Pre-U CIE Mathematics: Case Study Practical Walkthrough | Pre-U CIE 数学：案例分析实战演练

📚 Pre-U CIE Mathematics: Case Study Practical Walkthrough | Pre-U CIE 数学：案例分析实战演练

Case studies form the bridge between abstract mathematical techniques and real-world decision making. In this walkthrough, we tackle a classic inventory optimisation problem known as the newsvendor model. You will see how probability distributions, expected values, and differential calculus combine to give a clear optimal policy. The approach mirrors what examiners expect in Pre-U CIE Mathematics: a logical structure, careful notation, and insightful interpretation of results.

案例分析是将抽象数学技巧与现实决策连接起来的桥梁。在本次实战演练中，我们将处理一个经典的库存优化问题——报童模型。你将看到概率分布、期望值以及微分学如何共同给出明确的最优策略。整个推演过程完全符合 CIE Pre-U 数学考纲的要求：逻辑结构清晰、符号使用严谨、结果解读深刻。

1. Problem Scenario | 问题情境

A shop sells a seasonal product with a short selling window. The unit purchase cost is c, the unit selling price is p, and any unsold item at the end of the season can be salvaged for a unit value s. The demand D is a continuous random variable with probability density function f(x) and cumulative distribution function F(x). The decision maker must choose a single order quantity Q before observing demand, aiming to maximise the expected profit.

某商店销售一种销售窗口很短的季节性商品。单位进货成本为 c，单位售价为 p，季末未售出的商品只能以单位残值 s 处理。需求 D 是一个连续型随机变量，概率密度函数为 f(x)，累积分布函数为 F(x)。决策者必须在观察到实际需求之前确定唯一的订货量 Q，目标是最大化期望利润。

2. Defining Parameters and Assumptions | 定义参数与假设

We assume p > c > s > 0. There is no backordering; unsatisfied demand is lost. The probability distribution of D is stationary and fully known. The objective function is the expected profit per selling season, denoted by E[Π(Q)].

我们假设 p > c > s > 0。未满足的需求直接丢失，不允许延期交货。需求 D 的概率分布是平稳且已知的。目标函数是每个销售季度的期望利润，记作 E[Π(Q)]。

c = unit cost / 单位成本
p = selling price / 售价
s = salvage value per unit / 单位残值
Q = order quantity (decision variable) / 订货量（决策变量）
D = random demand / 随机需求
f(x) = pdf of D / D 的概率密度函数
F(x) = cdf of D / D 的累积分布函数

3. Constructing the Profit Function | 构建利润函数

For any realisation of demand x, the profit depends on whether demand exceeds Q or not. If x ≤ Q, the shop sells x units at price p, salvages Q – x units, and pays cQ. The profit is π₁(x) = p x + s(Q – x) – cQ. If x > Q, the shop sells all Q units at price p and pays cQ, giving π₂(x) = p Q – cQ.

对于任意实际需求 x，利润取决于需求是否超过订货量。若 x ≤ Q，店铺以单价 p 售出 x 件，剩余 Q – x 件以残值处理，总成本为 cQ，利润为 π₁(x) = p x + s(Q – x) – cQ。若 x > Q，则全部 Q 件以 p 售出，利润为 π₂(x) = p Q – cQ。

π(x, Q) = { p x + s(Q – x) – cQ, x ≤ Q; (p – c)Q, x > Q }

利润函数：当需求 x ≤ Q 时，π = p x + s(Q – x) – cQ；当 x > Q 时，π = (p – c)Q

4. Expected Profit Expression | 期望利润表达式

The expected profit E[Π(Q)] is obtained by integrating the profit function over the domain of D with respect to its probability density.

期望利润 E[Π(Q)] 是通过对利润函数在整个需求范围内关于概率密度积分得出的。

E[Π(Q)] = ∫₀^Q [p x + s(Q – x) – cQ] f(x) dx + ∫_Q^∞ [(p – c)Q] f(x) dx

Note that demand is non-negative, so the lower limit is 0. The integrals separate the cases of surplus and shortage.

注意需求为非负，故积分下限为 0。两个积分分别对应供过于求和供不应求的情形。

5. Simplifying the Expected Profit | 简化期望利润

We can rewrite the expression in a more compact form using the complementary cdf. Observe that ∫_Q^∞ f(x) dx = 1 – F(Q). After algebraic manipulation, we obtain:

我们可以利用互补累积分布函数将其写成更紧凑的形式。注意到 ∫_Q^∞ f(x) dx = 1 – F(Q)，经过代数整理后得到：

E[Π(Q)] = (p – c)Q – (p – s) ∫₀^Q (Q – x) f(x) dx

This form reveals that expected profit equals the profit if all Q units were sold minus an expected overage cost proportional to the expected leftover inventory.

此形式揭示出，期望利润等于假设 Q 件全部售出时的利润减去与期望剩余库存成正比的期望超储成本。

6. Differentiation for Optimisation | 求导优化

To find the optimal Q*, we differentiate E[Π(Q)] with respect to Q. We apply the Leibniz integral rule to the term containing the integral.

为求得最优订货量 Q*，我们对 E[Π(Q)] 关于 Q 求导，并对含积分项使用莱布尼茨积分法则。

d/dQ ∫₀^Q (Q – x) f(x) dx = ∫₀^Q f(x) dx + (Q – Q)f(Q)·1 = F(Q)

dE[Π]/dQ = (p – c) – (p – s) F(Q)

7. The First-Order Condition and Critical Fractile | 一阶条件与临界分位数

Setting the derivative to zero gives the necessary condition for an interior optimum. Since F(Q) is monotone increasing, the solution is unique and globally optimal.

令导数为零即得内点最优的必要条件。由于 F(Q) 单调递增，该解唯一且为全局最优。

F(Q*) = (p – c) / (p – s)

The ratio (p – c)/(p – s) is known as the critical fractile. It represents the probability that demand will not exceed the order quantity, balancing the cost of underordering (lost profit) against the cost of overordering (excess inventory).

比值 (p – c)/(p – s) 被称为临界分位数。它代表需求不超过订货量的概率，精准平衡了少订（损失利润）与多订（库存积压）的成本。

8. Worked Example with Normal Demand | 正态需求实例演算

Suppose weekly demand for a festival item is normally distributed with mean μ = 200 and standard deviation σ = 40. The unit cost c = £12, selling price p = £25, salvage value s = £5. Compute the optimal Q*.

假设某节日商品的周需求量服从正态分布，均值 μ = 200，标准差 σ = 40。成本 c = £12，售价 p = £25，残值 s = £5。试计算最优 Q*。

First, find the critical fractile: (25 – 12) / (25 – 5) = 13 / 20 = 0.65.

首先，计算临界分位数：(25 – 12) / (25 – 5) = 13/20 = 0.65。

We need Q* such that F(Q*) = 0.65. For a normal distribution, standardise: z = (Q* – μ) / σ. From standard normal tables, Φ(z) = 0.65 gives z ≈ 0.385. Hence Q* = 200 + 0.385 × 40 = 215.4. Rounding to the nearest integer, Q* ≈ 215 units.

需求满足 F(Q*) = 0.65。对正态分布标准化：z = (Q* – μ) / σ。查标准正态表，Φ(z)=0.65 对应 z ≈ 0.385。因此 Q* = 200 + 0.385 × 40 = 215.4。取整后最优订货量约为 215 件。

9. Interpretation of the Critical Fractile | 临界分位数的解读

The optimal service level—the probability of not stocking out—equals 65%. This means that in the long run, 65% of seasons will see no stockout, while 35% of seasons will run out of stock before all demand is met. The value of 65% reflects the relative profitability: a high profit margin (p – c) relative to the loss from unsold units (c – s) leads to a higher service level.

最优服务水平——即不发生缺货的概率——为 65%。这意味着长期来看，65% 的销售季不会缺货，而 35% 的销售季会在满足全部需求前售罄。65% 的数值反映了相对盈利水平：高利润率 (p – c) 相对于未售出损失 (c – s) 越大，服务水平就越高。

Parameter	Value	Meaning
(p – c)	£13	Profit margin per unit sold
(c – s)	£7	Loss per unit unsold
Critical ratio	0.65	Optimal in-stock probability

10. Sensitivity Analysis | 敏感性分析

How does Q* change if the standard deviation of demand increases to 60? The z-value remains 0.385, so Q* = 200 + 0.385 × 60 = 223.1 ≈ 223. A larger demand uncertainty increases the optimal order quantity because the overage risk is outweighed by the high profit margin.

若需求标准差增大至 60，Q* 会如何变化？z 值保持 0.385，则 Q* = 200 + 0.385 × 60 = 223.1 ≈ 223。更大的需求不确定性会提高最优订货量，因为高利润率带来的收益超过了超储风险。

If the salvage value s rises to £8, then critical fractile becomes (25 – 12)/(25 – 8) = 13/17 ≈ 0.7647, giving z ≈ 0.72 and Q* = 200 + 0.72 × 40 = 228.8 ≈ 229. Higher salvage value reduces the penalty for overstocking, so ordering more becomes attractive.

若残值 s 上升至 £8，临界分位数变为 (25-12)/(25-8)=13/17≈0.7647，对应 z≈0.72，Q* = 200+0.72×40=228.8≈229。更高的残值降低了积压惩罚，因此增加订货更具吸引力。

11. Extensions and Real-World Considerations | 拓展与实际考量

The basic model can be extended to incorporate a shortage penalty cost v per unit of unmet demand. The critical fractile then becomes (p – c + v) / (p – s + v). The logic remains identical: the ratio balances total underage and overage costs. Furthermore, when demand is discrete, the optimal Q is the smallest integer such that F(Q) ≥ critical ratio.

基本模型可拓展为包含单位缺货惩罚成本 v 的情形。此时临界分位数变为 (p – c + v) / (p – s + v)。逻辑完全一致：该比值平衡了总缺货成本与总超储成本。此外，若需求是离散型，最优 Q 即为满足 F(Q) ≥ 临界比率的最小整数。

In practice, demand distributions are often estimated from historical data, and the normality assumption must be verified. Pre-U CIE exam questions frequently ask candidates to comment on the suitability of the model and suggest improvements, so always link mathematics back to context.

实际应用中，需求分布通常由历史数据估计得出，并需检验正态性假设的合理性。CIE Pre-U 考试经常要求考生评价模型的适用性并提出改进建议，因此始终要将数学分析与实际情境相联系。

12. Summary of the Walkthrough | 演练总结

This case study demonstrated how to move from a verbal business problem to a formal optimisation model using expected values and calculus. The key takeaway is the critical fractile formula F(Q*) = (p – c)/(p – s), which elegantly ties the optimal decision to the cost structure and demand uncertainty. Mastering such models sharpens both your algebraic fluency and your ability to reason probabilistically.

本案例展示了如何将文字型商业问题转化为基于期望值与微积分的形式化优化模型。核心要点是临界分位数公式 F(Q*) = (p – c)/(p – s)，它将最优决策与成本结构和需求不确定性优雅地联系在一起。掌握此类模型既能增强代数运算的流畅度，也能提升概率推理的能力。

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Pre-U CIE Mathematics: Case Study Practical Walkthrough | Pre-U CIE 数学：案例分析实战演练