📚 Early and Late Event Times | 最早与最晚事件时间
In project management and Decision Mathematics, understanding early and late event times is essential for constructing and analysing activity networks. These times allow you to determine the minimum project duration, identify critical activities, and calculate floats. This article covers the forward and backward pass techniques required by the Edexcel A-Level specification, with a step-by-step worked example.
在项目管理和决策数学中，理解最早与最晚事件时间对于构建和分析活动网络至关重要。这些时间能让你确定最短项目工期，识别关键活动并计算浮动时间。本文涵盖 Edexcel A-Level 考试所要求的正向与反向推算方法，并配有分步计算示例。
1. Event Times in Activity Networks | 活动网络中的事件时间
In an activity-on-node network, each node represents an event – a point in time when one or more activities begin or finish. Every event is labelled with two numbers: its earliest event time (EET) and its latest event time (LET). These times, calculated by forward and backward passes, form the basis of critical path analysis.
在节点表示活动的网络中，每个节点代表一个事件——表明一项或多项活动开始或完成的时刻。每个事件标注有两个数字：最早事件时间（EET）和最晚事件时间（LET）。这些时间通过正向与反向推算得出，构成了关键路径分析的基础。
The forward pass moves from the start node to the end node, establishing the earliest possible moment each event can occur. The backward pass then works in reverse, determining the latest time an event can occur without delaying the whole project.
正向推算从开始节点向结束节点进行，确定每个事件可能发生的最早时刻。反向推算则逆向进行，确定在不延迟整个项目的前提下事件可以发生的最晚时间。
2. Forward Pass: Calculating Early Event Times | 正向推算：计算最早事件时间
We begin by setting the earliest event time of the source node to zero: EETsource ry activity i → j that enters j. The earliest event time of j is the maximum value of (EETi + duration of activity i → j) over all preceding activities.
我们首先将源节点的最早事件时间设为零：EET源 = 0。对于任何其他节点 j，检查所有进入 j 的活动 i → j。节点 j 的最早事件时间取所有前驱活动的 (EETi + 活动 i → j 的持续时间) 中的最大值。
EETj = max { EETi + dij } for all activities i → j
This 'max' rule ensures that an event cannot occur until every predecessor activity has been completed. If a node has only one incoming activity, the EET is simply the predecessor's EET plus the activity's duration.
这个“最大值”规则确保了在所有前驱活动都完成后事件才能发生。如果节点只有一条进入活动，则其最早事件时间直接等于前驱节点的 EET 加上该活动的持续时间。
Always work through the network from left to right, processing nodes in an order that guarantees all predecessors have been dealt with. Record each EET inside the top-right quadrant of the corresponding node.
始终从左向右处理网络，按照保证所有前驱节点都已处理完毕的顺序进行。将每个 EET 记录在对应节点的右上象限中。
3. Worked Example for Early Event Times | 最早事件时间计算实例
Consider the following project activities with their durations (in days):
考虑以下项目活动及其持续时间（单位：天）：
| Activity | Preceding event | Following event | Duration |
|---|---|---|---|
| A | 1 | 2 | 3 |
| B | 1 | 3 | 4 |
| C | 2 | 4 | 2 |
| D | 3 | 4 | 5 |
| E | 2 | 5 | 1 |
| F | 4 | 5 | 3 |
| G | 3 | 5 | 2 |
The network has five events, numbered 1 to 5. Event 1 is the start, event 5 is the end. Now perform the forward pass:
该网络包含五个事件，编号为 1 至 5。事件 1 为开始，事件 5 为结束。现在进行正向推算：
- EET(1) = 0.
- Event 2: only one incoming activity A from event 1 → EET(2) = 0 + 3 = 3.
- Event 3: only incoming activity B from event 1 → EET(3) = 0 + 4 = 4.
- Event 4: C from event 2 (3 + 2 = 5) and D from event 3 (4 + 5 = 9). Take the maximum → EET(4) = 9.
- Event 5: from event 4 (9 + 3 = 12) and G from event 3 (4 + 2 = 6). The maximum is 12 → EET(5) = 12.
因此，项目的最早完成时间为 12 天。
Hence the earliest project completion time is 12 days.
4. Backward Pass: Calculating Late Event Times | 反向推算：计算最晚事件时间
time (the project duration). Then work backwards through the network: for each node i, consider all activities i → j leaving i. The latest event time of i is the minimum value of (LETj – dij) over all outgoing activities.
正向推算完成后，我们从终点节点开始反向推算。将最终事件的最晚事件时间设定等于其最早事件时间（即项目工期）。然后逆向处理网络：对于每个节点 i，考虑所有离开 i 的活动 i → j。节点 i 的最晚事件时间等于所有 (LETj – dij) 中的最小值。
LETi = min { LETj – dij } for all activities i → j
The minimum rule reflects the tightest constraint: the event i must occur early enough to allow all its subsequent activities to finish by their late times. If an event had a single outgoing activity, its LET would simply be LETj – duration.
最小值规则反映了最严格的约束：事件 i 必须足够早地发生，以便其所有后续活动能在各自的最晚时间前完成。如果某个事件只有一条离开活动，则其 LET 直接等于 LETj – 持续时间。
Record each LET inside the bottom-right quadrant of the event node. The backward pass reveals which events have no slack and therefore lie on the critical path.
将每个 LET 记录在事件节点的右下象限中。反向推算揭示了哪些事件没有时差，从而位于关键路径上。
5. Worked Example for Late Event Times | 最晚事件时间计算实例
, we already know the project duration is 12, so LET(5) = 12. Now apply the backward pass:
继续使用同一网络，已知项目工期为 12，因此 LET(5) = 12。现在进行反向推算：
- Event 4: only one outgoing activity F to event 5 → LET(4) = 12 – 3 = 9.
- Event 3: two outgoing activities: D to event 4 (9 – 5 = 4) and G to event 5 (12 – 2 = 10). Take the minimum → LET(3) = 4.
- Event 2: two outgoing activities: C to event 4 (9 – 2 = 7) and E to event 5 (12 – 1 = 11). Minimum is 7 → LET(2) = 7.
- Event 1: outgoing activities A to event 2 (7 – 3 = 4) and B to event 3 (4 – 4 = 0). Minimum is 0 → LET(1) = 0.
ing that LET(1) = 0 confirms the arithmetic is consistent with a start event at time zero.
这些数值现在与 EET 一起写在节点中。验证 LET(1) = 0 可确认计算与起点时间为零是一致的。
6. Critical Path Identification | 关键路径的确定
An event is called critical if its earliest and latest event times are equal: EET = LET. A critical event has no wiggle room — oject. The critical path is formed by linking critical events through critical activities.
如果一个事件的最早和最晚事件时间相等，即 EET = LET，则该事件称为关键事件。关键事件没有任何回旋余地——对这一事件的任何延误都会直接推迟整个项目。关键路径由关键事件通过关键活动连接而成。
An activity i → j is critical if all the following hold:
- EETi = LETi
- EETj = LETj
- EETj – EETi = dij
一项活动 i → j 若同时满足以下条件即为关键活动：
- EETi = LETi
- EETj = LETj
- EETj – EETi = dij
l duration is 4 + 5 + 3 = 12 days. Any delay to these activities will increase the project's overall duration.
在我们的示例中，事件 1、3、4 和 5 是关键事件（EET = LET）。活动 B、D 和 F 满足条件，因此关键路径为 1 → 3 → 4 → 5。总工期为 4 + 5 + 3 = 12 天。对这些活动的任何延误都会增加项目的总工期。
7. Total Float and Event Times | 总浮动与事件时间
Total float is the amount of time an activity can be delayed without affecting the project completion date, assuming all other activities start at their earliest times. It can be calculated directly from the event times of its endpoints:
总浮动是指在不影响项目完成日期的条件下，某项活动可以延迟的时间量，假设所有其他活动都按最早时间开始。它可以直接由其端点的事件时间计算得出：
Total Float = LETj – EETi – Duration
e, activity A (1 → 2) has LET(2) = 7, EET(1) = 0 and duration 3, giving a total float of 7 – 0 – 3 = 4 days. Activity C (2 → 4) has total float 9 – 3 – well. Non-critical activities often share float, so careful resource scheduling is needed.
例如，活动 A (1 → 2) 的 LET(2) = 7，EET(1) = 0，持续时间为 3，总浮动为 7 – 0 – 3 = 4 天。活动 C (2 → 4) 的总浮动也是 9 – 3 – 2 = 4 天。非关键活动经常共享浮动，因此需要仔细安排资源调度。
Understanding the link between event times and floats helps you quickly verify calculations: a critical activity will always have total float equal to zero, and this matches the fact that for critical events EET and LET are identical.
理解事件时间与浮动之间的联系有助于快速验算：关键活动的总浮动始终为零，这与关键事件上 EET 和 LET 相等这一事实相符。
8. Avoiding Common Mistakes | 常见错误避免
Many errors occur during the forward pass when candidates forget to consider all incoming activities. Always list every predecessor and pick the maximum. Similarly, in the backward pass, the rule is minimum, not maximum. Start the backward pass with LET of the sink node equal to its EET – never zero unless the project duration is zero.
许多错误发生在正向推算时，考生忘记考虑所有进入活动。务必列出每个前驱并取最大值。同样，在反向推算中，使用最小值规则而非最大值。开始反向推算时，终点节点的 LET 应取为其 EET —— 除非项目工期为零，否则绝不要设为零。
Another frequent slip is assuming a network is automatically simple; always draw the activity network clearly, label events and durations, and record EET and LET step by step. Mixing up events and activities is common – remember that event times belong to nodes, while durations belong to the arcs connecting them.
另一个常见失误是以为网络会自动简化；始终清晰地画出活动网络，标出事件和持续时间，并逐步记录 EET 和 LET。事件与活动混淆也很常见——请记住事件时间属于节点，而持续时间属于连接它们的弧。
Finally, check that LET(source) equals 0. If it does not, a miscalculation has occurred in the backward pass or the network might contain a dangling activity that was missed.
最后，检查 LET（源节点）是否等于 0。如果不等于，说明反向推算中有误算，或者网络中可能遗漏了某项悬挂活动。
9. Interpreting Event Times in Precedence Networks | 在优先网络中诠释事件时间
In many Edexcel exam questions, you are given an activity-on-node diagram with the events already numbered. You must complete the EET and LET boxes and use them to answer questions about the critical path and floats. The event times directly tell you how much scheduling flexibility exists around each milestone.
在许多 Edexcel 考试题目中，你会拿到一张已标好事件编号的节点活动图。你需要填完 EET 和 LET 框，并用它们回答关于关键路径和浮动的问题。事件时间直接告诉你围绕每个里程碑存在多少调度灵活性。
If an event has EET = LET, that milestone is 'fixed' – n early and late times, the gap gives the total slack available before that event begins to affect the project end.
如果某个事件的 EET = LET，则该里程碑是“固定”的——任何结束于该点或从该点开始的活动，若同时满足持续时间条件，也将是关键活动。当事件的最早与最晚时间之间存在差异时，该差值表示在事件开始影响项目结束之前可用的总时差。
Practising the interpretation of these numbers is vital: a large gap at a non-critical event can suggest an opportunity to delay an entire chain of activities, which is useful when levelling resources.
练习解读这些数字至关重要：非关键事件上的较大差值提示有机会推迟整条活动链，这在均衡资源时非常有用。
10. Practice Tips for Exam | 考试练习建议
In the exam, show your working methodically. Draw a table with events 1 to n, list incoming/outgoing activities, and apply the max/min rules in order. Even if the diagram already exists, rewriting key calculations helps avoid careless slips.
在考试中，要有条理地展示步骤。绘制一个包含事件 1 至 n 的表格，列出进入/离开的活动，并依次应用最大/最小值规则。即使题目已给出网络图，重写关键计算过程也有助于避免粗心失误。
al activities, verify that total float is exactly zero. If time permits, re-calculate a sample node to ensure consistency.
始终检查两次推算中都纳入了所有活动。完成反向推算后，确认 LET（源节点）= 0。对于关键活动，核对其总浮动恰好为零。如果时间允许，重新计算一个样本节点以确保一致性。
Familiarity with common network shapes — such as multiple parallel paths and merging nodes — will speed up your analysis. Work through past paper questions repeatedly; the process of finding early and late event times becomes second nature with practice.
熟悉常见的网络结构——如多条平行路径和合并节点——将加快你的分析速度。反复练习历年真题；通过练习，寻找最早与最晚事件时间的过程会变成你的下意识动作。
Published by TutorHao | Decision Mathematics Revision Series | aleveler.com
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