IPA FE Subject A 2023 Question 13
A question IPA Japan published in 令和５年度 基本情報技術者試験 科目A 公開問題 問13, translated into English. ITPEC is based on the same exams.
The activities were scheduled to be performed as shown in the figure, but a delay of one day occurred in activity A. When the cost increase rate of each activity is the value in the table, which activity should be shortened to minimize the additional cost required to finish in the originally scheduled number of days? Here, the cost increase rate is the additional cost required to shorten an activity by one day.
Explanation
E — the one-day delay matters only on the critical path, so shorten the cheapest activity on it by one day.
The three paths through the diagram:
- •A → B → E → G: 4 + 6 + 5 + 5 = 20 days (critical path)
- •A → C → G: 4 + 8 + 5 = 17 days
- •A → D → F → G: 4 + 4 + 4 + 5 = 17 days
With A one day late the project takes 21 days. To finish in the planned 20, one activity on A → B → E → G after A must lose a day: B (6), E (2.5) or G (5). E is the cheapest. Afterwards the critical path is 20 days again and the other paths, 18 days each, stay shorter, so no second activity needs shortening.
Answer (d)
Why not others:
- •(a) B — on the critical path, so it would work, but at 6 per day it is the most expensive choice
- •(b) C and (c) D — not on the critical path; with the delay their paths take 18 days, so shortening them leaves the project at 21 days, however low their rates
Key rule: When crashing a schedule, shorten only critical-path activities, choose the lowest cost increase rate first, and check that no other path becomes longer than the target.
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.