A summary of all notes (the video also has a portal): University of Science and Technology of China-convex optimization
原问题
min f 0 ( x ) s . t . f i ( x ) ≤ 0 , i = 1 , ⋯ , m h i ( x ) = 0 , i = 1 , ⋯ , P \min f_0(x)\\s.t.\;f_i(x)\le0,i=1,\cdots,m\\h_i(x)=0,i=1,\cdots,P minf0(x)s.t.fi(x)≤0,i=1,⋯,mhi(x)=0,i=1,⋯,P
对偶函数 g ( λ , v ) = inf x { f 0 ( x ) + ∑ i = 1 m λ i f i ( x ) + ∑ i = 1 P v i h i ( x ) } g(\lambda,v)=\inf_x\{f_0(x)+\sum_{i=1}^m\lambda_if_i(x)+\sum_{i=1}^Pv_ih_i(x)\} g ( λ ,v )=xinf{ f0(x)+i=1∑mλifi(x)+i=1∑Pvihi(x)}
Dual problem max g (λ, v) s. T. Λ ≥ 0 \max\;g(\lambda,v)\\st\;\lambda\ge0maxg ( λ ,v )s.t.λ≥0
Hypothesis
- P ∗ = d ∗ P^*=d^* P∗=d∗
- All functions are differentiable
x ∗ , λ ∗ , v ∗ f i ( x ∗ ) ≤ 0 , i = 1 , ⋯ , m h i ( x ∗ ) = 0 , i = 1 , ⋯ , P λ ∗ ≥ 0 x^*,\lambda^*,v^*\\f_i(x^*)\le0,i=1,\cdots,m\\h_i(x^*)=0,i=1,\cdots,P\\\lambda^*\ge0 x∗,λ∗,v∗fi(x∗)≤0,i=1,⋯,mhi(x∗)=0,i=1,⋯,Pλ∗≥0
KKT conditions
- primal feasibility
- dual feasibility
- complementary slackness
- stationary
If the original problem is a convex problem, each function is differentiable, and the dual gap is zero, then the KKT condition is a necessary and sufficient condition
Next Chapter Portal: University of Science and Technology of China-Convex Optimization Notes (lec37)-KKT conditions for some problems