IPA IP 2024 Question 41
A question IPA Japan published in 令和6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.
The relationships between the activities of a project and their required times are shown in an arrow diagram. Among the four (4) nodes B through E in this arrow diagram, which of the following is the node at which the start of all the activities that begin from that node can be delayed the most without affecting the completion time of the entire process? Here, an activity that begins from a node cannot start until all the activities that lead to that node are completed, and the setup time from one activity to the next activity is not considered.
Explanation
C — node C has the most slack: the activity that starts there can begin up to 15 later without delaying the end of the project.
The slack of a node is its latest start time minus its earliest start time. A forward pass gives the earliest times (take the largest arrival), a backward pass from the end gives the latest times (take the smallest).
- •Forward: B = 15, C = 10, D = 15 + 10 = 25, E = max(15 + 25, 10 + 15, 25 + 10) = 40, F = max(25 + 20, 40 + 15) = 55
- •Backward from 55: E = 55 − 15 = 40, D = min(55 − 20, 40 − 10) = 30, C = 40 − 15 = 25, B = min(30 − 10, 40 − 25) = 15
The critical path is A→B→E→F (15 + 25 + 15 = 55).
Answer (b)
Why not others:
- •(a) B and (d) E — both lie on the critical path, so they have no slack at all
- •(c) D — only 5 of slack: the activity D→E (10) must reach E by 40; looking at D→F alone would wrongly give 10
Key rule: Node slack = latest time − earliest time; the forward pass takes the maximum of the incoming paths, the backward pass the minimum of the outgoing ones.
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.