📚 Edexcel A-Level Further Mathematics Decision 2 Key Concepts | Edexcel A-Level 进阶数学决策数学2 知识点精讲
Decision Mathematics 2 extends the ideas from D1 by introducing more powerful algorithms for resource management, optimisation, and game-like situations. This article walks through the core topics, pairing clear English explanations with Chinese translations to help bilingual learners master the techniques needed for the Edexcel exam.
决策数学2在D1的基础上引入了更强大的算法，用于资源管理、优化以及博弈情境。本文逐项讲解核心知识点，用清晰的英文解释配合中文翻译，帮助双语学习者掌握Edexcel考试所需的方法。
1. Critical Path Analysis with Resource Levelling | 资源均衡的关键路径分析
In D1 we constructed activity networks and found earliest and latest event times. D2 adds resource considerations: activities require workers or equipment, and we often have limited availability. A Gantt (cascade) chart shows each activity's time window based on total float. By sliding non‑critical activities within their float, we can produce a resource histogram and attempt to smooth demand so that resource usage stays below the maximum available. The goal is to minimise peaks without delaying the project — this is called resource levelling.
在D1中我们构建了活动网络并求出最早和最晚事件时间。D2加入了资源因素：活动需要工人或设备，而我们通常受到资源限制。甘特图（阶梯图）显示每个活动基于总浮时的时间窗口。通过在其浮时范围内移动非关键活动，我们可以生成资源直方图，并尝试平滑资源需求，使资源使用量始终低于最大可用量。目标是在不延误工期的前提下最小化使用峰值——这就是资源均衡。
A resource histogram itself is a bar chart showing the number of resources required on each day if all activities start at their earliest start times. When peaks exceed the limit, we must reschedule non‑critical activities, prioritising those with the highest float. The resulting diagram is a revised cascade chart and a levelled histogram. Exam questions often ask you to complete a given Gantt chart or draw a new one after levelling.
资源直方图本身是一个条形图，显示如果所有活动都在最早开始时间开始时每天所需的资源数量。当峰值超过限制时，我们必须重新排程非关键活动，优先考虑浮时最多的活动。最后得到的是一张修订后的阶梯图和一个均衡后的直方图。考试题经常要求你补全给定的甘特图或在均衡后绘制新的图表。
2. Allocation (Assignment) Problems – Hungarian Algorithm | 指派问题——匈牙利算法
An allocation problem pairs n tasks with n workers (or machines) to minimise total cost or time. The Hungarian algorithm transforms a square cost matrix so that an optimal assignment can be read off when a complete set of independent zeros is found. The steps are: reduce rows by subtracting the row minimum; reduce columns similarly; cover all zeros with the minimum number of lines; if lines equal n, an optimum exists; if not, find the smallest uncovered element, subtract it from uncovered entries, add it to doubly covered entries, then repeat covering. This algorithm guarantees the minimum cost.
指派问题将n项任务与n个工人（或机器）配对，以最小化总成本或总时间。匈牙利算法通过转换正方形的费用矩阵，使得当能够找到一组完全独立的零元素时，就可以直接读出最优指派。步骤是：行归约（每行减去该行最小值）；列归约；用最少数量的线覆盖所有零；如果线数等于n，则最优解已存在；否则找出未被覆盖的最小元素，从未被覆盖元素中减去它，并在双重覆盖处加上它，然后重复覆盖步骤。该算法保证找到最小成本。
Maximisation problems are handled by subtracting every entry from a large number (e.g. the largest entry) to convert into an equivalent minimisation. If the matrix is unbalanced (more tasks than workers), add dummy rows or columns with zero costs. The algorithm also works for assigning salesmen to territories or machines to jobs, always producing the least total cost or maximum profit.
最大化问题可以这样处理：用一个大的数（例如矩阵中的最大元素）减去每个元素，转化为等价的最小化问题。如果矩阵不平衡（任务多于工人），则添加成本为零的虚拟行或列。该算法同样适用于将销售员分配至区域、机器分配至工作等，总能得出最小总成本或最大利润。
3. Transportation Problems – Finding an Initial Solution | 运输问题——初始解求法
Transportation problems deal with distributing goods from several sources to several destinations at minimum cost. An initial feasible solution is required before optimisation. Three methods are commonly tested: the North‑West Corner method (simplest, but ignores costs), the Least Cost method (chooses the cell with smallest unit cost each time), and Vogel's Approximation Method (VAM). VAM uses penalty costs — the difference between the two smallest costs in each row and column — to guide allocation, usually producing a solution closer to the optimum.
运输问题处理的是以最低成本将货物从多个供应地配送到多个需求地。在优化之前需要一个初始可行解。考试中通常考察三种方法：西北角法（最简单但忽略成本）、最小成本法（每次选择单位成本最小的格子）以及沃格尔近似法（VAM）。VAM利用惩罚成本——每行和每列中两个最小成本的差值——来指导分配，通常能产生更接近最优解的方案。
To apply VAM: calculate row and column penalties; select the row or column with the largest penalty; allocate as much as possible to the lowest‑cost cell in that row/column; adjust supply and demand, delete the satisfied row or column; recalculate penalties until all supply and demand are exhausted. This gives a basic feasible solution with exactly (m + n − 1) occupied cells, where m and n are the numbers of sources and destinations.
应用VAM：计算行列惩罚值；选择惩罚值最大的行或列；在该行或列中成本最低的格子尽可能多地分配；调整供需量，划掉已满足的行或列；重新计算惩罚值，直至所有供需归零。这会产生一个基本可行解，恰好有 (m + n − 1) 个占用格子，其中m和n分别为供应地和需求地的数量。
4. Transportation Problems – Optimisation with Stepping Stone and MODI | 运输问题——优化：踏脚石法与MODI法
Once an initial solution is found, we test whether it is optimal. The Stepping Stone method evaluates each empty cell by creating a closed path through occupied cells, alternating plus and minus signs. The total change in cost for shifting one unit into that empty cell is the shadow cost (or improvement index). If all shadow costs are non‑negative (for minimisation), the solution is optimal. Otherwise, the most negative shadow cost cell is selected, and as many units as possible are shifted along its loop to obtain a new solution.
找到初始解后，我们需要检验它是否最优。踏脚石法通过占用格子为每个空格子构建一条闭合路径，路径上交替标正负号。将一单位货物移入该空格所引起的总成本变化即为影子成本（改进指数）。若所有影子成本皆非负（最小化问题），解为最优；否则，选出影子成本最负的格子，沿其回路尽可能多地移动单位，得到新解。
dified Distribution) is more systematic. Assign dual variables uᵢ to each source row and vⱼ to each destination column such that uᵢ + vⱼ = cᵢⱼ for every occupied cell. Then compute the opportunity cost for each empty cell as cᵢⱼ − (uᵢ + vⱼ er matrices, and exam questions often require computing shadow costs using this technique.
MODI法（修正分配法）更系统化。为每个供应行分配对偶变量uᵢ，每个需求列分配vⱼ，使得对所有占用格子满足uᵢ + vⱼ = cᵢⱼ。然后计算每个空格的机会成本：cᵢⱼ − (uᵢ + vⱼ)。这实际上与踏脚石法的影子成本相同。对于更大的矩阵，MODI更快捷，考试题通常要求用此方法计算影子成本。
5. Game Theory – Pure and Mixed Strategies | 博弈论——纯策略与混合策略
Two‑person zero‑sum games are represented by a pay‑off matrix, where one player's gain is the other' olution with pure strategies. The row player chooses the row that maximises the minimum pay‑off, and the column player chooses the column that minimises the maximum pay‑off. When these values coincide, no player can improve unilaterally.
两人零和博弈用一个支付矩阵表示，其中一方的收益就是另一方的损失。如果存在鞍点（最大值中的最小值等于最小值中的最大值），博弈就有稳定的纯策略解。行玩家选择能使最小收益最大化的行，列玩家选择能使最大损失最小化的列。当这两个值相等时，任何一方都无法单方面改善结果。
When no saddle point exists, players must randomise — e without a saddle point, the optimal mixed strategy for each player can be found by equating the expected pay‑offs of the opponent's pure strategies. If the row player's strategy is (p, 1−p) and the column player' we set up two equalities and solve for p and q. The value of the game is the expected pay‑off when both play optimally.
没有鞍点时，玩家必须随机化——这就是混合策略。对于没有鞍点的2×2博弈，令对手纯策略的期望支付相等，即可求出各方的最优混合策略。如果行玩家的策略为(p, 1−p)，列玩家为(q, 1−q)，我们建立两个等式解出p和q。博弈的值就是双方都按最优策略时的期望支付。
6. Game Theory – Graphical Method and Linear Programming | 博弈论——图解法与线性规划
2×n or m×2 games, the graphical method reduces the problem to two dimensions. Plot the pay‑off lines for the row player against the probability p. The lower envelope shows the worst‑case pay‑ ersection of active lines gives the optimal p, and the value v can be read from the graph.
对于2×n或m×2的博弈，图解法将问题降为二维。绘制行玩家支付线关于概率p的图像。下包络线显示了每个p对应的最差支付；行玩家选择能让下包络最大化的p。活跃线的交点给出最优p，博弈值v可从图中读出。
Larger games require linear programming. The optimal strategies can be found by formulating LPs for each player. For the row player, maximise v subject to constraints that each column pay‑off is at least v and probabilities sum to 1. The column player's LP minimises w subject to each row pay‑off not exceeding w. Since the two LPs are dual, they yield the same value. Exam questions may ask you to formulate the LP and interpret the final tableau.
更大的博弈需要线性规划。可以为每个玩家建立线性规划模型求最优化策略。行玩家在满足每一列支付至少为v、概率和为1的约束下最大化v；列玩家的LP则是在每行支付不超过w的约束下最小化w。这两个LP互为对偶，给出相同的值。考试题可能要求建立LP并解释最终表格。
7. Dynamic Programming – Principle of Optimality | 动态规划——最优化原理
Dynamic programming solves multi‑stage decision problems by breaking them into smaller sub‑problems. The principle of optimality states that an optimal policy has the property that, whatever the initial state and decisions, the remaining decisions must form an optimal policy with regard to the state resulting from the first decision. This allows us to work backwards from the last stage, determining optimal actions and value functions.
动态规划通过将多阶段决策问题分解为较小的子问题来求解。最优化原理指出：无论初始状态和当前决策如何，最优策略都具备这样的性质——剩下的决策对由第一个决策产生的状态而言必须是最优的。这使得我们可以从最后一个阶段开始逆向递推，确定最优行动和价值函数。
d a table showing the immediate reward plus the future value from the optimal policy of the next stage. This recursion is expressed as: fₙ(s) = max or min { rₙ(s, a) + fₙ₊₁(s') } where s' is the new state after action a.
我们通常定义阶段（如天、机器分配）、状态（如剩余资源、库存水平）和行动。对于每个阶段，我们建立一个表格，显示即时收益加上下一阶段最优策略的未来价值。这个递推关系表示为：fₙ(s) = max 或 min { rₙ(s, a) + fₙ₊₁(s') }，其中s'是采取行动a后的新状态。
8. Dynamic Programming – Table Method for Deterministic Problems | 动态规划——确定性问题的表格法
Common D2 applications include minimising cost over a production schedule or maximising profit from selling assets. A popular exam problem is "allocate x machines to three factories so that total profit is maximised". We set up stages as factories, state as machines remaining, decision as how many to give to the current factory. Starting from the last factory, we tabulate its profit for each possible number of machines; then move to the second, combining its profit with the optimal future profit already found. The process yields the optimal allocation and total maximum profit.
D2中常见的应用包括在排产计划中最小化成本或出售资产最大化利润。一个常考题是“将x台机器分配给三个工厂使得总利润最大”。我们以工厂为阶段，状态为剩余机器数，决策为分配多少台给当前工厂。从最后一个工厂开始，把每种可能的机器数量对应的利润制成表格；然后回溯到第二个工厂，将其利润与已求出的未来最优利润相加。这个过程最终给出最优分配方案和总最大利润。
Tabulation needs to cover every feasible state‑decision combination. State values are usually integers, and interpolation is not needed if all returns are given in tables. The final answer is found in the first stage's table for the full initial state. Students should be comfortable reading and completing dynamic programming tables, as well as interpreting the optimal decision sequence.
列表法需要覆盖所有可行的状态-决策组合。状态值通常是整数，如果所有收益都以表格给出，则无需插值。最终答案在全初始状态下的第一个阶段表中读出。学生应熟练阅读和填写动态规划表格，并能解释最优决策序列。
9. Flows in Networks – Maximum Flow Algorithm | 网络流——最大流算法
A flow network has a source, a sink, and directed arcs with capacities. The maximum flow problem asks how much flow can be sent from source to sink without exceeding any arc capacity. The labelling procedure (Ford‑Fulkerson) finds augmenting paths by assigning each node a label (predecessor, remaining capacity) starting from the source. When a path to the sink is found, we augment flow by the minimum residual capacity along the path, update forward and backward arcs, and relabel. This repeats until no augmenting path exists.
一个流网络具有源点、汇点和带容量的有向弧。最大流问题询问在不超出任何弧容量的情况下，能从源点输送多少流量到汇点。标号法（Ford‑Fulkerson）通过从源点开始为每个节点分配标号（前驱节点，剩余容量）来寻找增广路。当找到一条到达汇点的路径时，我们沿着该路径以最小的剩余容量增加流量，更新正向弧和反向弧，并重新标号。重复此过程直到不存在增广路。
The max‑flow min‑cut theorem states that the maximum flow equals the minimum cut capacity. A cut is a partition of vertices into two sets S and T with source in S and sink in T; its capacity is the sum of capacities of arcs from S to T. At the end of the labelling procedure, the set of labelled nodes defines a minimum cut. This theorem is often used to verify the maximum flow or to identify bottlenecks.
最大流-最小割定理指出，最大流等于最小割的容量。割是将顶点划分为两个集合S和T，源点在S中、汇点在T中；其容量为从S到T的所有弧的容量之和。在标号过程结束时，被标号的节点集合就定义了一个最小割。这个定理常用于验证最大流或识别瓶颈。
10. Network Flow Extensions – Feasible Flows and Multiple Sources | 网络流扩展——可行流与多源点
In some problems, arcs have lower bounds as well as upper bounds. A feasible flow must satisfy both capacity constraints and flow conservation at all nodes (except source and sink). To find an initial feasible circulation, we can introduce a super‑source and super‑sink, send the required lower‑bound flows, and then check whether all extra demands can be met. The concept of excess and deficit nodes is key. If a feasible flow exists, we then maximise flow by applying the standard labelling algorithm on the residual network.
有些问题中，弧不仅有容量上限还有下限。一个可行流必须同时满足容量约束和所有节点（除源汇外）的流量守恒。为了找到一个初始可行环流，我们可以引入一个超级源和一个超级汇，发送所要求的下限流，然后检查所有额外需求是否能满足。盈余节点和赤字节点的概念是关键。如果存在可行流，我们就可以在剩余网络上应用标准标号法最大化流量。
Networks with several supply nodes and several demand nodes are transformed by adding a single super‑source connected to all supplies with arcs of capacity equal to each supply, and a super‑sink from all demands with arcs of capacity equal to each demand. Then a maximum flow from super‑source to super‑sink solves the transshipment problem. This unifies transportation with network flow theory and is a neat D2 topic.
具有多个供应节点和多个需求节点的网络，可以通过添加一个超级源（连接所有供应点，弧容量等于各供应量）和一个超级汇（从所有需求点连入，弧容量等于各需求量）来转换。然后从超级源到超级汇的最大流就解决了转运问题。这将运输问题与网络流理论统一起来，是D2中一个简洁而优美的主题。
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