Urban Waste and its Disposal: Mathematical Modelling and Statistical Analysis | 城市废弃物及其处置：数学建模与统计分析

📚 Urban Waste and its Disposal: Mathematical Modelling and Statistical Analysis | 城市废弃物及其处置：数学建模与统计分析

Urban waste management is often presented as a geography or environmental science topic, but it provides a rich context for applying A-Level Mathematics. From forecasting waste tonnage to optimising collection routes and comparing disposal technologies, mathematical techniques help planners make evidence-based decisions. This article links key Edexcel A-Level Maths methods to the urban waste problem.

城市垃圾管理通常作为地理或环境科学主题呈现，但它为运用 A-Level 数学提供了丰富情境。从预测垃圾吨数、优化收运路线到比较处置技术，数学方法帮助规划者做出基于证据的决策。本文将 Edexcel A-Level 数学的关键方法与城市垃圾问题联系起来。

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1. Waste Generation as a Time Series | 垃圾产生量的时间序列

Urban waste generation often follows weekly and seasonal patterns. A local council records monthly residual waste tonnage for 24 months. Plotting the data reveals trend, seasonality and random noise. Moving averages smooth the series and help identify underlying growth in waste per household.

城市垃圾产生量常呈周度和季节性规律。某市政部门连续 24 个月记录每月残余垃圾吨数。绘制数据可见趋势、季节性和随机噪声。移动平均可平滑序列，并有助于识别每户垃圾量的潜在增长。

For quarterly data, a 3-point moving average at time t is calculated as the mean of the current, previous and next values. This is a standard technique in the Edexcel Statistics component for analysing time series data.

对于季度数据，时间 t 的 3 点移动平均为当前值、前一值和后一值的平均数。这是 Edexcel 统计学部分分析时间序列数据的标准方法。

M(t) = [x(t−1) + x(t) + x(t+1)] ÷ 3

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2. Exponential Growth and Decay in Waste | 垃圾的指数增长与衰减

Waste generation in rapidly growing cities may increase approximately exponentially. If the annual growth rate is k, the total waste tonnage after t years is P(t) = P₀ × exp(kt), where P₀ is the initial tonnage. Doubling time is found by solving exp(kt) = 2, giving t = ln 2 ÷ k.

快速扩张的城市中，垃圾产生量可能近似指数增长。若年增长率为 k，则 t 年后的垃圾总吨数为 P(t) = P₀ × exp(kt)，其中 P₀ 为初始吨数。加倍时间通过解 exp(kt) = 2 得到，即 t = ln 2 ÷ k。

Conversely, biodegradable waste in a landfill decays exponentially. The mass remaining after time t is M(t) = M₀ × exp(−λt), where λ is the decay constant. The half-life is t½ = ln 2 ÷ λ. This directly applies Pure Maths modelling with exponentials and logarithms.

相反，填埋场中的可生物降解垃圾呈指数衰减。t 时间后剩余质量为 M(t) = M₀ × exp(−λt)，其中 λ 为衰减常数。半衰期为 t½ = ln 2 ÷ λ。这直接应用了纯数学中指数与对数的建模。

P(t) = P₀ × exp(kt)

M(t) = M₀ × exp(−λt), t½ = ln 2 ÷ λ

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3. Linear Regression for Recycling Rates | 回收率的线性回归

A local authority records household recycling rate y against public awareness campaign spending x. The least-squares regression line is y = a + bx. The gradient b is calculated from the sums Sxy and Sxx, where Sxy = Σ (xᵢ − x_mean)(yᵢ − y_mean) and Sxx = Σ (xᵢ − x_mean)².

某地方当局记录家庭回收率 y 与公众意识宣传支出 x 的关系。最小二乘回归线为 y = a + bx。斜率 b 由 Sxy 和 Sxx 计算，其中 Sxy = Σ (xᵢ − x_mean)(yᵢ − y_mean)，Sxx = Σ (xᵢ − x_mean)²。

The product moment correlation coefficient r measures the strength of the linear relationship, and r² gives the proportion of variation in y explained by x. Interpolation within the data range is valid, but extrapolation beyond the range should be treated with caution.

积矩相关系数 r 衡量线性关系的强度，r² 给出 y 的变异中可由 x 解释的比例。在数据范围内插值是有效的，但超出范围的外推应谨慎处理。

b = Sxy ÷ Sxx, a = mean(y) − b × mean(x)

r = Sxy ÷ √(Sxx × Syy)

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4. Probability Distributions for Waste Composition | 废弃物组成的概率分布

When sampling 100 items from mixed household waste, the number of PET plastic bottles can be modelled by a binomial distribution X ~ B(100, p). If p = 0.25, the expected number is np = 25 and the variance is np(1 − p) = 18.75. This allows planners to predict variation between samples.

从混合家庭垃圾中抽取 100 件物品时，PET 塑料瓶数量可用二项分布 X ~ B(100, p) 建模。若 p = 0.25，则期望数量为 np = 25，方差为 np(1 − p) = 18.75。这使规划者能预测样本间的变异。

For rare items such as batteries per kilogram of waste, a Poisson distribution with mean λ is appropriate. If λ is small, the Poisson distribution gives P(X = r) = λʳ e⁻λ ÷ r!. Normal approximations may be used for large n with continuity correction.

对于每公斤垃圾中的电池等稀有物品，均值为 λ 的泊松分布是合适的。若 λ 很小，泊松分布给出 P(X = r) = λʳ e⁻λ ÷ r!。对于大 n，可在连续性校正下使用正态近似。

X ~ B(n, p), mean = np, variance = np(1−p)

X ~ Po(λ), P(X = r) = λʳ e⁻λ ÷ r!

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5. Hypothesis Testing: Comparing Disposal Methods | 假设检验：比较不同处置方法

Two incinerators report mean dioxin emissions. A two-sample t-test can test the null hypothesis H₀: μ₁ = μ₂ against the alternative H₁: μ₁ ≠ μ₂ at the 5% significance level. The test statistic is calculated from sample means, variances and sizes.

两个焚烧厂报告平均二噁英排放量。双样本 t 检验可在 5% 显著性水平下检验零假设 H₀：μ₁ = μ₂ 与备择假设 H₁：μ₁ ≠ μ₂。检验统计量由样本均值、方差和样本量计算。

If the p-value is less than 0.05, we reject H₀ and conclude there is sufficient evidence of a difference in mean emissions. Type I error occurs when rejecting a true H₀, and Type II error occurs when failing to reject a false H₀.

若 p 值小于 0.05，则拒绝 H₀，并认为有充分证据表明平均排放量存在差异。第一类错误是拒绝真实的 H₀，第二类错误是未能拒绝错误的 H₀。

t = (mean₁ − mean₂) ÷ √(s₁²/n₁ + s₂²/n₂)

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6. Optimisation of Waste Collection Routes | 垃圾收运路线的优化

A waste collection route can be modelled as a graph with nodes representing bins and weighted edges representing road distances. Finding the shortest route that visits all bins once and returns to the depot is a travelling salesman problem. For small networks, systematic listing or brute force finds the exact optimum.

垃圾收运路线可建模为图，节点代表垃圾桶，加权边代表道路距离。寻找访问所有垃圾桶一次并返回车场的最短路线是旅行商问题。对于小网络，系统列举或穷举法可找到精确最优解。

The nearest neighbour algorithm gives a fast heuristic: start at the depot, repeatedly visit the closest unvisited node, then return to the depot. While not always optimal, it provides a feasible route for real-time planning. Total distance is the sum of edge weights.

最近邻算法给出一种快速启发式方法：从车场出发，重复访问最近的未访问节点，然后返回车场。虽然不一定最优，但它为实时规划提供了可行路线。总距离为边权之和。

D = Σ dᵢⱼ for edges in route

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7. Linear Programming for Waste Facility Costs | 垃圾设施成本的线性规划

A council must decide how many tonnes x to send to recycling and y to landfill each day. Constraints include x + y = 1000, recycling capacity x ≤ 400, landfill capacity y ≤ 800, and x, y ≥ 0. The cost function is C = 40x + 90y.

某议会须决定每天将多少吨垃圾 x 送去回收、多少吨 y 送去填埋。约束条件包括 x + y = 1000、回收能力 x ≤ 400、填埋能力 y ≤ 800，以及 x、y ≥ 0。成本函数为 C = 40x + 90y。

The graphical method plots the feasible region and evaluates C at each vertex. The minimum cost occurs at a corner point of the feasible region. This application links linear inequality constraints to real-world waste management decisions.

图解法绘制可行域并在每个顶点处计算 C。最小成本出现在可行域的角点处。这一应用将线性不等式约束与现实垃圾管理决策联系起来。

Minimise C = 40x + 90y subject to x + y = 1000, x ≤ 400, y ≤ 800, x ≥ 0, y ≥ 0

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8. Statistical Sampling of Household Waste | 家庭垃圾的统计抽样

To estimate mean household waste per week, a random sample is required. The sample size n needed for a given margin of error E is found using n = (z × σ ÷ E)². For 95% confidence, z = 1.96. If the standard deviation σ is 2.5 kg and E is 0.2 kg, then n = (1.96 × 2.5 ÷ 0.2)² ≈ 600.

为估计每户每周平均垃圾量，需要随机样本。给定误差幅度 E 所需样本量 n 由 n = (z × σ ÷ E)² 计算。对于 95% 置信水平，z = 1.96。若标准差 σ 为 2.5 kg、E 为 0.2 kg，则 n = (1.96 × 2.5 ÷ 0.2)² ≈ 600。

Stratified sampling by housing type can reduce bias and improve precision. The sample size for each stratum is proportional to the stratum size in the population, ensuring all housing types are represented.

按住房类型分层抽样可减少偏差并提高精度。每层样本量与总体中层的大小成比例，确保所有住房类型都被代表。

n = (z × σ ÷ E)²

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9. Modelling Landfill Gas Production | 垃圾填埋气体产生的建模

Anaerobic decomposition in landfills produces methane. Cumulative gas yield can be modelled by bounded exponential growth: G(t) = G_max × (1 − exp(−kt)), where G_max is the ultimate gas yield and k is the rate constant. This is a standard Pure Maths model.

填埋场中的厌氧分解产生甲烷。累计气体产量可用有界指数增长模型描述：G(t) = G_max × (1 − exp(−kt))，其中 G_max 为最终气体产量，k 为速率常数。这是纯数学中的标准模型。

Differentiating gives the gas production rate dG/dt = k × G_max × exp(−kt). The rate is highest at t = 0 and decreases over time. Integrating the rate gives the cumulative yield, linking differentiation and integration in context.

求导得到气体产生速率 dG/dt = k × G_max × exp(−kt)。速率在 t = 0 时最大，并随时间递减。对速率积分得到累计产量，将微分与积分应用在情境中联系起来。

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