CPM network, 9 activities (old course)

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(2)--> (2)
(1) --B(4)--> (3)
(2) --C(5)--> (6)
(3) --D(5)--> (4)
(3) --E(3)--> (5)
(3) - - dummy - -> (6)
(4) --G(2)--> (7)
(5) --H(5)--> (7)
(6) --F(4)--> (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 + 2 = 2
• B: EST = 0; EFT = 0 + 4 = 4
• C: EST = EFT of A = 2; EFT = 2 + 5 = 7
• D: EST = EFT of B = 4; EFT = 4 + 5 = 9
• E: EST = EFT of B = 4; EFT = 4 + 3 = 7
• F: EST = max(B:4, C:7) = 7; EFT = 7 + 4 = 11
• H: EST = EFT of E = 7; EFT = 7 + 5 = 12
• G: EST = EFT of D = 9; EFT = 9 + 2 = 11
• I: EST = max(F:11, G:11, H:12) = 12; EFT = 12 + 3 = 15

Project duration = largest EFT = 15 time units.

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 = 15; LST = 15 - 3 = 12
• G: LFT = LST of I = 12; LST = 12 - 2 = 10
• H: LFT = LST of I = 12; LST = 12 - 5 = 7
• F: LFT = LST of I = 12; LST = 12 - 4 = 8
• E: LFT = LST of H = 7; LST = 7 - 3 = 4
• D: LFT = LST of G = 10; LST = 10 - 5 = 5
• C: LFT = LST of F = 8; LST = 8 - 5 = 3
• B: LFT = min(D:5, E:4, F:8) = 4; LST = 4 - 4 = 0
• A: LFT = LST of C = 3; LST = 3 - 2 = 1

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): B, E, H, I
• Critical path: B - E - H - I (= 4+3+5+3 = 15 time units)

Answer: project duration = 15 time units; critical path = B - E - H - I