增广拉格朗日方法在拉格朗日方法的基础上添加了二次惩罚项，从而使得转换后的问题能够更容易求解，不至于因条件数变大不好求。则转换后的问题为 Ψ ( x , λ , ν ) = L ( x , λ , ν ) + α 2 ∑ j = 1 m ( λ j g j ( x ) ) 2 + β 2 ∑ i = 1 q ( ν i h i ( x ) ) 2 \Psi(x,\lambda,\nu)=L(x,\lambda,\nu)+\frac{\alpha}{2}\sum_{j=1}^{m}({\lambda_jg_j(x)})^2+\frac{\beta}{2}\sum_{i=1}^{q}{(\nu_ih_i(x))^2} Ψ(x,λ,ν)=L(x,λ,ν)+2αj=1∑m(λjgj(x))2+2βi=1∑q(νihi(x))2 其中 α \alpha α 和 β \beta β 是两个非负的参数， Ψ ( x , λ , ν ) \Psi(x,\lambda,\nu) Ψ(x,λ,ν)的梯度为 ∇ x Ψ ( x , λ , ν ) = ∇ x L ( x , λ , ν ) + α ∑ j = 1 m λ j 2 g j ( x ) ∇ g j ( x ) + β ∑ i = 1 q ν i 2 h i ( x ) ∇ h i ( x ) \nabla_x\Psi(x,\lambda,\nu)=\nabla_xL(x,\lambda,\nu)+\alpha\sum_{j=1}^{m}{\lambda_j^2g_j(x)\nabla g_j(x)}+\beta\sum_{i=1}^{q}{\nu_i^2h_i(x)\nabla h_i(x)} ∇xΨ(x,λ,ν)=∇xL(x,λ,ν)+αj=1∑mλj2gj(x)∇gj(x)+βi=1∑qνi2hi(x)∇hi(x) 根据KKT条件 ∇ x Ψ ( x ∗ , λ ∗ , ν ∗ ) = ∇ x L ( x ∗ , λ ∗ , ν ∗ ) \nabla_x\Psi(x^*,\lambda^*,\nu^*)=\nabla_xL(x^*,\lambda^*,\nu^*) ∇xΨ(x∗,λ∗,ν∗)=∇xL(x∗,λ∗,ν∗)
