Critical Path Analysis for IGCSE CCEA Maths | IGCSE CCEA 数学：关键路径分析考点精讲

📚 Critical Path Analysis for IGCSE CCEA Maths | IGCSE CCEA 数学：关键路径分析考点精讲

Critical path analysis is a powerful decision-making tool used to plan and manage complex projects. In the CCEA IGCSE Mathematics syllabus, you are expected to construct activity networks, perform forward and backward passes, calculate floats, and identify the critical path. This article breaks down every step of the process, providing clear explanations and worked examples that align with exam-style questions.

关键路径分析是一种强大的决策工具，用于规划和管理复杂的项目。在 CCEA IGCSE 数学考纲中，你需要能够构建活动网络图、进行前向遍历和后向遍历、计算浮动时间并识别关键路径。本文逐一拆解该过程的每一步，提供清晰的解释和与考题风格一致的详细示例，帮助你掌握这一重要专题。

1. What Is Critical Path Analysis? | 什么是关键路径分析？

Critical path analysis (CPA) is a method of scheduling a set of project activities. It shows which tasks can be delayed without affecting the overall project completion time, and which tasks are critical – meaning any delay in them will delay the entire project.

关键路径分析是一种安排一系列项目活动的方法。它能够显示哪些任务可以延迟而不影响整个项目的完成时间，而哪些任务是关键的——这意味着这些任务的任何延迟都会导致整个项目的延误。

The technique is often applied in construction, software development, event planning, and logistics. It helps project managers allocate resources efficiently, avoid bottlenecks, and meet deadlines.

该技术常被应用于建筑、软件开发、活动策划和物流等领域。它帮助项目经理高效分配资源、避免瓶颈并按期完成任务。

2. Activity-On-Node Representation | 节点活动表示法

In CCEA IGCSE, we use the activity-on-node (AON) convention. Each node represents an activity, and the node is divided into sections displaying the activity’s duration, earliest start time, latest start time, and earliest finish time.

在 CCEA IGCSE 考试中，我们使用节点活动表示法。每个节点代表一个活动，节点被分割成几个部分，分别显示活动的持续时间、最早开始时间、最晚开始时间和最早完成时间。

A typical node layout looks like this:

一个典型的节点布局如下：

┌─────────────┐
│EST Duration│
│ Activity│
│LST Float │
└─────────────┘

The arrows (or directed edges) between nodes indicate dependencies – an activity cannot start until all its immediate predecessors are finished.

节点之间的箭头（或有向边）表示依赖关系——一个活动必须在其所有直接前驱完成后才能开始。

Make sure you are comfortable drawing and labelling these nodes accurately; small mistakes in layout can lead to lost marks in the exam.

确保你能准确画出并标注这些节点；布局中的小错误可能会导致考试失分。

3. Drawing an Activity Network from a Precedence Table | 根据前驱关系表绘制活动网络图

Exam questions will typically provide a table listing activities, their durations, and their immediate predecessors. Your first task is to construct the network diagram correctly.

考试题目通常会提供一个表格，列出活动、持续时间和直接前驱。你的首要任务是正确构建网络图。

Follow these steps: start with activities that have no predecessors. Draw them as separate nodes placed side by side. Then add successor activities, linking them with arrows. Always work from left to right, ensuring the dependencies are respected. A common approach is to sketch a rough version, check all dependencies, and then draw a neat final version.

遵循以下步骤：从没有前驱的活动开始，将它们作为独立的节点并排绘制。然后添加后续活动，用箭头连接。始终从左到右进行，确保所有依赖关系都得到满足。一种常见的做法是先画草图，检查所有依赖关系，再画出整洁的最终版本。

Do not forget to number the nodes or label them clearly. In CCEA questions, nodes may be represented by letters, and you are usually asked to complete a partially drawn network or start from scratch.

别忘了为节点编号或清晰标注。在 CCEA 问题中，节点可能用字母表示，通常要求你补全部分绘制的网络图或从头开始绘制。

4. Forward Pass: Earliest Start and Earliest Finish Times | 前向遍历：最早开始时间和最早完成时间

The forward pass calculates the earliest possible time each activity can start and finish, assuming the project begins at time 0.

前向遍历计算每个活动可能的最早开始和最早完成时间，假设项目从时间 0 开始。

For the initial activities, the earliest start time (EST) is 0. The earliest finish time (EFT) is EST + duration. For any subsequent activity, its EST is the maximum of the EFTs of all its immediate predecessors.

对于初始活动，最早开始时间为 0。最早完成时间为 EST + 持续时间。对于任何后续活动，其 EST 等于其所有直接前驱的 EFT 的最大值。

Mathematically, if an activity has predecessors P₁, P₂, …, Pₙ, then:
EST = max{EFT(P₁), EFT(P₂), …, EFT(Pₙ)}
EFT = EST + duration

数学表达为：若某活动有前驱 P₁, P₂, …, Pₙ，则：
EST = max{EFT(P₁), EFT(P₂), …, EFT(Pₙ)}
EFT = EST + 持续时间

Always work from left to right across the network, filling in each node’s top-left (EST) and top-right (EFT) sections as you go.

始终从左到右遍历网络，依次填入每个节点左上角（EST）和右上角（EFT）的数据。

5. Backward Pass: Latest Start and Latest Finish Times | 后向遍历：最晚开始时间和最晚完成时间

Once the minimum project duration is known from the forward pass, the backward pass determines the latest time each activity can start and finish without delaying the whole project.

从正向遍历得出最短项目工期后，反向遍历确定每个活动在不延误整个项目的情况下可以开始和完成的最晚时间。

Start from the final activity (or the end node). Its latest finish time (LFT) is set equal to the project’s minimum completion time (the maximum EFT from the forward pass). Its latest start time (LST) = LFT – duration.

从最终活动（或结束节点）开始。其最晚完成时间设等于项目最短工期（即正向遍历中的最大 EFT）。其最晚开始时间 LST = LFT – 持续时间。

For an earlier activity, its LFT is the minimum of the LSTs of all activities that immediately follow it. Then LST = LFT – duration.

对于更早的活动，其 LFT 为其所有直接后继活动的 LST 中的最小值。然后 LST = LFT – 持续时间。

Work from right to left, filling the bottom-left (LST) and bottom-right (LFT) sections of each node. Care with the minimum rule is essential; using the maximum here is a common mistake.

从右向左操作，填入每个节点左下角（LST）和右下角（LFT）的数据。务必小心最小值规则；这里错误地使用最大值是一个常见错误。

6. Calculating Total Float | 计算总浮动时间

Total float is the amount of time an activity can be delayed without affecting the overall project duration. It is calculated using:

总浮动时间是指一个活动可以延迟的时间量，而不会影响整个项目的工期。其计算公式为：

Total Float = LST – EST = LFT – EFT

Both formulas give the same result. If the float is zero, the activity is critical; if it is positive, there is some slack.

两个公式给出相同的结果。若浮动时间为零，则该活动是关键活动；若为正数，则表示存在一定的松弛时间。

When filling in the node, the float is often written in the bottom-right inner section or placed below the activity label, depending on the style used in the exam paper. CCEA questions may ask you to state the float explicitly or find all critical activities.

在填充节点时，浮动时间通常写在右下角内部区域或活动标签的下方，具体取决于试卷使用的风格。CCEA 问题可能会要求你明确写出浮动时间，或找出所有关键活动。

7. Identifying the Critical Path | 识别关键路径

The critical path is the longest path through the network in terms of duration. It consists of activities that have zero total float. Any delay on a critical activity will cause a delay in the whole project.

关键路径是网络图中持续时间最长的一条路径。它由总浮动时间为零的活动组成。任何关键活动的延迟都将导致整个项目延误。

To identify it, trace all activities with total float = 0 from the start to the end. Usually you state the critical path as a sequence of activities, e.g. A → C → F → H. There may be more than one critical path. If there are multiple critical paths, all must be given for full marks.

要识别它，从起点到终点追踪所有总浮动时间为零的活动。通常你将关键路径表述为活动序列，例如 A → C → F → H。可能存在多条关键路径。若存在多条，则必须全部列出才能得满分。

In exams, always explicitly state the path and its total duration. The total duration of the critical path equals the minimum project completion time.

在考试中，务必明确写出路径及其总工期。关键路径的总工期等于项目的最短完成时间。

8. Interpreting a Cascade Chart (Gantt Chart) | 解释阶梯图（甘特图）

CCEA may also test your ability to read or draw a cascade chart (bar chart) based on the activity network. Each activity is represented by a horizontal bar, with its start and finish times plotted on a timeline.

CCEA 可能还会考查你阅读或绘制基于活动网络图的阶梯图（条形图）的能力。每个活动由一条水平长条表示，其开始和结束时间绘制在时间轴上。

Activities are typically scheduled to start at their earliest start time, and the float is shown as a shaded extension or a separate dashed bar. The cascade chart helps visualise where slack exists and when resources might be over-allocated.

活动通常安排在其最早开始时间启动，浮动时间用阴影延伸或单独的虚线条形表示。阶梯图有助于直观地看出松弛时间存在的位置以及资源可能在何时被过度使用。

When drawing, label axes clearly: ‘Time’ on the horizontal axis and ‘Activities’ on the vertical axis. Use a ruler for neatness; messy diagrams may lose marks.

绘制时，清楚标注坐标轴：横轴为“时间”，纵轴为“活动”。使用尺子保持整洁；凌乱的图表可能导致失分。

9. Common CCEA Exam Pitfalls and How to Avoid Them | 常见 CCEA 考试陷阱及如何避免

Many students lose marks not because they do not understand the method, but due to small errors. Here are some pitfalls to watch out for:

许多学生失分并非因为不理解方法，而是由于小的错误。以下是需要注意的一些陷阱：

Skipping dependencies: Always double-check that every immediate predecessor is linked correctly. Drawing a rough draft first can prevent this.
Forgetting to start: 总是再次核对每个直接前驱是否正确连接。先画草图可以避免这一点。
Using max instead of min in backward pass: The LFT of an activity is the minimum LST of its successors, not the maximum. Think of it as pulling the activity as late as possible without delaying the earliest starting follower.
后向遍历中用最大值代替最小值： 活动的 LFT 是其所有后继 LST 的最小值，而不是最大值。可以理解为在不延迟最早开始的后续活动的前提下，尽可能地将此活动推迟。
Incorrect node layout: Make sure you are drawing nodes in the format expected by CCEA. If the exam provides a blank node template, copy it exactly.
节点布局错误： 确保你按照 CCEA 期望的格式绘制节点。如果试卷提供了空白的节点模板，请精确复制。
Mistaking total float for free float: CCEA normally asks for total float. Free float, which is the delay possible without affecting any successor’s EST, is a different concept and not always required. Confirm what the question is asking.
混淆总浮动时间与自由浮动时间： CCEA 通常要求总浮动时间。自由浮动时间是指在不影响任何后继活动最早开始时间的前提下可延迟的时间，是另一个概念，不常考。明确题目要求的是什么。

Carefully reading the question and showing your working in a structured way can help you avoid these errors.

仔细阅读题目并以结构化的方式展示解答过程，有助于避免这些错误。

10. Worked Example: From Precedence Table to Critical Path | 实例解析：从前驱关系表到关键路径

Let’s apply the steps to a typical exam-style problem. Consider a small project with the following activities:

让我们将步骤应用于一道典型的考试题。考虑一个具有以下活动的小型项目：

Activity	Duration (hours)	Predecessors
A	4	—
B	5	A
C	3	A
D	6	B
E	2	B, C
F	3	D, E

Draw the network, perform forward and backward passes, find the total project duration, identify the critical path(s), and calculate the float for non-critical activities.

绘制网络图，执行前向与后向遍历，计算总项目工期，确定关键路径，并计算非关键活动的浮动时间。

Solution:

解答：

Network order: A (start) → B, C. Then B → D and B, C → E. Finally D, E → F. Forward pass gives: A: EST=0, EFT=4. B: EST=4, EFT=9. C: EST=4, EFT=7. D: EST=9, EFT=15. E: EST=max(9,7)=9, EFT=11. F: EST=max(15,11)=15, EFT=18. Minimum project duration = 18 hours.

网络顺序：A（开始）→ B, C。然后 B → D 且 B, C → E。最后 D, E → F。正向遍历得出：A: EST=0, EFT=4。B: EST=4, EFT=9。C: EST=4, EFT=7。D: EST=9, EFT=15。E: EST=max(9,7)=9, EFT=11。F: EST=max(15,11)=15, EFT=18。最短项目工期 = 18 小时。

Backward pass: F: LFT=18, LST=15. D: LFT=15, LST=9. E: LFT=15, LST=13. B: LFT=min(LST D,LST E)=min(9,13)=9, LST=4. C: LFT=min(LST E)=13, LST=10. A: LFT=min(LST B,LST C)=min(4,10)=4, LST=0.

后向遍历：F: LFT=18, LST=15。D: LFT=15, LST=9。E: LFT=15, LST=13。B: LFT=min(LST D,LST E)=min(9,13)=9, LST=4。C: LFT=min(LST E)=13, LST=10。A: LFT=min(LST B,LST C)=min(4,10)=4, LST=0。

Floats: A:0; B:0; C: LFT-EFT=13-7=6 or LST-EST=10-4=6; D:0; E:13-11=2; F:0. Critical activities: A, B, D, F. Critical path: A → B → D → F with duration 18 hours. Alternatively, you can check path durations: A-B-D-F = 4+5+6+3=18; A-B-E-F = 4+5+2+3=14; A-C-E-F = 4+3+2+3=12. The longest is indeed A-B-D-F.

浮动时间：A:0；B:0；C: LFT-EFT=13-7=6 或 LST-EST=10-4=6；D:0；E:13-11=2；F:0。关键活动：A, B, D, F。关键路径：A → B → D → F，工期 18 小时。或者，你可以检查各路径长度：A-B-D-F=18；A-B-E-F=14；A-C-E-F=12。最长的确实是 A-B-D-F。

11. Quick Tips for Success in CCEA Exams | CCEA 考试高分速成技巧

Always label each node clearly with the activity letter, EST, EFT, LST, and LFT. Use the same format throughout the network.
始终清晰地在每个节点上标注活动字母、EST、EFT、LST 和 LFT。整个网络使用相同的格式。
When checking your work, verify that the float calculation (LST–EST) equals (LFT–EFT) for every activity. An inequality indicates an arithmetic error.
检查时，核实每个活动的浮动时间（LST–EST）等于（LFT–EFT）。不相等即表明存在计算错误。
If you have spare time, re-calculate the project duration by adding durations along the critical path to confirm it matches the terminal node’s EFT.
如有余裕，沿着关键路径将持续时间相加，核实其与终端节点 EFT 一致，以此重新计算项目工期。
Be careful with activities that share successors – the backward pass demands finding the smallest LST. Circle or highlight those numbers on your diagram to avoid oversight.
小心处理共享后继的活动——后向遍历要求找出最小的 LST。在图上圈出或突出显示这些数字以避免疏忽。
Remember that the critical path can change if durations are altered. Some questions may ask you to consider the effect of a delay in one activity on the whole project; refer to the float of that activity.
记住，如果持续时间改变，关键路径可能会转移。有些问题可能要求你考虑某项活动延误对整个项目的影响；此时应参考该活动的浮动时间。

12. Summary and Final Check | 总结与最后核查

Critical path analysis is a structured, logical topic that rewards careful step-by-step working. Once you master the forward pass (max of predecessors’ EFT), the backward pass (min of successors’ LST), and float computation, most exam questions become a matter of applying the same procedure accurately.

关键路径分析是一个结构化、逻辑性强的专题，稳步推进即可得分。一旦你掌握了前向遍历（取前驱 EFT 的最大值）、后向遍历（取后继 LST 的最小值）和浮动时间的计算，大多数考题都只是准确应用相同步骤的问题。

Practice drawing networks from various precedence tables, and time yourself to ensure you can complete a full question within the allocated minutes. With consistent practice, you will find that critical path analysis becomes one of the most straightforward and high-scoring topics on the CCEA IGCSE Mathematics paper.

多练习从前驱关系表绘制网络图，并计时以确保能在规定时间内完整作答。通过持续练习，你会发现关键路径分析成为 CCEA IGCSE 数学试卷中最直接且容易拿高分的专题之一。

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Critical Path Analysis for IGCSE CCEA Maths | IGCSE CCEA 数学：关键路径分析考点精讲