📚 GCSE Edexcel Maths: Critical Path Analysis - Full Revision Guide | 关键路径分析 考点精讲完整指南
Critical Path Analysis (CPA) is a core topic in the Edexcel GCSE Mathematics specification, sitting inside the Decision Mathematics strand. It gives you a systematic method for modelling a project as a network of activities, working out the shortest possible time in which the whole project can be completed, and identifying exactly which tasks are allowed to slip and which are not. be delayed. This revision guide walks through every stage you need for the exam: drawing activity networks, using dummy activities correctly, carrying out the forward and backward passes, calculating float, identifying the critical path, drawing cascade charts, and avoiding the mistakes that cost marks most often.
关键路径分析（CPA）是 Edexcel GCSE 数学大纲中的核心主题，属于决策数学（Decision Mathematics）板块。它为你提供了一套系统方法：把项目建模成活动网络，计算出整个项目可能的最短完工时间，并准确识别哪些任务可以延误、哪些绝对不能延误。本复习指南将完整讲解考试所需的每一个环节：绘制活动网络图、正确使用虚活动、执行正向与反向遍历、计算时差、识别关键路径、绘制级联图，以及规避最常见的失分错误。
1. Introduction to Critical Path Analysis | 关键路径分析简介
Critical Path Analysis is a project management technique developed in the 1950s to schedule and coordinate large-scale engineering and construction projects. In GCSE Edexcel Maths you apply it to model a project as a network of activities. The network shows the order in which tasks must be completed, and the analysis reveals the critical path - the sequence of activities that determines the overall project duration.
关键路径分析是一种项目管理技术，起源于20世纪50年代，用于安排和协调大型工程与建筑项目。在 GCSE Edexcel 数学中，你用它把项目建模为活动网络。该网络展示任务必须完成的先后顺序，而分析的结果会揭示关键路径——即决定整个项目工期的活动序列。
Questions on CPA almost always ask you to answer one of three things: what is the earliest time the project can finish; which activities can be delayed without affecting that finishing time; and how the project can be scheduled to make the best use of resources. Once you can complete the forward pass, the backward pass and the float calculation reliably, every question in this topic becomes routine.
CPA 的考题几乎总是在问三件事之一：项目最早何时能完成；哪些活动可以延迟而不影响这个完工时间；以及如何安排项目才能最有效地利用资源。一旦你能稳定地完成正向遍历、反向遍历和时差计算，这一专题的所有题目都会变得套路化。
The three exam skills you must be able to demonstrate are summarised below.
你必须能够展示的三项考试技能总结如下。
- Construction: draw a correct activity network from a precedence table, including any dummy activities.
构建：根据前序关系表绘制正确的活动网络图，包括所需的虚活动。 - Calculation: complete the forward and backward passes and find the minimum completion time.
计算：完成正向与反向遍历，求出最短完工时间。 - Interpretation: state the critical path, calculate float, and comment on scheduling in context.
解读：写出关键路径，计算时差，并结合实际情境对排程作出说明。
2. Activities, Durations and Precedence Tables | 活动、工期与前序关系表
An activity is any task that consumes time and resources. "Paint the walls" and "lay the carpet" are activities; the duration is how long each one takes. Activities are usually labelled with capital letters, and each activity is assumed to be indivisible - once started it runs to completion.
活动是指任何消耗时间和资源的任务。"粉刷墙壁"和"铺设地毯"都是活动；工期是指每项活动需要多长时间。活动通常用大写字母标注，并且假定每项活动不可分割——一旦开始就会一直做完。
A precedence table (also called a dependence table) summarises all the activities, their durations and which other activities must be finished before each one can start. Every CPA question begins with a table of this kind, so the first thing to do is read it carefully.
前序关系表（也称依赖表）汇总了所有活动、它们的工期，以及每项活动开始前必须先完成哪些活动。每一道 CPA 题目都从这样一张表开始，所以第一件事就是仔细阅读它。
| Activity | Duration (days) | Immediate predecessors |
| A | 3 | - |
| B | 4 | - |
| C | 2 | A |
| D | 6 | A |
| E | 3 | B, C |
| F | 2 | D, E |
Activities with no immediate predecessor can start at time zero. Activities that share the same predecessors can run in parallel, which is exactly what makes CPA useful - if it were not for parallel working, the minimum completion time would simply be the sum of all durations.
没有前序活动的任务可以在时间零点开始。拥有相同前序活动的任务可以并行进行，而这正是 CPA 的价值所在——如果没有并行作业，最短完工时间就只是所有工期之和。
3. Drawing Activity Networks | 绘制活动网络图
Edexcel uses the activity-on-arc convention. Each activity is drawn as an arrow, the length of which carries no meaning, and each event (the start or finish of one or more activities) is drawn as a numbered node.
Edexcel 采用箭线式（activity-on-arc）表示法。每项活动用一条箭头表示，箭头长度没有任何含义；每个事件（一项或多项活动的开始或结束）用一个带编号的节点表示。
The rules below must be followed exactly, because examiners award marks specifically for network structure.
以下规则必须严格遵守，因为考官会针对网络结构单独给分。
- Every network has a single start node and a single finish node.
每个网络只有一个起始节点和一个终止节点。 - Arrows must always point from a lower-numbered node to a higher-numbered node.
箭头必须始终从小号节点指向大号节点。 - Two nodes may be joined by at most one arrow - if two activities share both a start and a finish event, one of them needs a dummy.
两个节点之间最多只能有一条箭头——如果两项活动共享同一个起始事件和同一个终止事件，其中一项就需要借助虚活动。 - Numbers are allocated so that nodes are numbered in a logical sequence from left to right.
节点编号按从左到右的逻辑顺序依次分配。
Inside each node you record the event number together with two times: the earliest time the event can occur and the latest time it can occur without delaying the project. These two values are the whole point of the forward and backward passes that follow.
在每个节点内部，你记录事件编号以及两个时间：该事件最早可以发生的时间，以及在不延误项目的前提下最迟可以发生的时间。这两个值正是接下来正向遍历与反向遍历的全部意义所在。
4. Dummy Activities | 虚活动
A dummy activity is an arrow with a duration of zero. It consumes no time and no resources; it exists purely to express a logical dependency that cannot otherwise be drawn correctly. Dummies are always drawn as dashed arrows and are never given a letter in the precedence table.
虚活动是工期为零的箭头。它不消耗时间，也不消耗资源；它存在的唯一目的是表达一个无法用其他方式正确绘制的逻辑依赖关系。虚活动始终画成虚线箭头，并且在前序关系表中绝不会被赋予字母。
Consider this classic situation: A and B have no predecessors, C depends on A only, and D depends on both A and B. Activity A finishes at node 2 and activity B finishes at node 3. If D started from node 2, it would appear to depend only on A, which is wrong. The fix is to draw a dummy from node 2 to node 3, so that node 3 now represents both A and B being complete, and D can start from node 3.
考虑这个经典情形：A 和 B 没有前序活动，C 只依赖 A，而 D 同时依赖 A 和 B。活动 A 在节点 2 结束，活动 B 在节点 3 结束。如果 D 从节点 2 出发，看上去就只依赖 A，这是错误的。解决办法是从节点 2 向节点 3 画一条虚活动，这样节点 3 就同时代表 A 和 B 均已完成，D 也就可以从节点 3 开始。
Remember that a dummy arrow still has a direction, and that direction matters. Reversing it changes the logic of the network and will lose marks. Also remember that a dummy occupies no time, so it contributes zero to any forward or backward pass calculation.
记住，虚活动箭头仍然有方向，而且方向很重要。方向画反会改变网络的逻辑并导致失分。同样要记住，虚活动不占用时间，因此在任何正向或反向遍历计算中它贡献的都是零。
5. Forward Pass: Earliest Start and Finish Times | 正向遍历：最早开始与最早完成时间
The forward pass works from left to right through the network and produces the earliest time at which each event can occur. If an event has more than one incoming activity, the earliest time is the maximum of the candidate values, because the event cannot happen until every preceding activity has finished.
正向遍历从网络左端向右端推进，求出每个事件最早可以发生的时间。如果某个事件有多条进入的活动，最早时间取所有候选值中的最大值，因为只有当所有前置活动都完成后，该事件才能发生。
EST(event) = max over all incoming activities of (EST(start of activity) + duration)
tion, so EFT = EST + duration. The earliest time recorded at the finish node is the minimum completion time of the whole project.
对每项活动而言，最早开始时间（EST）就是其起始事件的最早时间，最早完成时间（EFT）等于该值加上工期，即 EFT = EST + 工期。记录在终止节点上的最早时间就是整个项目的最短完工时间。
Using the table from Section 2, the finish node takes the value 11 days: A starts at 0 and finishes at 3, C then runs from 3 to 5, E can only start once both B (finishing at 4) and C (finishing at 5) are done, so E starts at 5, and F needs D (finishing at 9) and E (finishing at 8), so F starts at 9 and finishes at 11.
使用第 2 节的表格，终止节点取值为 11 天：A 在 0 开始、3 完成，C 随后从 3 进行到 5，E 必须在 B（4 完成）和 C（5 完成）都完成之后才能开始，所以 E 从 5 开始；F 需要 D（9 完成）和 E（8 完成），因此 F 从 9 开始，到 11 结束。
6. Backward Pass: Latest Start and Finish Times | 反向遍历：最迟开始与最迟完成时间
The backward pass works from right to left. You begin by writing the minimum completion time into the finish node as its latest time, because there is no slack at the end of the project. At every other node, the latest time is the minimum of the candidate values, since the event must happen early enough for every activity that leaves it.
反向遍历从右向左进行。首先把最短完工时间写入终止节点，作为它的最迟时间，因为项目终点没有任何松动余地。对于其余每个节点，最迟时间取所有候选值中的最小值，因为该事件必须足够早地发生，才能让从它出发的每一项活动都来得及完成。
LFT(event) = min over all outgoing activities of (LFT(end of activity) - duration)
n, so LST = LFT - duration. A useful check is that the start node must end up with a latest time of zero; if it does not, you have made an arithmetic error somewhere.
对每项活动而言，最迟完成时间（LFT）等于其结束事件的最迟时间，最迟开始时间（LST）等于该值减去工期，即 LST = LFT - 工期。一个实用的检查方法是：起始节点的最迟时间必须等于零；如果不是，说明你在某处算错了。
Applying this to the example: the finish node has value 11, so E has LFT 9 and LST 6, while D has LFT 9 and LST 3. Node 2 takes the minimum of 4 (from C) and 3 (from D), giving 3, and node 1 then takes the minimum of 0 and 2, correctly giving 0.
把这个方法用到例题中：终止节点取值为 11，因此 E 的 LFT 为 9、LST 为 6，而 D 的 LFT 为 9、LST 为 3。节点 2 取 4（来自 C）与 3（来自 D）中的最小值，得到 3；节点 1 再取 0 与 2 中的最小值，正确得到
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