CPM schedule, critical path and float

(a) Network (activity-on-node)

 A(4) --+----------> C(5) ---------+
        |                          v
        +----------> D(6) --+---> F(3) ----+
                     ^      |              v
 B(3) --+------------+      +---> G(5) --> H(2)
        |                         ^
        +----------> E(4) --------+

A and B start the project. C follows A; D needs both A and B; E follows B; F needs C and D; G needs D and E; H needs F and G.

(b) Forward and backward pass

Forward pass: $EF = ES + t$, where ES is the largest EF of the predecessors. Backward pass: $LS = LF - t$, where LF is the smallest LS of the successors. Project duration = largest EF = 17.

Sample: ES of D = max(EF of A, EF of B) = max(4, 3) = 4. ES of F = max(9, 10) = 10. LF of D = min(LS of F, LS of G) = min(12, 10) = 10.

(c) Critical path

Activities with zero total float are critical: A, D, G, H.

Check: $4 + 6 + 5 + 2 = 17$ days. The other paths are shorter (e.g. B-D-G-H = 16, A-D-F-H = 15, A-C-F-H = 14).

Answer: Critical path A-D-G-H; project duration 17 days; floats: B = 1, C = 3, E = 3, F = 2 days.