📚 Decision Mathematics 2 Exam Question Types Analysis | 决策数学2 题型解析
Decision Mathematics 2 (D2) is a demanding module in A-Level Further Mathematics that builds on the algorithms introduced in D1 and extends them into more sophisticated optimisation problems. Typical exam questions require not only accurate execution of algorithms but also careful interpretation of results, often within real-world contexts. In this article we analyse the major question types you will encounter, highlighting common pitfalls and efficient solution strategies. Mastery of these topics will significantly boost your confidence in tackling any D2 paper.
决策数学 2（D2）是 A-Level 进阶数学中要求较高的模块，它在 D1 所介绍算法的基础上，进一步拓展到更为复杂的优化问题。典型的考题不仅要求精确执行算法，还需要结合现实情境仔细解读结果。本文将分析你会遇到的主要题型，指出常见失分点并给出高效的解题策略。掌握这些专题将极大提升你应对任何 D2 试卷的信心。
1. Game Theory: Payoff Matrices and Dominance | 博弈论：支付矩阵与劣势策略
Zero-sum game questions almost always begin with a payoff matrix for player A. The first step is to check for a saddle point using the standard maximin-minimax approach. If the row maximin equals the column minimax, the game is strictly determined and you simply write down the value and optimal pure strategies. However, most D2 questions lead to a mixed-strategy solution. Before solving a 2×2 or 3×3 game, you are expected to reduce the matrix using dominance arguments. Remember: from A’s perspective, a row is dominated if every entry is less than or equal to the corresponding entry in another row. From B’s perspective, a column is dominated if every entry is greater than or equal to the corresponding entry in another column – because B wants to minimise A’s payoff.
零和博弈题几乎总是从玩家 A 的支付矩阵开始。第一步是利用标准的最大最小-最小最大方法检查鞍点。如果行最大最小等于列最小最大，博弈就被严格确定，你直接写出博弈值和最优纯策略即可。然而，大多数 D2 题目都会导向混合策略解。在求解 2×2 或 3×3 博弈之前，你应当用劣势策略原则化简矩阵。请记住：从 A 的角度，若某行的每个元素都小于或等于另一行的对应元素，则该行被占优；从 B 的角度，若某列的每个元素都大于或等于另一列的对应元素，则该列被占优——因为 B 希望最小化 A 的收益。
Once the matrix is reduced to 2×2, the standard formulas for the mixed strategies and game value should be second nature. For a matrix [[a, b], [c, d]], player A’s probability for row 1 is p = (d – c) / (a – b – c + d). Never forget to check that the calculated probabilities lie strictly between 0 and 1 – if not, you have made a reduction or calculation error. In a 3×3 scenario without dominance, you may be required to set up and solve simultaneous equations, but this is less common in D2 exams.
一旦矩阵化简为 2×2，混合策略和博弈值的标准公式就应成为第二本能。对于矩阵 [[a, b], [c, d]]，玩家 A 选择行 1 的概率为 p = (d – c) / (a – b – c + d)。永远不要忘记检查算出的概率是否严格介于 0 和 1 之间——若否，说明你的化简或计算有误。在无法使用劣势策略化简的 3×3 情形中，你可能需要建立并求解联立方程组，但这在 D2 考试中比较少见。
A typical exam twist asks you to modify the matrix slightly (e.g., a change in one payoff) and then discuss how the strategies alter. Always connect your answer to the dominance relationships rather than recomputing everything from scratch.
常见的考题变化是让你对矩阵稍作修改（例如改变一个收益值），然后讨论策略如何变化。永远要将你的答案与优劣关系联系起来，而不是从头再算一遍。
2. Simplex Method for Linear Programming | 线性规划的单纯形法
Standard simplex tableaux form the backbone of many D2 questions. You must be able to convert a worded linear programming problem into the canonical form with slack variables, then set up the initial tableau. The objective is to maximise Z, and all constraints should be of the ≤ type when using the standard maximisation simplex. If a constraint is ≥, you must multiply through by -1 (and of course reverse the inequality) before adding a slack variable – a very common source of sign errors.
标准单纯形表是许多 D2 题目的核心。你必须能够将文字描述的线性规划问题转化为带有松弛变量的标准形式，随后建立初始单纯形表。目标是最大化 Z，并且在使用标准最大化单纯形法时，所有约束应为 ≤ 类型。若约束为 ≥，则必须先将两边乘以 -1（当然不等号也要反向）再加松弛变量——这是一个非常常见的符号错误来源。
The simplex iterations require you to pick the most negative entry in the objective row to identify the pivot column, then compute the ratios (RHS divided by positive entries in the pivot column) to find the pivot row. Pay careful attention to degenerate cases where a ratio is zero or multiple rows tie; the exam may ask you to explain the significance of degeneracy or to apply the smallest-index rule.
每次单纯形迭代需要你选取目标行中最负的项以确定主元列，然后计算比率（右端项除以主元列中的正项）来找主元行。要特别注意退化情形，即某个比率为零或多行并列最小；考试可能要求你解释退化的意义或应用最小下标规则。
The tableau format itself needs to be perfectly legible. Always label your basic variables in the leftmost column and keep the objective row as the bottom row. When reading the final tableau, remember that only basic variables have values equal to the RHS; non-basic variables are zero. The shadow prices are found in the objective row under the slack variable columns of the optimal tableau.
单纯形表本身的格式必须完全清晰无误。总是把基变量写在最左边一列，并使目标行保持在最下行。阅读最终单纯形表时，请记住只有基变量的值等于右端项；非基变量均为零。影子价格可以从最优表中松弛变量列下方的目标行找到。
After obtaining the optimal solution, questions often ask for an interpretation of shadow prices. A shadow price of 2 for a resource means that if the resource availability increases by 1 unit, Z would increase by 2, provided the current basis remains optimal.
在得到最优解后，题目常会要求解释影子价格。若某种资源的影子价格为 2，则意味着在保持当前基仍为最优的条件下，若资源可用量增加 1 单位，Z 将增加 2。
3. Two-Stage Simplex and Big M Method | 两阶段单纯形法与大 M 法
When constraints involve ≥ or = , artificial variables must be introduced. In the two-stage method, you first minimise the sum of artificial variables (or maximise its negative) to find a feasible starting basis. Only if the optimal value of this Phase I objective is zero can you proceed to Phase II with the original objective. In the Big M method, you assign a huge penalty M to artificial variables in the objective function and perform a single sequence of iterations. D2 exams often favour the two-stage approach because it avoids the ambiguity of ‘big enough’ M.
当约束条件包含 ≥ 或 = 时，必须引入人工变量。在两阶段法中，首先最小化所有人工变量之和（或最大化其相反数）以找到一个可行的初始基。仅当第一阶段目标的最优值为零时，才能进入带有原目标的第二阶段。在大 M 法中，你给目标函数中的人工变量分配一个极大的罚值 M，并执行单轮迭代。D2 考试通常更偏爱两阶段法，因为它避免了 M “足够大”的模糊性。
A tricky exam question might present a final tableau and ask you to identify which stage it belongs to, or to deduce whether the original problem is infeasible. If any artificial variable remains basic with a positive value at the end of Phase I, the problem has no feasible solution.
一道有难度的考题可能会给出最终单纯形表，并要求你判断它属于哪一阶段，或推断原问题是否无可行解。如果在第一阶段结束时仍有任何人工变量作为基变量且取正值，则该问题无可行解。
Always write a clear dictionary or tableau template before starting, and double-check that all artificial variables are initially included with coefficient -1 in the Phase I objective. A single sign slip can make the entire table false.
开始解题前，一定要写出清晰的字典或表格模板，并仔细检查第一阶段目标中所有人工变量是否初始都以系数 -1 出现。一个符号的纰漏就可能导致整个表格出错。
4. Transportation Problem | 运输问题
Transportation problems involve distributing goods from several sources to several destinations at minimum cost. A D2 question typically gives a cost matrix and supply/demand values, and asks for an initial feasible solution using the North-West corner rule or the least-cost method, followed by the stepping-stone or MODI (modified distribution) method to find the optimal solution. The exam often requires you demonstrate both phases clearly.
运输问题涉及以最低成本将货物从多个供应点运送到多个目的地。D2 考题通常会给出一个成本矩阵以及供应/需求值，要求使用西北角法则或最小成本法给出初始可行解，随后采用踏脚石法或改进分布法（MODI 法）寻找最优解。考试往往要求你清晰地展示这两个阶段。
When applying the stepping-stone method, always draw a closed loop for the entering cell and ensure each corner alternates between adding and subtracting one unit. The improvement index must be negative for minimisation problems; if all indices are non-negative, the current solution is optimal. However, if the index is zero, multiple optimal solutions exist – the examiner may ask for one alternative optimum.
应用踏脚石法时，一定要为入基格画出闭回路，并确保每个顶点交替增加和减去一个单位。对于最小化问题，改进指数必须为负；若所有指数均已非负，当前解即为最优。但若指数为零，则存在多重最优解——考官可能要求你给出一个替代最优解。
Degeneracy in transportation problems (fewer than m+n-1 occupied cells) must be handled by inserting a zero shipment in an appropriately chosen empty cell, marked with an epsilon or zero. Never forget to check the number of occupied cells after each iteration.
运输问题中的退化（占据格少于 m+n-1 个）必须通过在适当选择的空单元格中插入零运量来处理，并标记为 ε 或 0。每次迭代后都别忘了检查占据格的数量。
If the problem is unbalanced – total supply does not equal total demand – you must introduce a dummy source or dummy destination with zero cost, then solve as standard. Missing this step is a classic error.
如果问题不平衡——总供应不等于总需求——你必须引入一个虚拟供应点或虚拟目的地，其成本为零，然后按标准方法求解。遗漏这一步是典型的错误。
5. Assignment Problem (Hungarian Algorithm) | 指派问题（匈牙利算法）
The assignment problem asks you to pair n workers to n jobs at minimum total cost, or maximum total profit. The Hungarian algorithm has well-defined steps: subtract row minima, then column minima; cover all zeros with the minimum number of lines; if the number of lines equals n, an optimal assignment is found; otherwise, subtract the smallest uncovered element and add it to double-covered elements, and repeat. An exam question may present a benefit matrix and require conversion to a regret matrix before applying the algorithm if it is a maximisation problem.
指派问题要求你将 n 个工人与 n 项工作配对，使得总成本最小或总收益最大。匈牙利算法有明确的步骤：先减去行最小值，再减去列最小值；用最少的直线覆盖所有零；若直线条数等于 n，则已找到最优指派；否则减去未被覆盖的最小元素，并将其加回被双重覆盖的元素，然后重复。如果问题是最大化形式，考题可能会给出收益矩阵，并要求在应用算法前先将其转换为后悔值矩阵。
After finding the optimal assignment, always restate the solution clearly: Worker 1 → Job 3, etc., and compute the total cost. Occasionally the exam requires you to adapt the algorithm for a non-square matrix, which involves adding dummy rows or columns with zero entries.
找到最优指派后，一定清晰重述解答：工人 1 → 工作 3 等，并计算总成本。考试偶尔会要求你为非方阵调整算法，这时需要添加带有零项的虚拟行或列。
One subtlety is when there are multiple zero entries and more than one possible covering. The Hungarian method will still lead to the optimum, but be systematic in your choice of lines to avoid getting stuck in a loop.
一个微妙之处在于当有多个零格且存在不止一种可能的覆盖方式时。匈牙利算法仍将通向最优解，但在选取直线时要保持系统性，以免陷入循环。
6. Critical Path Analysis (CPA) | 关键路径分析
CPA questions in D2 often move beyond simple project networks to include dummy activities, multiple dependencies, and resource leveling. You must be able to draw the activity-on-arc network from a precedence table, perform forward and backward passes to obtain earliest and latest event times, and identify the critical path(s). The float of an activity – total, free, and independent – is frequently examined. Remember: total float = latest finish time – earliest start time – duration; free float = earliest start of all following activities – earliest finish; independent float may be negative and is set to zero if so.
D2 中的关键路径分析题目常常超越简单的项目网络，包含虚工作、多重依赖关系以及资源平衡。你必须能够根据前导表绘制活动-箭线网络，执行正推与逆推以获得事件的最早和最晚时间，并找出关键路径。活动的时差——总时差、自由时差和独立时差——是常考内容。请记住：总时差 = 最晚结束时间 – 最早开始时间 – 持续时间；自由时差 = 所有后续活动的最早开始时间 – 最早完成时间；独立时差可能为负，若为负则取零。
Resource histograms and scheduling are a hallmark of D2. Given a list of activities, their durations, and resource requirements, you must construct a resource histogram for the earliest start schedule, and then smooth the resources within the available float so that the maximum resource demand is minimised. This often requires a trial-and-improvement approach within a Gantt chart.
资源直方图与调度是 D2 的标志性内容。给出活动列表、持续时间及资源需求，你必须为最早开始时间表构建资源直方图，然后在可用时差内平滑资源，使得资源需求的最大值最小化。这通常需要在甘特图中采用试凑法进行优化。
Don’t forget that if a project is delayed, you can use the updated completion time to calculate the cost of crashing certain activities. crashing involves reducing duration at extra cost; D2 questions often ask you to find the cheapest way to achieve a specified reduction in project length.
不要忘记，如果项目发生延误，你可以利用更新后的完成时间来赶工某些活动。赶工涉及以额外成本缩短工期；D2 题目常要求你找出在指定缩短项目工期条件下成本最低的方案。
7. Resource Histograms and Scheduling | 资源直方图与调度
This topic is closely related to CPA but merits its own section because questions often isolate the scheduling process. You will be given a precedence table and a resource (e.g., number of workers) required per activity. You first build a Gantt chart using earliest start times, then plot a resource histogram showing the total resource requirement per time unit. The goal is to reschedule non-critical activities within their float to reduce peaks. The exam expects you to draw a smoothed histogram and state the maximum resource level after levelling.
该专题与关键路径分析密切相关，但因其问题常会分离出调度过程，故单独列为一节。题目会给出前导表和每项活动所需的资源（例如工人数）。你首先用最早开始时间画出甘特图，然后绘制资源直方图，展示每时间单位的总资源需求。目标是在浮动时间内重新安排非关键活动以降低峰值。考试要求你画出平滑后的直方图，并说明平衡后的最大资源水平。
A useful technique is to list activities by float descending, and try to delay those with the largest total float first. Always check that any delay respects both the preceding and succeeding dependencies. Clearly state your final schedule and the revised resource peak; marks are often awarded for the clarity of your reasoning as much as the final histogram.
一个实用的技巧是按总时差降序列出活动，并优先尝试推迟总时差最大的活动。每次延迟都要检查是否遵守前后依赖关系。清晰地写出最终调度和修改后的资源峰值；你的推理清晰度往往与最终的直方图一样能拿到分值。
A second, related question type is ‘smoothing with a fixed resource limit’. If the total available resource is capped, you may need to delay some activities beyond their total float, which extends the project duration. The exam will ask you to find the minimum project duration subject to the resource constraint.
第二类相关问题类型是“在固定资源限制下的平滑”。如果可用资源总量有上限，你可能需要将某些活动延迟至超出其总时差，这会导致项目工期延长。考试会要求你找出在该资源约束下的最短项目工期。
8. Network Flows and Max-Flow Min-Cut Theorem | 网络流与最大流最小割定理
Flow network problems in D2 involve directed graphs with capacities on arcs, and you must find the maximum flow from a source S to a sink T. The labelling procedure – systematically finding augmenting paths and updating flows – is the standard algorithm. Each time you label a node, record the preceding node and the potential additional flow. Backtrack from T to S and add the flow along the path, adjusting residual capacities. Stop when no more augmenting paths exist.
D2 中的流网络问题涉及带有弧容量的有向图，你必须找出从源点 S 到汇点 T 的最大流。标号法——系统地寻找增广路径并更新流量——是标准算法。每次为节点标号时，记录前驱节点和潜在增加的流量。从 T 回溯至 S，并沿路径增加流量，同时调整剩余容量。当再也找不到增广路径时停止。
The max-flow min-cut theorem states that the maximum flow equals the minimum cut capacity. A cut partitions the vertices into two sets containing S and T respectively, and its capacity is the sum of capacities of arcs going from the S-set to the T-set. Exam questions frequently ask you to verify the theorem by stating both the max flow and a corresponding minimum cut.
最大流最小割定理指出，最大流量等于最小割的容量。割将顶点划分为分别包含 S 和 T 的两个集合，其容量是所有从 S-集指向 T-集的弧的容量之和。考试题常要求你通过同时陈述最大流值和相应的最小割来验证该定理。
A tricky aspect is when there are multiple sources or sinks. You must introduce a super-source and a super-sink with infinite capacities. Also be aware of capacity constraints on nodes rather than arcs; you can split a node into two with a new arc carrying the node’s capacity.
一个棘手之处是存在多个源点或多个汇点的情况。此时必须引入一个超级源点和一个超级汇点，并以无穷大容量连接。还要注意节点容量限制而非弧容量限制的情况；你可以将节点分裂为两个，并用一条容量为该节点容量限制的新弧相连。
After finding the maximum flow, the exam often probes your understanding by asking: ‘If the capacity of arc AB were increased by 1, would the max flow increase?’ You must check whether that arc lies in every minimum cut. If yes, then flow can increase.
在找出最大流后，考试常常会这样检验你的理解：“如果将弧 AB 的容量增加 1，最大流会增加吗？”你必须检查该弧是否位于每一个最小割中。如果是，则最大流可以增加。
9. Dynamic Programming | 动态规划
Dynamic programming (DP) in D2 is characterised by finding optimal decisions in a multi-stage process. Typical applications include shortest-path problems through a network of stages, allocation of resources, or knapsack-type problems. You must define the state variables and the stage, and then write down the recurrence relation. For a deterministic DP, the backward recursion fn(s) = opt { decision cost + fn+1(next state) } is fundamental. Always build a table for each stage showing all possible states and their optimal values.
D2 中的动态规划围绕在多阶段过程中寻找最优决策。典型应用包括穿越分层网络的最短路径问题、资源分配问题或背包类问题。你必须定义状态变量和阶段，然后写出递推关系。对于确定性 DP，反向递推公式 fn(s) = opt { 决策成本 + fn+1(下一状态) } 是最基本的。始终要为每个阶段建表，列出所有可能状态及其最优值。
When tackling a shortest-path on a directed acyclic network, label the nodes with backward costs. Start from the sink with value 0, and work backwards. The optimal path is traced forward by following the minimising decisions. This is extremely algorithmic, so present your working in neat columns to avoid confusion.
在处理有向无环网络的最短路径时，用逆向成本为节点标号。从汇点标 0 开始，向后递推。最优路径通过正向追踪最小化决策来重构。这个过程算法性极强，因此用整洁的列来展示你的计算以避免混乱。
Another popular DP problem is the ‘stagecoach’ or ‘optimal investment’ type where you have to allocate a resource (e.g., money, machines) among several projects. The stage is the project, the state is the amount left to allocate. Draw a grid and fill in returns for each state-decision combination.
另一种常见的 DP 问题是“驿站马车”或“最优投资”类型，你需要将一种资源（如资金、机器）在若干项目间分配。阶段是项目，状态是待分配的资源数量。画一个网格，并为每个状态-决策组合填入回报值。
A subtle point: when the DP involves probabilities (stochastic DP), you must use expected values. The recurrence becomes fn(s) = opt { immediate expected return + Σ pk fn+1(state reached) }. The exam might present a decision tree and ask you to fold back to find the optimal policy.
一个需要注意的地方：当 DP 包含概率（随机 DP）时，你必须使用期望值。递推关系变为 fn(s) = opt { 即时期望回报 + Σ pk fn+1(达到的状态) }。考题可能呈现一个决策树，并要求你进行折叠回推以找出最优策略。
10. Recurrence Relations and Solving Techniques | 递推关系及其求解技巧
D2 often includes questions on first and second order linear recurrence relations. A first-order recurrence un+1 = a un + b can be solved by finding the steady-state value L = b/(1 – a) and expressing un = L + (u₀ – L) an, provided a ≠ 1. You must be able to interpret long-term behaviour: if |a| < 1, the sequence converges to L; if |a| > 1, it diverges (unless the starting point is exactly L).
D2 常会出关于一阶和二阶线性递推关系的题目。一阶递推 un+1 = a un + b 可通过求稳态值 L = b/(1 – a) 并表达为 un = L + (u₀ – L) an 来求解，前提是 a ≠ 1。你必须能够解释长期行为：若 |a| < 1，序列收敛到 L；若 |a| > 1，则发散（除非起始值恰好为 L）。
For second-order homogeneous recurrences un+2 = p un+1 + q un, you form the characteristic equation r² – p r – q = 0. If the roots r₁ and r₂ are distinct, the general solution is un = A r₁n + B r₂n. If there is a repeated root r, the solution is un = (A + B n) rn. The constants A and B are determined by the given initial values. Later the question may ask you to find the limit of un as n → ∞ or to apply the model to a real situation, such as population growth or financial loans.
对于二阶齐次递推 un+2 = p un+1 + q un，你需要构造特征方程 r² – p r – q = 0。若根 r₁ 和 r₂ 不相等，通解为 un = A r₁n + B r₂n。若有重根 r，解为 un = (A + B n) rn。常数 A 和 B 由给定的初值确定。随后题目可能要求你求 n → ∞ 时 un 的极限，或将模型应用于实际情境，如人口增长或金融贷款。
A common pitfall is forgetting to check whether the recurrence is homogeneous.
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导