s02lec08

# s02lec08 - 15.053 Bounds Tuesday March 5 Duality One of the...

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1 15.053 Tuesday, March 5 z Duality The art of obtaining bounds weak and strong duality z Handouts: Lecture Notes 2 Bounds z One of the great contributions of optimization theory (and math programming) is the providing of upper bounds for maximization problems z We can prove that solutions are optimal z For other problems, we can bound the distance from optimality 3 A 4-variable linear program maximize z = 3x 1 + 4x 2 +6x 3 + 8x 4 subject to x 1 + x 2 + x 3 + x 4 = 1 2x 1 + 3x 2 +4x 3 + 5x 4 = 3 x 1 , x 2 , x 3 , x 4 0 David has minerals that he will mix together and sell for profit. The minerals all contain some gold content, and he wants to ensure that the mixture has 3% gold, and each bag will weigh 1 kilogram. Mineral 1: 2% gold, \$3 profit/kilo Mineral 2: 3% gold, \$4 profit/kilo Mineral 3: 4% gold, \$6 profit/kilo Mineral 4: 5% gold, \$8 profit/kilo 4 A 4-variable linear program 1 = = 1 1 3 1 1 x 1 x 2 x 4 x 3 2 3 4 5 3 4 8 6 -z 0 0 1 = 0 maximize z = 3x 1 + 4x 2 +6x 3 + 8x 4 subject to x 1 + x 2 + x 3 + x 4 = 1 2x 1 + 3x 2 +4x 3 + 5x 4 = 3 x 1 , x 2 , x 3 , x 4 0 5 Obtaining a Bound x 1 x 2 x 4 x 3 -z Subtract 8 times constraint 1 from the objective function. -z – 5 x 1 – 4 x 2 – 2 x 3 = -8 Does this show that z 8? YES! z + 5 x 1 + 4x 2 + 2 x 3 = 8 1 = = 1 1 3 1 1 2 3 4 5 0 0 1 = -5 -4 0 -2 -8 6 Obtaining a Second Bound: Treat the operation as pricing out x 1 x 2 x 4 x 3 -z Subtract 3 * constraint 1 and subtract constraint 2 from the objective function. -z – 2 x 1 – 2 x 2 – 1 x 3 = -6 Thus z 6! Which bound is better: 6 or 8? z + 2 x 1 + 2x 2 + 1 x 3 = 6 1 = = 1 1 3 1 1 2 3 4 5 0 0 1 = 1 3 Prices -2 -2 -1 0 -6

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7 Obtaining the Best Bound: Formulate the problem as an LP x 1 x 2 x 4 x 3 -z 1 = = 1 1 3 1 1 2 3 4 5 0 0 1 = y 2 y 1 Prices A A: 3 - y 1 - 2y 2 0 Î y 1 + 2y 2 3 C C: 6 – y 1 –4y 2 0 Î y 1 + 4y 2 6 B B: 4 - y 1 - 3y 2 0 Î y 1 + 3y 2 4 -y 1 -3y 2 minimize y 1 + 3y 2 D D: 8 – y 1 –5y 2 0 Î y 1 + 5y 2 8 8 The problem that we formed is called the dual problem Subject to y 1 + 2y 2 3 y 1 + 4y 2 6 y 1 + 3y 2 4 y 1 + 5y 2 8 minimize y 1 + 3y 2 y 1 and y 2 are unconstrained in sign 9 Summary of previous slides z
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