📚 A-Level Further Maths Unit 5 Mark Scheme (Jun19) Key Concepts | A-Level进阶数学单元5 2019年6月评分方案知识点精讲
This article breaks down the essential knowledge points from the June 2019 A-Level Further Mathematics Unit 5 mark scheme, focusing on Decision Mathematics. Whether you are revising linear programming, critical path analysis, or network flows, understanding how examiners award method and accuracy marks can dramatically boost your performance. We explore typical question styles, common pitfalls, and the precise wording required to secure full marks.
本文深度剖析2019年6月A-Level进阶数学第五单元（决策数学）评分方案中的核心知识点。无论你正在复习线性规划、关键路径分析还是网络流，了解阅卷员如何给方法分和准确度分可以显著提升你的考试成绩。我们将逐一分析典型题型、常见错误，以及获得满分所需的关键表达。
1. Formulating a Linear Programming Problem | 线性规划问题的构建
ectly identifying decision variables and writing the objective function. Candidates needed to define variables such as 'let x be the number of type A cabinets' and 'let y be the number of type B cabinets' 40x + 65y, and a full set of constraints including non-negativity. A method mark (M1) was given for any two correct constraints, while all constraints correct earned accuracy marks (A1).
在2019年6月的试卷中，前几分通常奖励给正确定义决策变量并写出目标函数的考生。考生需要定义变量，如“设 x 为 A 型柜子的数量”、“设 y 为 B 型柜子的数量”。评分方案明确要求清晰的利润表达式，例如 P = 40x + 65y，以及包含非负约束在内的完整约束组。任何两个正确约束条件即可获得方法分（M1），全部约束正确则拿到准确度分（A1）。
0 into equations with slack variables. The mark scheme insisted on 2x + 3y + s₁ = 60 with s₁ ≥ 0. Examiners looked for the correct introduction of surplus variables for ≥ inequalities, which often caused confusion in minimisation problems.
当将约束条件整理为单纯形法的标准形式时，许多学生因为未能把像 2x + 3y ≤ 60 这样的不等式转化为带松弛变量的方程而丢分。评分方案要求写成 2x + 3y + s₁ = 60，且 s₁ ≥ 0。阅卷员会检查是否正确引入剩余变量来处理 ≥ 不等式，这在极小化问题中经常造成混淆。
2. Simplex Tableau and Optimality Checking | 单纯形表与最优性检验
rded the construction of the initial simplex tableau with the objective row expressed as P – 40x – objective coefficients with wrong signs. M1 was awarded for a tableau with correctly placed coefficients, and A1 for the fully correct initial tableau. The pivot selection step required identifying the most negative indicator in the P‑row, which determined the entering variable.
2019年评分方案奖励能够正确构建初始单纯形表的考生，要求目标行写成 P – 40x – 65y = 0 的形式。常见错误是目标函数系数的符号写反。正确放置系数可获得方法分（M1），完全正确的初始表得准确度分（A1）。选择主元的步骤要求找到 P 行中最负的指示数，由此确定入基变量。
next tableau, and for testing optimality by checking that all indicators in the objective row were non‑ th the corresponding values read from the final tableau. Losing a mark for not giving the value of the objective function was a recurring theme in examiner reports.
后续分数分配给行变换得到下一单纯形表，以及通过检查目标行所有指示数是否非负来检验最优性。最终答案必须明确写明 x = a, y = b, P = c，并且这些值应从最终的表中读出。阅卷报告反复提到，未能给出目标函数值会导致丢分。
3. Critical Path Analysis – Activity on Arc | 关键路径分析 – 箭线图
June 2019 Unit 5 featured a precedence table for a construction project. Candidates were required to draw an activity-on-arc network, ensuring that dummy activities were used correctly to preserve logical dependencies. The mark scheme gave M1 for a network that respected most of the precedence relationships, and A1 for a fully correct diagram. Common mistakes included missing dummies between activities that shared some but not all predecessors, or creating redundant dummies that inflated the number of events.
2019年6月的单元5考了一道建筑工程的先后关系表。考生需要画出箭线图（activity on arc），并正确使用虚拟活动以保持逻辑依赖关系。评分方案对大部分先后关系正确的网络图给方法分（M1），完全正确的图则给准确度分（A1）。常见错误包括在部分共祖但非全同的活动之间遗漏了虚拟活动，或者添加了多余的虚拟活动而增加了事件数量。
Once the network was drawn, a forward pass calculated earliest start times (EST), and a backward pass gave latest finish times (LFT). A mark was awarded specifically for correctly computing the earliest event times, and another for the latest event times. The critical path was identified as the path with zero total float on all its activities. Writing only the critical activities without stating them as a path (e.g., A–C–G) sometimes lost the final communication mark.
画出网络图后，前推法计算最早开始时间（EST），后推法得到最迟完成时间（LFT）。准确计算最早事件时间可得一个分数，准确计算最迟事件时间再得另一个分数。关键路径被定义为所有活动总时差为零的路径。仅写出关键活动而不以路径形式（如 A–C–G）呈现，有时会导致丢掉最后的表达分。
4. Total Float and Scheduling | 总时差与排程
Total float was calculated as LFT – EST – duration for each activity. The June 2019 mark scheme awarded B1 for giving the correct total float of at least three non‑critical activities. A scheduling diagram (Gantt chart) was then requested, where examiners looked for the correct positioning of bars according to EST and the use of floats to smooth resource usage. It was essential to label each bar clearly, and to indicate the time axis with consistent units.
每个活动的总时差计算公式为 LFT – EST – 持续时间。2019年6月的评分方案对至少三个非关键活动的正确总时差给出 B1 分。随后要求画出排程图（甘特图），阅卷员关注的是根据最早开始时间正确放置条形，并利用时差平衡资源使用。关键是要清晰标注每条，并标明时间轴及一致的单位。
A typical pitfall was confusing independent float with total float; the mark scheme only required total float for scheduling. When drawing the Gantt chart, some candidates placed activities at their latest start times without reference to resource constraints, which lost marks if the question specifically asked for a resource‑levelled schedule.
典型的误区是将独立时差与总时差混淆；排程只需要使用总时差。绘制甘特图时，有些考生在不考虑资源约束的情况下将活动放在最迟开始时间处，如果题目明确要求资源均衡排程，就会导致失分。
5. Hungarian Algorithm for Allocation Problems | 指派问题的匈牙利算法
An allocation problem in the 2019 paper required finding the minimum total cost of assigning four workers to four tasks. The mark scheme expected the cost matrix to be reduced row‑by‑row, then column‑by‑column. M1 was awarded for a correct row reduction, showing the subtraction of the smallest element in each row. A1 followed for a complete reduced cost matrix. The augmenting path method was then used to identify a minimum vertex cover and update the matrix until a complete matching was found.
2019年试卷中的指派问题要求找出将四名工人分派给四项任务的最小总成本。评分方案要求对成本矩阵先逐行削减，再逐列削减。正确进行行削减（逐行减去该行最小元素）可得 M1 分。得到完全化简的成本矩阵则得 A1 分。随后使用增广路径方法找出最小顶点覆盖，并更新矩阵，直至找到完全匹配。
Task A, etc.), and the minimum cost stated as an integer. Many candidates lost the final A‑mark because they forgot to extract the original costs and sum them – they only provided the pattern of zeros. Examiners advised double‑ e number of lines used to cover zeros equals the matrix order before declaring optimality.
在评分方案中，最终分配需要清晰列出（如工人 1→任务 C、工人 2→任务 A 等），并给出整数形式的最小成本。许多考生丢掉了最后准确度分，因为他们忘了提取原始成本并求和——只给出了零元素的模式。阅卷员建议，在宣布最优解之前，务必检查覆盖零元素所用的线条数等于矩阵阶数。
6. Game Theory – Two‑Person Zero‑Sum Games | 对策论 – 两人零和对策
mark scheme tested a 3×3 payoff matrix for player A. Candidates had to determine the play‑ ction using dominance arguments, where a row or column could be removed if its payoffs were uniformly worse for the respective player. M marks were given for correctly identifying and eliminating a dominated row or column, with a clear justification.
2019年6月的评分方案考了一道3×3支付矩阵，求局中人 A 的稳妥策略。考生需要通过求行最小值和列最大值来检查鞍点。没有鞍点时，需要用优势策略进行约简——如果某行或某列对相应局中人一致较差，则可以删除。正确识别并删除一个劣势行或列，并给出明确理由，即可获得方法分。
lity calculations. The expected payoffs were expressed as linear equations, e.g., 5p + 2(1 – p) = 3p + 4(1 – ated. A common error was mixing up the probabilities for the row player and column player.
约简后，对2×2对策用混合策略概率计算求解。期望收益表示为线性方程，例如 5p + 2(1 – p) = 3p + 4(1 – p)。评分方案要求局中人 A 的最终概率以精确分数给出，并写明对策的值。常见错误是把行局中人和列局中人的概率弄混。
7. Maximum Flow and Minimum Cut | 最大流与最小割
Network flow questions appeared in the 2019 Unit 5 paper with a directed graph showing capacities on arcs. Candidates had to find the maximum feasible flow from source S to sink T by augmenting along paths with available capacity. The mark scheme awarded M1 for each correctly identified flow‑augmenting path and the updated flow values. The final flow had to be verified by a cut whose capacity equals the total flow; stating the cut as a set of arcs with their capacities was essential for the full A1 mark.
2019年单元5试卷中的网络流题给出了一张有向图，弧上标有容量。考生需要通过沿有剩余容量的路径增流，找出从源 S 到汇 T 的最大可行流。评分方案对每条正确识别的增流路径及更新的流量值都给出 M1 分。最终流量必须用容量等于总流量的割来验证；将割表示为一组弧及其容量对拿到完整 A1 分至关重要。
In many mark schemes, a common mistake was failing to label both directions of flow after augmenting, especially the back‑flow on reversed arcs, which allowed subsequent paths to be found. Candidates were advised to maintain a clear flow‑capacity diagram, noting that the maximum flow was unique even though different sequences of augmenting paths might lead to it.
在许多评分方案中，常见错误是增流后忘记标注双向流量，特别是反向弧上的回流量，这会影响后续路径的发现。建议考生维护一张清晰的流量-容量图，注意最大流是唯一的，即便达到它的增流路径顺序可能不同。
8. Route Inspection (Chinese Postman Problem) | 路径检查（中国邮递员问题）
For a route inspection problem, the 2019 mark scheme required stating the condition for a closed trail that traverses every edge at least once: the graph must be Eulerian, i.e., all vertex degrees even. When the graph had exactly two odd vertices, the problem reduced to finding the shortest path between them and duplicating that path. Marks were given for correctly listing all odd vertices, finding shortest distances using Dijkstra, and calculating the new minimum route length.
针对路径检查问题，2019年评分方案要求写出能够遍历每条边至少一次的闭合迹的条件：图必须是欧拉图，即所有顶点度数均为偶数。当图恰好有两个奇顶点时，问题简化为求出它们之间的最短路径并复制该路径。正确列出所有奇顶点、用 Dijkstra 算法求出最短距离，并计算新的最小路线长度，均能获得分数。
The mark scheme frequently allocated a method mark for the pairing of odd vertices and a separate A1 for the total mew length. Candidates who forgot to add the original total weight of all edges often lost the final accuracy mark. Furthermore, when more than two odd vertices existed, the weighing of possible pairings needed systematic listing; the marking scheme rewarded candidates who showed a table of all possible complete pairings and their costs.
评分方案通常对奇顶点配对给出方法分，对新的总路线长度给出单独的 A1 分。忘记加上所有边的原始总权重，往往导致失去最后的准确度分。此外，当存在两个以上奇顶点时，需要对可能的配对组合进行系统列举；展示所有可能完整配对及其成本表格的考生，可获得奖励分。
9. Dynamic Programming – Maximising Returns | 动态规划 – 收益最大化
The June 2019 dynamic programming question involved allocating a limited resource (e.g., £ investments to maximise total return. The mark scheme credited the construction of a stage‑by‑ ion and cumulative return had to be recorded. A typical form was f₃(s) = max{r₃(x₃) + f₄(s – x₃)}, where f₄ represented the final stage with no further decisions.
2019年6月的动态规划题涉及将有限资源（如 50,000 英镑）分配到三项投资，以最大化总回报。评分方案认定分阶段表格的建立值得给分，从最后一个阶段开始逆向递推给 M1。对于每个状态，必须记录最优决策和累积回报。常见的形式为 f₃(s) = max{r₃(x₃) + f₄(s – x₃)}，这里 f₄ 代表无后续决策的最终阶段。
Each value in the table demanded a precise calculation, and the mark scheme gave A1 for each correct row of optimal values. The final answer was presented as the optimal allocation, e.g., £10k to A, £25k to B, £15k to C, and the maximum return. Omitting the reasoning trail, such as the recurrence relations, could lose method marks even if the final numbers were correct.
表格中的每个值都需要精确计算，评分方案对每行正确的最优值给出 A1。最终答案呈现为最优分配，例如 A 项目 1 万镑、B 项目 2.5 万镑、C 项目 1.5 万镑，以及最大回报。即便最终数字正确，遗漏推理过程（如递推关系）仍可能丢掉方法分。
10. Matchings in Bipartite Graphs | 二分图匹配
Questions on bipartite graphs appeared in the 2019 Unit 5 exam, requiring the construction of an initial matching and then improving it using an alternating path. The mark scheme awarded M1 for finding any feasible initial matching. An alternating path had to start at an unmatched vertex on one side and end at an unmatched vertex on the other side; the symmetric difference then produced a larger matching. Indicating the alternating path by a sequence of vertices was the expected notation.
2019年单元5考试中出现了二分图匹配的题目，要求构造一个初始匹配，然后利用交错路径改进它。评分方案对任何一个可行的初始匹配给予 M1 分。交错路径必须从一侧未匹配的顶点开始，结束于另一侧未匹配的顶点；利用对称差便得到一个更大的匹配。题目期望的表示法是通过顶点序列指示交错路径。
The final maximum matching had to be clearly stated, and the total size noted. In some cases, the exam required using the matching to solve an allocation problem where the bipartite graph represented suitability. Marks were lost when candidates did not explicitly link the matching back to the original problem, e.g., 'Person 1 does Job X, Person 2 does Job Y'.
最终的最大匹配必须清晰写出，并注明匹配总数。在一些情形中，考试要求用匹配解决一个分配问题，其中二分图代表适合性。如果考生没有将匹配明确关联回原问题，比如“人员 1 做工作 X，人员 2 做工作 Y”，则会失分。
11. Algorithms and Sorting – Efficiency and Tracing | 算法与排序 – 效率与追踪
The 2019 mark scheme included a question on sorting algorithms, typically asking how many comparisons and swaps would occur when applying bubble sort to a list of numbers. Candidates were expected to trace the algorithm, showing the state of the list after each pass. M1 was given for a correct first pass; full A1 for the completely sorted list with the correct number of passes. Some mark schemes also asked for the order of complexity, often 'O(n²) for bubble sort'.
2019年评分方案中包含一道排序算法题，通常要求说明对一个数字列表进行冒泡排序时会发生多少次比较和交换。考生需要追踪算法，展示每一遍之后列表的状态。正确的第一遍可得 M1；完全排好序的列表及正确的遍数可得完整的 A1。有些评分方案还要求写出复杂度阶，通常是“冒泡排序 O(n²)”。
For further mathematics, candidates might also face questions on the first‑fit decreasing algorithm for bin packing, or Dijkstra's shortest path algorithm. In all cases, the mark scheme stressed the importance of clear working, such as a table of permanent labels for Dijkstra. Skipping intermediate steps or failing to label vertices as the algorithm progresses often resulted in lost marks even if the final answer was correct.
在进阶数学中，考生还可能遇到装箱问题的首次适应递减算法，或 Dijkstra 最短路径算法。评分方案始终强调清晰呈现过程的重要性，例如 Dijkstra 的永久标号表。跳过中间步骤或者在算法推进中没有标注顶点，即使最终答案正确，也屡屡导致丢分。
Published by TutorHao | Further Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图