hw7sol_fa09

hw7sol_fa09 - UC Berkeley Department of Electrical...

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UC Berkeley Department of Electrical Engineering and Computer Science EECS 227A Nonlinear and Convex Optimization Solutions 7 Fall 2009 Problem 7.1 (a) The derivatives are f 0 ( x ) = - 1 /x 2 ,f 00 ( x ) = 2 /x 3 ,f 000 ( x ) = - 6 /x 4 so the self-concordance condition is 6 x 4 2 ± 2 x 3 ² 3 / 2 = 4 2 x 4 x which holds if x 4 2 / 6 = p 8 / 9 (b) If we make an affine change of variables y i = 8( b i - a T i x ) / (9 α ), then y i < 8 / 9 for all x dom f . The function f reduces to m i =1 1 /y i , which is self concordant by the result in (a). Problem 7.2 (a) We have d dx f 00 ( x ) - 1 / 2 = ( - 1 / 2) f 000 ( x ) f 00 ( x ) 3 / 2 Integrating d dx f 00 ( x ) - 1 / 2 = 1 gives f ( x ) = - log( x + c 0 ) + c 1 x + c 2 . Integrating d dx f 00 ( x ) - 1 / 2 = - 1 gives f ( x ) = - log( - x + c 0 ) + c 1 x + c 2 (b) Suppose f 00 (0) > 0 ,f 00 x ) = 0 for ¯ x > 0, and f 00 ( x ) > 0 on the interval between 0 and ¯ x . The inequality - 1 d dx f 00 ( x ) - 1 / 2 1 holds for x between 0 and ¯ x . Integrating gives f 00 x ) - 1 / 2 - f 00 (0) - 1 / 2 ¯ x which contradicts f 00 x ) = 0 1
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Problem 7.3 (a) The composition rules show that tf 0 ( x )+ φ h ( x ) is convex in x , since h is increasing and convex, and f i are convex. (b) The minimizer of tf 0 ( x ) + φ h ( x ) ,z = x * ( t ), satisfies t f 0 ( z ) + φ ( z ) = 0. Expanding this we get t f 0 ( z ) + m X i =1 h 0 ( f i ( z )) f i ( z ) = 0 This shows that z minimizes the Lagrangian f 0 ( z ) + m i =1 λ i f i ( z ), for λ i = h 0 ( f i ( z )) /t,i
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hw7sol_fa09 - UC Berkeley Department of Electrical...

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