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# 数学代写|凸优化代写Convex Optimization代考|LAGRANGE DUALITY

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## 数学代写|凸优化代写Convex Optimization代考|LAGRANGE DUALITY

We start our overview of Lagrange duality with the basic case of nonlinear inequality constraints, and then consider extensions involving linear inequality and equality constraints. Consider the problem $\neq$
$$\begin{array}{ll} \operatorname{minimize} & f(x) \ \text { subject to } & x \in X, \quad g(x) \leq 0, \end{array}$$
where $X$ is a nonempty set,
$$g(x)=\left(g_1(x), \ldots, g_r(x)\right)^{\prime},$$
and $f: X \mapsto \Re$ and $g_j: X \mapsto \Re, j=1, \ldots, r$, are given functions. We refer to this as the primal problem, and we denote its optimal value by $f^*$. A vector $x$ satisfying the constraints of the problem is referred to as feasible. The dual of problem (1.1) is given by
$$\begin{array}{ll} \operatorname{maximize} & q(\mu) \ \text { subject to } & \mu \in \Re^r, \end{array}$$

where the dual function $q$ is
$$q(\mu)= \begin{cases}\inf _{x \in X} L(x, \mu) & \text { if } \mu \geq 0, \ -\infty & \text { otherwise, }\end{cases}$$
and $L$ is the Lagrangian function defined by
$$L(x, \mu)=f(x)+\mu^{\prime} g(x), \quad x \in X, \mu \in R^r ;$$
(cf. Section 5.3 of Appendix B).

## 数学代写|凸优化代写Convex Optimization代考|Convex Programming with Inequality and Equality Constraints

Let us consider an extension of problem (1.1), with additional linear equality constraints. It is our principal constrained optimization model under convexity assumptions, and it will be referred to as the convex programming problem. It is given by
$$\begin{array}{ll} \operatorname{minimize} & f(x) \ \text { subject to } & x \in X, \quad g(x) \leq 0, \quad A x=b, \end{array}$$
where $X$ is a convex set, $g(x)=\left(g_1(x), \ldots, g_r(x)\right)^{\prime}, f: X \mapsto \Re$ and $g_j: X \mapsto \Re, j=1, \ldots, r$, are given convex functions, $A$ is an $m \times n$ matrix, and $b \in \Re m$.

The preceding duality framework may be applied to this problem by converting the constraint $A x=b$ to the equivalent set of linear inequality constraints
$$A x \leq b, \quad-A x \leq-b,$$
with corresponding dual variables $\lambda^{+} \geq 0$ and $\lambda^{-} \geq 0$. The Lagrangian function is
$$f(x)+\mu^{\prime} g(x)+\left(\lambda^{+}-\lambda^{-}\right)^{\prime}(A x-b)$$
and by introducing a dual variable
$$\lambda=\lambda^{+}-\lambda^{-}$$

with no sign restriction, it can be written as
$$L(x, \mu, \lambda)=f(x)+\mu^{\prime} g(x)+\lambda^{\prime}(A x-b) .$$
The dual problem is
$$\begin{array}{ll} \operatorname{maximize} & \inf _{x \in X} L(x, \mu, \lambda) \ \text { subject to } & \mu \geq 0, \lambda \in \Re^m . \end{array}$$

## 数学代写|凸优化代写Convex Optimization代考|LAGRANGE DUALITY

$$\begin{array}{ll} \operatorname{minimize} & f(x) \ \text { subject to } & x \in X, \quad g(x) \leq 0, \end{array}$$

$$g(x)=\left(g_1(x), \ldots, g_r(x)\right)^{\prime},$$
$f: X \mapsto \Re$和$g_j: X \mapsto \Re, j=1, \ldots, r$是给定的函数。我们称其为原始问题，并用$f^*$表示其最优值。满足问题约束的向量$x$称为可行向量。(1.1)问题的对偶由
$$\begin{array}{ll} \operatorname{maximize} & q(\mu) \ \text { subject to } & \mu \in \Re^r, \end{array}$$

$$q(\mu)= \begin{cases}\inf _{x \in X} L(x, \mu) & \text { if } \mu \geq 0, \ -\infty & \text { otherwise, }\end{cases}$$
$L$为拉格朗日函数，定义为
$$L(x, \mu)=f(x)+\mu^{\prime} g(x), \quad x \in X, \mu \in R^r ;$$
(参见附录B第5.3节)。

## 数学代写|凸优化代写Convex Optimization代考|Convex Programming with Inequality and Equality Constraints

$$\begin{array}{ll} \operatorname{minimize} & f(x) \ \text { subject to } & x \in X, \quad g(x) \leq 0, \quad A x=b, \end{array}$$

$$A x \leq b, \quad-A x \leq-b,$$

$$f(x)+\mu^{\prime} g(x)+\left(\lambda^{+}-\lambda^{-}\right)^{\prime}(A x-b)$$

$$\lambda=\lambda^{+}-\lambda^{-}$$

$$L(x, \mu, \lambda)=f(x)+\mu^{\prime} g(x)+\lambda^{\prime}(A x-b) .$$

$$\begin{array}{ll} \operatorname{maximize} & \inf _{x \in X} L(x, \mu, \lambda) \ \text { subject to } & \mu \geq 0, \lambda \in \Re^m . \end{array}$$

## MATLAB代写

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