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ee227a_review_soln

# ee227a_review_soln - 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 1 Fall 2009 Review 1.1 For future reference, the function f has the following first and second derivatives: ∂f ∂x ( x, y ) = 4 x 3 4 yx 20 x ∂f ∂y ( x, y ) = 2 x 2 + 2 y 2 f ∂y 2 ( x, y ) = 4 x 2 f ∂x∂y ( x, y ) = 2 (a) Note that f (1 , 0) = f ( 1 , 0) = 9, so that 1 2 f (1 , 0) + 1 2 f ( 1 , 0) = 9. On the other hand, we have f (0 , 0) = 0 > 9, showing that f is not convex. (b) We need to solve f ( x, y ) = 0 for ( x, y ) R 2 . Taking derivatives yields f ( x, y ) = bracketleftbig 4 x ( x 2 y ) 20 x 2( x 2 y ) bracketrightbig T By inspection, ( x, y ) = (0 , 0) is the only stationary point. It is neither a local maximum nor a local minimum, since f ( δ, 0) = δ 4 10 δ 2 < 0 for δ small enough, and f (0 , ǫ ) = ǫ 2 > 0 for ǫ negationslash = 0. Alternatively, we may compute the Hessian matrix 2 f (0 , 0) = parenleftbigg 20 0 0 2 parenrightbigg which is indefinite, showing (0 , 0) is neither a local minimum nor a local maximum. (c) The function f is convex. The set C is bounded in the y -direction, and the function value tends to infinity as | x | increases, so that a global minimum exists and is attained. By necessary optimality condition for constrained optimization to solve this problem, if ( x , y ) is global minimum it needs to satisfy f ( x , y ) T bracketleftbiggparenleftbigg x y parenrightbigg parenleftbigg x y parenrightbiggbracketrightbigg 0 , ( x, y ) X In part (b), we showed that f has only one stationary point in the interior of the given set, and it is not a minimum. Hence, the minimum must occur on the boundary of the set— i.e., either at some point ( x , 1) or some point ( x , 0), and we must have ∂f ∂x ( x , 1) = 0 or ∂f ∂x ( x , 0) = 0. For y = 0, we have f ( x, 0) = x 4 10 x 2 . Since ∂f ∂x ( x, 0) = 4 x 3 20 x 2 , the relevant stationary points are at x = ± 5, with f ( ± 5 , 0) = 25 50 = 25. Otherwise, for y = 1, we have f ( x, 1) = x 4 12 x 2 + 1. We have ∂f ∂x ( x, 1) = 4 x 3 24 x , so stationary points are found at x = ± 6 with f ( ± 6 , 0) = 35. Checking second derivatives shows that this is the global optimum. 1

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Review 1.2 Using KKT conditions, we have g j ( x ) 0 , j = 1 , . . . , m λ j 0 , j = 1 , . . . , m λ j g j ( x ) = 0 , j = 1 , . . . , m f ( x ) + m summationdisplay j =1 λ j g j ( x ) = 0 So f ( x ) T ( x x ) = m summationdisplay j =1 λ j g j ( x ) T ( x x ) Since g j , j = 1 , . . . , m are convex functions: g j ( x ) T ( x x ) g j ( x ) g j ( x ), therefore f ( x ) T ( x x ) ≥ − m summationdisplay j =1 λ j ( g j ( x ) g j ( x )) = m summationdisplay j =1 λ j g j ( x ) + m summationdisplay j =1 λ j g j ( x ) = m summationdisplay j =1
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ee227a_review_soln - UC Berkeley Department of Electrical...

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