問52
頻出度★★★
次の表に示す作業全体の最短の所要日数を増やすことなく作業Eの所要日数を増やしたい。最大何日増やすことができるか。
| 作業 | 前提作業 | 所要日数 |
|---|---|---|
| A | — | 7 |
| B | — | 4 |
| C | — | 2 |
| D | C | 1 |
| E | B，D | 1 |
平成28年度 春期 ITパスポート試験 問52
ストラテジ系企業活動 › 業務分析・データ利活用
This question's explanation and your answers so far.
解説
問 52
Incorrectア
This is the value you get if you mistakenly think E has no float (slack) at all. In reality E is not on the critical path formed by A (7 days) but belongs to the B–D chain; it can start on day 4 and finish on day 5. Against the total of 7 days it has 2 days of float, so the number of days it can be increased is not 0.
Incorrectイ
This is the value you get if you underestimate the float as 1 day. The time window available to E is 7−4=3 days, and subtracting its current duration of 1 day means it can be increased by 2 days. Getting E’s start day or the total duration wrong tends to give 1, but the correct value is 2 days.
- ウ
正解 First find the earliest start and earliest finish for each activity. A, B, and C have no predecessor activities, so they can all start on day 0; A finishes on day 7, B on day 4, and C on day 2. D comes after C, so it starts on day 2 and finishes on day 3. E can start only after both B and D are finished, so its earliest start is max(4,3)=4, that is day 4, and with its current duration of 1 day it finishes on day 5. The completion of all the activities is set by the latest finish time, and A’s 7 days is the longest path (the critical path), so the shortest total duration is 7 days. E can start on day 4, and to keep the total at 7 days it only needs to finish by day 7 at the latest, so the time window available to E is 7−4=3 days. Since its current duration is 1 day, it can be increased by at most 3−1=2 days, and the correct answer is choice ウ, 2 days.
Incorrectエ
This is the mistake of answering with the time window available to E itself (7−4=3 days). The question asks for the number of days it can be increased, so you have to subtract its current duration of 1 day. That gives 3−1=2 days, not 3.
ここがポイント
- The total duration of all the activities is set by the longest path (the critical path). In this problem, A’s 7 days, which has no predecessor, is the longest, so the total is 7 days.
- An activity that is not on the critical path has float (total float = latest finish − earliest finish), and lengthening its duration by that amount does not change the total duration. E has 2 days of float.
用語