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CPM network, 13 activities (A=2 to M=7)

Short answer

Forward/backward pass yields project duration 25 days. Critical path is B-E-I-K-L; activities B, E, I, K, L have zero total float.

Draw the CPM network diagram and compute EST, EFT, LST, LFT, TF, FF, Int.F and Ind.F from the information given below. Also compute the project duration and mark the critical path.

ActivityDuration (days)Predecessor
A2-
B4-
C3A
D5A, B
E5B
F1C
G4D
H5D
I7E
J2D, F
K3H, I
L6K
M7G

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)
(1) --B(4)--> (3)
(2) --C(3)--> (4)
(2) - - dummy - -> (6)
(3) --E(5)--> (5)
(3) - - dummy - -> (6)
(4) --F(1)--> (10)
(5) --I(7)--> (9)
(6) --D(5)--> (7)
(7) --G(4)--> (8)
(7) --H(5)--> (9)
(7) - - dummy - -> (10)
(8) --M(7)--> (12)
(9) --K(3)--> (11)
(10) --J(2)--> (12)
(11) --L(6)--> (12)

The network has 12 events and 3 dummy activities. Start event is (1) and the end event is (12).

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 = 0; EFT = 0 + 4 = 4
  • C: EST = EFT of A = 2; EFT = 2 + 3 = 5
  • D: EST = max(A:2, B:4) = 4; EFT = 4 + 5 = 9
  • E: EST = EFT of B = 4; EFT = 4 + 5 = 9
  • F: EST = EFT of C = 5; EFT = 5 + 1 = 6
  • G: EST = EFT of D = 9; EFT = 9 + 4 = 13
  • H: EST = EFT of D = 9; EFT = 9 + 5 = 14
  • I: EST = EFT of E = 9; EFT = 9 + 7 = 16
  • J: EST = max(D:9, F:6) = 9; EFT = 9 + 2 = 11
  • M: EST = EFT of G = 13; EFT = 13 + 7 = 20
  • K: EST = max(H:14, I:16) = 16; EFT = 16 + 3 = 19
  • L: EST = EFT of K = 19; EFT = 19 + 6 = 25

Project duration = largest EFT = 25 days.

Backward pass (LFT and LST)

Rule: LFT = smallest LST of the succeeding activities (project duration for end activities); LST = LFT - D.

  • L: LFT = project duration = 25; LST = 25 - 6 = 19
  • K: LFT = LST of L = 19; LST = 19 - 3 = 16
  • M: LFT = project duration = 25; LST = 25 - 7 = 18
  • J: LFT = project duration = 25; LST = 25 - 2 = 23
  • I: LFT = LST of K = 16; LST = 16 - 7 = 9
  • H: LFT = LST of K = 16; LST = 16 - 5 = 11
  • G: LFT = LST of M = 18; LST = 18 - 4 = 14
  • F: LFT = LST of J = 23; LST = 23 - 1 = 22
  • E: LFT = LST of I = 9; LST = 9 - 5 = 4
  • D: LFT = min(G:14, H:11, J:23) = 11; LST = 11 - 5 = 6
  • C: LFT = LST of F = 22; LST = 22 - 3 = 19
  • B: LFT = min(D:6, E:4) = 4; LST = 4 - 4 = 0
  • A: LFT = min(C:19, D:6) = 6; LST = 6 - 2 = 4

Floats

  • Total float TF=LST−EST=LFT−EFTTF = LST - EST = LFT - EFT
  • Free float FF=min⁡(EST of successors)−EFTFF = \min(\text{EST of successors}) - EFT
  • Interfering float IntF=TF−FFIntF = TF - FF
  • Independent float IndF=max⁡(0, min⁡(EST of successors)−max⁡(LFT of predecessors)−D)IndF = \max\left(0,\ \min(\text{EST of successors}) - \max(\text{LFT of predecessors}) - D\right)

Result table

ActivityDESTEFTLSTLFTTFFFInt.FInd.F
A202464040
B404040000
C3251922170170
D5496112020
E549490000
F1562223173140
G491314185050
H591411162200
I79169160000
J29112325141400
K3161916190000
L6192519250000
M7132018255500

Event times (TE from the forward pass, TL from the backward pass):

EventTE (earliest)TL (latest)Slack
1000
2264
3440
452217
5990
6462
79112
813185
916160
1092314
1119190
1225250

Critical path and project duration

  • Critical activities (TF = 0): B, E, I, K, L
  • Critical path: B - E - I - K - L (= 4+5+7+3+6 = 25 days)

Answer: project duration = 25 days; critical path = B - E - I - K - L

✓ Answer checked by Claude Opus 5.5 (high effort)

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