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Lecture4

# Lecture4 - C&O 355 Lecture 4 N Harvey...

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C&O 355 Lecture 4 N. Harvey http://www.math.uwaterloo.ca/~harvey/

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Outline Equational form of LPs Basic Feasible Solutions for Equational form LPs Bases and Feasible Bases Brute-Force Algorithm Neighboring Bases
Local-Search Algorithm: Pitfalls & Details 1. What is a corner point? 2. What if there are no corner points? 3. What are the “neighboring” corner points? 4. What if there are no neighboring corner points? 5. How can I find a starting corner point? 6. Does the algorithm terminate? 7. Does it produce the right answer? Algorithm Let x be any corner point For each corner point y that is a neighbor of x If c T y>c T x then set x=y Halt

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Pitfall #2: No corner points? This is possible Case 1: LP infeasible Case 2: Not enough constraints x 1 x 2 x 2 · 2 x 2 ¸ 0 This is unavoidable. Algorithm must detect this case. A Fix! We avoid this case by using equational form. x 1 x 2 x 2 - x 1 ¸ 1 x 1 + 6x 2 · 15 4x 1 - x 2 ¸ 10 (0,0) x 1 ¸ 0 x 2 ¸ 0
Converting to Equational Form “Inequality form” Equational form” Claim : These two forms of LPs are equivalent. x 1 x 2 x 2 - x 1 · 1 x 1 + 6x 2 · 15 4x 1 - x 2 · 10 (0,0) x 1 ¸ 0 x 2 ¸ 0 (3,2) x 1 x 2 Solutions of Ax=b Feasible region x 3 A · x b A = x b Tall, skinny A Short, wide A

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Easy: Just use “Simple LP Manipulations” from Lecture 2 Trick 1: ¸ ” instead of “ · Trick 2: “=” instead of “ · This shows P={ x : Ax=b, x ¸ 0 } is a polyhedron . “Inequality form” Equational form”
Trick 1: x 2 R can be written x=y-z where y,z ¸ 0 So Trick 2: For u,v 2 R , u · v , 9 w ¸ 0 s.t. u+w=v So Rewrite it: Then “Inequality form” Equational form” ´ ´ ~ A = [ A; ¡ A;I ] ~ c = [ c; ¡ c; 0] ~ x = [ y;z;w ] ´ “slack variable” This is already in equational form!

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Pitfall #2: No corner points?
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