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Lecture6 - C&O 355 Lecture 6 N Harvey...

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C&O 355 Lecture 6 N. Harvey http://www.math.uwaterloo.ca/~harvey/
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Outline Proof of Optimality Does the algorithm terminate? Bland’s Rule Corollaries
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1. What is a corner point? (BFS and bases) 2. What if there are no corner points? (Infeasible) 3. What are the “neighboring” bases? (Increase one coordinate) 4. What if no neighbors are strictly better? (Might move to a basis that isn’t strictly better (if ± =0), but whenever x changes it’s strictly better) 5. How can I find a starting feasible basis? 6. Does the algorithm terminate? 7. Does it produce the right answer? Local-Search Algorithm Let B be a feasible basis (If none, Halt: LP is infeasible) For each entering coordinate k B If “benefit” of coordinate k is > 0 Compute y( ± ) (If ± = 1 , Halt: LP is unbounded) Find leaving coordinate h 2 B (y( ± ) h =0) Set x=y( ± ) and B’= B n {h} [ {k} Halt: return x
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The Benefits Vector Recall: Benefit of coordinate k is c k - c B T A B -1 A k Let’s define a benefits vector r to record all the benefits Define: Note: For k B, r k is benefit of coordinate k For k 2 B, r k =0 Claim: For any point z, we have c T z = c T x + r B T z B Proof:
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Quick Optimality Proof Suppose algorithm terminates with BFS x and basis B Claim: x is optimal. Proof: Loop terminates ) every k B has benefit · 0 ) r · 0 For any feasible point z, we have: c T z = c T x + r B T z B · c T x ) x is optimal. ¥ ¸ 0 since z feasible · 0 since loop terminated
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1. What is a corner point? (BFS and bases) 2. What if there are no corner points? (Infeasible) 3. What are the “neighboring” bases? (Increase one coordinate) 4. What if no neighbors are strictly better? (Might move to a basis that isn’t strictly better (if ± =0), but whenever x changes it’s strictly better) 5. How can I find a starting feasible basis? 6. Does the algorithm terminate? 7. Does it produce the right answer? (Yes) Local-Search Algorithm Let B be a feasible basis (If none, Halt: LP is infeasible) For each entering coordinate k B If “benefit” of coordinate k is > 0 Compute y( ± ) (If ± = 1 , Halt: LP is unbounded) Find leaving coordinate h 2 B (y( ± ) h =0) Set x=y( ± ) and B’= B n {h} [ {k} Halt: return x
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Does Algorithm Terminate?
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