📚 Critical Path Analysis in IGCSE CIE Mathematics | IGCSE CIE 数学：关键路径分析考点精讲
Critical Path Analysis (CPA) is a project management technique used to identify the minimum time needed to complete a project and to determine which activities are critical - meaning any delay in them will delay the whole project. In the IGCSE CIE Mathematics syllabus (0580/0980), CPA appears in the Statistics and probability strand, often tested through activity networks, precedence tables, forward and backward passes, float calculations and identifying the critical path. This article covers every essential concept you need to master for your exam.
关键路径分析是一种项目管理技术，用于确定完成项目所需的最短时间，并找出哪些活动是“关键”的——即这些活动一旦延误，整个项目就会被推迟。在 IGCSE CIE 数学大纲（0580/0980）中，关键路径分析属于统计与概率板块，常考题包括活动网络图、前导表、正向和反向推算、浮动时间计算以及识别关键路径。本文覆盖了你必须掌握的每一个核心概念。
1. Understanding Activity Networks and Nodes | 理解活动网络图与结点
An activity network (also called a precedence network) is a diagram that shows the order in which activities must be carried out. Activities are represented by arcs (arrows) or by nodes depending on the convention used. The most common convention in CIE IGCSE is Activity-on-Arc (AOA), where each directed edge is an activity and each node (circle) represents an event - the start or finish of one or more activities. Each activity is labelled with a letter and its duration in square brackets, e.g., A[5] means activity A takes 5 units of time.
活动网络图（也叫前导网络图）展示了活动必须执行的顺序。在 CIE IGCSE 中最常见的是弧表示活动（AOA），即每条有向边代表一个活动，每个结点（圆圈）代表一个事件——一个或多个活动的开始或结束。每个活动用一个字母标注，并在方括号内注明持续时间，例如 A[5] 表示活动 A 耗时 5 个单位时间。
The network must have exactly one start node (often numbered 0 or 1) and one end node. Dependencies are read from the diagram: an activity cannot begin until all preceding activities leading into its start node are complete. A common mistake is to create redundant dummies; remember that a dummy activity (drawn as a dashed arrow, duration 0) is used only to maintain logical dependencies or to avoid two activities having the same start and end nodes.
网络图必须只有一个起始结点（通常编号为 0 或 1）和一个终止结点。依赖关系需要从图中读取：一个活动必须等所有指向其起始结点的前序活动都完成后才能开始。常见的错误是画了多余的虚拟活动；请记住，虚拟活动（虚线箭头，持续时间为 0）仅用于保持逻辑依赖或避免两个活动有相同的起点和终点。
When constructing a network from a precedence table, first identify which activities have no predecessors - they start at the source node. Next, draw activities one by one following the dependencies, introducing a dummy only when necessary. For example, if activity C depends on A and B, but D depends only on B, then after B you would split using a dummy to show that C needs both A and B while D only needs B.
从优先表构建网络图时，首先找出没有前驱的活动——它们从源结点开始。然后按照依赖关系依次画出活动，必要时才引入虚拟活动。例如，若活动 C 依赖 A 和 B，而活动 D 只依赖 B，则在 B 后需要用虚拟活动分叉，以表示 C 需要 A 和 B，而 D 只需要 B。
2. Precedence Tables: Reading and Interpreting Dependencies | 前导表：读取与解释依赖关系
Exam questions often provide a precedence table listing each activity, its duration, and its immediate predecessors (the activities that must be completed just before it). You must be able to translate this table into a network diagram and vice versa. A typical row looks like: Activity: E, Duration: 7, Predecessors: B, C. This means E can start only after both B and C are finished.
考试题目通常会提供一个前导表，列出每个活动、持续时间以及它的直接前驱（即必须在该活动之前完成的活动）。你必须能够将该表转换成网络图，反之亦然。典型的行如下：活动 E，持续时间 7，前驱 B, C。这意味着 E 只能在 B 和 C 都完成后开始。
Some tables might use dashes or 'None' to indicate no predecessors. The first step in any CPA question is to list all activities and check if there is a unique start and finish. If there are multiple finishing activities, they must all feed into the final end node. When reading the table, look out for 'dummy' needs: if two activities share some but not all predecessors, dummies are required to preserve the exact logic.
有些表格可能用短横线或“None”表示无前驱。任何关键路径分析题的第一步都是列出所有活动，并检查是否有唯一的起点和终点。如果有多个结束活动，它们都必须汇入最终的终止结点。阅读表格时，注意是否需要虚拟活动：如果两个活动共享部分但不是全部前驱，就需要用虚拟活动来保持精确的逻辑关系。
3. The Forward Pass: Calculating Earliest Start Times (EST) | 正向推算：计算最早开始时间 (EST)
The forward pass works from left to right across the network, assigning the earliest time each event (node) can be reached. At the start node, the earliest event time is 0. To find the earliest time for the next event, we take the maximum of (earliest time of preceding event + duration of connecting activity). This gives the earliest start time (EST) of any activity leaving that node.
正向推算从网络图的左端向右端进行，为每个事件（结点）分配可以到达的最早时间。起始结点的最早事件时间为 0。要找到下一个事件的最早时间，我们取（前驱事件的最早时间 + 连接活动的持续时间）的最大值。这给出了离开该结点的任何活动的最早开始时间 (EST)。
(which is at 5), then earliest for node 3 = max(0+4, 5+6) = max(4,11) = 11. Always choose the largest value to ensure all preceding activities have finished.
考虑一个简单网络：结点 1（时间 0）通过活动 A[5] 到结点 2，则结点 2 最早时间 = 0 + 5 = 5。如果结点 3 从结点 1 接收 A[4]，从结点 2（时间 5）接收 B[6]，则结点 3 最早时间 = max(0+4, 5+6) = max(4,11) = 11。一定要选最大值，以确保所有前驱活动都已完成。
Write the earliest event time clearly inside the left half of each node or above the node as specified in the question. The EST of an activity is simply the earliest time of the node from which it originates.
按要求将最早事件时间清楚地写在每个结点的左半部分或结点上方。活动的 EST 就是其起始结点的最早时间。
4. The Backward Pass: Calculating Latest Start Times (LST) | 反向推算：计算最晚开始时间 (LST)
After completing the forward pass and recording the project's minimum duration (the earliest time at the end node), we perform a backward pass from right to left. At the end node, set the latest event time equal to its earliest time (the project duration). Then for each preceding node, subtract activity durations and take the minimum. This finds the latest time each event can occur without delaying the whole project.
完成正向推算并记录项目的最短工期（终止结点的最早时间）之后，我们从右向左进行反向推算。在终止结点，设最晚事件时间等于其最早时间（项目工期）。然后对每个前一个结点，减去活动持续时间并取最小值。这样能找出每个事件在不延误整个项目的前提下可以发生的最晚时间。
re than one activity leaves node 4, we take the minimum of the computed values to maintain the tightest constraint. The latest start time (LST) for an activity is the latest time of its start node.
假设结点 5 是终点，最早时间为 20，则其最晚时间 = 20。如果活动 G[3] 从结点 4 指向结点 5，则结点 4 的最晚时间 = 20 - 3 = 17。如果从结点 4 出发的活动不止一个，我们取计算值的最小值以保持最严格的约束。活动的最晚开始时间 (LST) 就是其开始结点的最晚时间。
alculations: the start node must have latest time = 0 after the backward pass. If not, you've made an error in arithmetic or missed a path.
一定要验算：反向推算后起始结点的最晚时间必须为 0。如果不是，就说明计算有误或漏算了一条路径。
5. Float: Total Float and Free Float | 浮动时间：总浮动与自由浮动
atest Start Time (LST) - Earliest Start Time (EST). Equivalently, you can use Latest Finish Time - Earliest Finish Time. Both give the same result.
活动的总浮动时间是指在不影响整个项目工期的前提下，该活动可以延迟的时间量。计算公式为：总浮动 = 最晚开始时间 (LST) - 最早开始时间 (EST)。或者等价地，使用最晚完成时间 - 最早完成时间，两者结果相同。
loat = (earliest time of its end node) - (earliest time of its start node + duration). This tells you if it can slip without pushing the next activity. Free float is always less than or equal to total float.
自由浮动时间是指在不影响任何后续活动最早开始的前提下，活动可以延迟的时间量。对于弧上的活动，自由浮动 =（其终止结点的最早时间）-（其开始结点的最早时间 + 持续时间）。这表示在不推迟下一个活动的情况下可以滑动的余地。自由浮动总是小于或等于总浮动。
ivity; if its free float is also 0, then any delay will directly affect the next activity.
考试中可能会要求计算某个特定活动的浮动时间并解释其含义。总浮动为 0 的活动是关键活动；如果其自由浮动也是 0，那么任何延误都会直接影响后续活动。
6. Identifying the Critical Path | 识别关键路径
ted. All activities on the critical path have total float = 0. There can be more than one critical path; if so, any delay on any of those paths will delay the project.
关键路径是网络图中持续总时间最长的路径，它决定了项目可以完成的最短时间。关键路径上的所有活动总浮动时间均为 0。可能存在多条关键路径；如果是这样，任何一条关键路径上的延误都会导致项目延期。
To identify the critical path, list all activities with zero total float and trace a continuous path from start to end. Write the path as a sequence of activities or events as required. For instance, Start - A - C - F - Finish. Double-check that the sum of durations along this path equals the project duration.
要识别关键路径，列出所有总浮动为 0 的活动，并从起点到终点连续追踪。按考试要求将路径写成活动序列或事件序列，例如：开始 - A - C - F - 结束。要核对该路径上各活动的持续时间之和等于项目总工期。
ivities or visually shortest line - always use total float = 0 as the objective criterion.
一个常见误区是选择活动数目最少或直观上线条最短的路径——始终要以总浮动 = 0 作为客观判据。
7. Dummy Activities in Depth | 深入理解虚拟活动
A dummy activity is drawn as a dashed arrow with duration 0. Its sole purpose is to represent a dependency that cannot be shown by parallel arcs. For example, if activity D depends on A only, but activity E depends on both A and B, a dummy from the end of B to the start of E might be needed to prevent D from incorrectly depending on B. Dummies can also be used to give two activities separate end events when they share the same start and end, which is not allowed in AOA networks.
虚拟活动用虚线箭头表示，持续时间为 0。它的唯一目的是表示无法用平行弧线表示的依赖关系。例如，如果活动 D 只依赖 A，而活动 E 同时依赖 A 和 B，那么可能需要从 B 的终点画一条虚拟活动到 E 的起点，以避免 D 错误地依赖 B。虚拟活动还可用于当两个活动有相同的起点和终点时（AOA 网络不允许），给它们分开的结束事件。
When performing forward and backward passes, treat dummies as any other activity but with duration 0. They do not change times but ensure logical flow. A common exam mistake is to forget that a dummy can be part of the critical path if it lies between two activities that are critical.
进行正向和反向推算时，将虚拟活动像其他活动一样处理，但持续时间为 0。它们不改变时间，但保证逻辑流程。一个常见的考试错误是忘记虚拟活动如果位于两个关键活动之间，也可能成为关键路径的一部分。
8. Gantt Charts (Cascade Charts) | 甘特图 (级联图)
Some IGCSE questions ask you to draw a Gantt chart (also called a cascade chart) based on the results of forward and backward passes. A Gantt chart is a horizontal bar chart where each activity is represented by a bar starting at its EST and ending at its earliest finish time (EST + duration). The total float is often shown as a dotted extension from the end of the bar up to the LST.
有些 IGCSE 题目会要求你根据正向和反向推算的结果绘制甘特图（也称级联图）。甘特图是一种水平条形图，每个活动由一个条形表示，其起点为 EST，终点为最早完成时间（EST + 持续时间）。总浮动通常用从条形末端到 LST 的虚线延伸来表示。
l representation helps to see which activities have flexibility and when resources might be leveled. In the exam, you may need to read scheduling constraints from a given Gantt chart, such as the earliest day a particular worker is needed or when an overlap occurs.
关键活动将没有虚线延伸（浮动 = 0）。这种可视化表示有助于看清哪些活动具有灵活性，以及何时可以进行资源均衡。考试中你可能需要从给定的甘特图中读出排程约束，例如某个工人最早需要到岗的日期，或何时发生重叠。
When constructing a Gantt chart, always use a consistent time scale. Label axes clearly and mark the critical path activities differently, e.g., with a solid fill or a darker colour.
绘制甘特图时，务必使用一致的时间刻度。清晰标记坐标轴，并用不同方式标记关键路径活动，例如实心填充或较深的颜色。
9. Scheduling with Limited Resources | 资源受限的排程
An extension topic within CPA considers the availability of resources (e.g., workers, machines). The question might give the number of workers required for each activity and a limit on total workers per day. Using the float information, you may need to reschedule activities within their float to smooth the resource usage over time, a process called resource levelling.
关键路径分析的拓展内容之一考虑资源可用性（如工人、机器）。题目可能给出每个活动所需的工人数量以及每日总工人数限制。利用浮动信息，你可能需要在活动容许的浮动范围内重新安排活动，以使资源使用随时间变得平滑，这一过程称为资源均衡。
Typically, you produce a resource histogram (a bar chart showing number of workers used each day if all activities start at EST). Then, if any day's total exceeds the limit, you delay non-critical activities (using their total float) to shift the demand to later days. You must ensure no critical activity is delayed, as that would extend the project.
通常你会制作一个资源直方图（显示若所有活动都在最早开始时间启动时每天使用的工人数量的柱状图）。然后，如果某一天的总量超出限制，你就利用非关键活动的总浮动将它们推迟，把需求转移到后续日期。你必须确保没有关键活动被延误，否则会延长项目工期。
Exam problems often ask for the minimum number of workers needed to complete the project in the critical time, or to create a worker schedule. Remember to state the final resource profile and any changes from the original earliest start schedule.
考题经常要求找出在关键时间内完成项目所需的最少工人数，或制定工人排程。请记得说明最终的资源概况，以及相比原始最早开始排程有哪些变化。
10. Common Exam Mistakes and How to Avoid Them | 常见考试错误及避免方法
Students often lose marks by misreading the precedence table, especially when an activity depends on multiple predecessors. List dependencies carefully and double-check each connection. Another frequent error is incorrect arithmetic in the forward and backward passes, such as forgetting to take the maximum or minimum. Always work systematically, node by node, writing down each calculation.
学生常因误读前导表而丢分，特别是当某个活动依赖于多个前驱时。应仔细列出依赖关系，并核对每一条连线。另一个常见错误是在正向和反向推算中计算错误，例如忘记取最大值或最小值。始终按结点系统推算，写下每一步计算过程。
ads to wrong answers. Remember: total float = LST - EST; free float involves the next node' the diagram clear. If the question provides a grid for a Gantt chart, use it properly; don't misalign the bars.
混淆总浮动和自由浮动会导致错误答案。记住：总浮动 = LST - EST；自由浮动涉及下一个结点的最早时间。此外，画网络图时尽量避免不必要的线条交叉——利用虚拟活动使图清晰。如果题目提供了甘特图的网格，要正确使用，不要将条形错位。
Finally, always show your working. Even if the final path is wrong, clear calculation steps can earn method marks.
最后，一定要展示解题过程。即使最终的关键路径错误，清晰的计算步骤也能得到方法分。
11. Worked Example: From Table to Critical Path | 完整示例：从表格到关键路径
Let's consider a typical IGCSE question: Activity table – A[3] (no predecessors), B[4] (after A), C[2] (after A), D[5] (after B), E[3] (after B and C), F[2] (after D and E). Draw the network, find EST/LST, floats and the critical path.
考虑一个典型的 IGCSE 问题：活动表 – A[3]（无前驱），B[4]（后于 A），C[2]（后于 A），D[5]（后于 B），E[3]（后于 B 和 C），F[2]（后于 D 和 E）。画出网络图，计算 EST/LST、浮动及关键路径。
T = 3+2=5. Node 4 combines B and C for E: we need a dummy from end of B (node 2) to node 4? h B and C, so node 4 receives E from node 2 and from node 3. Node 2 EST=7, node 3 EST=5. E's duration 3, so node 5 (end of E) EST = max(7+3, 5+3? and C, so we merge B and C into a single node 4 using a dummy from the end of C to node 4? ode 3 to node 2, giving node 2 now EST = max(7, 5+0)=7. Then E[3] from node 2 to node 4, giving EST = 7+3=10. Meanwhile D[5] from node 2 to node 5 (EST=7+5=12). Finally F[2] from node 4 (10) and node 5 (12) to end node 6: we need dependencies from both D and E, so node 6 EST = max(10+2, 12+2) = max(12,14)=14. Project duration 14.
开始结点 0。A 到结点 1（EST=3）。从结点 1，B 到结点 2：EST = 3+4=7；C 到结点 3：EST = 3+2=5。E 需要 B 和 C，因此我们需要把 B 和 C 汇合。从结点 3 画一条虚拟活动到结点 2，这样结点 2 的 EST = max(7, 5+0)=7。然后 E[3] 从结点 2 到结点 4，EST = 7+3=10。同时 D[5] 从结点 2 到结点 5，EST = 7+5=12。最后 F[2] 从结点 4（10）和结点 5（12）到终止结点 6：因为 F 依赖 D 和 E，我们需要把两者合并。结点 6 的 EST = max(10+2, 12+2) = max(12,14)=14。项目工期 14。
7-4=3, via C to node 3: 7-2=5, min=3. Node 0 = 3-3=0. Total floats: A: 0-0=0, B: 0-0=0, C: LST(3? float=0, E: EST 7, LFT_node4=12, LST=12-3=9 → float=2, F: EST max(10,12)=12 for its start? Actually F starts at a node after merging D and E, but we used separate nodes 4 and 5. Let's design properly: A standard solution would merge D and E into a single node before F. I' A-B-D-F (duration 3+4+5+2=14). C and E each have float of 2. This illustrates the process.
反向推算：结点 6 最晚时间=14。结点 5 = 14-2=12。结点 4 = 14-2=12。结点 2 通过 D：12-5=7；通过 E：12-3=9；取最小=7。结点 3（虚拟活动到结点 2）：7-0=7。结点 1 通过 B：7-4=3，通过 C 到结点 3：7-2=5，取最小=3。结点 0 = 3-3=0。总浮动：A：0-0=0，B：0-0=0，C（起始于结点 1 EST=3，结点 1 最晚=3，C 终止结点 3 最晚=7，LST=7-2=5，EST=3，浮动=2），D：EST 7，结点 2 最晚=7，结点 5 最晚=12，LST=12-5=7，浮动=0，E：EST 7，结点 4 最晚=12，LST=12-3=9，浮动=2，F：从合并结点开始……关键路径为 A-B-D-F（持续时间 3+4+5+2=14）。C 和 E 各有浮动 2。这个例子说明了操作流程。
12. Preparing for the Exam: Tips and Revision | 备考建议与复习要点
Practice drawing networks from precedence tables with various dependency patterns, including those requiring dummies. Focus on accurate forward and backward passes; remember that these are often worth several marks and are straightforward if you work methodically. Revise float formulas and be able to interpret the meaning of a float value in the context of project management.
多练习从优先表绘制网络图，包括需要虚拟活动的各种依赖模式。注重正向和反向推算的准确性；记住这部分通常占好几分，只要按部就班就能轻松拿到。复习浮动公式，并能在项目管理情境中解释浮动值的含义。
Use past papers to get familiar with the phrasing of scheduling and resource questions. Pay attention to units (hours, days, weeks) and ensure your Gantt chart reflects the scale correctly. When identifying the critical path, always state it explicitly as the path of activities or events, not just the duration.
利用历年真题熟悉排程和资源问题的表述方式。注意单位（小时、天、周），并确保甘特图正确反映时间刻度。在识别关键路径时，务必明确写出活动或事件路径，而不只是写出总工期。
Finally, manage your time in the exam. CPA questions can be lengthy, so practice completing a full CPA problem in under 15 minutes. With structured revision, you can turn this topic into an easy high-scorer.
最后，考试中要管理好时间。关键路径分析题可能篇幅较长，所以要练习在 15 分钟内完整解答一道 CPA 题。通过有结构的复习，你完全可以把本专题变成轻松拿分的高分项。
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