CPM network, road project (days)

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

The network has 9 events and 4 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 + 6 = 6
• B: EST = 0; EFT = 0 + 4 = 4
• C: EST = 0; EFT = 0 + 5 = 5
• D: EST = max(A:6, B:4) = 6; EFT = 6 + 2 = 8
• E: EST = max(C:5, D:8) = 8; EFT = 8 + 4 = 12
• F: EST = EFT of D = 8; EFT = 8 + 2 = 10
• G: EST = EFT of D = 8; EFT = 8 + 5 = 13
• H: EST = EFT of D = 8; EFT = 8 + 8 = 16
• I: EST = EFT of G = 13; EFT = 13 + 5 = 18
• J: EST = max(E:12, G:13) = 13; EFT = 13 + 6 = 19

Project duration = largest EFT = 19 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 = 19; LST = 19 - 6 = 13
• I: LFT = project duration = 19; LST = 19 - 5 = 14
• H: LFT = project duration = 19; LST = 19 - 8 = 11
• G: LFT = min(I:14, J:13) = 13; LST = 13 - 5 = 8
• F: LFT = project duration = 19; LST = 19 - 2 = 17
• E: LFT = LST of J = 13; LST = 13 - 4 = 9
• D: LFT = min(E:9, F:17, G:8, H:11) = 8; LST = 8 - 2 = 6
• C: LFT = LST of E = 9; LST = 9 - 5 = 4
• B: LFT = LST of D = 6; LST = 6 - 4 = 2
• A: LFT = LST of D = 6; LST = 6 - 6 = 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, J
• Critical path: A - D - G - J (= 6+2+5+6 = 19 days)

Answer: project duration = 19 days; critical path = A - D - G - J

Analysis of the project situation

• Project completion time = 19 days.
• Critical path = A - D - G - J (6 + 2 + 5 + 6 = 19 days); critical activities = A, D, G and J (zero float). A delay in any of them delays the road project by the same number of days.
• Non-critical activities: B, C, E, F, H and I. E and I have only 1 day of float, so they are near-critical and must be watched; F (9 days) and H (3 days) have the most float.
• Management should give priority in resources and supervision to the critical activities, and use the float of F, H, B and C to level labour and equipment.