CPM network, 10 activities (A=5 to J=5)

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(5)--> (2)
(1) --B(6)--> (3)
(1) --C(7)--> (4)
(1) --D(8)--> (5)
(2) --E(9)--> (6)
(3) --F(7)--> (6)
(4) --G(5)--> (6)
(5) --H(3)--> (6)
(6) --J(5)--> (7)
(6) --I(4)--> (8)
(7) - - dummy - -> (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 + 5 = 5
• B: EST = 0; EFT = 0 + 6 = 6
• C: EST = 0; EFT = 0 + 7 = 7
• D: EST = 0; EFT = 0 + 8 = 8
• E: EST = EFT of A = 5; EFT = 5 + 9 = 14
• F: EST = EFT of B = 6; EFT = 6 + 7 = 13
• G: EST = EFT of C = 7; EFT = 7 + 5 = 12
• H: EST = EFT of D = 8; EFT = 8 + 3 = 11
• I: EST = max(E:14, F:13, G:12, H:11) = 14; EFT = 14 + 4 = 18
• J: EST = max(E:14, F:13, G:12, H:11) = 14; EFT = 14 + 5 = 19

Project duration = largest EFT = 19 time units.

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 = 19; LST = 19 - 5 = 14
• I: LFT = project duration = 19; LST = 19 - 4 = 15
• H: LFT = min(I:15, J:14) = 14; LST = 14 - 3 = 11
• G: LFT = min(I:15, J:14) = 14; LST = 14 - 5 = 9
• F: LFT = min(I:15, J:14) = 14; LST = 14 - 7 = 7
• E: LFT = min(I:15, J:14) = 14; LST = 14 - 9 = 5
• D: LFT = LST of H = 11; LST = 11 - 8 = 3
• C: LFT = LST of G = 9; LST = 9 - 7 = 2
• B: LFT = LST of F = 7; LST = 7 - 6 = 1
• A: LFT = LST of E = 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, E, J
• Critical path: A - E - J (= 5+9+5 = 19 time units)

Answer: project duration = 19 time units; critical path = A - E - J