SACE General Mathematics Practice Questions

Forward scan (earliest starting times, in days): EST(A) = 0 and EST(B) = 0; EST(C) = 0 + 5 = 5; EST(D) = max(5, 3) = 5 because D waits for both A and B; EST(E) = 5 + 6 = 11; EST(F) = 5 + 4 = 9; EST(G) = max(11 + 2, 9 + 7) = max(13, 16) = 16. The project finishes at 16 + 3 = 19, so the minimum completion time is 19 days. Checking the paths confirms this: A–C–E–G = 5 + 6 + 2 + 3 = 16, A–D–F–G = 5 + 4 + 7 + 3 = 19 and B–D–F–G = 3 + 4 + 7 + 3 = 17. The critical path is the longest path, A–D–F–G. Backward scan (latest completion times, in days), starting from 19: LCT(G) = 19; LCT(E) = LCT(F) = 19 − 3 = 16; LCT(C) = 16 − 2 = 14; LCT(D) = 16 − 7 = 9; LCT(B) = 9 − 4 = 5. Slack time = latest completion time − (task time + earliest starting time). Task B: 5 − (3 + 0) = 2 days. Task C: 14 − (6 + 5) = 3 days. Task E: 16 − (2 + 11) = 3 days. So B may be delayed by up to 2 days, and C and E share 3 days of slack along the path A–C–E, without delaying the 19-day completion. Note that C and E do not each have an independent 3 days: they lie on the same non-critical path, so using C's slack removes E's.