📚 IGCSE Maths: Critical Path Analysis — Key Concepts and Exam Tips | IGCSE 数学：关键路径分析考点精讲
Critical path analysis is a powerful decision‑mathematics tool used to plan and manage complex projects. By breaking down a project into individual tasks, determining their dependencies, and identifying the longest path through the network, you can find the minimum time needed to finish the entire project and spot which activities must not be delayed. For IGCSE Maths, you need to be confident drawing activity networks, calculating earliest and latest start times, finding total float, and clearly stating the critical path. This article will walk you through every concept with clear explanations, worked examples, and exam‑focused tips.
关键路径分析是决策数学中一个非常实用的工具，用来规划和管控复杂的项目。通过把项目拆分成单个任务、确定任务之间的依赖关系，并找出网络图中最长的路径，我们就能求得完成整个项目所需的最短时间，同时确定哪些活动绝对不能拖延。对于IGCSE数学考试，你必须熟练掌握绘制活动网络图、计算最早和最晚开始时间、求总浮动时间，并能清晰地写出关键路径。本文将通过清晰的讲解、例题和应试技巧，带你吃透所有考点。
1. Activity Networks and Precedence Tables | 活动网络与前导表
An activity network is a visual representation of the tasks (activities) in a project and their logical order. Each activity has a duration, and some activities must be completed before others can start. These relationships are usually given in a precedence table, which lists each activity, its duration, and its immediate predecessors.
活动网络是将项目中的任务（活动）及其逻辑顺序用图形直观表示出来的一种工具。每个活动都有一个持续时间，有些活动必须完成后，另一些活动才能开始。这些依赖关系通常由一张前导表给出，表中列出每个活动的编号、持续时间以及它的紧前活动。
For example, consider a simple project with activities A (3 days), B (4 days) depending on A, C (2 days) depending on A, and D (1 day) depending on both B and C. The precedence table might look like this:
例如，某项目包含活动 A（3 天）、B（4 天，依赖 A）、C（2 天，依赖 A）、D（1 天，依赖 B 和 C）。前导表可以这样表示：
| Activity | Duration (days) | Predecessors |
| A | 3 | – |
| B | 4 | A |
| C | 2 | A |
| D | 1 | B, C |
Reading precedence tables accurately is the first step towards drawing a correct network diagram. An exam question will often provide such a table and ask you to construct the network.
准确读懂前导表是画出正确网络图的第一步。考试题常常直接给出这样一张表，然后让你构造网络图。
2. Activity‑on‑Node Representation | 活动在节点表示法
IGCSE syllabuses almost exclusively use the activity‑on‑node (AoN) convention. Here, each activity is represented by a node (a box or circle), and arrows show the dependencies between activities. A node typically contains the activity label, its duration, the earliest start time (EST), and the latest start time (LST). You will fill in EST and LST as you perform forward and backward passes.
IGCSE 大纲几乎全部采用活动在节点（AoN）表示法。在这种方法里，每个活动用一个节点（方框或圆圈）表示，箭头表示活动之间的依赖关系。一个节点通常包含活动标签、持续时间、最早开始时间（EST）和最晚开始时间（LST）。你会通过前向计算和后向计算逐步填入 EST 和 LST。
The standard node layout you may encounter is:
你可能会遇到的节点标准布局如下：
| EST | LST | Activity (duration) | ||
You do not need to memorise a specific design, but you must be consistent and label all values clearly. Always check which notation your exam board expects; most will accept a simple box with four compartments.
你不必死记某一种节点设计，但必须保持一致并清楚标注所有数值。记得确认你的考试局偏好的格式；大多数都接受分成四格的简单方框。
3. Constructing a Network Diagram | 构建网络图
To draw an AoN network from a precedence table, follow these steps: first, identify the activities with no predecessors — they will start the project. Draw a node for each such activity and connect them to a common start node (if needed). Then, for each subsequent activity, draw its node and draw arrows from all of its predecessor nodes. Continue until all activities are placed. Finally, connect any activities with no successors to a common end node.
根据前导表绘制 AoN 网络图时，遵循以下步骤：首先找出没有紧前活动的任务——它们将是项目开始的起点。为每个这样的活动画一个节点，必要时用一个共同的起始节点连接它们。然后，按照顺序为每个后续活动画节点，并从它的所有紧前活动节点向它画箭头。一直进行到所有活动都就位。最后，将所有没有后继活动的节点连接到一个共同的终止节点。
It is vital to include every dependency. If two activities share the same predecessors but are not dependent on each other, they will simply both have arrows coming from the same predecessor nodes. A common mistake is to omit an arrow, which changes the logic entirely.
画出每一条依赖关系至关重要。如果两个活动共享相同的紧前活动但彼此之间不依赖，它们就从相同的紧前节点各自引出一条箭头。常见的错误是漏掉箭头，这会完全改变逻辑关系。
4. Dummy Activities | 虚拟活动
Sometimes, the precedence relationships cannot be represented cleanly with direct arrows, especially when activities have different combinations of predecessors. Dummy activities are artificial tasks with zero duration; they exist solely to maintain correct dependency logic without creating unintended links. In AoN networks, dummies are often not needed as frequently as in activity‑on‑arc diagrams, but they can still appear when you must show that one activity depends on two separate chains with a different third dependency.
有时，仅靠直接箭头无法干净地表达依赖关系，尤其是当几项活动有着不同的紧前活动组合时。虚拟活动是持续时间为零的人造任务，存在的唯一目的就是维持正确的依赖逻辑，避免产生多余的连接。在 AoN 网络中，虚拟活动不像在活动在弧网络里那样频繁出现，但当需要表达一个活动既依赖于两条独立链路又受另一个依赖约束时，虚拟活动仍然会出现。
When you use a dummy, label it with a dashed arrow or a distinct identifier (like D₁) and note its duration as 0. It does not affect the project’s total time but helps you calculate ESTs and LSTs correctly.
当你使用虚拟活动时，用虚线箭头或特殊标识（如 D₁）标注，并把持续时间记为 0。它不影响项目总工期，但能帮助你正确计算 EST 和 LST。
5. Forward Pass: Earliest Start Times (EST) | 前向计算：最早开始时间
The forward pass determines the soonest each activity can begin, given the network of predecessors. Start by setting the EST of the first activity (or start node) to 0. For any other activity, its EST is the maximum of (EST of predecessor + duration of predecessor) over all its immediate predecessors.
前向计算用来求出在网络中各活动最早何时能够开始。先将第一个活动（或起始节点）的 EST 设为 0。对于任何其它活动，其 EST 等于所有紧前活动中 (紧前活动的 EST + 紧前活动的持续时间) 的最大值。
ESTₐ = max { ESTₚ + dₚ } for all predecessors p of a
Work through the network from left to right, filling in the EST for each node. The EST of the final node gives the minimum project duration. In our earlier example, EST of A = 0; B and C both have EST = 0 + 3 = 3; D’s predecessors are B (EST 3 + 4 = 7) and C (EST 3 + 2 = 5), so EST of D = max(7,5) = 7.
从网络图左端向右推进，逐个填入节点的 EST。最终节点的 EST 就是项目的最短工期。在前面的例子中，A 的 EST = 0；B 和 C 的 EST 均为 0 + 3 = 3；D 的紧前活动是 B（EST 3 + 4 = 7）和 C（EST 3 + 2 = 5），因此 D 的 EST = max(7,5) = 7。
Always double‑check that you take the maximum; picking a smaller value would mean starting before a needed predecessor has finished, which is impossible.
一定要核对是否取了最大值；如果取了较小的值，就意味着在某个必需的紧前活动完成之前就开始，这在实际中是不可能的。
6. Backward Pass: Latest Start Times (LST) | 后向计算：最晚开始时间
The backward pass calculates the latest time each activity can start without delaying the whole project. Begin at the end node: set its LST equal to its EST (i.e., the project duration). Then move backwards through the network. For an activity, its LST is the minimum of (LST of successor − duration of this activity) over all its immediate successors.
后向计算用来求出每个活动在不延误整个项目的前提下最晚何时必须开始。从终止节点开始：将其 LST 设为和它的 EST 相等（即项目总工期）。然后反向遍历网络。对于一个活动，其 LST 等于所有紧后活动中 (紧后活动的 LST − 本活动的持续时间) 的最小值。
LSTₐ = min { LSTₛ − dₐ } for all successors s of a
In the example, the project duration is 7 + 1 = 8 days (if D lasts 1). So the end node representing project finish has EST = LST = 8. For D, its LST = 8 − 1 = 7. For B, its successor is D, so LST₆ = 7 − 4 = 3. For C, successor is D, so LSTₜ = 7 − 2 = 5. For A, successors B and C have LST₆ = 3 and LSTₜ = 5, so LSTₐ = min(3 − 3, 5 − 3) = min(0,2) = 0.
在此例中，项目工期为 7 + 1 = 8 天（若 D 持续 1 天）。因此代表项目完成的终点节点的 EST = LST = 8。对于 D，LST = 8 − 1 = 7。对于 B，其紧后活动是 D，LST₆ = 7 − 4 = 3。对于 C，紧后活动是 D，LSTₜ = 7 − 2 = 5。对于 A，紧后活动 B 和 C 的 LST 分别为 3 和 5，所以 LSTₐ = min(3 − 3, 5 − 3) = min(0,2) = 0。
Remember, you are subtracting the duration of the activity you are calculating, not the successor’s duration. Practise this step carefully, as it is a frequent source of errors.
请记住，你减去的是正在计算的那个活动的持续时间，而不是紧后活动的持续时间。这一步需要仔细练习，因为它是常见的出错点。
7. Total Float and Critical Activities | 总浮动时间与关键活动
Total float is the amount of time an activity can be delayed without affecting the overall project completion time. It is found by subtracting the earliest start time from the latest start time (or equivalently, the earliest finish from the latest finish).
总浮动时间是指在不影响整个项目完工时间的前提下，一项活动可以被推迟的时间量。它由最晚开始时间减去最早开始时间求得（或者等价地，最晚完成时间减去最早完成时间）。
Total float = LST − EST
An activity with zero total float is called a critical activity. Any delay to a critical activity will directly delay the whole project. In the example, activities A, B, and D all have float 0 (A: 0−0=0, B: 3−3=0, D: 7−7=0), while C has float 5−3=2 days. Thus C can be delayed by up to 2 days without affecting the finish date.
总浮动时间为零的活动称为关键活动。任何关键活动的延误都会直接导致整个项目延误。在例子中，活动 A、B 和 D 的浮动时间均为 0（A：0−0=0，B：3−3=0，D：7−7=0），而活动 C 的浮动时间为 5−3=2 天。因此 C 最多可以延误 2 天而不影响完工日期。
Sometimes a question may ask for ‘free float’ as well, but IGCSE typically focuses on total float. Free float is the delay possible without affecting the early start of any successor. Be aware of the distinction, but concentrate on total float unless your syllabus specifically asks otherwise.
有时题目也会问及“自由浮动时间”，但 IGCSE 通常侧重总浮动时间。自由浮动时间是指不推迟任何紧后活动最早开始时间的前提下可以延误的时间。你需要了解这一区别，但除非大纲明确要求，否则重点掌握总浮动时间即可。
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 project. To determine it, highlight all activities where total float = 0 and trace a path from the first node to the last. You must list the activities in order, e.g., A → B → D. The duration of the critical path equals the project duration. If more than one path has zero float, all such paths are critical; the project can then be delayed by a holdup on any of them.
关键路径是从项目开始到结束、由关键活动组成的连续链路。要确定它，把所有总浮动时间为 0 的活动标出来，然后从起始节点到终止节点画出一条路径。你必须按顺序列出这些活动，如 A → B → D。关键路径的持续时间就等于项目的总工期。如果存在不止一条零浮动路径，那么所有这些路径都是关键的；此时任何一条链路上的延误都会拖后整个项目。
In exam answers, always give the critical path as a sequence of activity letters, and state the project duration clearly. For example: ‘Critical path: A – B – D, project duration 8 days.’
在考试作答时，总是用活动字母序列来表示关键路径，并清楚地写出项目工期。例如：“关键路径：A – B – D，项目工期 8 天。”
9. Gantt Charts (Cascade Charts) | 甘特图（级联图）
Some IGCSE specifications also require you to draw a Gantt chart (sometimes called a cascade chart) based on the ESTs and the critical path. A Gantt chart displays each activity as a horizontal bar along a time axis. Critical activities are typically drawn in one colour or shaded, and floats are shown as a dashed extension after the bar. This visual representation helps project managers schedule resources.
部分 IGCSE 大纲还要求根据 EST 和关键路径绘制甘特图（有时亦称级联图）。甘特图将每个活动沿时间轴表示为水平条。关键活动通常用某种颜色或阴影表示，浮动时间则用条棒后的虚线延伸展示。这种可视化方式有助于项目经理安排资源。
When drawing a Gantt chart, use the EST to position the start of each bar, and draw it for the activity’s duration. If an activity has float, extend a dashed line from the end of the bar to show the latest finish. Always label your axes (time on horizontal axis, activities on vertical) and indicate the critical activities with a key.
绘制甘特图时，根据 EST 确定每个条棒的起点，并按活动的持续时间绘制长度。如果活动有浮动时间，从条棒尾部画一条虚线延伸到最晚完成时刻。务必标注坐标轴（横轴为时间，纵轴为活动），并用图例标出关键活动。
10. Common Pitfalls and Exam Strategies | 常见错误与应试策略
The most frequent mistake in critical path analysis is miscalculating the forward pass by taking the minimum instead of the maximum, or vice versa for the backward pass. Another error is forgetting that the EST of a node with multiple predecessors must consider all of them.
关键路径分析中最常见的错误是在前向计算时取了最小值而不是最大值，或者后向计算时相反。另一个常见错误是忘记对于有多个紧前活动的节点，必须综合考虑所有紧前活动来计算 EST。
Students also sometimes misinterpret the precedence table, sketching arrows incorrectly. Always work step‑by‑step: read the table carefully, draw the nodes and arrows, label durations, then perform the forward pass, then the backward pass. Only after that should you calculate floats and highlight the critical path.
学生有时也会误解前导表，画错箭头。一定要一步一步来：仔细阅读表格，画出节点和箭头，标注持续时间，然后进行前向计算，再后向计算。只有在完成这些步骤之后，才去计算浮动时间并标出关键路径。
In exam questions, even if you cannot completely solve the critical path, you can still earn method marks for correct nodes, arrows, and durations. Write neatly and show your working, including the EST and LST values in the nodes.
在考试中，即使你没能完全解出关键路径，只要节点、箭头和持续时间正确，仍然能拿到方法分。书写要整洁，在节点中展示你的 EST 和 LST 值及计算过程。
11. Worked Example: Full Calculation | 完整例题演示
Let us work through a slightly larger project to consolidate all steps. The precedence table is:
我们来演示一个稍大的项目，以巩固所有步骤。前导表如下：
| Activity | Duration | Predecessors |
| P | 5 | – |
| Q | 3 | P |
| R | 4 | P |
| S | 2 | Q, R |
| T | 6 | R |
| U | 1 | S, T |
Step 1: Draw the AoN network. P is first. Arrows from P to Q and P to R. Q and R lead to S, and R also leads to T. Both S and T lead to U. Add a finish node after U.
第 1 步：绘制 AoN 网络。P 是第一个活动。从 P 画箭头到 Q 和 R。Q 和 R 都指向 S，R 也指向 T。S 和 T 都指向 U。在 U 之后加一个结束节点。
Step 2: Forward pass. EST(P)=0. EST(Q)=5, EST(R)=5. EST(S)=max(5+3,5+4)=max(8,9)=9. EST(T)=5+4=9. EST(U)=max(9+2,9+6)=max(11,15)=15. So project duration = 15 + 1 = 16.
第 2 步：前向计算。EST(P)=0。EST(Q)=5，EST(R)=5。EST(S)=max(5+3,5+4)=max(8,9)=9。EST(T)=5+4=9。EST(U)=max(9+2,9+6)=max(11,15)=15。因此项目工期 = 15 + 1 = 16。
Step 3: Backward pass. Finish node LST=16. LST(U)=16−1=15. LST(S)=15−2=13. LST(T)=15−6=9. LST(R)=min(13−4, 9−4)=min(9,5)=5. LST(Q)=13−3=10. LST(P)=min(5−5, 10−5)=min(0,5)=0.
第 3 步：后向计算。结束节点 LST=16。LST(U)=16−1=15。LST(S)=15−2=13。LST(T)=15−6=9。LST(R)=min(13−4, 9−4)=min(9,5)=5。LST(Q)=13−3=10。LST(P)=min(5−5, 10−5)=min(0,5)=0。
Step 4: Total floats. P: 0−0=0; Q: 10−5=5; R: 5−5=0; S: 13−9=4; T: 9−9=0; U: 15−15=0.
第 4 步：总浮动时间。P：0−0=0；Q：10−5=5；R：5−5=0；S：13−9=4；T：9−9=0；U：15−15=0。
Step 5: Critical path. Activities with float 0 are P, R, T, U. The sequence is P → R → T → U. Project duration is 16 days.
第 5 步：关键路径。浮动时间为 0 的活动为 P、R、T、U。路径序列为 P → R → T → U。项目工期为 16 天。
Practice such examples until you can perform the forward and backward passes fluently, as speed matters in the IGCSE exam.
反复练习类似的例子，直到你能流畅地进行前向和后向计算，因为在 IGCSE 考试中速度也很关键。
12. Exam‑Style Practice and Final Tips | 考试风格练习与最后建议
IGCSE questions often ask you to (a) draw the network from a precedence table, (b) complete the forward and backward passes, (c) state the critical path and project duration, and (d) discuss the effect of a delay in a specific activity. Be methodical: label everything, show the EST and LST in each node as you go, and write the float calculations alongside the network.
IGCSE 考题通常会要求你 (a) 根据前导表画出网络图，(b) 完成前向和后向计算，(c) 写出关键路径和项目工期，(d) 讨论某项具体活动延误带来的影响。要有条理：标注一切信息，在节点中逐步显示 EST 和 LST，并在网络图旁边写出浮动时间的计算。
When discussing delays, use the float: if the delay is less than or equal to the total float, the project finish is unaffected; if it exceeds the float, the project will be delayed by (delay minus float) days. Always refer back to the critical activities.
在讨论延误影响时，利用浮动时间：如果延误小于或等于总浮动时间，项目完工不受影响；如果超过浮动时间，项目将延误（延误天数减去浮动天数）天。一定要联系关键活动来回答。
Finally, remember that neat drawings and clear sequences earn valuable marks. Even if you struggle with the backward pass, a correct forward pass and correct float calculation for at least some activities can still secure a good grade. Good luck with your revision!
最后，请记住整洁的图形和清晰的序列能为你赢得宝贵的分数。即使你在后向计算上遇到困难，一个正确的前向计算和部分活动的正确浮动时间仍然可以确保不错的分数。祝复习顺利！
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