📚 Critical Path Analysis: AQA GCSE Maths Revision | 关键路径分析 AQA GCSE 数学考点精讲
Critical path analysis is a powerful decision‑making tool used to plan and manage complex projects. In AQA GCSE Mathematics, you will learn to model projects using activity networks, calculate earliest and latest start times, determine floats, and identify the critical path – the sequence of activities that directly impacts the overall project duration. Mastering this topic not only prepares you for exam questions but also builds valuable skills for real‑world scheduling problems.
关键路径分析是一种用于规划和管理复杂项目的强大决策工具。在 AQA GCSE 数学中，你将学习如何使用活动网络对项目建模，计算最早开始时间和最迟开始时间，确定浮动时间，并找出关键路径——即直接影响整个项目工期的活动序列。掌握这一考点不仅能帮助你应对考试题目，还能培养解决现实排程问题的宝贵技能。
1. What Is Critical Path Analysis? | 什么是关键路径分析？
Critical path analysis (CPA) is a method for identifying the minimum time needed to complete a project and pinpointing which tasks must not be delayed. It was developed in the 1950s for large engineering projects and is now widely used in construction, software development and event planning. For GCSE, we break a project into individual activities, each with a duration, and use a network to show the order in which they must be performed.
关键路径分析（CPA）是一种用来确定完成项目所需的最短时间，并找出哪些任务不能有任何延误的方法。它于 20 世纪 50 年代为大型工程项目而开发，如今广泛应用于建筑、软件开发和活动策划等领域。在 GCSE 阶段，我们将一个项目分解为若干独立的作业（活动），每个作业都有一个持续时间，并用网络图表示它们的执行顺序。
2. Key Terminology in CPA | 关键路径分析的核心术语
Before drawing networks, it is essential to understand the key terms used in CPA. An activity is a task that takes time and resources. Its duration indicates how long it takes, usually given in hours, days or weeks. Predecessors (or preceding activities) are those that must be finished before the activity can start. The terms earliest start time (EST), latest start time (LST) and total float will be defined formally when we perform the forward and backward passes.
在绘制网络图之前，理解 CPA 的核心术语非常重要。作业（活动）是需要耗费时间和资源的任务。其持续时间表示完成任务需要多久，通常以小时、天或周为单位。前驱作业（前置活动）是指在某一作业开始之前必须完成的那些作业。在进行前向扫描和后向扫描时，我们会正式定义最早开始时间（EST）、最迟开始时间（LST）和总浮动时间这些术语。
3. Activity‑on‑Node Diagrams | 节点式活动网络图
In AQA GCSE Mathematics, we typically use activity‑on‑node (AON) diagrams. Each node represents one activity and is drawn as a rectangle divided into sections. The most common layout shows the activity identifier (e.g. A, B, C) and its duration inside the node, while the EST and LST are recorded in the top corners of the node box. Arrows between nodes indicate dependencies; an arrow from Activity P to Activity Q means Q can only start after P has finished.
在 AQA GCSE 数学中，我们通常使用节点式活动网络图（AON）。每个节点代表一个作业，绘制成一个分成若干部分的矩形。最常见的布局是在节点内部标明作业代号（如 A、B、C）及其持续时间，而将 EST 和 LST 记录在节点框的顶角处。节点之间的箭头表示依赖关系；从作业 P 指向作业 Q 的箭头意味着 Q 只能在 P 完成后才能开始。
4. Precedence Tables | 前驱关系表
Exam questions often provide a precedence table listing activities, their durations and the activities that must be completed immediately before them. Your first task is to interpret this table and construct the network correctly. A typical table might look like the following:
考试题目通常会给出一个前驱关系表，列出各项作业、它们的持续时间以及必须在其之前立即完成的作业。你的首要任务是正确解读该表格并构建网络图。一个典型的表格可能如下所示：
| Activity | Duration | Preceding activities |
|---|---|---|
| A | 4 | – |
| B | 5 | A |
| C | 3 | A |
| D | 6 | B, C |
Notice that Activity A has no predecessors and can begin at time 0, while Activity D requires both B and C to be finished before it can start.
请注意，作业 A 没有前驱作业，可以从时刻 0 开始，而作业 D 要求 B 和 C 都完成后才能开始。
5. Forward Pass: Calculating Earliest Start Times (EST) | 前向扫描：计算最早开始时间
The forward pass moves from the start of the network to the end. We set the EST of the first activity to 0. For every other activity, the EST is calculated as the maximum of (EST of predecessor + predecessor’s duration) over all its immediate predecessors. The logic is simple: an activity cannot begin until every predecessor has finished, so its earliest possible start is the latest finish time of its predecessors.
前向扫描从网络的起始端走向终止端。我们将第一个作业的 EST 设为 0。对于其他每个作业，EST 的计算方式为：遍历该作业的所有直接前驱，取最大值（前驱的 EST + 前驱的持续时间）。所依据的逻辑很简单：一个作业必须等所有前驱都完成后才能开始，因此它最早可能开始的时间就是其前驱完成时间的最大值。
ESTj = max { ESTi + di } for all immediate predecessors i of j
Using the table above, we would obtain ESTA = 0, ESTB = 0 + 4 = 4, ESTC = 0 + 4 = 4, and ESTD = max(4 + 5, 4 + 3) = max(9, 7) = 9.
使用上表，我们可以得到 ESTA = 0，ESTB = 0 + 4 = 4，ESTC = 0 + 4 = 4，ESTD = max(4 + 5, 4 + 3) = max(9, 7) = 9。
6. Backward Pass: Finding Latest Start Times (LST) | 后向扫描：求最迟开始时间
The backward pass works from the end of the network back to the start. We begin by setting the LST of the final activity equal to its EST (assuming we want the minimum project duration). For activities before the finish, the LST is the minimum of (LST of successor − duration of the current activity) over all immediate successors. This ensures that if an activity starts at its LST, it will finish just in time for the succeeding activities to start at their LSTs.
后向扫描从网络末端逆向回到起点。开始时，我们将终点作业的 LST 设为其 EST（假设我们希望得到最短项目工期）。对于终点之前的作业，LST 的计算方式为：遍历该作业的所有直接后继，取最小值（后继的 LST − 当前作业的持续时间）。这可以保证如果一个作业在其 LST 开始，它刚好能在后续作业的 LST 时刻完工。
LSTi = min { LSTj − di } for all immediate successors j of i
In our example, the project ends after D finishes, so the project duration is 9 + 6 = 15. We set LSTD = 9. Then LSTB = 9 − 5 = 4, LSTC = 9 − 3 = 6, and LSTA = min(4 − 4, 6 − 4) = min(0, 2) = 0.
在我们的例子中，项目在 D 完成时结束，因此项目工期为 9 + 6 = 15。我们设 LSTD = 9。然后 LSTB = 9 − 5 = 4，LSTC = 9 − 3 = 6，而 LSTA = min(4 − 4, 6 − 4) = min(0, 2) = 0。
7. Total Float and Its Meaning | 总浮动时间及其含义
The total float of an activity is the amount of time it can be delayed without affecting the overall project duration. It is calculated by the simple formula:
一个作业的总浮动时间是指在不影响整个项目工期的前提下，该作业可以延误的时间量。它由以下简单公式计算：
Total Float = LST − EST
An activity with zero total float is critical; any delay to it will directly delay the entire project. Activities with positive float have some scheduling flexibility. In our example, activities A, B and D have zero float (EST = LST), while activity C has a float of 6 − 4 = 2 time units. This means C could be delayed by up to 2 time units without pushing back the project finish.
总浮动时间为零的作业是关键作业；对它任何延误都将直接拖延整个项目。浮动时间为正的作业具有一定的排程弹性。在我们的例子中，作业 A、B 和 D 的浮动时间为零（EST = LST），而作业 C 的浮动时间为 6 − 4 = 2 个时间单位。这意味着 C 最多可以延误 2 个时间单位而不推迟项目完工。
8. Identifying the Critical Path | 识别关键路径
The critical path is the continuous chain of critical activities from the start to the end of the network. In the AON diagram, it is simply the path where every node has a total float of 0. Because all activities on the critical path are urgent, the length of this path determines the minimum project duration. In our worked scenario, the critical path is A → B → D, giving a total project length of 4 + 5 + 6 = 15.
关键路径是从网络起点到终点的一条由关键作业连接而成的连续链。在 AON 图中，它就是每个节点的总浮动时间均为 0 的那条路径。由于关键路径上的所有作业都不可延误，所以这条路径的长度决定了最短项目工期。在示例情景中，关键路径为 A → B → D，总工期为 4 + 5 + 6 = 15。
Sometimes there is more than one critical path. In that case, you must list all of them, and the total duration remains the same. AQA questions may ask you to state the critical path(s) and the minimum completion time.
有时会出现多条关键路径。此时你需要列出所有关键路径，而总工期保持不变。AQA 的题目可能会要求你写出关键路径和最短完成时间。
9. Cascade Charts (Gantt Charts) | 级联图（甘特图）
A cascade chart, also known as a Gantt chart, is a horizontal bar chart that shows each activity’s earliest start time, duration and float. Activities are listed down the vertical axis, while the horizontal axis represents time. A bar is drawn from the EST for the length of the duration. The float is often shown as a dotted extension of the bar, indicating how far the activity can slide without delaying the project. Exam questions may ask you to draw a cascade chart from the information in your network.
级联图（也称甘特图）是一种水平条形图，展示每个作业的最早开始时间、持续时间和浮动时间。纵轴列出各项作业，横轴表示时间。从 EST 位置开始绘制一条长度等于持续时间的实心横条。浮动时间通常用虚线延长部分表示，表明该作业可以滑移的范围而不至于拖延项目。考试题目可能会要求你根据网络图中的信息绘制级联图。
When drawing a cascade chart, ensure you use a sensible scale, label activities clearly, and distinguish between the fixed bar (solid) and the float (dashed). The critical activities will appear with no dashed extension because their float is zero.
绘制级联图时，务必使用合适的比例，清晰标注各项作业，并用实线表示固定作业条、虚线表示浮动部分。关键作业将没有虚线延长部分，因为它们的浮动时间为零。
10. Dummy Activities in Arrow Diagrams | 箭线图中的虚作业
Some textbooks introduce CPA using activity‑on‑arrow (AOA) networks, where arrows represent activities and nodes represent events. In that format, dummy activities – shown as dashed arrows with zero duration – are used to preserve logical dependencies without implying extra work. However, the AQA GCSE specification focuses on activity‑on‑node diagrams, so you will rarely encounter dummies in the exam. The key is to interpret the precedence table correctly and build a clean AON network.
一些教材使用箭线式活动网络图（AOA）来介绍 CPA，其中箭头表示作业，节点表示事件。在这种格式中，会使用虚作业——以零持续时间的虚线箭线表示——来保持正确的依赖逻辑，而不暗示额外的工作量。不过，AQA GCSE 的考试纲要侧重于节点式活动网络图，因此你在考试中很少会遇到虚作业。关键是正确解读前驱关系表，并绘制出清晰的 AON 网络。
11. Common Mistakes and How to Avoid Them | 常见错误与避免方法
Mistake 1: Forgetting to take the maximum during the forward pass. An activity cannot start until all predecessors finish, so you must compare all incoming paths and pick the largest value.
Mistake 2: Using addition instead of subtraction in the backward pass. Always calculate LST of a predecessor as LSTsuccessor minus the predecessor’s duration.
Mistake 3: Confusing EST with duration. An activity with a long duration does not automatically have a large EST; its EST is governed by predecessors.
Mistake 4: Incorrectly reading the precedence table, especially when multiple dependencies exist. Draw a rough sketch first.
错误 1：在前向扫描时忘记取最大值。一个作业必须等所有前驱完成后才能开始，因此你必须比较所有传入路径并选取最大值。
错误 2：在后向扫描时误用加法而非减法。务必用后继的 LST 减去前驱的持续时间来计算前驱的 LST。
错误 3：混淆 EST 与持续时间。持续时间长的作业，其 EST 并不一定大；EST 由前驱决定。
错误 4：错误解读前驱关系表，尤其是存在多重依赖时。先画一个粗略的草图会很有帮助。
12. Worked Example – Full Analysis | 例题详解 – 完整分析
Question: A project consists of activities P, Q, R, S, T with the following data. Draw the activity‑on‑node network, calculate EST, LST and total float for each activity, identify the critical path and state the minimum project duration.
题目：某项目包含作业 P、Q、R、S、T，数据如下。请绘制节点式活动网络图，计算每项作业的 EST、LST 和总浮动时间，找出关键路径，并写出最短项目工期。
| Activity | Duration | Predecessors |
|---|---|---|
| P | 5 | – |
| Q | 7 | P |
| R | 4 | P |
| S | 6 | Q, R |
| T | 3 | S |
Solution: Start with P (no predecessors). ESTP = 0. Q and R both depend on P, so ESTQ = 0 + 5 = 5, ESTR = 0 + 5 = 5. S requires both Q and R, therefore ESTS = max(5 + 7, 5 + 4) = max(12, 9) = 12. T depends on S, so ESTT = 12 + 6 = 18. The project finishes at 18 + 3 = 21. So the minimum project duration is 21.
解答：从 P（无前驱）开始。ESTP = 0。Q 和 R 都依赖于 P，因此 ESTQ = 0 + 5 = 5，ESTR = 0 + 5 = 5。S 同时需要 Q 和 R，故 ESTS = max(5 + 7, 5 + 4) = max(12, 9) = 12。T 依赖于 S，因此 ESTT = 12 + 6 = 18。项目在 18 + 3 = 21 时结束。故最短项目工期为 21。
Now backward pass: Begin with T. LSTT = ESTT = 18. LSTS = 18 − 6 = 12. LSTQ = 12 − 7 = 5, LSTR = 12 − 4 = 8. LSTP = min(5 − 5, 8 − 5) = min(0, 3) = 0.
现在进行后向扫描：从 T 开始。LSTT = ESTT = 18。LSTS = 18 − 6 = 12。LSTQ = 12 − 7 = 5，LSTR = 12 − 4 = 8。LSTP = min(5 − 5, 8 − 5) = min(0, 3) = 0。
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