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handout4 - Math 171A: Linear Programming Lecture 4...

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Unformatted text preview: Math 171A: Linear Programming Lecture 4 Properties of the Objective Function Philip E. Gill c 2011 http://ccom.ucsd.edu/~peg/math171a Monday, January 10th, 2011 Recap: Properties of linear constraints The constraints may be infeasible constraint #1: x 1 + x 2 ≥ 1 , a T 1 = ( 1 1 ) , b 1 = 1 constraint #2: x 1 ≥ , a T 2 = ( 1 ) , b 2 = constraint #3:- x 1 ≥ - 2 , a T 3 = (- 1 ) , b 3 =- 2 constraint #4: x 2 ≥ , a T 4 = ( 1 ) , b 4 = constraint #5: x 1 + 2 x 2 ≥ 1 , a T 5 = ( 1 2 ) , b 5 = 1 constraint #6:- 2 x 1- 4 x 2 ≥ - 1 , a T 6 = (- 2- 4 ) , b 6 =- 1 A = 1 1 1- 1 1 1 2- 2- 4 , b = 1- 2 1- 1 UCSD Center for Computational Mathematics Slide 2/63, Monday, January 10th, 2011 x 1 x 2 a 1 # 1 #2 #3 #4 a 4 #5 a 5 a 2 #6 If constraint #6 is present, there are no feasible points! Definition A set of constraints is feasible (aka consistent ) if there exists at least one feasible point. Otherwise, the constraints are infeasible (aka inconsistent ). UCSD Center for Computational Mathematics Slide 4/63, Monday, January 10th, 2011 Shift constraint # 6: 2 x 1 + 4 x 2 ≤ 2 . constraint #1: x 1 + x 2 ≥ 1 , a T 1 = ( 1 1 ) , b 1 = 1 constraint #2: x 1 ≥ , a T 2 = ( 1 ) , b 2 = constraint #3:- x 1 ≥ - 2 , a T 3 = (- 1 ) , b 3 =- 2 constraint #4: x 2 ≥ , a T 4 = ( 1 ) , b 4 = constraint #5: x 1 + 2 x 2 ≥ 1 , a T 5 = ( 1 2 ) , b 5 = 1 constraint #6:- 2 x 1- 4 x 2 ≥- 2 , a T 6 = (- 2- 4 ) , b 6 =- 2 A = 1 1 1- 1 1 1 2- 2- 4 b = 1- 2 1- 2 UCSD Center for Computational Mathematics Slide 5/63, Monday, January 10th, 2011 x 1 x 2 a 1 # 1 #2 #3 #4 a 4 #5 a 5 a 2 #6 If the shifted constraint #6 is present, there is just one feasible point! A linear program can have a unique feasible point (necessarily optimal!) UCSD Center for Computational Mathematics Slide 7/63, Monday, January 10th, 2011 An important feature of the feasible region Result Every point on the boundary of FF lies on the intersection of a subset of the constraint hyperplanes . The number of intersecting hyperplanes depends on the type of boundary point. UCSD Center for Computational Mathematics Slide 8/63, Monday, January 10th, 2011 x 1 x 2 a 1 # 1 #2 #3 #4 a 4 #5 a 5 a 2 x 1 x 2 ¯ x = 1 x 1 + x 2 = 1 x 1 = 0 The boundary point ¯ x = 1 ! lies on 2 hyperplanes: H 1 = { x : x 1 + x 2 = 1 } H 2 = { x : x 1 = 0 } ⇒ ¯ x satisfies the square nonsingular system of equations 1 1 1 0 !...
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This note was uploaded on 03/19/2012 for the course MATH 171a taught by Professor Staff during the Spring '08 term at UCSD.

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handout4 - Math 171A: Linear Programming Lecture 4...

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