問35
頻出度★★★
図のアローダイアグラムで，A から G に至る全体の作業日数に影響を与えないことを条件に，C→F の作業の遅れは最大何日間まで許容できるか。
| 作業 | 作業日数 |
|---|---|
| A→B | 3 |
| B→D | 4 |
| A→C | 3 |
| C→D | 3 |
| D→F | 6 |
| C→F | 6 |
| F→G | 4 |
| A→E | 7 |
| E→F | 4 |
| E→G | 7 |
〔凡例〕○ —作業日数→ ○
平成22年度 春期 ITパスポート試験 問35
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This question's explanation and your answers so far.
解説
問 35
Incorrectア
1 day is the slack of the event C (latest event time 4 − earliest event time 3 = 1 day), not the float of the activity C→F. The latest time 4 for C is fixed by the C→D route (7−3=4), while the C→F side still has more room. The slack of an event and the float of each activity leaving that event are different quantities, and that is the trap in this question.
Incorrectイ
2 days is what you get if you fix the time of F at day 11 via E (7+4=11) and then compute 11−9=2. The earliest event time of F must be the maximum of the three incoming routes, so day 13 via D is the one that applies. Using the value of an intermediate route instead of the maximum makes you underestimate the float.
Incorrectウ
3 days is what you get if you mistake the earliest event time of D for day 6 via C (3+3=6), make F day 12 (6+6=12) and then compute 12−9=3. D must be the larger of 3+4=7 via B and 6 via C, that is day 7. The earliest event time of an event is the moment when all of its preceding activities have finished, so it is always the maximum.
- エ
正解 First, find the earliest event time of each node. Starting from A = day 0, we get B=3, C=3 and E=7. For D, take the larger of 3+4=7 via B and 3+3=6 via C, so D=7. For F, take the largest of 7+6=13 via D, 3+6=9 via C and 7+4=11 via E, so F=13. For G, take the larger of 13+4=17 via F and 7+7=14 via E, so G=17. In other words the whole project takes 17 days, and the critical path is A→B→D→F→G. Next, look at the float of the activity C→F. C→F can start at the earliest on day 3 and takes 6 days, so it finishes on day 9. On the other hand F still needs 4 days for F→G, so the latest event time of F that keeps the total at 17 days is 17−4=13 (day 13); since F lies on the critical path, this matches its earliest time of day 13. Therefore the delay allowed for C→F is 13−9=4 days. As a check: even with a delay of 4 days, C→F still finishes on day 13, F stays at day 13 and G stays at day 17, so the total number of days does not change. With a delay of 5 days, C→F finishes on day 14, F slips to day 14 and G to day 18, and the whole project is extended. Hence the maximum allowable delay is 4 days, and the correct answer is “エ”.
ここがポイント
- Basics of the arrow diagram (PERT chart): the earliest event time of each node is the maximum of the arrival times of the routes coming into it. The earliest event time of the goal node is the total number of days required, and the longest path that determines that number is the critical path. Here it is 17 days, along A→B→D→F→G.
- The delay allowed for an activity (its float, that is the total float) = latest event time of the following node − (earliest start time of the activity + its duration). Latest event times are obtained by working backward from the number of days at the goal. Activities on the critical path have zero float; float appears only on the other activities. For C→F here, F is on the critical path, so its latest time = earliest time = day 13, giving 13 − (3+6) = 4 days.
- Do not confuse the slack of an event (latest event time − earliest event time) with the float of an activity. The slack of event C is 4−3=1 day, but that value is fixed by the constraint on the C→D side, and it is a different quantity from the 4 days of delay allowed for the activity C→F.
用語