Lagrange Duality

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SoHardToNamed

最新推荐文章于 2025-03-04 22:54:15 发布

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minimize $f(x)$

subject to $x\in X$ ， $g(x)\leq 0$

where $X$ is a nonempty set, $g(x)=(g_{1}(x), ... , g_{r}(x))^{'}$ and $f : X \mapsto \Re$ and $g_{j}:Xx \mapsto \Re ,j=1,...,r$ , are given functions.

$f^{*}$ : optimal value.

feasible $x$ (vecter) : satisfying the CP

maximize $q(\mu )$

subject to $\mu \in \Re ^{r}$

where the dual function $q$ is:

$q(\mu )=\left\{\begin{matrix} \inf _{x\in X}\pounds (x,\mu ) &if\mu \geq 0\\ -\infty &otherwise \end{matrix}\right.$

and $\pounds$ is the Lagrangian function defined by $\pounds (x,\mu )=f(x) + \mu ^{'}g(x)$ , $x\in X,\mu \in \Re ^{r}$ ;

the effective constraint set of the dual problem is :

$\left \{ \mu\geqslant 0|\inf_{x\in X}\pounds (x,\mu)> -\infty \right \}$

$q^{*}$ : optimal value of the DP.

weak duality relation : $q^{*}\leqslant f^{*}$

as well as:

for all $\mu\geqslant 0$ , and $x\in X$ with $g(x)\leqslant 0$ ,

$q(\mu) = \inf_{z\in X}\pounds (z,\mu)\leqslant \pounds (x,\mu)= f(x)+\sum_{j=1}^{r}\mu_{j}g_{j}\leqslant f(x)$

so that:

$q^{*}=\sup_{\mu\in \Re ^{r}}q(\mu)=\sup_{\mu\geqslant 0}q(\mu)\leqslant \inf_{x\in X,g(x) \leqslant 0}f(x)=f^{*}$

we state this formally as follows.

Proposition

1.Deak Duality Theorem:

For any feasible solution $x$ and any $\mu \in \Re^{r}$ , we have $q(\mu )\leqslant f(\mu)$ . Moreover, $q^{*}\leqslant f^{*}$ .

2.Optimality Conditions:

There holds $q^{*}=f^{*}$ , and $(x^{*}, \mu ^{*})$ are a primal and dual optimal solution pair if and only if $x^{*}$ is feasible, $\mu ^{*}\geqslant 0$ , and

$x^{*} \in \arg \min_{x\in X}\pounds (x, \mu^{*}), \mu_{j}^{*}g_{j}(x^{*})=0,j=1,...,r$ .

3.Strong Duality -Existence of Dual optimal solutions:

Assume the set $X$ is convex, and the functions $f$ , and $g_{1}$ , ..., $g_{r}$ are convex. Assume further that $f^{*}$ is finite, and that one of the following twoconditions holds: