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Lecture 9 - Feasible Directions for Inequality Constraints

# Lecture 9 - Feasible Directions for Inequality Constraints...

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Lecture 9 Feasible Directions for Inequality Constraints UCSD Math 171A: Numerical Optimization Philip E. Gill ( [email protected] ) Monday, January 26, 2009 Recap: Linear programming with equality constraints LP’s with equality constraints ELP minimize x R n c T x subject to Ax = b , UCSD Center for Computational Mathematics Slide 2/34, Monday, January 26, 2009 Optimality conditions for ELP Result Consider minimizing ( x ) = c T x subject to Ax = b . (a) If Ax = b is incompatible, no solution exists; (b) If Ax = b is compatible and c does not lie in the range of A T , the objective function is unbounded below at feasible points; (c) If Ax = b is compatible and c lies in the range of A T (so that c = A T λ * for some vector λ * ), then: (i) * , the optimal value of , is finite and unique; (ii) Every feasible point is a minimizer x * ; (iii) x * is unique if and only if the columns of A are linearly independent; (iv) λ * is unique if and only if the rows of A are linearly independent. UCSD Center for Computational Mathematics Slide 3/34, Monday, January 26, 2009 x = solve(A,b) If incompatible, no feasible point exists If compatible, x is feasible lambda = solve(A’,c) If incompatible, no bounded solution If compatible, x is a minimizer UCSD Center for Computational Mathematics Slide 4/34, Monday, January 26, 2009

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Example 1: A = - 1 5 0 1 1 3 - 1 1 4 2 ! , b = 4 - 5 ! , c = - 4 6 - 1 - 3 - 1 Feasibility? x * = solve(A,b) x * = 0 1 0 - 1 0 , which is a basic solution of Ax = b UCSD Center for Computational Mathematics Slide 5/34, Monday, January 26, 2009 Example 1: (continued) A = - 1 5 0 1 1 3 - 1 1 4 2 ! , b = 4 - 5 ! , c = - 4 6 - 1 - 3 - 1 Optimality? λ * = solve(A’,c) λ * = 1 - 1 ! , which are unique Lagrange multipliers x * is optimal, with ( x * ) = c T x * = 9. UCSD Center for Computational Mathematics Slide 6/34, Monday, January 26, 2009 Example 1: (continued) A = - 1 5 0 1 1 3 - 1 1 4 2 ! , b = 4 - 5 ! , c = - 4 6 - 1 - 3 - 1 Note that b x = - 3 2 1 2 0 0 0 is another basic solution, with ( b x ) = c T b x = 9. UCSD Center for Computational Mathematics Slide 7/34, Monday, January 26, 2009 Example 2: A = - 1 5 0 1 1 3 - 1 1 4 2 ! , b = 4 - 5 ! , c = - 8 - 2 6 3 3 Feasibility? x * = solve(A,b) x * = 0 1 0 - 1 0 , which is a basic solution of Ax = b UCSD Center for Computational Mathematics Slide 8/34, Monday, January 26, 2009
Example 2: (continued) A = - 1 5 0 1 1 3 - 1 1 4 2 !

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