ITPEC IP April 2025 Question 53

1 day — the project lasts as long as its longest path, so it slips only by the amount the delayed path now exceeds that.

The three paths from start to end:

• •A → D: 2 + 4 = 6 days
• •B → E: 4 + 3 = 7 days
• •C → F: 7 + 1 = 8 days, the critical path

Path A–D has 8 − 6 = 2 days of float. Delaying A and D by 3 days in total makes it 6 + 3 = 9 days, now longer than C–F, so the project takes 9 days instead of 8: a delay of 9 − 8 = 1 day.

Answer (a)

Why not others:

• •(b) 2 — measures the new length of A–D against B–E, 9 − 7, instead of against the critical path C–F
• •(c) 3 — passes the whole delay on to the project, ignoring the 2 days of float on A–D
• •(d) 4 — reads "three days in total" as 3 days each on A and D: 6 + 6 = 12 and 12 − 8 = 4

Key rule: A delay on a non-critical path costs the project only what exceeds that path's float; recompute the longest path rather than adding the delay to the old duration.

Project Management · AI-written, may contain errors