# 数学代写|线性规划作业代写Linear Programming代考|Projective Transformation

## 数学代写|线性规划作业代写Linear Programming代考|Projective Transformation

Let $\bar{x}$ be an interior point, satisfying $A \bar{x}=0, e^{\mathrm{T}} \bar{x}=1$ and $\bar{x}>0$. Denote by $\bar{X}$ the diagonal matrix, whose diagonals are components of $\bar{x}$, i.e.,
$$\bar{X}=\operatorname{diag}\left(\bar{x}1, \ldots, \bar{x}_n\right)$$ Consider transformation $$x^{\prime}=\frac{\bar{X}^{-1} x}{e^{\mathrm{T}} \bar{X}^{-1} x} \triangleq T(x),$$ whose inverse transformation is $$x=T^{-1}\left(x^{\prime}\right)=\frac{\bar{X} x^{\prime}}{e^{\mathrm{T}} \bar{X} x^{\prime}} .$$ $T(x)$, termed projective transformation, is a 1-1 mapping from $\Gamma$ to $\Gamma$ itself. In fact, for any $x \in \Gamma$, it holds that $x^{\prime}=T(x) \in \Gamma$; conversely, for any $x^{\prime} \in \Gamma$, it holds that $x=T^{-1}\left(x^{\prime}\right) \in \Gamma$. Under $T$, in particular, each vertex $e_j(j=1, \ldots, n)$ of $\Gamma$ corresponds to itself, and so does each edge; most importantly, $\bar{x}$ corresponds to the center $e / n$ of $\Gamma$, whereas any interior point corresponds to interior point. Using $T,(10.1)$ is transformed to problem \begin{aligned} & \min f=\frac{c^{\mathrm{T}} \bar{X} x^{\prime}}{e^{\mathrm{T}} \bar{X} x^{\prime}}, \ & \text { s.t. } A \bar{X} x^{\prime}=0 \text {, } \ & e^{\mathrm{T}} x^{\prime}=1, \quad x^{\prime} \geq 0 \text {. } \ & \end{aligned} which is no longer a LP problem though, because its objective function is not linear. However, it is known from (10.2) and $e^{\mathrm{T}} x=1$ that when $x$ is close to $\bar{x}$, the denominator in the objective function can approximately be regarded as a positive constant, i.e., $$e^{\mathrm{T}} \bar{X} x^{\prime}=\frac{e^{\mathrm{T}} x}{e^{\mathrm{T}} \bar{X}^{-1} x}=1 / \sum{j=1}^n x_j / \bar{x}_j \approx 1 / n$$

## 数学代写|线性规划作业代写Linear Programming代考|Karmarkar Algorithm

Assume that a descent direction has been determined. If some components of $\bar{x}$ are close to zero, then the stepsize along the direction could be very small, and the associated improvement in objective function would be negligible. Problem (10.5) helps avoid this situation, to some extent, since $\bar{x}^{\prime}$ ‘s image, $\bar{x}^{\prime}=T(\bar{x})=e / n$, in $x^{\prime}$ space is now at the center of the simplex, the distance from which to each coordinate plane is the same. Of course, we do not really want to solve (10.5) itself, but only use it as a subproblem to determine a “good” search direction and associated stepsize.

To this end, denote the coefficient matrix of subproblem (10.5) by:
$$F=\left(\begin{array}{c} A \bar{X} \ \vdots \ e^{\mathrm{T}} \end{array}\right)$$
Then the orthogonal projection matrix from $x^{\prime}$ space to the null of $F$ is
$$P=I-F^{\mathrm{T}}\left(F F^{\mathrm{T}}\right)^{-1} F$$
Thus, the orthogonal projection of the objective gradient $\bar{X} c$ is
$$\Delta x=P \bar{X} c=\left(I-F^{\mathrm{T}}\left(F F^{\mathrm{T}}\right)^{-1} F\right) \bar{X} c$$
Proposition 10.1.2 Vector $\Delta x$ is nonzero, satisfying $F \Delta x=0$ and $(\bar{X} c)^{\mathrm{T}} \Delta x>0$. Proof Assume $\Delta x=0$. It is known from (10.8) that there is $h \in \mathcal{R}^m, h_{m+1}$ such that
$$\bar{X}c=F^{\mathrm{T}}\left(h^{\mathrm{T}}, h{m+1}\right)^{\mathrm{T}}$$

# 线性规划代考

## 数学代写|线性规划作业代写Linear Programming代考|Projective Transformation

$$\bar{X}=\operatorname{diag}\left(\bar{x} 1, \ldots, \bar{x}_n\right)$$

$$x^{\prime}=\frac{\bar{X}^{-1} x}{e^{\mathrm{T}} \bar{X}^{-1} x} \triangleq T(x)$$

$$x=T^{-1}\left(x^{\prime}\right)=\frac{\bar{X} x^{\prime}}{e^{\mathrm{T}} \bar{X} x^{\prime}} .$$
$T(x)$ ，称为投影变换，是从 $\Gamma$ 到 $\Gamma$ 本身。事实上，对于任何 $x \in \Gamma$ ，它认为 $x^{\prime}=T(x) \in \Gamma$; 反之，对于任何 $x^{\prime} \in \Gamma$ ，它认为 $x=T^{-1}\left(x^{\prime}\right) \in \Gamma$. 在下面 $T$ ，特别 是每个顶点 $e_j(j=1, \ldots, n)$ 的 $\Gamma$ 对应于自身，每条边也是如此；最重要的是， $\bar{x}$ 对应 于中心 $e / n$ 的 $\Gamma$ ，而任何内点对应于内点。使用 $T,(10.1)$ 转化为问题
$$\min f=\frac{c^{\mathrm{T}} \bar{X} x^{\prime}}{e^{\mathrm{T}} \bar{X} x^{\prime}}, \quad \text { s.t. } A \bar{X} x^{\prime}=0, e^{\mathrm{T}} x^{\prime}=1, \quad x^{\prime} \geq 0 .$$

$$e^{\mathrm{T}} \bar{X} x^{\prime}=\frac{e^{\mathrm{T}} x}{e^{\mathrm{T}} \bar{X}^{-1} x}=1 / \sum j=1^n x_j / \bar{x}_j \approx 1 / n$$

## 数学代写|线性规划作业代写Linear Programming代考|Karmarkar Algorithm

$$F=\left(A \bar{X} \vdots e^{\mathrm{T}}\right)$$

$$P=I-F^{\mathrm{T}}\left(F F^{\mathrm{T}}\right)^{-1} F$$

$$\Delta x=P \bar{X} c=\left(I-F^{\mathrm{T}}\left(F F^{\mathrm{T}}\right)^{-1} F\right) \bar{X} c$$

$$\bar{X}_c=F^{\mathrm{T}}\left(h^{\mathrm{T}}, h m+1\right)^{\mathrm{T}}$$

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