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Unformatted text preview: x 12 x 2 = 4, and the vector [2 , 1] T is a basis for the nullspace of the matrix A = [1 ,2]. (These choices are not unique, so there are many correct answers) Using our choices, any solution to the equality constraint can be expressed as x = • 6 1 ‚ + • 2 1 ‚ v = • 6 + 2 v 1 + v ‚ . Therefore, our problem is equivalent to min v 5(6 + 2 v ) 4 + (6 + 2 v )(1 + v ) + 6(1 + v ) 2 subject to 6 + 2 v ≥ , 1 + v ≥ . Using a log barrier function for these constraints, we obtain the unconstrained problem min v B μ ( v ) where B μ ( v ) = 5(6+2 v ) 4 +(6+2 v )(1+ v )+6(1+ v ) 2μ log(6+2 v )μ log(1+ v ) . Notice that if 1 + v ≥ 0, then 6 + 2 v ≥ 0. Therefore, the ±rst log term can be dropped from B μ ( v ). 2...
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 Fall '06
 oleary
 Derivative, Optimization, 2 Pt, Bμ, Pt Hx, pT Hp

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