L10 Optimal Design L.Multipliers Homework Review Meaning & of Lagrange Multiplier Summary

Homework 4.44 Now is a “minimize” We have only used “necessary conditions” We cannot yet conclude that the pt is a MIN!

4.44 cont’d H(x) is negative definite, therefore the candidate pt is not a local min. (therefore Pt A is NOT a max of the original F(x)). Unbounded?

Prob 4.54

Gaussian Elimination Case 2 x R1 by -1 + to R2 x R3 by -13 + to R2

Prob 4.57

Gaussian Elmination 4.57 Case 1 u=0 +R1 to R2 x R2 by -1/2 + to R3 Check feasibility Backsub Using R2

Gaussian Elmination 4.57 Case 2 s=0 +R3 to R4 Check feasibility Backsub Using R3

Prob 4.57

Prob 4.59

Prob 4.59

Prob 4.59

Gaussian Elmination 4.59 Case 4 s1,2=0 +R3 to R4 Backsub Using R3 Check feasibility Both s1 and s2 =0

Prob 4.59 Where is: Case 1 Case 2 Case 3 Case 4

MV Optimization Inequality & Equality Constrained

KKT Necessary Conditions for Min Regularity check - gradients of active inequality constraints are linearly independent

Relax both constraints (Prob 4.59)

Constraint Variation Sensitivity Theorem The instantaneous rate of change in the objective function with respect to relaxing a constraint IS the LaGrange multiplier!

Practical Use of Multipliers in 4.59 The first-order approximation on f(x), of relaxing a constraint is obtained from a Taylor Series expansion: f(actual)=1 versus f(approx)=0

Summary Min =-Max, i.e. f(x)=-F(x) Necessary Conditions for Min KKT point is a CANDIDATE min! (need sufficient conditions for proof) Use switching conditions, Gaussian Elimination to find KKT pts LaGrange multipliers are the instantaneous rate of change in f(x) w.r.t. change in constraint relaxation.