This question's explanation and your answers so far.
システム開発を示した図のアローダイアグラムにおいて，工程 A と D が合わせて 3 日遅れると，全体では何日遅れるか。 図のアローダイアグラム（凡例: 円が結合点，矢線上が「工程名」，矢線下が「所要日数」）: - 開始結合点から 3 本の矢線が出て，それぞれ別の経路を通り，最後に同じ終了結合点に集まる。 - 経路1（上）: 開始 →[工程 A，所要日数 2]→ 結合点 →[工程 D，所要日数 4]→ 終了 - 経路2（中）: 開始 →[工程 B，所要日数 4]→ 結合点 →[工程 E，所要日数 3]→ 終了 - 経路3（下）: 開始 →[工程 C，所要日数 7]→ 結合点 →[工程 F，所要日数 1]→ 終了
正解 In an arrow diagram, the longest path from start to end (the critical path) determines the total duration of the whole project. The durations of the three paths in the figure are A→D = 2 + 4 = 6 days, B→E = 4 + 3 = 7 days, and C→F = 7 + 1 = 8 days. So the original total duration is the critical path C→F, which is 8 days. When processes A and D are delayed by a total of 3 days, the A→D path becomes 6 + 3 = 9 days. The new total duration is the maximum of the three paths = max(9, 7, 8) = 9 days. Therefore the delay is 9 − 8 = 1 day, and the correct answer is “1” (ア).
“2” is wrong. This answer mistakes the baseline: it subtracts the second-longest path B→E (7 days) from the delayed A→D path (9 days) to get 9 − 7 = 2. The baseline for the overall delay should be the original critical path C→F at 8 days, so 9 − 8 = 1 day is correct.
“3” is wrong. This simply treats the delay of 3 days as the overall delay. The A→D path was originally 6 days, giving it 2 days of slack (float) against the 8-day critical path. Of the 3-day delay, 2 days are absorbed by the slack, so only 1 day propagates to the whole project.
“4” is wrong. This misreads “a total of 3 days” as A and D each being delayed by 3 days (6 days in total), making the A→D path 6 + 6 = 12 days and giving 12 − 8 = 4. The question states 3 days in total for A and D combined, so the A→D path only reaches 9 days.