CPM network, 9 activities with successors

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 (days). Events are circled numbers; dashed arrows are dummy activities (zero duration) that show dependencies only.

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

Project duration = largest EFT = 13 days.

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 - 2 = 11
• H: LFT = project duration = 13; LST = 13 - 1 = 12
• E: LFT = LST of H = 12; LST = 12 - 2 = 10
• G: LFT = LST of I = 11; LST = 11 - 3 = 8
• F: LFT = LST of I = 11; LST = 11 - 1 = 10
• D: LFT = LST of G = 8; LST = 8 - 3 = 5
• C: LFT = min(F:10, H:12) = 10; LST = 10 - 2 = 8
• B: LFT = LST of E = 10; LST = 10 - 4 = 6
• A: LFT = min(B:6, C:8, D:5) = 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, D, G, I
• Critical path: A - D - G - I (= 5+3+3+2 = 13 days)

Answer: project duration = 13 days; critical path = A - D - G - I

Significance of total float

• Activities with TF = 0 are critical; they decide the project duration, so they get the closest control and priority in resources.
• TF tells how long a non-critical activity may be delayed without delaying the project, which helps to decide its start date.
• It is used to level and smooth resources, shifting activities inside their float to cut peaks.
• It shows which activities can release men and machines to the critical activities when the project must be speeded up.
• It is the cushion against delay; when TF becomes negative the project is behind schedule and corrective action is needed.
• Float is shared along a path, so using it on one activity reduces the float of the others on the same path.