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# 数学代考|Relations to Convex Geometry 运筹学代写

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## 运筹学代写

Our development to this point, including the above proof of the fundamental theorem, has been based only on elementary properties of systems of linear equations. These results, however, have interesting interpretations in terms of the theory of convex sets that can lead not only to an alternative derivation of the The main link between the algebraic and geometric theories is the formal relation between basic feasible solutions of linear inequalities in standard form and extreme points of polytopes. We establish this correspondence as follows. The reader is referred to Appendix B for a more complete summary of concepts related to convexity, but the definition of an extreme point is stated here.

Definition A point $\mathbf{x}$ in a convex set $C$ is said to be an extreme point of $C$ if there are no two distinct points $\mathbf{x}{1}$ and $\mathbf{x}{2}$ in $C$ such that $\mathbf{x}=\alpha \mathbf{x}{1}+(1-\alpha) \mathbf{x}{2}$ for some $\alpha, 0<\alpha<1$.
$2.5$ Relations to Convex Geometry
29
An extreme point is thus a point that does not lie strictly within a line segment connecting two other points of the set. The extreme points of a triangle, for example, are its three vertices.
Theorem (Equivalence of Extreme Points and Basic Solutions) Let $\mathbf{A}$ be an $m \times n$ matrix of rank $m$ and $\mathbf{b}$ an $m$-vector. Let $K$ be the convex polytope consisting of all $n$-vectors $\mathbf{x}$ satisfying
$(2.19)$
A vector $\mathbf{x}$ is an extreme point of $K$ if and only if $\mathbf{x}$ is a basic feasible solution to $(2,19)$.
Proof Suppose first that $\mathbf{x}=\left(x_{1}, x_{2}, \ldots, x_{m}, 0,0, \ldots, 0\right)$ is a basic feasible solution to $(2.19)$. Then
$$x_{1} \mathbf{a}{1}+x{2} \mathbf{a}{2}+\cdots+x{m} \mathbf{a}{m}=\mathbf{b}$$ where $\mathbf{a}{1}, \mathbf{a}{2}, \ldots, \mathbf{a}{m}$, the first $m$ columns of $\mathbf{A}$, are linearly independent. Suppose that $\mathbf{x}$ could be expressed as a convex combination of two other points in $K$; say, $\mathbf{x}=\alpha \mathbf{y}+(1-\alpha) \mathbf{z}, 0<\alpha<1, \mathbf{y} \neq \mathbf{z}$. Since all components of $\mathbf{x}, \mathbf{y}, \mathbf{z}$ are nonnegative and since $0<\alpha<1$, it follows immediately components of $\mathbf{y}$ and $\mathbf{z}$ are zero. Thus, in particular, we have
$$y_{1} \mathbf{a}{1}+y{2} \mathbf{a}{2}+\cdots+y{m} \mathbf{a}{m}=\mathbf{b}$$ and $$z{1} \mathbf{a}{1}+z{2} \mathbf{a}{2}+\cdots+z{m} \mathbf{a}{m}=\mathbf{b}$$ Since the vectors $\mathbf{a}{1}, \mathbf{a}{2}, \ldots, \mathbf{a}{m}$ are linearly independent, however, it follows that $\mathbf{x}=\mathbf{y}=\mathbf{z}$ and hence $\mathbf{x}$ is an extreme point of $K .$

Conversely, assume that $\mathbf{x}$ is an extreme point of $K$. Let us assume that the nonzero components of $\mathbf{x}$ are the first $k$ components. Then
$$x_{1} \mathbf{a}{1}+x{2} \mathbf{a}{2}+\cdots+x{k} \mathbf{a}{k}=\mathbf{b}$$ with $x{i}>0, i=1,2, \ldots, k$. To show that $\mathbf{x}$ is a basic feasible solution it must be shown that the vectors $\mathbf{a}{1}, \mathbf{a}{2}, \ldots, \mathbf{a}{k}$ are linearly independent. We do this by contradiction. Suppose $\mathbf{a}{1}, \mathbf{a}{2}, \ldots, \mathbf{a}{k}$ are linearly dependent. Then there is a nontrivial linear combination that is zero:
$$y_{1} \mathbf{a}{1}+y{2} \mathbf{a}{2}+\cdots+y{k} \mathbf{a}{k}=0$$ Define the $n$-vector $\mathbf{y}=\left(y{1}, y_{2}, \ldots, y_{k}, 0,0, \ldots, 0\right)$. Since $x_{i}>0,1 \leqslant i \leqslant k$, it is possible to select $\varepsilon$ such that
$$\mathbf{x}+\varepsilon \mathbf{y} \geqslant 0, \quad \mathbf{x}-\varepsilon \mathbf{y} \geqslant 0$$
30
2 Basic Properties of Linear Programs
We then have $\mathbf{x}=\frac{1}{2}(\mathbf{x}+\varepsilon \mathbf{y})+\frac{1}{2}(\mathbf{x}-\varepsilon \mathbf{y})$ which expresses $\mathbf{x}$ as a convex combination of two distinct vectors in $K$. This cannot occur, since $\mathbf{x}$ is an extreme point of $K$. Thus $\mathbf{a}{1}, \mathbf{a}{2}, \ldots, \mathbf{a}_{k}$ are linearly independent and $\mathbf{x}$ is a basic feasible solution. (Although if $k<m$, it is a degenerate basic feasible solution.)

y_{1} \mathbf{a}{1}+y{2} \mathbf{a}{2}+\cdots+y{k} \mathbf{a}{k}=0 $$定义n-向量\mathbf{y}=\left(y{1}, y_{2}, \ldots, y_{k}, 0,0, \ldots, 0\right)。由于 x_{i}>0,1 \leqslant i \leqslant k，因此可以选择 \varepsilon，使得$$
\mathbf{x}+\varepsilon \mathbf{y} \geqslant 0, \quad \mathbf{x}-\varepsilon \mathbf{y} \geqslant 0

$30 \quad 2$ 线性规划的基本性质

## 什么是运筹学代写

• 确定需要解决的问题。
• 围绕问题构建一个类似于现实世界和变量的模型。
• 使用模型得出问题的解决方案。
• 在模型上测试每个解决方案并分析其成功。
• 实施解决实际问题的方法。

## 运筹学代写的三个特点

• 优化——运筹学的目的是在给定的条件下达到某一机器或者模型的最佳性能。优化还涉及比较不同选项和缩小潜在最佳选项的范围。
• 模拟—— 这涉及构建模型，以便在应用解决方案刀具体的复杂大规模问题之前之前尝试和测试简单模型的解决方案。
• 概率和统计——这包括使用数学算法和数据挖掘来发现有用的信息和潜在的风险，做出有效的预测并测试可能的解决方法。