hw5sol_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 5 Fall 2009 Reading: Boyd and Vandenberghe, Chapter 5 Solution 5.1 Note that we have f ( x ) = 1 for any x R n . Since for any feasible point h ( x ) = 2 x 6 = 0, any local minimum will be regular. From the Lagrange multiplier theorem, there exists λ * such that f ( x * ) + λ * h ( x * ) = 0 , or equivalently 1 + 2 λ * x * = 0. Using the requirement that h ( x * ) = k x * k 2 - 1 = 0, this yields the two possibilities ( x * i * i ) = ± ( 1 n , - n 2 ) . By using the second order optimality conditions, we have y 0 ( 2 f ( x * ) + λ * 2 h ( x * ) ) y 0 for all y V ( x * ) : = { y | 2( x * ) T y = 0 } . Since 2 f ( x * ) = 0 and 2 h ( x * ) = 2 I , the above condition is equivalent to 2 λ * y T Iy 0 for all y V ( x * ), which can hold only for non-negative λ * . Hence the only point satisfying the first and second order conditions for a local minimum is x * = - 1 n 1 . Since the constraint set is compact and f is continuous, an optimal point x * exists, and the above point must be the optimum. One geometric interpretation of the Lagrange multiplier conditions is that the gradient direction f ( x * ) = 1 is parallel with the gradient h ( x * ) = 2 x * associated with the constraint h ( x ) = k x k 2 - 1 = 0, which defines the boundary of a circle. Solution 5.2 (a) Note that the constraint set for k th problem is C k = { x R n : k x k 2 = 1 ,e T i x = 0 ,i = 1 ,...,k } satisfies C 1 C 2 ... C n . Since all these problems have the same objective, we must have the optimal value satisfies λ 1 λ 2 ... λ n (b) Suppose α 1 ,...,α n satisfies n i =1 α i e i = 0. By multiplying both sides with e T k , note that e T k e i = 0 , i 6 = k and e T k e k = 1, we have
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hw5sol_fa09 - UC Berkeley Department of Electrical...

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