:7, D:5) = 7; EFT = 7 + 2 = 9
• I: EST = max(F:10, G:9, H:10) = 10; EFT = 10 + 3 = 13

Project duration = largest EFT = 13 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 = 13; LST = 13 - 3 = 10
• G: LFT = LST of I = 10; LST = 10 - 2 = 8
• F: LFT = LST of I = 10; LST = 10 - 3 = 7
• H: LFT = LST of I = 10; LST = 10 - 5 = 5
• C: LFT = min(F:7, G:8) = 7; LST = 7 - 4 = 3
• E: LFT = LST of H = 5; LST = 5 - 3 = 2
• D: LFT = LST of G = 8; LST = 8 - 3 = 5
• B: LFT = min(D:5, E:2) = 2; LST = 2 - 2 = 0
• A: LFT = LST of C = 3; LST = 3 - 3 = 0

Floats

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

Result table

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

Critical path and project duration

• Critical activities (TF = 0): A, B, C, E, F, H, I
• Critical path: A - C - F - I (= 3+4+3+3 = 13 weeks)
• Critical path: B - E - H - I (= 2+3+5+3 = 13 weeks)
• There are two critical paths of the same length, so a delay on either one delays the project.

Answer: project duration = 13 weeks; critical paths = A - C - F - I and B - E - H - I