CPM precedence diagram, 9 activities

Total minimum project time, critical activities, the eight CPM values (ES, EF, LS, LF, TF, FF, IntF, IndF) and the critical path are found below.

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. Double arrows (==>) mark the critical path.

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

The network has 8 events and 2 dummy activities. The dummy 3-4 makes E depend on both B and C; the dummy 5-6 makes G depend on D while H follows D alone. The start event is (1) and the end event is (8).

Critical path on the network: (1) - A - (2) - C - (3) - F - (6) - G - (7) - I - (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 = EFT of A = 3; EFT = 3 + 5 = 8
• C: EST = EFT of A = 3; EFT = 3 + 5 = 8
• D: EST = EFT of A = 3; EFT = 3 + 6 = 9
• E: EST = max(B:8, C:8) = 8; EFT = 8 + 2 = 10
• F: EST = EFT of C = 8; EFT = 8 + 3 = 11
• H: EST = EFT of D = 9; EFT = 9 + 2 = 11
• G: EST = max(D:9, E:10, F:11) = 11; EFT = 11 + 4 = 15
• I: EST = max(G:15, H:11) = 15; EFT = 15 + 6 = 21

Project duration = largest EFT = 21 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 = 21; LST = 21 - 6 = 15
• G: LFT = LST of I = 15; LST = 15 - 4 = 11
• H: LFT = LST of I = 15; LST = 15 - 2 = 13
• F: LFT = LST of G = 11; LST = 11 - 3 = 8
• E: LFT = LST of G = 11; LST = 11 - 2 = 9
• D: LFT = min(G:11, H:13) = 11; LST = 11 - 6 = 5
• C: LFT = min(E:9, F:8) = 8; LST = 8 - 5 = 3
• B: LFT = LST of E = 9; LST = 9 - 5 = 4
• A: LFT = min(B:4, C:3, D:5) = 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, C, F, G, I
• Critical path: A - C - F - G - I (= 3+5+3+4+6 = 21 time units)

Answer: total minimum project time = 21 time units; critical path = A - C - F - G - I