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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|线性规划作业代写Linear Programming代考|Relations to Convex Geometry

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 fundamental theorem, but also to a clearer geometric understanding of the result. 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 $\alpha, 0<\alpha<1$.

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
$$\mathbf{A x}=\mathbf{b}, \mathbf{x} \geqslant 0$$
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 that the last $n-m$ 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 $$z{1} \mathbf{a}{1}+z{2} \mathbf{a}{2}+\cdots+z{m} \mathbf{a}_{m}=\mathbf{b}$$

## 数学代写|线性规划作业代写Linear Programming代考|Farkas’ Lemma and Alternative Systems

We now present a theorem to check whether or not a feasible solution exists for constraint system (2.19). If one can find a single solution to meet all the constraints, then it is a “positive” certificate to prove the system feasible. The question is: how could we construct a “negative” certificate to prove the system infeasible?
Theorem (Farkas’ Lemma) Let $\mathbf{A}$ be an $m \times n$ matrix and $\mathbf{b}$ an m-vector. The system of constraints
$$\mathbf{A x}=\mathbf{b}, \mathbf{x} \geqslant \mathbf{0}$$
has a feasible solution $\mathbf{x}$ if and only if the system of constraints
$$-\mathbf{y}^{T} \mathbf{A} \geqslant \mathbf{0}, \mathbf{y}^{T} \mathbf{b}=1(\text { or }>0)$$
has no feasible solution y. Therefore a single feasible solution y for system (2.21) establishes a certificate to prove system (2.20) infeasible.
The two systems, (2.20) and (2.21), are called alternative systems: one of them is feasible and the other is infeasible.

Example 1 Let $1 \times 2$ matrix $\mathbf{A}=(11)$ and scalar $b=-1$. Then, $y=-1$ is feasible for system (2.21), which proves that system (2.20) is infeasible.
Before we prove the theorem, we first present a lemma.
Lemma 1 Let $C$ be the cone generated by the columns of matrix $\mathbf{A}$, that is,
$$C=\left{\mathbf{A x} \in E^{m}: \mathbf{x} \geqslant \mathbf{0}\right} .$$
Then $C$ is a closed and convex set.
The definition of cone and conic combination can be found in Sect. A.3. We leave the proof of the lemma as an exercise, where the closeness proof needs to use Carathéodory’s theorem given in Sect. 2.4.

# 线性规划代写

## 数学代写|线性规划作业代写Linear Programming代考|Relations to Convex Geometry

$$\mathbf{A x}=\mathbf{b}, \mathbf{x} \geqslant 0$$

$$x_{1} \mathbf{a} 1+x 2 \mathbf{a} 2+\cdots+x m \mathbf{a} m=\mathbf{b}$$

$$y_{1} \mathbf{a} 1+y 2 \mathbf{a} 2+\cdots+y m \mathbf{a} m=\mathbf{b}$$

$$z \mathbf{1} \mathbf{a} 1+z 2 \mathbf{a} 2+\cdots+z m \mathbf{a}_{m}=\mathbf{b}$$

## 数学代写|线性规划作业代写Linear Programming代考|Farkas’ Lemma and Alternative Systems

$$\mathbf{A x}=\mathbf{b}, \mathbf{x} \geqslant \mathbf{0}$$

$$-\mathbf{y}^{T} \mathbf{A} \geqslant \mathbf{0}, \mathbf{y}^{T} \mathbf{b}=1(\text { or }>0)$$

(2.20) 和 (2.21) 这两个系统被称为替代系统: 一个是可行的，另一个是不可行的。

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## 有限元方法代写

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## MATLAB代写

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