CPM network, 10 activities with successors

Definitions

• Total float (TF): the time by which an activity can be delayed without delaying the project completion. $TF = LST - EST = LFT - EFT$.
• Free float (FF): the time by which an activity can be delayed without delaying the earliest start of any following activity. $FF = \min(EST\ \text{of successors}) - EFT$.
• Independent float (IndF): the time available to an activity when its predecessors finish at their latest and its successors start at their earliest; using it affects no other activity. $IndF = \max(0,\ \min(EST_{succ}) - \max(LFT_{pred}) - D)$.
• Interfering float $= TF - FF$ (the part of the total float that is shared with the following activities).

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(2)--> (2)
(2) --B(3)--> (3)
(2) --C(4)--> (4)
(2) --D(5)--> (5)
(3) --E(4)--> (6)
(4) - - dummy - -> (6)
(4) - - dummy - -> (7)
(4) --F(3)--> (8)
(5) --G(2)--> (7)
(6) --H(1)--> (8)
(7) --I(2)--> (8)
(8) --J(3)--> (9)

The network has 9 events and 2 dummy activities. Start event is (1) and the end event is (9).

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 = EFT of A = 2; EFT = 2 + 3 = 5
• C: EST = EFT of A = 2; EFT = 2 + 4 = 6
• D: EST = EFT of A = 2; EFT = 2 + 5 = 7
• E: EST = EFT of B = 5; EFT = 5 + 4 = 9
• F: EST = EFT of C = 6; EFT = 6 + 3 = 9
• G: EST = EFT of D = 7; EFT = 7 + 2 = 9
• H: EST = max(C:6, E:9) = 9; EFT = 9 + 1 = 10
• I: EST = max(C:6, G:9) = 9; EFT = 9 + 2 = 11
• J: EST = max(F:9, H:10, I:11) = 11; EFT = 11 + 3 = 14

Project duration = largest EFT = 14 days.

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 = 14; LST = 14 - 3 = 11
• I: LFT = LST of J = 11; LST = 11 - 2 = 9
• H: LFT = LST of J = 11; LST = 11 - 1 = 10
• G: LFT = LST of I = 9; LST = 9 - 2 = 7
• F: LFT = LST of J = 11; LST = 11 - 3 = 8
• E: LFT = LST of H = 10; LST = 10 - 4 = 6
• D: LFT = LST of G = 7; LST = 7 - 5 = 2
• C: LFT = min(F:8, H:10, I:9) = 8; LST = 8 - 4 = 4
• B: LFT = LST of E = 6; LST = 6 - 3 = 3
• A: LFT = min(B:3, C:4, D:2) = 2; LST = 2 - 2 = 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, J
• Critical path: A - D - G - I - J (= 2+5+2+2+3 = 14 days)

Answer: project duration = 14 days; critical path = A - D - G - I - J