This question's explanation and your answers so far.
機械メーカのS社では，製品Xに組み込むソフトウェア（以下，Xソフトという）の開発作業A～Hを表1のように計画した。ここで，前作業とは当該作業を開始する前に終了していなければならない作業のことであり，各作業は前作業が終了すればすぐに開始する。各作業に掛かった費用は，各作業完了時に計上する。 表1 Xソフト開発の作業
| 作業 | 日数 | 費用 | 前作業 |
|---|---|---|---|
| A | 2 | 5 | なし |
| B | 3 | 7 | A |
| C | 4 | 4 | B |
| D | 4 | 5 | A |
| E | 5 | 6 | D |
| F | 2 | 5 | A |
| G | 6 | 6 | D, F |
| H | 7 | 12 | C, E, G |
図1は，表1に基づいて作成したアローダイアグラムである。 Xソフトの開発が終了する日は, 開始から最短で何日目か。
※ Our check while writing this explanation could not fully confirm the published answer; scoring follows IPA's official key.
16 is the length of the path A→B→C→H, that is 2+3+4+7=16 days. C does indeed finish on day 9, but the predecessors of H are not only C: they also include E (finishing on day 11) and G (finishing on day 12), so H cannot be started on day 9. This is the mistake of lining up with the earliest-finishing predecessor at a merge point; the earliest event time must always be aligned with the latest predecessor.
A path that gives 17 does not exist under the dependencies shown in Table 1 and Figure 1. If you misread the predecessor of G as only F, you get A→F→G→H = 2+2+6+7=17 days, but in Table 1 the predecessors of G are both D and F, and in Figure 1 that relation is shown by a dummy activity (a dashed arrow) drawn from the end node of D to the end node of F. This is the typical mistake of ignoring a dummy activity just because its duration is 0.
18 is the length of the path A→D→E→H, that is 2+4+5+7=18 days. E finishes on day 11, but the predecessors of H include G, which finishes on day 12, so the start of H has to wait until day 12. This mistakes the path through E for the longest one, whereas that path is not the critical path.
正解 The correct answer is エ, “19”. To find the total duration from an arrow diagram, compute the earliest event time of each node (the earliest time by which all activities entering that node are finished) in order from the start side; at a node where several activities merge, take the latest of the finishing times of the preceding activities. Following Table 1 and Figure 1, the calculation goes as follows. Activity A (2 days) has no predecessor, so it finishes on day 2 from the start. After A the network splits into three directions: B, D and F. B (3 days) finishes on day 2+3=5, and the following C (4 days) finishes on day 5+4=9. D (4 days) finishes on day 2+4=6, and the following E (5 days) finishes on day 6+5=11. F (2 days) finishes on day 2+2=4. Here G (6 days) has both D and F as predecessors, so it can start only at the later of the two, namely D finishing on day 6, and G finishes on day 6+6=12. Finally H (7 days) has three predecessors, C, E and G, so it can start only at the latest of them, G finishing on day 12, and H finishes on day 12+7=19. That is the day the development of software X ends, so the answer is day 19. Put the other way round, the longest path (the critical path) is A→D→G→H, which is 2+4+6+7=19 days, and if any activity on this path slips even by one day, the overall finishing day slips by the same amount.