問48
頻出度★★★
図のアローダイアグラムにおいて, 作業 B が 3 日遅れて完了した。全体の遅れを 1 日にするためには, どの作業を何日短縮すればよいか。
平成23年度 特別試験 ITパスポート試験 問48
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This question's explanation and your answers so far.
解説
問 48
Incorrectア
Shortening task C by 1 day to 4 days makes path A→C take 3+4=7 days, but task B is late and takes 10 days, so the earliest node time of the middle node stays at max(7, 10)=10 days and does not change. Once B is late, it is the path on the B side that governs the merge point, and no matter how much the A→C path on the opposite side is shortened, the timing of the merge does not become earlier. As a result the total remains 10+12=22 days and the delay stays at 2 days with no improvement. This is the classic mistake of shortening a task that is off the critical path, which does not reduce the total duration.
Incorrectイ
Even if task D is shortened by 1 day to 9 days, the duration from the middle node to the end node is still max(D=9, E→F=12)=12 days and does not change. This is because the path through D originally takes 10 days, 2 days shorter than the 12 days of E→F, so it is a non-critical path that has 2 days of float. Shortening a task that has float only increases the float further and has no effect on the overall completion date. The total remains 10+12=22 days, so the delay stays at 2 days and does not become 1 day.
- ウ
正解 The correct answer is ウ. First, check the schedule before the delay occurs. In an arrow diagram, the earliest node time of each node is determined by “the latest arrival time among all paths entering that node”. At the node where task C and task B merge (the middle node in the figure), path A→C takes 3+5=8 days and path B takes 7 days, so the later value, 8 days, is used. From that node to the end node, compare D at 10 days with E→F at 7+5=12 days; 12 days dominates, so the total duration is 8+12=20 days and the critical path is A→C→E→F. Next, consider the case where task B finishes 3 days late, taking 7+3=10 days. The middle node becomes max(A+C=8, B=10)=10 days, so B being 3 days late pushes the merge point back by 2 days (B originally had 8-7=1 day of float, so 1 of the 3 delayed days is absorbed by that float and only 2 days actually take effect). The total becomes 10+max(D=10, E→F=12)=10+12=22 days, which is 2 days later than the original 20 days. To limit the overall delay to 1 day, that is, to finish in 21 days, you only need to shorten by 1 more day the path that is pushing the schedule out by 2 days. After the middle node, the path that governs the duration is E→F at 12 days, so a task on that path must be shortened. If task E is shortened by 1 day to 6 days, E→F becomes 6+5=11 days, which is still longer than D at 10 days and therefore still dominant, so the total is 10+11=21 days and the delay is exactly 1 day. Therefore the answer is to shorten task E by 1 day.
Incorrectエ
If nothing is shortened, 1 of the 3 days by which task B is late is absorbed by the float that B had (the difference between the 8 days of A→C and the 7 days of B), but the remaining 2 days cannot be absorbed, so the total is 10+12=22 days, that is, 2 days later than the original 20 days. This does not satisfy the requirement to limit the overall delay to 1 day, so it cannot be said that no shortening is needed. Note that even when a task with float is delayed, the portion that exceeds the float turns directly into a delay of the whole project.
ここがポイント
- How to read an arrow diagram (PERT chart) and the earliest node time — the task name is written above the arrow and its duration below it, and the earliest node time of each node is the maximum of the arrival times of all paths entering that node. The path with the longest duration from start to end is the critical path, and its length is the total duration of the project. In this question the original critical path is A→C→E→F at 20 days.
- Float and the idea behind schedule shortening — non-critical tasks have float; a delay within that float does not affect the whole project, but any portion beyond the float turns directly into a delay of the whole project. Also, when you want to reduce the total duration, shortening has no effect unless it is applied to a task on the path that is dominant at that moment (the critical path). Because a delay can shift the critical path, always decide what to shorten based on the state after the delay.
用語