ITPEC FE Morning April 2019 Question 51

Removing F ends the project 1 day earlier — F lies on the 12-day critical path, and without it the longest path is 11 days.

The diagram has three paths from start to finish:

• •A – C – G: 2 + 2 + 4 = 8 days
• •B – F – G: 4 + 4 + 4 = 12 days (critical path)
• •B – D – H: 4 + 2 + 5 = 11 days

The project takes 12 days. If F is no longer needed, path B – F – G disappears and the longest remaining path is B – D – H at 11 days, so the project finishes 1 day earlier.

Answer (c)

Why not others:

• •(a) — A – C – G is the shortest path at 8 days; the critical path is the longest one, B – F – G
• •(b) — 11 days is the length of B – D – H, not of the critical path; the project needs 12 days
• •(d) — H has 1 day of float (12 − 11); a 2-day delay makes B – D – H 13 days, so the project slips by only 1 day

Key rule: A change on a non-critical path affects the end date only by the amount it exceeds that path's float; shortening the critical path helps only until another path becomes critical.

Project Management · AI-written, may contain errors