拉格朗日乘子法简述 - A Brief Tutorial of Using Lagrange Multipliers

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 d variables and k constraints to one with d+k variables without constraint.

1. Equality Constraint
   Suppose x∈Rd , we would like to solve some optimal value x∗ s. t. minxf(x) and g(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 multiplier λ≠0 s.t.
   
   ∇f(x)+λ∇g(x)=0
   
   Define the corresponding Lagrange funcntion as
   
   L(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.
2. Inequality Constraint
   In the inequality constraint case, the optimal problem can be defined as
   
   minxf(x),s.t.g(x)≤0
   
   When g(x)<0 , the constraint g(x)≤0 makes no sense which means that we can take ∇f(x)=0 to solve the optimal problem. In addition, when g(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.
3. Multi-constraints
   Consider an optimal problem with m equality constraints and n inequality constraints. In addition there is a feasible region D⊂Rd :
   
   minxs.t.f(x)hi(x)=0,i=1,2,…,m,gj(x)≤0,j=1,2,…,n
   
   Define the lagrange function as
   
   L(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 function Γ:Rm×Rn→R is defined as
   
   Γ(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 . Let x~∈D , then
   
   Γ(λ,μ)=infx∈DL(x,λ,μ)≤L(x~,λ,μ)≤f(x~)
   That is to say, the dual function shows the lower bound of the value of the primal problem. Then the dual problem is given by
   
   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 p∗ be the optimal value of the primal problem, d∗ be the optimal value of the dual problem. It has been shown that d∗≤p∗ which is also known as weak duality. If d∗=p∗ , then strong duality holds. Generally, the strong duality does not always hold. However, when the primal problem is a convex optimal problem, i.e. f(x),gj(x) are convex functions, hi(x) are affine functions and there exists a least one x~ making the inequality constraints strictly hold, the strong duality holds. In this case, take the derivative of the Lagrange function over x , λ and μ and set it to be zero. Then we can solve the dual problem as well as the primal problem.