题目
| Activity | Immediately preceding activities |
|---|---|
| A | – |
| B | – |
| C | – |
| D | A |
| E | A |
| F | B, C, E |
| G | B, C, E |
| H | C |
| I | C |
| J | D, F, G, H, I |
| K | D, F, G, H, I |
| L | I |
(a) Draw the activity network described in the precedence table above, using activity on arc and the minimum number of dummies.
A project is modelled by the activity network drawn in (a). Each activity requires exactly one worker. The project is to be completed in the shortest possible time.
The table below gives the time, in hours, to complete three of the activities.
| Activity | Duration (in hours) |
|---|---|
| A | 10 |
| E | 7 |
| F | 8 |
The length of the critical path AEFK is 33 hours.
(b) Determine the range of possible values for the duration of activity J. You must make your method and working clear.
题目中文翻译
| 活动 | 紧前活动 |
|---|---|
| A | – |
| B | – |
| C | – |
| D | A |
| E | A |
| F | B, C, E |
| G | B, C, E |
| H | C |
| I | C |
| J | D, F, G, H, I |
| K | D, F, G, H, I |
| L | I |
(a) 使用弧表示活动和最少数量的虚活动，画出上面优先级表中描述的活动网络。
一个项目由 (a) 中画出的活动网络模拟。每项活动恰好需要一名工人。项目应在最短时间内完成。
下表给出了完成三项活动的时间（单位：小时）。
| 活动 | 持续时间（小时） |
|---|---|
| A | 10 |
| E | 7 |
| F | 8 |
关键路径 AEFK 的长度为 33 小时。
(b) 确定活动 J 持续时间的可能取值范围。必须清楚说明方法和计算过程。
解答
(a)
解法一
思路
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先让 A、B、C 从同一开始事件出发。F、G 同时依赖 B、C、E，因此用一条虚活动把 C 的完成状态并入 B、E 的汇合事件；J、K 同时依赖 D、F、G、H、I，再用两条虚活动合并这些分支。最后用一条虚活动把 J 并入唯一结束事件，共需四条虚活动。
答题过程
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One valid activity-on-arc network is specified below. The symbol denotes a dummy activity.
| Activity | Start event | Finish event |
|---|---|---|
| A | 1 | 2 |
| B | 1 | 3 |
| C | 1 | 4 |
| D | 2 | 5 |
| E | 2 | 3 |
| 4 | 3 | |
| F | 3 | 5 |
| G | 3 | 6 |
| H | 4 | 6 |
| I | 4 | 7 |
| 7 | 6 | |
| 6 | 5 | |
| J | 5 | 8 |
| K | 5 | 9 |
| 8 | 9 | |
| L | 7 | 9 |
This network has one start event, one finish event and four dummies. Its completed diagram is:
(b)
解法一
思路
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先利用关键路径 AEFK 的总时长求出 K 的持续时间。路径 AEFJ 与 AEFK 只有最后一项活动不同；为使 AEFK 保持为题目给定的关键路径，J 的持续时间必须小于 K 的持续时间。活动持续时间取正值，因此还要写出下界。
答题过程
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Since AEFK has duration hours,
so
The path AEFJ must be shorter than the critical path AEFK. Hence
which gives
As an activity duration is positive, the required range is