Critical paths and scheduling · 关键路径与调度
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| float/fləʊt/ | 浮动时间 | fú dòng shí jiān |
Must every activity wait for every other?
- A school event has activities that can run in parallel. Adding every duration overestimates the minimum completion time.
- This lesson studies float 浮动时间: The time an activity can be delayed without delaying completion under the network model.
是否每项活动都必须等待其他所有活动？
- 学校活动包含可并行进行的环节。将所有持续时间相加会高估最短完成时间。
- 本课研究浮动时间：在网络模型下，某项活动在不延误整体完工时间的前提下最多可延迟的时间。
Choose the mathematical structure
- For an activity network, calculate earliest event times forward and latest event times backward. Total float for activity i to j is L_j-E_i-duration. A critical activity has zero total float; more than one critical path may exist.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
选择数学结构
- 对于活动网络，从前向后计算最早事件时间，从后向前计算最迟事件时间。活动 i 到 j 的总浮动时间为 L_j - E_i - 持续时间。关键活动的总浮动时间为零；可能存在多条关键路径。
- 计算前请先明确允许的输入项和单位。方程应表达关系本身，而不仅仅是记录计算器按键过程。
Which description correctly defines float? · 哪项描述正确定义了浮动时间（Float）？
The time an activity can be delayed without delaying completion under the network model. · 在网络模型中，某活动在不延误整体完工时间的前提下可延迟的时间。
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
Activities A=3 and B=5 start together. C=4 follows A, and D=2 follows both B and C. The earliest completion of C is 7; D must wait until max(5,7)=7 and finishes at 9. Path A-C-D is critical; B has float 2.
通过验证案例进行推导
- 将结果与初始量进行比对。代入原始关系式，或在适当情况下对比图表与数值答案。
活动 A=3 和 B=5 同时开始。C=4 在 A 之后进行，D=2 需在 B 和 C 均完成后才能开始。C 的最早完工时间为 7；D 必须等待至 max(5,7)=7，最终完工于 9。路径 A-C-D 为关键路径；B 拥有浮动时间 2。
Critical paths and scheduling · 关键路径与调度
For an activity network, calculate earliest event times forward and latest event times backward · 对于活动网络，向前计算最早事件时间，向后计算最迟事件时间。
Compare the model with the worked case and explain one change. · 对比模型与已解案例，并说明其中一处变化。
For A=3,C=4 after A,D=2 after C, find total path duration. · 当A=3, C=4，且C后接D=2时，求总路径持续时间。
The path A-C-D takes 3+4+2=9 time units. · 路径A-C-D耗时3+4+2=9时间单位。
Test a tempting shortcut
- Do not add durations of independent parallel activities. A zero-float activity belongs to a critical path, but there may be several such paths. Resource limits can require a schedule longer than the network's theoretical minimum.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
The minimum project duration is always the sum of every activity duration. This claim is false. Explain which definition or assumption it violates.
检验一个诱人的捷径
- 不要将相互独立的并行活动的持续时间相加。总浮动时间为零的活动属于关键路径，但可能存在多条此类路径。资源限制可能导致实际排程长于网络理论上的最短工期。
- 当捷径失效时，找出其违背的假设。保留精确值直到题目要求的最终舍入步骤。
项目最短工期总是等于所有活动持续时间之和。此说法错误。请说明它违背了哪一定义或假设。
B=5 must finish before D starts at time 7. Find B's float. · B=5必须在时间7前完成，D方可开始。求B的浮动时间。
B normally finishes at 5 but can finish at 7: total float=7-5=2. · B正常完成时间为5，但最晚可完成于7：总浮动时间=7-5=2。
The minimum project duration is always the sum of every activity duration. · 最小项目工期总是所有活动工期的总和。
Do not add durations of independent parallel activities. A zero-float activity belongs to a critical path, but there may be several such paths. Resource limits can require a schedule longer than the network's theoretical minimum. · 不要累加独立并行活动的工期。零时差活动属于关键路径，但可能存在多条这样的路径。资源限制可能导致进度计划长于网络理论上的最小值。
Interpret a new situation
- Draw precedence relationships before assigning times. Distinguish activity duration from event time. Explain what a delay does to the completion date, and record any resource assumptions.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
解读新情境
- 在分配时间之前先绘制优先关系图。区分活动持续时间与事件时间。解释延误对完工日期的影响，并记录任何关于资源的假设。
- 完整解答需给出数学结果并解释其含义。检查其在所述背景下是否可行。
A project starts at time 0 and finishes at 9. Find its minimum duration. · 某项目始于时间0，止于时间9。求其最短工期。
Project duration=end time-start time=9-0=9. · 项目工期=结束时间-开始时间=9-0=9。
Match each part of a complete solution to its purpose. · 将完整解答的每个部分与其目的相匹配。
An assumption justifies the model; a check tests the result; interpretation connects it to the question. · 假设用于论证模型的合理性；检查用于验证结果；解释用于将其与问题建立联系。
Use this in your course
- edexcel IAL further mathematics; official unit D1. Other-unit enrichment is identified in the scope review; it is not an extra cash-in requirement.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The time an activity can be delayed without delaying completion under the network model. Choose the relationship, show the method, check its assumptions and interpret the result.
将其应用于你的课程
- Edexcel IAL高等数学；官方单元D1。其他单元的补充内容已在范围审查中标识；这并非额外的加分要求。
- 在给出最终答案前先展示解题方法，并遵循试卷的计算器及公式使用规则。通过定位第一个无效步骤来复盘错误答案。
在网络模型下，某项活动在不延误整体完工时间的前提下最多可延迟的时间。选择关系，展示方法，检查其假设并解释结果。