📚 Mastering Decision Mathematics 2: Top Tips for High Scores | 掌握决策数学2：高分技巧全攻略
Decision Mathematics 2 (D2) builds on the algorithms and logical structures encountered in D1, introducing more powerful optimisation techniques such as the simplex method, dynamic programming and game theory. To score highly, students must combine procedural fluency with deep conceptual understanding, and be able to interpret results in real-world contexts. This guide presents the essential strategies and common pitfalls to help you secure top marks in your D2 exam.
决策数学2（D2）建立在D1算法与逻辑结构的基础上，引入了简单形法、动态规划和博弈论等更强大的优化技术。要取得高分，学生必须将流程的熟练与深层概念理解相结合，并能在真实情境中解读结果。本指南提供了关键策略和常见错误分析，帮助你在D2考试中稳获高分。
1. Exam Overview and Key Topics | 考试概述与关键主题
The D2 module typically covers assignment problems, transportation problems, linear programming via the simplex method (including two-stage and big-M approaches), game theory for zero-sum games, dynamic programming for sequential decisions, network flow optimisation and advanced critical path analysis with crashing. High-scoring answers demand precise execution of algorithms, clear tabulation and the ability to explain the significance of optimal solutions.
D2模块通常涵盖指派问题、运输问题、用单纯形法求解线性规划（包括两阶段法与大M法）、零和博弈的博弈论、序贯决策的动态规划、网络流优化以及带赶工分析的高级关键路径。高分答案要求精确执行算法、清晰地制表并能解释最优解的意义。
Familiarity with the exam structure is vital: questions often combine multiple topics, such as a transportation problem followed by sensitivity analysis, or a game theory payoff matrix converted into a linear programming formulation. Time pressure is significant, so you must be able to move confidently between methods without hesitation.
熟悉考试结构至关重要：题目常常组合多个主题，例如运输问题后紧跟敏感性分析，或将博弈论收益矩阵转化为线性规划模型。时间压力很大，因此你必须能够自信地在不同方法间切换，毫不迟疑。
2. Mastering the Assignment Problem | 掌握指派问题
The Hungarian algorithm is the cornerstone. Remember the four key stages: subtract row minima, subtract column minima, cover all zeros with the minimum number of lines, and if the number of lines equals the order of the matrix you have an optimal assignment; otherwise, adjust the matrix by the smallest uncovered value and repeat. For unbalanced problems, always add dummy rows or columns with zero costs before applying the algorithm.
匈牙利算法是核心。记住四个关键步骤：减去行最小值、减去列最小值、用最少的直线覆盖所有零，如果直线数等于矩阵阶数则已得最优指派；否则用最小未被覆盖元素调整矩阵并重复。对于不平衡问题，务必在算法前加入成本为零的虚拟行或列。
When converting a maximisation assignment to a minimisation one, subtract all entries from the largest value in the matrix. After obtaining the optimal assignment, retrieve the original profit by referring back to the initial table. A common error is forgetting to re-substitute and consequently quoting the reduced-cost total as the final answer.
当把最大化指派问题转化为最小化问题时，用矩阵中的最大值减去所有元素。得到最优指派后，需回溯初始表格得出原始利润。常见错误是忘记回代，直接将缩简后的成本总和当作最终答案。
Consider the following cost matrix for three workers and three tasks:
| Worker \ Task | T₁ | T₂ | T₃ |
|---|---|---|---|
| W₁ | 18 | 26 | 21 |
| W₂ | 24 | 20 | 28 |
| W₃ | 27 | 22 | 19 |
After row and column reduction, you would obtain a reduced cost matrix and proceed to cover zeros. Practise drawing the minimum number of lines accurately; misidentifying the covering pattern is a frequent source of lost marks.
经过行列缩减，你将得到一个缩减成本矩阵并继续覆盖零元素。准确画出最少直线需要练习；错误判断覆盖模式是常见的失分点。
3. Transportation Problems: Methods and Optimality | 运输问题：方法与最优性
For an initial feasible solution, Vogel’s Approximation Method (VAM) usually gives a solution closer to optimal than the North-West Corner rule or the minimum cost method. VAM penalises the failure to use the cheapest routes by considering the difference between the two smallest costs in each row and column. Always attempt VAM first to reduce the number of iterations needed for optimality testing.
对于初始可行解，伏格尔近似法（VAM）通常比西北角法或最低成本法更接近最优解。VAM通过考虑每行和每列两个最小成本的差额，惩罚了未使用最便宜路径的情况。始终优先尝试VAM以减少最优性检验所需的迭代次数。
To test optimality, the modified distribution (MODI) method is efficient. Calculate dual variables uᵢ and vⱼ for occupied cells using uᵢ + vⱼ = cᵢⱼ, then evaluate the shadow costs for unoccupied cells: dᵢⱼ = cᵢⱼ − (uᵢ + vⱼ). If any dᵢⱼ is negative, the solution can be improved by introducing the cell with the most negative value. Beware of degeneracy: if an initial basic feasible solution has fewer than m+n−1 occupied cells, you must insert a zero allocation to maintain the stepping-stone path.
最优性检验可使用MODI法。对已占格用uᵢ + vⱼ = cᵢⱼ计算对偶变量uᵢ和vⱼ，再计算未占格的影子成本dᵢⱼ = cᵢⱼ − (uᵢ + vⱼ)。若任一dᵢⱼ为负，解可改进，引入最负值所在格。警惕退化问题：若初始基本可行解的已占格数少于m+n−1，必须插入一个虚分配0以维持踏石回路。
When forming the stepping-stone loop, alternate between adding and subtracting the re-allocated quantity θ. Keep adjustments clear using a labelled loop; many candidates lose accuracy by mis-tracing the loop. Choosing θ as the smallest allocation among the cells marked with a minus sign guarantees non-negativity.
构造踏石回路时，交替加上和减去重新分配量θ。使用带标注的回路保持清晰，很多考生因路线错误而失分。选择θ为标有减号的单元格中的最小分配量，能确保非负性。
4. Linear Programming with the Simplex Method | 线性规划与单纯形法
The simplex method solves maximisation and minimisation linear programmes in standard form. Convert inequalities to equalities using slack (≤), surplus (≥) and artificial (for ≥ and =) variables. For a two-stage approach, Stage I minimises the sum of artificial variables to find a feasible starting point; if the optimal value of the artificial objective is zero, a basic feasible solution exists. Stage II then optimises the original objective.
单纯形法求解标准形式下的最大化与最小化线性规划问题。用松弛变量（≤）、剩余变量（≥）和人工变量（≥与=）将不等式转换为等式。两阶段法中，阶段一最小化人工变量之和以寻找可行起点；若人工目标最优值为零，则存在基本可行解。阶段二接着优化原目标。
A typical simplex tableau includes the basic variable column, the coefficients matrix and the objective row. Select the pivot column by the most negative coefficient in the objective row (for maximisation) and the pivot row by the minimum ratio test: divide the right-hand side by the positive entries in the pivot column; the smallest non-negative ratio determines the departing variable. Update the tableau using row operations to make the pivot element 1 and all other entries in the column 0.
典型的单纯形表包含基变量列、系数矩阵和目标行。选择主元列：在最大化问题中，选目标行最负的系数；主元行由最小比值检验决定：右端项除以主元列中的正元素，最小的非负比值确定离开变量。通过使主元素为1且列中其它项为0的行变换更新表格。
After an optimal tableau is reached, interpret the final solution: the value of each basic variable appears in the RHS column, and the objective value in the bottom-right corner. Shadow prices are given by the final objective row coefficients of the slack/surplus variables. For sensitivity analysis, the allowable ranges for right-hand side values can be deduced from the final tableau ratios.
达到最优表后，解读最终解：每个基变量的值出现在右端列中，目标值在右下角。松弛/剩余变量在最终目标行的系数即为影子价格。对于敏感性分析，右端项系数的允许变化范围可从最终表的比值中推出。
| Basis | x₁ | x₂ | s₁ | s₂ | RHS |
|---|---|---|---|---|---|
| x₂ | 0 | 1 | ⅓ | -½ | 20 |
| x₁ | 1 | 0 | -¼ | ⅔ | 30 |
| Z | 0 | 0 | 5 | 8 | 310 |
In this optimal tableau, x₁ = 30, x₂ = 20, maximum Z = 310, and the shadow price of the first constraint is 5.
在此最优表中，x₁ = 30，x₂ = 20，最大Z = 310，第一个约束的影子价格为5。
5. Game Theory: Pure and Mixed Strategies | 博弈论：纯策略与混合策略
For two-person zero-sum games, first check for a saddle point: the maximin value equals the minimax value. If a saddle point exists, the game is stable and pure strategies are optimal. In the absence of a saddle point, you must determine mixed strategies, where players randomise over their available actions to maximise their worst-case payoff.
对于两人零和博弈，首先检查鞍点：最大最小值等于最小最大值。若鞍点存在，博弈稳定，纯策略即为最优。无鞍点时，需确定混合策略，玩家在其可用行动上随机化以最大化其最差收益。
Formulate the mixed-strategy problem as a linear programme. For the row player, let probabilities be p₁, p₂, … such that ≤ 1; they choose p to maximise the expected value V subject to the condition that against any column the expected payoff is at least V. This leads to dual linear programmes that can be solved by the simplex method. In a 2×2 game, the formula p₁ = (d − c) / (a + d − b − c) is quick to use, where the payoff matrix for row player is:
| a | b |
| c | d |
将混合策略问题构造为线性规划。对行玩家，设概率p₁, p₂, … 且 ≤ 1；他们选择p以最大化期望值V，条件是对任意列，期望收益至少为V。这导出对偶线性规划，可用单纯形法求解。在2×2博弈中，公式 p₁ = (d − c) / (a + d − b − c) 可快速使用，此处的行玩家收益矩阵如上。
For games with more than two strategies, apply dominance reduction first: a strategy that is never better than another can be deleted. Then, if the game reduces to 2×n or m×2, graphical method helps identify the active strategies and the optimal mixed strategy. Be careful with sign conventions when setting up the linear programmes: the row player’s problem is a maximisation, and the column player’s is a minimisation.
对于多于两个策略的博弈，首先应用优势约简：从不优于另一策略的可被删除。若博弈缩减为2×n或m×2，图解法可帮助识别有效策略和最优混合策略。建立线性规划时注意符号约定：行玩家问题是最大化，列玩家问题是最小化。
6. Dynamic Programming for Sequential Decisions | 动态规划：序贯决策
Dynamic programming decomposes a multistage decision problem into smaller subproblems, solving backwards from the final stage. Define the state variables clearly (e.g., remaining budget, number of items) and the decision variables that transform the state from one stage to the next. The recurrence relation connects the optimal value function fₙ(s) at stage n with state s to fₙ₊₁ at the next stage.
动态规划将多阶段决策问题分解为较小的子问题，从最后阶段逆向求解。清晰定义状态变量（如剩余预算、物品数量）和将状态从一阶段转换至下一阶段的决策变量。递推关系将阶段n状态s的最优值函数fₙ(s)与下一阶段的fₙ₊₁联系起来。
Typical applications include shortest-path networks, resource allocation and production/inventory planning. Always construct a tableau for each stage, listing feasible states, decisions, immediate reward and the resulting value. A common error is mislabelling the stages: in a shortest-path problem, the stages are often the number of arcs away from the destination, not the nodes themselves. Keep the direction consistent – backward induction is the norm.
典型应用包括最短路径网络、资源分配和生产/库存计划。始终为每个阶段构建表格，列出可行状态、决策、即时收益及结果值。常见错误是阶段标记不当：在最短路径问题中，阶段通常是距离终点的弧数，而非节点本身。保持方向一致——逆向归纳是常规做法。
For a resource allocation example, if a company has 5 units to allocate to three projects with returns given in a table, the DP table for stage 3 (last project) would list returns for each possible remaining resource 0 to 5; then for stage 2 combine with the optimal returns from stage 3. This recursive logic ensures the global optimum.
以资源分配为例，若公司有5单位资源分配到三个项目并有收益表，阶段3（最后一个项目）的DP表将列出每个可能剩余资源0到5的收益；然后阶段2再与阶段3的最优收益结合。这种递归逻辑确保全局最优。
7. Network Flow Optimisation | 网络流优化
The maximum flow through a capacitated network can be found using the Ford-Fulkerson method with flow augmentation. Start with zero flow and repeatedly find an augmenting path from source to sink where every forward arc has spare capacity or every backward arc has positive flow. Label each node with the predecessor and the additional flow that can be sent. When no more augmenting paths exist, the flow is maximum, and the minimum cut equals the maximum flow.
带容量限制网络的最大流可用福特-富尔克森流量增广方法求解。从零流开始，反复找到一条从源点到汇点的增广路，路径上每条正向弧有余量或每条反向弧有正流量。为每个节点标注前驱节点和可增流量。当不存在增广路时，流量即为最大流，且最小割等于最大流。
A powerful exam technique is to record the saturated arcs and then identify the cut by separating the nodes reachable from the source in the final residual network. The cut capacity must be computed from the original capacities on arcs crossing from the set S to the set T, ignoring backward arcs. Candidates often lose marks by incorrectly including backward arcs in the cut capacity or failing to list the cut clearly.
一个强大的考试技巧是记录饱和弧，然后通过最终残余网络中从源点可达的节点来识别割集。割容量必须根据横跨集合S到集合T的原始弧容量计算，忽略后向弧。考生常因错误包括后向弧在割容量内或未能清楚列出割集而失分。
When drawing flow diagrams, update the flows on each arc and mark backflows. If the problem involves multiple sources or sinks, introduce a super-source or super-sink with infinite capacity arcs. Treat the algorithm systematically and double-check that flow conservation holds at every intermediate node.
绘制流图时，更新每条弧上的流量并标注反向流。若问题涉及多源或多汇，引入超级源或超级汇并连接无限容量弧。系统执行算法并复核每个中间节点是否满足流量守恒。
8. Advanced Critical Path Analysis | 高级关键路径分析
In D2, critical path analysis extends to project crashing: the reduction of activity durations at extra cost to shorten the overall project. For each activity, you are normally given a normal time, crash time, normal cost and crash cost. Calculate the cost slope = (Crash cost − Normal cost) / (Normal time − Crash time). To reduce the project by one time unit, choose the critical activity with the smallest cost slope; as you crash, new critical paths may emerge, so you need to re-evaluate the network.
在D2中，关键路径分析扩展到项目赶工：付费缩短活动持续时间以缩短整体
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