优化-拉格朗日乘子法简述
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Lagrange Multipliers are used to solve the optimal value of multivariate functions under a group of constraints. By lagrange multipliers, we can convert an optimal problem with
- Equality Constraint
Supposex∈Rd , we would like to solve some optimal valuex∗ s. t.minxf(x) andg(x)=0 , i.e.minxf(x),s.t.g(x)=0
For simplicity, we omit the geometric explanation of the optimal problem. Define the Lagrange multiplierλ≠0 s.t.∇f(x)+λ∇g(x)=0
Define the corresponding Lagrange funcntion asL(x,λ)=f(x)+λg(x)
So∇xL(x,λ)=∇f(x)+λ∇g(x)=0 ∇λL(x,λ)=g(x)=0
Obviously, the original optimal problem is converted to the new optimal problem with no constraint. - Inequality Constraint
In the inequality constraint case, the optimal problem can be defined asminxf(x),s.t.g(x)≤0
Wheng(x)<0 , the constraintg(x)≤0 makes no sense which means that we can take∇f(x)=0 to solve the optimal problem. In addition, wheng(x)=0 , the inequality constraint degrades to the equality constraint.
In summary, the KKT (Karush-Kuhn-Tucker) conditions are always satisfied:⎧⎩⎨⎪⎪g(x)≤0;λ≥0;λg(x)=0
With efficiency concerned, we show the solution of the optimal problem together with next problem. Go on. - Multi-constraints
Consider an optimal problem withm equality constraints andn inequality constraints. In addition there is a feasible regionD⊂Rd :minxs.t.f(x)hi(x)=0,i=1,2,…,m,gj(x)≤0,j=1,2,…,n
Define the lagrange function asL(x,λ,μ)=f(x)+∑i=1mλihi(x)+∑j=1nμjgj(x)
Since there are inequality constraints, the KKT condition is followed:⎧⎩⎨⎪⎪⎪⎪gj(x)≤0;μj≥0;μjgj(x)=0.
Solving the original optimal problem, also known as primal problem, can be achieved by solving the corresponding dual problem. Then the Lagrange dual functionΓ:Rm×Rn→R is defined asΓ(λ,μ)=infx∈DL(x,λ,μ)=infx∈D⎛⎝f(x)+∑i=1mλihi(x)+∑j=1nμjgj(x)⎞⎠
Evidently,∑mi=1λihi(x)+∑nj=1μjgj(x)≤0 . Letx~∈D , thenThat is to say, the dual function shows the lower bound of the value of the primal problem. Then the dual problem is given byΓ(λ,μ)=infx∈DL(x,λ,μ)≤L(x~,λ,μ)≤f(x~) maxλ,μΓ(λ,μ)s.t.μ≥0
The dual problem is always a convex optimal problem, regardless of the convexity of the primal problem.
Let
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