Differential Equations Solutions 45

Differential Equations Solutions 45 - 55 Therefore, our...

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55 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 0 , 1+ v 0 . 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 v )(1 + v ) + 6(1 + v ) 2 μ log(6 + 2 v ) μ log(1 + v ) . Notice that if 1 + v 0, then 6 + 2 v 0. Therefore, the Frst log term can be dropped from B μ ( v ). CHALLENGE 10.4. (a) The central path is deFned by Ax = b , A w + s = c , μF ± ( x )+ s = 0 , where F ± ( x )= X 1 e , w is an
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.

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