CPM network, 9 activities (A=5 to I=2)

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(5)--> (2)
(1) --B(4)--> (3)
(2) --C(0)--> (3)
(2) --D(6)--> (4)
(3) --E(7)--> (4)
(3) --F(8)--> (5)
(4) --G(6)--> (6)
(5) --H(3)--> (6)
(6) --I(2)--> (7)

The network has 7 events and 0 dummy activities. Start event is (1) and the end event is (7).

Activity C has zero duration, so it is treated as a normal activity of 0 weeks (it only passes the logic on).

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 + 5 = 5
• B: EST = 0; EFT = 0 + 4 = 4
• C: EST = EFT of A = 5; EFT = 5 + 0 = 5
• D: EST = EFT of A = 5; EFT = 5 + 6 = 11
• E: EST = max(B:4, C:5) = 5; EFT = 5 + 7 = 12
• F: EST = max(B:4, C:5) = 5; EFT = 5 + 8 = 13
• G: EST = max(D:11, E:12) = 12; EFT = 12 + 6 = 18
• H: EST = EFT of F = 13; EFT = 13 + 3 = 16
• I: EST = max(G:18, H:16) = 18; EFT = 18 + 2 = 20

Project duration = largest EFT = 20 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 = 20; LST = 20 - 2 = 18
• H: LFT = LST of I = 18; LST = 18 - 3 = 15
• G: LFT = LST of I = 18; LST = 18 - 6 = 12
• F: LFT = LST of H = 15; LST = 15 - 8 = 7
• E: LFT = LST of G = 12; LST = 12 - 7 = 5
• D: LFT = LST of G = 12; LST = 12 - 6 = 6
• C: LFT = min(E:5, F:7) = 5; LST = 5 - 0 = 5
• B: LFT = min(E:5, F:7) = 5; LST = 5 - 4 = 1
• A: LFT = min(C:5, D:6) = 5; LST = 5 - 5 = 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, C, E, G, I
• Critical path: A - C - E - G - I (= 5+0+7+6+2 = 20 weeks)

Answer: project duration = 20 weeks; critical path = A - C - E - G - I