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Unformatted text preview: (1.2) Rewrite the question in the form min c x + 1 2 x Cx s.t. Ax = b (1) where C = diag (4 , 6 , 8) = 4 0 0 0 6 0 0 0 8 which is positive definite, c = (0 , , 0) , A = (2 , 3 , 4) , and b = 9 By Theorem 1.3 of the text (firstorder necessary and sufficient condition of optimality for problem (1)), the optimal solution can be found by solving the following system for the vector ( x * ; u * ) (where x * is the optimal solution and entries of u * are the multipliers corresponding to the constraints 1 ): C A A x * u * = c b In the given question, C is nonsingular (can you tell the inverse by inspection?) and A has only one row, so the matrix on the lefthand side of the above equation is invertible, and the solution can be computed easily: x * u * = 4 0 0 2 0 6 0 3 0 0 8 4 2 3 4 0  1 9 = 1 18 3 . 5 1 1 2 1 2 1 2 1 1 1 . 25 2 2 2 2 4 9 = 1 1 1 2 Therefore the optimal solution is x * = (1 , 1 , 1) and the optimal value is 9. 1 Note that dimension of the vector u * is equal to the number of constraints. 1 (1.4) Let f ( x ) = f ( x 1 ,...,x n ) := n X i =1 c i x i + 1 2 n X i =1 n X j =1 γ ij x i x j . For any k = 1 ,...,n , ∂f ( x ) ∂x k = ∂ ∂x k n X i =1 c i x i + 1 2 n X i =1 n X j =1 γ ij x i x j = c k + 1 2 n X i =1 ∂x i ∂x k n X j =1 γ ij x j + n X i =1 x i ∂ ∂x k n X j =1 γ ij x j = c k + 1 2 n X j =1 γ kj x j + n X i =1 γ ik...
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 Winter '09
 MICHAEL
 Trigraph, Hilbert space, positive definite

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