Development of Regional Organisations Excluding the EU: A-Level Mathematical Modelling | 区域组织发展（不含欧盟）：A-Level 数学建模

📚 Development of Regional Organisations Excluding the EU: A-Level Mathematical Modelling | 区域组织发展（不含欧盟）：A-Level 数学建模

This article applies core A-Level Mathematics techniques from the Edexcel specification to model the growth and development of regional organisations outside the European Union. We explore exponential and logistic growth, regression, probability, integration, hypothesis testing, sequences, differential equations, statistical distributions, and numerical methods, using realistic scenarios based on regional blocs such as ASEAN, Mercosur, and the African Union.

本文运用 Edexcel 考纲中的 A-Level 数学核心方法，对欧盟以外的区域组织发展进行建模分析。我们将探讨指数增长与逻辑斯蒂增长、回归分析、概率、积分、假设检验、数列、微分方程、统计分布和数值方法，并结合东盟、南方共同市场及非洲联盟等区域组织的实际情境。

---

1. Exponential Growth of Membership | 成员数量的指数增长

If a regional organisation starts with N₀ members and the number of member states grows at a continuous rate k per year, the membership size N(t) after t years is modelled by the exponential function N(t) = N₀ e^(kt). This model assumes no limiting factors and is useful for short-term projections.

若某区域组织初始有 N₀ 个成员国，且成员国数量以每年连续速率 k 增长，则 t 年后的成员规模 N(t) 可用指数函数 N(t) = N₀ e^(kt) 建模。该模型假设不存在限制因素，适用于短期预测。

N(t) = N₀ e^(kt)

For example, an organisation founded in 2000 with 5 members and an annual continuous growth rate of 8% would have approximately N(20) = 5 × e^(0.08 × 20) = 5 × e^1.6 ≈ 24.8 members by 2020, which rounds to 25 member states.

例如，某组织于 2000 年成立时有 5 个成员，年连续增长率为 8%，则到 2020 年其成员数约为 N(20) = 5 × e^(0.08 × 20) = 5 × e^1.6 ≈ 24.8，四舍五入为 25 个成员国。

• N₀ is the initial number of members at time t = 0.
• k is the continuous growth rate (positive for growth, negative for decline).
• The doubling time is given by ln 2 / k.

• N₀ 是 t = 0 时刻的初始成员国数量。
• k 是连续增长率（正值为增长，负值为下降）。
• 翻倍时间由 ln 2 / k 给出。

---

2. Logistic Growth Model and Carrying Capacity | 逻辑斯蒂增长模型与承载能力

Exponential growth cannot continue indefinitely because there is a maximum feasible number of member states K, known as the carrying capacity. The logistic differential equation dN/dt = rN(1 – N/K) captures this saturation effect, where r is the intrinsic growth rate.

指数增长不可能无限持续，因为成员国数量存在一个最大可行值 K，称为承载能力。逻辑斯蒂微分方程 dN/dt = rN(1 – N/K) 描述了这种饱和效应，其中 r 为内在增长率。

dN/dt = rN(1 – N/K)

The explicit solution is N(t) = K / [1 + ((K – N₀) / N₀) e^(-rt)]. For instance, if K = 50, N₀ = 5, and r = 0.1, then after 30 years we obtain N(30) = 50 / [1 + (45/5) e^(-3)] = 50 / [1 + 9 e^(-3)] ≈ 50 / 1.448 ≈ 34.5, so about 35 members.

其显式解为 N(t) = K / [1 + ((K – N₀) / N₀) e^(-rt)]。举例，若 K = 50，N₀ = 5，r = 0.1，则 30 年后 N(30) = 50 / [1 + (45/5) e^(-3)] = 50 / [1 + 9 e^(-3)] ≈ 50 / 1.448 ≈ 34.5，约为 35 个成员。

Parameter	Symbol	Meaning
Carrying capacity	K	Maximum possible members
Intrinsic growth rate	r	Growth rate when N is small
Initial size	N₀	Number of members at t = 0

---

3. Modelling Trade Flows with Linear Regression | 用线性回归模拟贸易流量

Suppose the total intra-regional trade volume Y (in billions of USD) depends linearly on the combined GDP X (in trillions of USD) of member states. Least squares regression gives a line of best fit Y = a + bX, where b = Sxy / Sxx and a = ȳ – b x̄.

假设区域内贸易总额 Y（单位：十亿美元）与成员国合计 GDP X（单位：万亿美元）呈线性关系。最小二乘回归给出最佳拟合直线 Y = a + bX，其中 b = Sxy / Sxx，a = ȳ – b x̄。

b = Σ(xᵢ – x̄)(yᵢ – ȳ) / Σ(xᵢ – x̄)²

If data for five years show X values 2, 3, 5, 6, 8 and corresponding Y values 20, 28, 42, 50, 66, then x̄ = 4.8, ȳ = 41.2, Sxy = 156.8, Sxx = 16.8, so b = 9.33 and a = 41.2 – 9.33 × 4.8 ≈ -3.58. The regression line is Y = -3.58 + 9.33X.

若五年数据中 X 值为 2、3、5、6、8，对应 Y 值为 20、28、42、50、66，则 x̄ = 4.8，ȳ = 41.2，Sxy = 156.8，Sxx = 16.8，因此 b = 9.33，a = 41.2 – 9.33 × 4.8 ≈ -3.58。回归直线为 Y = -3.58 + 9.33X。

• A positive slope b indicates that as GDP rises, trade volume increases.
• The coefficient of determination r² measures how well the line fits the data.

• 正斜率 b 表明 GDP 上升时贸易量增加。
• 决定系数 r² 衡量直线对数据的拟合优度。

---

4. Probability of New Member Accession | 新成员加入的概率

When several countries apply to join a regional organisation, each application may succeed independently with probability p. The number of successful applications X follows a binomial distribution B(n, p), with probability mass function P(X = k) = C(n, k) p^k (1 – p)^(n – k).

当多个国家申请加入某区域组织时，每个申请可能以概率 p 独立成功。成功申请的数量 X 服从二项分布 B(n, p)，概率质量函数为 P(X = k) = C(n, k) p^k (1 – p)^(n – k)。

P(X = k) = C(n, k) pᵏ (1 – p)ⁿ⁻ᵏ

If 5 countries apply and each has a 0.4 chance of being accepted, the probability that at least 2 are accepted is P(X ≥ 2) = 1 – P(X = 0) – P(X = 1) = 1 – 0.6⁵ – 5 × 0.4 × 0.6⁴ = 1 – 0.07776 – 0.2592 = 0.66304.

若 5 个国家申请且每个被接受的概率为 0.4，则至少 2 个被接受的概率为 P(X ≥ 2) = 1 – P(X = 0) – P(X = 1) = 1 – 0.6⁵ – 5 × 0.4 × 0.6⁴ = 1 – 0.07776 – 0.2592 = 0.66304。

Number accepted (k)	0	1	2	3	4	5
Probability	0.0778	0.2592	0.3456	0.2304	0.0768	0.0102

---

5. Cumulative Cooperation Projects: Integration | 合作项目累计数量：积分

The rate at which a regional organisation launches new cooperation projects can be expressed as a function R(t) of time t. The total number of projects completed between t = a and t = b is the definite integral ∫ₐᵇ R(t) dt.

某区域组织启动新合作项目的速率可用时间 t 的函数 R(t) 表示。在 t = a 到 t = b 之间完成的累计项目数量为定积分 ∫ₐᵇ R(t) dt。

Total = ∫₀¹⁰ (10 + 2t) dt

If R(t) = 10 + 2t projects per year, then between t = 0 and t = 10 the total is ∫₀¹⁰ (10 + 2t) dt = [10t + t²]₀¹⁰ = 100 + 100 = 200 projects.

若 R(t) = 10 + 2t 个项目/年，则在 t = 0 到 t = 10 之间的总数为 ∫₀¹⁰ (10 + 2t) dt = [10t + t²]₀¹⁰ = 100 + 100 = 200 个项目。

• Integration here corresponds to finding the area under a rate-of-change curve.
• This method can also model cumulative trade volume or total investment over time.

• 这里的积分对应求变化率曲线下的面积。
• 该方法还可用于模拟随时间累计的贸易量或总投资。

---

6. Hypothesis Testing for Regional GDP Differences | 区域GDP差异的假设检验

To test whether two regional organisations (excluding the EU) have significantly different mean GDP per capita, we use a two-sample t-test. The test statistic is t = (x̄₁ – x̄₂) / √(s₁²/n₁ + s₂²/n₂), compared with a critical value from the t-distribution.

为检验两个区域组织（不含欧盟）的人均 GDP 均值是否存在显著差异，我们使用双样本 t 检验。检验统计量为 t = (x̄₁ – x̄₂) / √(s₁²/n₁ + s₂²/n₂)，并与 t 分布的临界值比较。

t = (x̄₁ – x̄₂) / √(s₁²/n₁ + s₂²/n₂)

Suppose Organisation A has sample mean 18,000 USD, standard deviation 2,500, sample size 12; Organisation B has mean 15,500 USD, standard deviation 2,000, sample size 15. Then t = (18000 – 15500) / √(2500²/12 + 2000²/15) = 2500 / √(520833 + 266667) = 2500 / 887.4 ≈ 2.817.

假设组织 A 的样本均值为 18,000 美元，标准差 2,500，样本量 12；组织 B 的均值为 15,500 美元，标准差 2,000，样本量 15。则 t = (18000 – 15500) / √(2500²/12 + 2000²/15) = 2500 / √(520833 + 266667) = 2500 / 887.4 ≈ 2.817。

Using a 5% significance level and a two-tailed test with degrees of freedom approximately 24, the critical t-value is about 2.064. Since 2.817 > 2.064, we reject the null hypothesis and conclude a significant difference exists.

在 5% 显著性水平下，双尾检验、自由度约为 24 的临界 t 值约为 2.064。由于 2.817 > 2.064，我们拒绝原假设，认为存在显著差异。

---

7. Sequences and Series in Annual Contributions | 年度会费缴纳中的数列与级数

Member states often pay an annual contribution that increases by a fixed amount each year, forming an arithmetic sequence. The sum of the first n terms is Sₙ = n/2 [2a + (n – 1)d], where a is the first term and d is the common difference.

成员国缴纳的年度会费通常每年增加固定金额，形成等差数列。前 n 项之和为 Sₙ = n/2 [2a + (n – 1)d]，其中 a 为首项，d 为公差。

Sₙ = n/2 [2a + (n – 1)d]

If a country pays an initial contribution of 100 million USD, increasing by 20 million each year, the total paid over 10 years is S₁₀ = 10/2 [2 × 100 + 9 × 20] = 5 × (200 + 180) = 5 × 380 = 1900 million USD.

若某国初始会费为 1 亿美元，每年增加 2,000 万美元，则 10 年缴纳总额为 S₁₀ = 10/2 [2 × 100 + 9 × 20] = 5 × (200 + 180) = 5 × 380 = 1,900 百万美元。

• Arithmetic sequences model linear growth in payments or projects.
• Geometric sequences with ratio r would be used if contributions grow by a fixed percentage.

• 等差数列用于模拟付款或项目的线性增长。
• 如果会费按固定百分比增长，则使用公比为 r 的等比数列。

---

8. Differential Equations for Integration Depth | 一体化深度的微分方程

The depth of regional integration I(t), ranging from 0 (no integration) to 1 (full political and economic union), can be modelled by the differential equation dI/dt = k(1 – I), where k is a positive constant. This reflects that integration becomes slower as it approaches full union.

区域一体化深度 I(t)，从 0（无一体化）到 1（完全政治经济联盟），可用微分方程 dI/dt = k(1 – I) 建模，其中 k 为正常数。这反映了一体化在接近完全联盟时速度变慢。

dI/dt = k(1 – I)

Solving by separation of variables gives ∫ dI/(1 – I) = ∫ k dt, so -ln|1 – I| = kt + C. With I(0) = 0, we obtain I(t) = 1 – e^(-kt). If k = 0.15, then after 10 years, I(10) = 1 – e^(-1.5) ≈ 0.777, meaning about 78% integration depth.

分离变量得 ∫ dI/(1 – I) = ∫ k dt，因此 -ln|1 – I| = kt + C。由 I(0) = 0，得 I(t) = 1 – e^(-kt)。若 k = 0.15，则 10 年后 I(10) = 1 – e^(-1.5) ≈ 0.777，即一体化深度约为 78%。

• The solution approaches I = 1 asymptotically.
• This is an example of a first-order linear differential equation.

• 解渐近趋于 I = 1。
• 这是一阶线性微分方程的一个例子。

---

9. Statistical Distributions of Trade Volumes | 贸易额的概率分布

Annual intra-regional trade volume X (in billions USD) is often approximately normally distributed with mean μ and

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）