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IOE510+Lecture+03+10+10+Mon

# IOE510+Lecture+03+10+10+Mon - IOE 510 Linear Programming I...

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1 IOE 510 Linear Programming I Wednesday March 10, 2010 Professor Amy Cohn

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2 Announcements Problem set 8 due today; problem set 9 to be posted tomorrow
3 Re-cap from Last Time Duality – using penalties to enforce constraints

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4 Duality
5 Another Example Primal problem: Min 3x + 2y - z St x - y + z > 2 5y > 3 x, y, z > 0

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6 Example, cont. Given that we have two constraints, we’ll consider two different penalties: Min 3x + 2y - z St x - y + z > 2 (p 1 ) 5y > 3 (p 2 ) x, y, z > 0
7 Example, cont. D(p 1 , p 2 ) becomes: Min 3x + 2y – z + p 1 (2 – x + y - z) + p 2 (3 – 5y) St x, y, z > 0

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8 Example, cont. D(p 1 , p 2 ): can be re-written as Min 2p 1 + 3p 2 + (3 – p 1 ) x + (2 + p 1 – 5p 2 ) y + (-1 –p 1 ) z St x, y, z > 0 What can we say about the solution to this problem for given values of p 1 and p 2 ? Derive the dual problem…
9 Taking Duals for Generic LP Forms If a primal constraint is > , the dual variable is > 0 If a primal constraint is =, the dual variable is free If a primal constraint is < , the dual variable is < 0 If a primal variable is > 0 , the dual constraint is < If a primal variable is free, the dual constraint is = If a primal variable is < 0, the dual constraint is >

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10 Shortcut to Finding the Dual The dual of Min cx St Ax > b x > 0 is Max b T p St A T p < c T p > 0
11 Claim

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IOE510+Lecture+03+10+10+Mon - IOE 510 Linear Programming I...

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