IPA IP 2023 Question 41
A question IPA Japan published in 令和5年度 ITパスポート試験 公開問題 問41, translated into English. ITPEC is based on the same exams.
As a result of performing activities based on the following arrow diagram, activity D was delayed by 2 days, and activity F was completed 3 days ahead of schedule. How much did the number of days required for the entire set of activities change compared with the plan?
Explanation
2 days earlier — the project lasts as long as its longest path, so recompute every path with the new durations and compare the longest one with the plan.
Paths from start to end, as planned and as performed (D: 1 + 2 = 3 days, F: 5 − 3 = 2 days):
The plan's critical path A–C–F took 11 days. F finishing 3 days early shortens it to 8, and D's 2-day delay only uses up float on B–D–F, which also ends at 8. The longest path is now B–E–G at 9 days, untouched by either change, so the whole takes 9 days instead of 11: 11 − 9 = 2 days earlier.
Answer (c)
Why not others:
- •(a) 3 days later — reads F's change as a 3-day delay: A–C–F would become 2 + 4 + 8 = 14 days, 3 more than planned
- •(b) 1 day earlier — nets the two changes, 3 − 2 = 1, as if both lay on the same path
- •(d) 3 days earlier — passes the whole 3-day gain on F to the project, missing that B–E–G (9 days) becomes the new critical path
Key rule: Shortening a critical activity helps only until another path becomes the longest; after any change, recompute every path and take the new maximum.
Project Management · AI-written, may contain errors
This question comes from the past questions IPA Japan publishes, translated into English by ITPEC Practice. ITPEC Practice is an independent study tool and is not affiliated with IPA or ITPEC. See the original paper (PDF) or Report an issue.