{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Lecture3 - f we can consider the fol-lowing linear program...

This preview shows pages 1–9. Sign up to view the full content.

Lecture 3: Handling Nonlinearity August 27, 2010

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Nonlinear Programs Is this a linear program?
Converting nonlinear programs to linear programs This is an example of a nonlinear program. A mathematical program that is not linear is called a nonlinear program But some nonlinear programs can be converted to linear programs in terms of new variables

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Handling nonlinearity

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Why does this work?
Handling Minimax (Maximin) Objective Functions Consider the following mathematical program: min u = max { 3 x 1 + 4 x 2 , 5 x 1 + 2 x 2 } s.t. x 1 + x 2 4 2 x 1 + x 2 5 This is a Minimax problem. The objective function is a max function of some linear functions. This is a nonlinear program since the objective is not linear. However, we can convert it to a linear program by introducing a new decision variable f to represent the objective function value. 18

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Handling Minimax (Maximin) Objective Functions After introducing the new variable
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: f , we can consider the fol-lowing linear program: min u = f s.t. x 1 + x 2 ≥ 4 2 x 1 + x 2 ≤ 5 3 x 1 + 4 x 2 ≤ f 5 x 1 + 2 x 2 ≤ f If ( x * 1 ,x * 2 ,f * ) is an optimal solution of the linear program, then ( x * 1 ,x * 2 ) is an optimal solution of the minimax problem, and f * is the optimal value of the minimax problem. Also, we have f * = max { 3 x * 1 + 4 x * 2 , 5 x * 1 + 2 x * 2 } . All minimax problems can be converted to linear programs, it is similar for maximin problems. 19 Handling absolute values: another way ● Observe that |z| = max {z, – z} ● So the problem min |x 1 – x 2 | s.t. x 1 + x 2 ≥ 4 becomes min max {x 1 – x 2 , x 2 – x 1 } s.t. x 1 + x 2 ≥ 4 ● Now it can be handled as a minimax program ● This works when the original problem is a minimization...
View Full Document

{[ snackBarMessage ]}