📚 Critical Activities in Critical Path Analysis | 关键路径分析中的关键活动
In Edexcel A-Level Decision Mathematics, the concept of ‘critical activities’ lies at the heart of project planning and control. Knowing how to identify these activities allows you to determine the shortest possible completion time for a project and to recognise which tasks cannot be delayed without impacting the overall schedule. This article explains the essential theory, calculation methods, and exam techniques for critical activities in critical path analysis.
在 Edexcel A-Level 决策数学中，“关键活动”的概念是项目规划与控制的核心。掌握如何识别这些活动，可以帮助你确定项目的最短完成时间，并判断哪些任务不能延迟，否则会影响整体进度。本文阐述关键路径分析中关键活动的基本理论、计算方法和考试技巧。
1. Activity Networks and Dependencies | 活动网络与依赖关系
A project is modelled as a network of activities, where each activity is represented by a node (in activity-on-node notation, which Edexcel uses) or by an edge (in activity-on-arrow notation). Activities are connected by directed edges that show precedence constraints: an activity cannot start until all its immediate predecessors have been completed.
项目被建模为一个活动网络，其中每个活动用节点表示（Edexcel 使用活动节点表示法），或用边表示（活动箭线表示法）。活动之间用有向边连接，以显示先后约束：一个活动必须在其所有直接前导活动完成后才能开始。
- English: Each node typically contains the activity name and its duration.
- 中文：每个节点通常包含活动名称及其持续时间。
- English: A dummy activity may be used in activity-on-arrow networks to represent dependencies, but in activity-on-node notation dummies are not needed.
- 中文：在活动箭线网络中可能使用虚活动来表示依赖关系，但在活动节点表示法中不需要虚活动。
Understanding the network structure is the first step in finding critical activities, because the timing calculations depend entirely on the precedence relationships and durations.
理解网络结构是寻找关键活动的第一步，因为时间计算完全依赖于先后关系和持续时间。
2. Defining a Critical Activity | 关键活动的定义
A critical activity is an activity for which any delay in its start or finish will cause a delay in the completion of the entire project. In other words, a critical activity has zero total float. Total float is the amount of time an activity can be delayed without affecting the project’s minimum completion time.
关键活动是指其开始或完成时间的任何延迟都会导致整个项目完成时间延迟的活动。换句话说，关键活动的总浮动时间为零。总浮动时间是指活动可以延迟而不会影响项目最短完成时间的时间量。
Mathematically, an activity is critical if and only if its total float is zero. The chain of critical activities from the start node to the end node forms the critical path.
从数学上讲，一个活动是关键的当且仅当其总浮动时间为零。从起始节点到结束节点的关键活动链构成关键路径。
3. Forward Pass: Earliest Start and Earliest Finish Times | 前向扫描：最早开始与最早完成时间
The forward pass calculates the earliest start time (EST) and earliest finish time (EFT) for each activity. The earliest start time of the first activity is usually set to 0. For any other activity, the EST is the maximum of the EFTs of all its immediate predecessors.
前向扫描计算每个活动的最早开始时间（EST）和最早完成时间（EFT）。第一个活动的最早开始时间通常设为 0。对于任何其他活动，其 EST 是其所有直接前导活动 EFT 的最大值。
EFT = EST + Duration
The project’s minimum completion time is the maximum EFT of all terminal activities (those with no successors).
项目的最短完成时间是所有终止活动（没有后继活动的活动）EFT 的最大值。
4. Backward Pass: Latest Finish and Latest Start Times | 后向扫描：最晚完成与最晚开始时间
The backward pass begins at the terminal activity or activities. The latest finish time (LFT) of a terminal activity is set equal to the project’s minimum completion time. For any other activity, the LFT is the minimum of the LSTs of all its immediate successors.
后向扫描从终止活动开始。终止活动的最晚完成时间（LFT）设为项目的最短完成时间。对于任何其他活动，其 LFT 是其所有直接后继活动 LST 的最小值。
LST = LFT − Duration
The latest start time (LST) is therefore the latest time an activity can start without delaying the project.
因此，最晚开始时间（LST）是活动在不延迟项目的前提下可以开始的最晚时间。
5. Calculating Total Float | 计算总浮动时间
Total float (TF) for an activity is the maximum amount of time the activity can be delayed beyond its earliest start without delaying the project completion. It can be calculated in two equivalent ways:
活动的总浮动时间（TF）是指活动可以比最早开始时间延迟的最大时间量，而不会延迟项目完成。它有两种等价的计算方式：
Total Float = LST − EST
Total Float = LFT − EFT
These two differences are always equal for a valid network. If an activity has zero total float, it is critical.
对于一个有效的网络，这两个差值总是相等的。如果一个活动的总浮动时间为零，则它是关键活动。
6. Identifying Critical Activities and the Critical Path | 识别关键活动与关键路径
Once the forward and backward passes are complete, you can identify critical activities by checking which activities have total float equal to zero. The sequence of critical activities forms one or more continuous paths from the start to the end. Every critical path must have the same total duration, equal to the project’s minimum completion time.
完成前向和后向扫描后，你可以通过检查哪些活动的总浮动时间为零来识别关键活动。关键活动的序列形成从起点到终点的一条或多条连续路径。每条关键路径的总持续时间必须相同，等于项目的最短完成时间。
In exam questions, you may be asked to state the critical path and its length. Always write the path as a sequence of activity letters, e.g. A → C → E → G, and give the total duration.
在考试题目中，你可能被要求说明关键路径及其长度。始终将路径写为活动字母序列，例如 A → C → E → G，并给出总持续时间。
7. Worked Example: A Simple Project | 示例：一个简单项目
Consider a project with the following activities and durations (in days):
考虑一个具有以下活动和持续时间（以天为单位）的项目：
| Activity | Duration | Predecessors |
| A | 3 | – |
| B | 4 | A |
| C | 2 | A |
| D | 5 | B, C |
| E | 1 | C |
| F | 3 | D, E |
Perform a forward pass. Set EST(A) = 0, so EFT(A) = 0 + 3 = 3. Then EST(B) = EST(C) = 3. EFT(B) = 3 + 4 = 7; EFT(C) = 3 + 2 = 5. Activity D has predecessors B and C, so EST(D) = max(7,5) = 7, EFT(D) = 7 + 5 = 12. Activity E has predecessor C, so EST(E) = 5, EFT(E) = 5 + 1 = 6. Activity F has predecessors D and E, so EST(F) = max(12,6) = 12, EFT(F) = 12 + 3 = 15. The minimum project completion time is 15 days.
进行前向扫描。设 EST(A) = 0，则 EFT(A) = 0 + 3 = 3。然后 EST(B) = EST(C) = 3。EFT(B) = 3 + 4 = 7；EFT(C) = 3 + 2 = 5。活动 D 的前导是 B 和 C，因此 EST(D) = max(7,5) = 7，EFT(D) = 7 + 5 = 12。活动 E 的前导是 C，因此 EST(E) = 5，EFT(E) = 5 + 1 = 6。活动 F 的前导是 D 和 E，所以 EST(F) = max(12,6) = 12，EFT(F) = 12 + 3 = 15。项目的最短完成时间为 15 天。
Now perform a backward pass. Set LFT(F) = 15, so LST(F) = 15 − 3 = 12. Activity D has successor F, so LFT(D) = 12, LST(D) = 12 − 5 = 7. Activity E has successor F, so LFT(E) = 12, LST(E) = 12 − 1 = 11. Activity C has successors D and E, so LFT(C) = min(7,11) = 7, LST(C) = 7 − 2 = 5. Activity B has successor D, so LFT(B) = 7, LST(B) = 7 − 4 = 3. Activity A has successors B and C, so LFT(A) = min(3,5) = 3, LST(A) = 3 − 3 = 0.
现在进行后向扫描。设 LFT(F) = 15，则 LST(F) = 15 − 3 = 12。活动 D 的后继是 F，所以 LFT(D) = 12，LST(D) = 12 − 5 = 7。活动 E 的后继是 F，所以 LFT(E) = 12，LST(E) = 12 − 1 = 11。活动 C 的后继是 D 和 E，因此 LFT(C) = min(7,11) = 7，LST(C) = 7 − 2 = 5。活动 B 的后继是 D，所以 LFT(B) = 7，LST(B) = 7 − 4 = 3。活动 A 的后继是 B 和 C，因此 LFT(A) = min(3,5) = 3，LST(A) = 3 − 3 = 0。
Finally, calculate total floats: A: 0 − 0 = 0; B: 3 − 3 = 0; C: 5 − 3 = 2; D: 7 − 7 = 0; E: 11 − 5 = 6; F: 12 − 12 = 0. The critical activities are A, B, D, F. The critical path is A → B → D → F, with length 15 days.
最后计算总浮动时间：A：0 − 0 = 0；B：3 − 3 = 0；C：5 − 3 = 2；D：7 − 7 = 0；E：11 − 5 = 6；F：12 − 12 = 0。关键活动为 A、B、D、F。关键路径为 A → B → D → F，长度为 15 天。
8. Properties of the Critical Path | 关键路径的性质
The critical path is the longest path through the network in terms of total duration. It determines the minimum time needed to complete the project. If any critical activity is delayed by a certain amount of time, the project completion will be delayed by the same amount, unless corrective action is taken.
关键路径是网络中总持续时间最长的路径。它决定了完成项目所需的最短时间。如果任何关键活动被延迟一定时间，项目完成时间将延迟相同时间，除非采取纠正措施。
There may be more than one critical path in a network. When multiple critical paths exist, all of them must have the same total duration. Activities that are on at least one critical path are critical; activities that are on no critical path have positive float and are non-critical.
网络中可能存在多条关键路径。当存在多条关键路径时，它们必须具有相同的总持续时间。位于至少一条关键路径上的活动是关键活动；不在任何关键路径上的活动具有正浮动时间，是非关键活动。
9. Common Mistakes and Exam Tips | 常见错误与考试提示
Students often confuse earliest and latest times, especially in the backward pass. Remember that the backward pass works from the end to the start, and you must take the minimum of successors’ LSTs when calculating an activity’s LFT. A common error is to use the maximum instead.
学生经常混淆最早和最晚时间，尤其是在后向扫描中。请记住，后向扫描是从结束到开始进行的，并且在计算活动的 LFT 时必须取后继活动 LST 的最小值。一个常见错误是使用了最大值。
Another frequent mistake is forgetting that total float can be calculated either as LST − EST or LFT − EFT. Always check both values; if they are not equal, you have made an arithmetical error.
另一个常见错误是忘记总浮动时间可以通过 LST − EST 或 LFT − EFT 计算。始终检查两个值；如果它们不相等，则说明存在算术错误。
When writing the critical path in the exam, always list activities in their correct sequence and include arrows between them. Also, clearly state the total duration, as this is often worth a mark.
在考试中书写关键路径时，始终按正确顺序列出活动并在它们之间加上箭头。同时，清楚地说明总持续时间，因为这通常值得一个分数。
10. The Role of Critical Activities in Resource Management | 关键活动在资源管理中的作用
Identifying critical activities is not only an exam skill; it is also essential for real-world project management. Project managers focus most attention on critical activities because any delay in these tasks directly extends the project deadline. Non-critical activities, with their positive float, can be scheduled flexibly within their float windows, allowing for better resource levelling.
识别关键活动不仅是一项考试技能，它对现实世界的项目管理也至关重要。项目管理者最关注关键活动，因为这些任务的任何延误都会直接延长项目截止日期。具有正浮动时间的非关键活动可以在其浮动窗口内灵活安排，从而更好地进行资源平衡。
If resources are over-allocated, a manager may decide to delay a non-critical activity to reduce peak demand. However, delaying a critical activity is never an option without extending the whole project. Understanding this distinction is a key learning outcome in Decision Mathematics.
如果资源分配过度，管理者可以决定延迟非关键活动以减少峰值需求。然而，延迟关键活动从来不是一个可选项，因为它会延长整个项目。理解这一区别是决策数学中的一个关键学习成果。
11. Summary and Key Formulae | 总结与关键公式
To summarise, the steps to find critical activities are:
总结起来，寻找关键活动的步骤是：
- English: Draw the activity network using precedence information.
- 中文：使用先后关系信息绘制活动网络。
- English: Perform a forward pass to calculate EST and EFT.
- 中文：执行前向扫描来计算 EST 和 EFT。
- English: Perform a backward pass to calculate LFT and LST.
- 中文：执行后向扫描来计算 LFT 和 LST。
- English: Compute total float for each activity.
- 中文：计算每个活动的总浮动时间。
- English: Identify activities with zero total float as critical.
- 中文：将总浮动时间为零的活动识别为关键活动。
- English: Trace the critical path(s) along these activities.
- 中文：沿着这些活动追踪关键路径。
Total Float = LST − EST = LFT − EFT
Mastering these calculations unlocks a significant portion of the Edexcel Decision Mathematics examination, and provides a powerful tool for analysing real-life projects.
掌握这些计算可以解锁 Edexcel 决策数学考试的一个重要部分，并为分析现实生活中的项目提供强大的工具。
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