📚 A-Level CCEA Mathematics: Critical Path Analysis – Key Exam Points | A-Level CCEA 数学：关键路径分析 考点精讲
Critical Path Analysis (CPA) is a fundamental topic in CCEA A-Level Decision Mathematics. It helps project managers schedule activities so that a project is completed in the shortest possible time, identifying which tasks must not be delayed. This guide covers the key exam techniques for constructing activity-on-node networks, performing forward and backward passes, calculating floats, drawing cascade (Gantt) charts, and applying resource levelling.
关键路径分析是 CCEA A-Level 决策数学的核心考点。它帮助项目经理合理安排活动，使项目在最短时间内完成，并识别出绝对不能延误的任务。本文梳理了构建节点活动网络、进行前向与后向传递、计算浮动时间、绘制级联图（甘特图）以及资源平衡等关键考试技巧。
1. Introduction to Critical Path Analysis | 关键路径分析简介
Critical Path Analysis models a project as a set of activities that have durations and precedence constraints. Each activity must be completed before its successors can begin. The aim is to find the minimum project completion time and the activities that govern it.
关键路径分析将一个项目建模为一组带有持续时间和先后约束的活动。每个活动必须在其后续活动开始前完成。目标是找到最短的项目完成时间以及支配该时间的活动。
The critical path is the longest path through the network in terms of total duration. Any delay on this path directly delays the whole project. Activities not on the critical path have some flexibility, known as float.
关键路径是网络图中总持续时间最长的一条路径。该路径上的任何延误都会直接推迟整个项目。不在关键路径上的活动拥有一定的灵活性，称为浮动时间。
2. Activity Networks – Activity on Node | 活动网络——节点活动图
CCEA uses activity-on-node networks. Each activity is represented by a box (node) usually divided into cells: top cell for the activity label, lower-left for EST, lower-right for LST. The duration is written next to the node or inside a separate cell. Arrows show immediate predecessors.
CCEA 考试采用节点活动图。每个活动用一个方框（节点）表示，通常划分为几个单元格：上方写活动代号，左下为最早开始时间（EST），右下为最晚开始时间（LST）。持续时间可写在节点旁或单独的单元格内。箭头表示紧前活动。
A precedence table is often given. From it we draw the network, ensuring no dangling activities and that all dependencies are respected.
题目常会给出一个依赖关系表。我们据此绘制网络图，确保没有悬空的活动，并且所有依赖关系都被满足。
| Activity | Predecessors | Duration (days) |
|---|---|---|
| A | – | 4 |
| B | A | 3 |
| C | A | 5 |
| D | B, C | 2 |
3. Forward Pass – Finding Earliest Start Times | 前向传递——求最早开始时间
We perform a forward pass to calculate the earliest start time (EST) for each activity. The EST of the start node (the first activity) is 0. For any other activity, its EST is the maximum of the earliest finish times of all its immediate predecessors.
通过前向传递计算每个活动的最早开始时间（EST）。起始活动（第一个节点）的 EST 为 0。对于其他任何活动，其 EST 等于其所有紧前活动的最早完成时间的最大值。
The earliest finish time (EFT) is EST + duration. We record EST in the lower-left cell of each node. The project completion time is the maximum EFT among all final activities.
最早完成时间（EFT）= EST + 持续时间。我们将 EST 填入每个节点左下角单元格。项目的完成时间就是所有最终活动中最大的 EFT。
EST(current) = max{ EFT(all predecessors) }
For the sample table: EST(A)=0, EFT(A)=4. Then EST(B)=4, EST(C)=4. Then D depends on B and C, so EST(D)=max{4+3, 4+5}=max{7,9}=9.
对于上表示例：EST(A)=0，EFT(A)=4。于是 EST(B)=4，EST(C)=4。而 D 依赖 B 和 C，因此 EST(D)=max{4+3, 4+5}=max{7,9}=9。
4. Backward Pass – Latest Start & Finish Times | 后向传递——求最晚开始与完成时间
Next, we conduct a backward pass to find the latest start time (LST) and latest finish time (LFT) for each activity without delaying the project. For the final activity, its LFT equals the project completion time.
紧接着进行后向传递，求出每个活动在不延误项目前提下的最晚开始时间（LST）和最晚完成时间（LFT）。对于最后的活动，其 LFT 等于项目完成时间。
The LST of an activity is its LFT minus its duration. For a non-final activity, its LFT is the minimum LST of all its immediate successors. We then record LST in the lower-right cell.
某活动的 LST = LFT − 持续时间。对于非最终活动，其 LFT 等于其所有紧后活动中最小的 LST。最后将 LST 填入右下角单元格。
LFT(current) = min{ LST(all successors) }
In our example, project completion = 11. So LFT(D)=11, LST(D)=11−2=9. D has no successor, so LFT(D) set to 11. Then LFT(B)=LFT(C)=LST(D)=9. Therefore LST(B)=9−3=6, LST(C)=9−5=4. Finally LFT(A)=min{LST(B), LST(C)}=min{6,4}=4, so LST(A)=4−4=0.
在我们的例子中，项目完成时间为 11。因此 LFT(D)=11，LST(D)=11−2=9。D 无后续活动，所以 LFT(D) 设为 11。然后 LFT(B)=LFT(C)=LST(D)=9。于是 LST(B)=9−3=6，LST(C)=9−5=4。最后 LFT(A)=min{LST(B), LST(C)}=min{6,4}=4，所以 LST(A)=4−4=0。
5. Total Float and Free Float | 总浮动时间与自由浮动时间
Total float is the amount of time an activity can be delayed without affecting the overall project completion. It is calculated as:
总浮动时间指一个活动可以延误但不会影响整个项目完成时间的时长。计算公式为：
Total Float = LST − EST
Alternatively, Total Float = LFT − EFT. Activities with total float = 0 are critical. Free float is the delay allowed without affecting the EST of any successor; it is often not required in CCEA but useful for resource scheduling.
也可写作 总浮动 = LFT − EFT。总浮动为 0 的活动即为关键活动。自由浮动是指在不影响任何后续活动 EST 前提下可延误的时长；CCEA 考试不一定要求计算，但它对资源调度有帮助。
For our network: A:0, B:2, C:0, D:0. Thus B has 2 days of total float.
就我们的网络而言：A 0，B 2，C 0，D 0。所以活动 B 拥有 2 天的总浮动。
6. Identifying the Critical Path | 识别关键路径
The critical path consists of all activities with zero total float. In our example, the critical path is A-C-D with a duration of 4+5+2 = 11 days. Any delay on A, C, or D will delay the entire project.
关键路径由所有总浮动为 0 的活动构成。本例中，关键路径为 A-C-D，总时长 4+5+2 = 11 天。A、C、D 中任何一个延误都会推迟整个项目。
In exam, you must state the critical path(s) clearly, show the critical activities and the minimum project duration. Sometimes more than one critical path exists.
在考试中，必须清晰地写出关键路径，列出关键活动以及最短项目工期。有时可能存在不止一条关键路径。
7. Cascade (Gantt) Charts | 级联图（甘特图）
A cascade chart, also called a Gantt chart, visualises the schedule. Activities are drawn as horizontal bars, with length proportional to duration. The bars are positioned according to their earliest start times on the time axis. Critical activities are often shown in a different colour or filled.
级联图（也称甘特图）用于可视化进度。活动用水平条形表示，长度与持续时间成比例。条形按照最早开始时间放置在时间轴上。关键活动通常用不同颜色或填充表示。
The chart also displays float as a shaded or empty extension after the bar, showing the latest possible finish. This helps in resource smoothing.
级联图还在条形末端以阴影或空白延伸的方式显示浮动时间，表示最晚可能完成时间。这有助于资源平滑。
You are typically asked to draw a cascade chart directly from the EST/LFT values and the activity durations. Make sure axes are labelled and the scale is accurate.
考试中通常要求根据 EST/LFT 值和活动持续时间直接画出级联图。务必标注坐标轴，保持比例准确。
8. Resource Histograms and Resource Levelling | 资源直方图与资源平衡
When resources (e.g. workers) are limited, we construct a resource histogram to show the number of resources required per time unit when each activity starts at its EST. The histogram often shows peaks that exceed the available resources.
当资源（如工人）有限时，我们绘制资源直方图来展示每个活动按最早开始时间启动时，每单位时间所需资源的数量。直方图常出现超过可用资源的高峰。
Resource levelling aims to reduce the peak resource usage by delaying non-critical activities within their float. The cascade chart is very useful here: you slide non-critical bars as far as their float allows to minimise the maximum resource demand.
资源平衡旨在利用非关键活动的浮动时间，将其推迟以降低资源使用高峰。级联图在此非常有用：在浮动允许范围内滑动非关键条形，将最大资源需求最小化。
In exam questions, you may be asked to schedule activities to meet a given resource limit, often following a priority rule (e.g. shortest duration first). Show your working clearly and state the new completion time if it changes.
考题中可能会要求你在给定资源限制下调度活动，通常遵循某个优先规则（例如最短工期优先）。要清晰展示调度过程，若完成时间发生变化也需说明。
9. Scheduling Using a Priority List | 利用优先列表进行调度
When multiple activities are available, a priority list determines the order. A common list is by critical path or by ascending float. CCEA may provide a predetermined list and ask you to produce a schedule showing when each activity can start given limited resources.
当多个活动可供选择时，优先列表决定执行顺序。常见的列表依据是关键路径或浮动时间递增。CCEA 可能会给出一个预设列表，然后要求你根据有限资源生成时间表，说明每个活动何时开始。
Construct a table or Gantt chart showing day-by-day allocation. If an activity cannot start due to resource shortage, it is delayed until resources become free. The final completion time may be longer than the critical path length.
可建立一个表格或甘特图来展示每天的分配情况。如果某项活动因资源不足无法开始，就推迟到资源空闲为止。最终的完成时间可能大于关键路径长度。
10. Interpreting Float in Context | 浮动时间在场景中的理解
Total float shows how much scheduling flexibility exists. However, using float in one activity may reduce float for others that share the same slack. Be careful when multiple non-critical activities lie on the same branch.
总浮动展示了进度安排的弹性。然而，某一活动使用浮动时间后，可能会减少共享同一松弛时间的其他活动的浮动量。当多个非关键活动位于同一条分支时需特别小心。
Free float, if considered, is the delay possible before affecting the EST of any successor. It is calculated as EST(successor) − EFT(current). This value is always less than or equal to total float.
若考虑自由浮动，它是指在影响任何后续活动 EST 之前可延误的时间，计算方法为 EST(后续) − EFT(当前)。该值总小于等于总浮动。
11. Worked Example – CCEA Style | 真题示例解析
Consider a project with activities and dependencies: A(–,2), B(A,4), C(A,3), D(B,1), E(B,5), F(C,2), G(D,E,3), H(F,G,1). All times in hours.
考虑一个项目，其活动与依赖关系如下：A(–,2)，B(A,4)，C(A,3)，D(B,1)，E(B,5)，F(C,2)，G(D,E,3)，H(F,G,1)。时间单位均为小时。
Forward pass: EST A=0, EFT=2. B: EST=2, EFT=6; C: EST=2, EFT=5. D: EST=6, EFT=7; E: EST=6, EFT=11; F: EST=5, EFT=7. G: max(7,11)=11, EFT=14. H: max(7,14)=14, EFT=15. Project duration=15.
前向传递：EST A=0，EFT=2。B: EST=2，EFT=6；C: EST=2，EFT=5。D: EST=6，EFT=7；E: EST=6，EFT=11；F: EST=5，EFT=7。G: max(7,11)=11，EFT=14。H: max(7,14)=14，EFT=15。项目工期 15 小时。
Backward pass: LFT H=15, LST=14. G: LFT=14, LST=11; F: LFT=14, LST=12; E: LFT=11, LST=6; D: LFT=11, LST=10; C: LFT=12, LST=9; B: min{LST D, LST E}=min{10,6}=6, LST=2; A: min{LST B, LST C}=min{2,9}=2, LST=0.
后向传递：LFT H=15，LST=14。G: LFT=14，LST=11；F: LFT=14，LST=12；E: LFT=11，LST=6；D: LFT=11，LST=10；C: LFT=12，LST=9；B: min{LST D, LST E}=min{10,6}=6，LST=2；A: min{LST B, LST C}=min{2,9}=2，LST=0。
Floats: A:0, B:0, C:7, D:4, E:0, F:7, G:0, H:0. Critical path: A-B-E-G-H with total 15 h. A cascade chart would show C and D and F with large floats that can be used for resource levelling.
浮动时间：A 0，B 0，C 7，D 4，E 0，F 7，G 0，H 0。关键路径：A-B-E-G-H，总计 15 小时。级联图中 C、D、F 拥有较大浮动量，可用于资源平衡。
12. Common Exam Pitfalls & Tips | 常见考试陷阱与技巧
Mistake 1: Forgetting to use the maximum rule in the forward pass when multiple predecessors exist. Always pick the largest EFT.
错误 1：当存在多个紧前活动时忘记前向传递中的最大值规则。务必选取最大的 EFT。
Mistake 2: In the backward pass, using the maximum instead of minimum LST for the LFT of a predecessor. Always use the smallest LST from the successors.
错误 2：后向传递中为前驱活动的 LFT 错误地使用了最大值而非最小值。必须从后续活动中取最小的 LST。
Mistake 3: Misinterpreting float — double-check subtraction and ensure you use the correct values for EST and LST from the activity’s own node.
错误 3：浮动计算有误——复核减法，并确保使用该活动自身节点的 EST 和 LST 值。
Tip: Check that the forward and backward pass values are consistent; the EST and LST of the start node should be 0, and the final node’s EFT and LFT should equal the project duration.
技巧：检查前后传递数值是否一致；起始节点的 EST 和 LST 都应等于 0，最终节点的 EFT 和 LFT 都应等于项目工期。
Tip: When drawing cascade charts, label the float extension clearly and indicate the resource usage per day. A ruler and careful scaling win exam marks.
技巧：绘制级联图时，清楚标注浮动延伸区间并标明每天的资源使用量。借助尺子和精确的比例能赢得考试分数。
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