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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 ( v ) where B ( v ) = 5(6+2 v ) 4 +(6+2 v )(1+ v )+6(1+ v ) 2 log(6+2 v ) log(1+ v ) . Notice that if 1 + v 0, then 6 + 2 v 0. Therefore, the rst log term can be dropped from B ( v ). 2...
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This homework help was uploaded on 02/05/2008 for the course CMSC 660 taught by Professor Oleary during the Fall '06 term at Maryland.
 Fall '06
 oleary

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