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Lecture 20 - Duality Theory I

# Lecture 20 - Duality Theory I - Recap Simplex method for...

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Lecture 20 Duality Theory I UCSD Math 171A: Numerical Optimization Philip E. Gill http://ccom.ucsd.edu/~peg/math171 Friday, February 27th, 2009 Recap: Simplex method for standard form Solve B T π = c B ; z N = c N - N T π ; ( z N ) s = min( z N ); if ( z N ) s 0 then stop ; Solve Bp B = - a ν s ; σ i = ( x B ) i - ( p B ) i if ( p B ) i < 0; + if ( p B ) i 0; σ t = min { σ i } ; α = σ t ; if α = + then stop ; x B x B + α p B ; ( x B ) t α ; Exchange index β t of B with index ν s of N ; UCSD Center for Computational Mathematics Slide 2/29, Friday, February 27th, 2009 Getting feasible How do we get a feasible basic solution of Ax = b ? Example: A = 2 3 1 1 0 1 2 1 0 - 1 ! , b = 5 3 ! UCSD Center for Computational Mathematics Slide 3/29, Friday, February 27th, 2009 2 x 1 + 3 x 2 + x 3 + x 4 = 5 x 1 + 2 x 2 + x 3 - x 5 = 3 x 1 , x 2 , x 3 , x 4 , x 5 0 Add positive shifts x 6 and x 7 : 2 x 1 + 3 x 2 + x 3 + x 4 + x 6 = 5 x 1 + 2 x 2 + x 3 - x 5 + x 7 = 3 x 1 , x 2 , x 3 , x 4 , x 5 , x 6 , x 7 0 The basis B = { 6 , 7 } defines the basic solution x B = 5 3 . UCSD Center for Computational Mathematics Slide 4/29, Friday, February 27th, 2009

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Minimize the “sum of infeasibilities” minimize x R 7 x 6 + x 7 2 x 1 + 3 x 2 + x 3 + x 4 + x 6 = 5 x 1 + 2 x 2 + x 3 - x 5 + x 7 = 3 x 1 , x 2 , x 3 , x 4 , x 5 , x 6 , x 7 0 If a feasible point exists, both x 6 and x 7 will be zero (i.e., nonbasic) at the end of phase 1. two other variables will be basic phase 1 solution is an initial feasible basic solution for phase 2 UCSD Center for Computational Mathematics Slide 5/29, Friday, February 27th, 2009 If the first constraint had been of the form 2 x 1 + 3 x 2 + x 3 + x 4 = - 5 we would have included the shift x 6 as 2 x 1 + 3 x 2 + x 3 + x 4 - x 6 = - 5 but we still use + x 6 in the sum of infeasibilities. UCSD Center for Computational Mathematics Slide 6/29, Friday, February 27th, 2009 General case To find an initial vertex for Ax = b , x 0 we define shifts v = x n +1 x n +2 . . . x n + m UCSD Center for Computational Mathematics Slide 7/29, Friday, February 27th, 2009 General case Define shifted constraints A V x v ! = b with V = sign( b 1 ) sign( b 2 ) . . . sign( b m ) sign( b i ) = - 1 if b i < 0 1 if b i 0 The initial basis is B = { n + 1, n + 2, . . . , n + m } .
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Lecture 20 - Duality Theory I - Recap Simplex method for...

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