对于问题： m i n f ( x ) s . t . g ( x ) ≤ 0 h ( x ) = 0 l ≤ x ≤ u \begin{aligned} min \quad &f(x)\\ s.t.\quad &g(x)\le 0\\ &h(x)=0\\ &l\le x\le u \end{aligned} mins.t.f(x)g(x)≤0h(x)=0l≤x≤u 与问题关联的朗格朗日函数： L ( x , λ , ν ) = f ( x ) + λ T g ( x ) + ν T h ( x ) L(x, \lambda, \nu)=f(x)+\lambda^T g(x)+\nu^Th(x) L(x,λ,ν)=f(x)+λTg(x)+νTh(x) KKT条件为： ∇ f ( x ) + ∇ g ( x ) λ + ∇ h ( x ) ν = 0 , l ≤ x ≤ u g ( x ) ≤ 0 λ ≥ 0 λ T g ( x ) = 0 h ( x ) = 0 \begin{aligned} &\nabla f(x)+\nabla g(x)\lambda+\nabla h(x)\nu =0,\quad l\le x\le u\\ &g(x)\leq0 \\ &\lambda \ge0\\ &\lambda^T g(x)=0\\ &h(x)=0 \\ \end{aligned} ∇f(x)+∇g(x)λ+∇h(x)ν=0,l≤x≤ug(x)≤0λ≥0λTg(x)=0h(x)=0
根据投影定理，KKT条件可以等价为： { P Ω ( x − ( ∇ f ( x ) + ∇ g ( x ) λ + ∇ h ( x ) ν ) ) = x ( λ + g ( x ) ) + = λ h ( x ) = 0 \left\{ \begin{aligned} &P_\Omega(x-(\nabla f(x)+\nabla g(x)\lambda+\nabla h(x)\nu)) =x\\ &(\lambda+g(x))^+=\lambda\\ &h(x)=0 \end{aligned} \right. ⎩⎪⎨⎪⎧PΩ(x−(∇f(x)+∇g(x)λ+∇h(x)ν))=x(λ+g(x))+=λh(x)=0 其中 ( ξ ) + = m a x { 0 , ξ } P Ω ( ξ ) = { u i , ξ > u i ξ , l i ≤ ξ ≤ u i l i , ξ < l i (\xi)^+=max\{0, \xi\} \\ P_\Omega(\xi)=\left\{\begin{aligned} &u_i,\qquad \xi>u_i\\&\xi,\qquad l_i\le\xi\le u_i\\&l_i,\qquad \xi<l_i\end{aligned}\right. (ξ)+=max{0,ξ}PΩ(ξ)=⎩⎪⎨⎪⎧ui,ξ>uiξ,li≤ξ≤uili,ξ<li
