mthsc810-lecture20-1x2

mthsc810-lecture20-1x2 - MthSc 810 Mathematical Programming...

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MthSc 810: Mathematical Programming Lecture 20 Pietro Belotti Dept. of Mathematical Sciences Clemson University December 6, 2011 Reading for today: Sections 6.1, 6.2 Reading for Thursday: Sections 6.3, 6.4. Recap: Lagrangian relaxation Consider a problem z OPT = min c x s . t . A x = b d x = f x 0 . Lagrangian relaxation applied to the last constraint yields min c x + λ ( d x f ) s . t . A x = b x 0 , and a lower bound on z for any λ R .

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Lagrangian function Consider the function L ( λ ) = min { c x + λ ( d x f ) : A x = b , x 0 } = λ f + min { ( c + λ d ) x : A x = b , x 0 } , which gives a lower bound on z OPT for any λ R . What does it look like? For variable c , G ( c ) = min { c x : A x = b , x 0 } is a concave function. It is the objective function value of an LP whose objective function has parametric coef±cients, just like L ( λ ) . Similarly, it can be proved that L ( λ ) is concave. A tight lower bound Because L ( λ ) z for any λ R , the tightest lower bound is max {L ( λ ) : λ R } . If the relaxed constraint were d x f , then λ would be constrained in sign Anyway, max λ R L ( λ ) is a maximization problem of a concave function Easy! Equivalent to min {−L ( λ ) : λ R } , where −L ( λ ) is convex
Minimizing −L ( λ ) −L ( λ ) is a convex , piece-wise linear function Piece-wise linear there exists no gradient However, it admits a subgradient . Consider λ = ¯ λ : −L ( ¯ λ ) = f ¯ λ min { ( c + ¯ λ d ) x : A x = b , x 0 } −L ( ¯ λ ) = f ¯ λ ( c + ¯ λ d ) x , where x is the optimal solution of the relaxation with λ = ¯ λ A subgradient of −L ( λ ) at λ = ¯ λ is f d x If f d x > 0, the minimum is at λ ¯ λ If f d x < 0, the minimum is at λ ¯ λ If f d x = 0, STOP : we found a minimum of −L ( λ ) , i.e., a good lower bound on z OPT .

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This note was uploaded on 03/14/2012 for the course MTHSC 810 taught by Professor Staff during the Fall '08 term at Clemson.

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mthsc810-lecture20-1x2 - MthSc 810 Mathematical Programming...

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