📚 Decision Mathematics 2: Optimisation, Flows and Games | 决策数学2：优化、流与博弈
Decision Mathematics 2 builds on D1 by applying algorithms to larger optimisation, network and strategic problems. This revision guide covers transportation and assignment models, dynamic programming, network flows, game theory, simplex extensions and critical path analysis. Each section pairs the key English explanation with a Chinese summary for effective bilingual revision.
决策数学2在D1的基础上，将算法应用于更大规模的优化、网络与策略问题。本复习指南涵盖运输与指派模型、动态规划、网络流、博弈论、单纯形扩展以及关键路径分析。每一节将英文要点与中文总结配对，便于双语复习。
1. What Decision Mathematics 2 Covers | 决策数学2的考查内容
In D1 you were usually told which algorithm to use. In D2 the marks often come from choosing the right model, setting it up correctly in a table or a network, and then interpreting the final numbers in the context of the original question. A transportation problem is answered on a cost table, a game is answered on a payoff matrix, and a project is answered on an activity network.
在D1中，题目通常会直接告诉你要用哪个算法；而在D2中，得分点常常在于选择正确的模型、在表格或网络中正确建模，并把最终数值放回原题情境中解释。运输问题在成本表上作答，博弈问题在支付矩阵上作答，项目问题则在活动网络上作答。
The table below summarises each model, the standard method and the form of answer an examiner expects.
下表总结了各模型、标准方法以及考官期望的答案形式。
| Topic | Model 模型 | Core method 核心方法 |
| Transportation 运输 | Cost table, m sources × n destinations | North-west corner, least cost, VAM, MODI |
| Assignment 指派 | n × n cost matrix, one-to-one | Hungarian algorithm |
| Dynamic programming 动态规划 | Stages, states, decisions | Backward recursion tables |
| Network flow 网络流 | Directed arcs with capacities | Labelling, augmenting paths, cuts |
| Game theory 博弈论 | Two-person zero-sum payoff matrix | Dominance, maximin, mixed strategies |
| Linear programming 线性规划 | Objective plus constraints | Big-M, two-phase, duality |
| Critical path analysis 关键路径 | Activity network with precedences | Forward and backward passes, float |
Boards differ in what they place where: some examine critical path analysis or the graphical simplex in D1, and some treat resource levelling or integer programming as extension material only. Always check your own specification and past papers before deciding how deep to go.
各考试局的安排有所不同：有的把关键路径分析或图解法单纯形放在D1，有的只把资源平衡或整数规划作为拓展内容。务必先核对自己考试局的大纲与真题，再决定学习的深度与广度。
2. Modelling Transportation Problems | 运输问题建模
A transportation problem is a special linear model that moves units from m supply sources to n demand destinations at minimum total cost. The standard assumption is balanced: total supply equals total demand.
运输问题是一类特殊的线性模型，将货物从 m 个供应地运往 n 个需求地，使总成本最小。标准假设是供需平衡：总供应等于总需求。
Let aᵢ be the supply at source i, bⱼ the demand at destination j, and cᵢⱼ the unit cost of moving goods from source i to destination j. The decision variable xᵢⱼ is the quantity sent along that route.
设 aᵢ 为供应地 i 的供应量，bⱼ 为需求地 j 的需求量，cᵢⱼ 为从供应地 i 运往需求地 j 的单位成本，决策变量 xᵢⱼ 为该路线上的运量。
minimise z = Σᵢ Σⱼ cᵢⱼ xᵢⱼ subject to Σⱼ xᵢⱼ = aᵢ, Σᵢ xᵢⱼ = bⱼ, xᵢⱼ ≥ 0
Notice the equality signs. In the classic formulation every source is emptied and every destination is filled exactly, which is precisely why a dummy row or column is needed whenever the given numbers do not balance.
注意这里用的是等号。经典模型要求每个供应地的货全部运出、每个需求地的需求全部被满足，这正是当题目给出的数字不平衡时必须引入虚拟行或虚拟列的原因。
If the problem is unbalanced, introduce a dummy source or destination with zero transport cost, or with a penalty cost if shortages or stockouts are not allowed.
若问题不平衡，可引入一个虚拟供应地或需求地，运输成本设为零；若不允许缺货，则设为惩罚成本。
Interpretation carries marks. A dummy source receiving a positive allocation means that destination's demand cannot be met from real supply, while a dummy destination absorbing units means that source has surplus stock left over.
解释部分同样有分数。虚拟供应地被分配到正数，说明该需求地的需求无法由真实供应满足；虚拟需求地吸收了货物，则说明该供应地有剩余库存。
3. Finding an Initial Solution | 求初始基本可行解
Three standard methods produce an initial basic feasible solution: the north-west corner rule, the least cost (matrix minimum) method, and Vogel's approximation method, usually abbreviated VAM.
常用的三种方法可以给出初始基本可行解：西北角法、最小成本法（矩阵最小元素法）以及沃格尔近似法，通常简写为 VAM。
- North-west corner rule — start at the top-left cell, allocate as much as the smaller of the remaining supply and demand allows, then move right or down. It is quick but ignores costs, so it rarely gives a good starting solution. 西北角法——从左上角单元格开始，按剩余供应量与需求量的较小值分配，然后向右或向下移动；速度快但完全忽略成本，因此起始解通常较差。
- Least cost method — repeatedly choose the cheapest available cell in the whole table and allocate as much as possible. 最小成本法——反复在整张表中挑选成本最低的可用单元格并尽量多分配。
- Vogel's approximation method — for every row and column compute the penalty, the difference between the two smallest remaining costs, then allocate to the cheapest cell in the row or column with the largest penalty. 沃格尔近似法——对每一行与每一列计算惩罚值，即剩余成本中最小两个之差的绝对值，然后在惩罚值最大的行或列中，把货分配给成本最低的单元格。
A basic feasible solution in a transportation table must occupy exactly m + n − 1 cells, where m is the number of sources and n the number of destinations.
运输表的基可行解必须恰好占用 m + n − 1 个单元格，其中 m 为供应地数，n 为需求地数。
number of occupied cells = m + n − 1
VAM generally gives the best start and often the optimal solution outright, which saves time in an exam. Show the penalty table clearly: examiners award method marks for the penalties themselves, not only for the final allocation.
VAM 通常给出最好的初始解，甚至常常直接就是最优解，这在考试中能节省大量时间。要把惩罚值表写清楚：考官对惩罚值本身也给方法分，而不是只看最终分配方案。
4. Testing Optimality: Stepping Stone and MODI | 最优性检验：踏脚石法与MODI法
Once you have a basic feasible solution you must prove it is optimal, or improve it. Two equivalent techniques are used: the stepping stone method and the MODI (modified distribution, or u–v) method.
得到基可行解后，
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