4 54 given f x c x 1 x cx where 2 c 4 4 c 3

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Unformatted text preview: his becomes −2 u1 + 2u2 −4 = u1 , −5 u1 + 3u2 which has no solution. Therefore, x∗ is not optimal for this problem either. Thus, this QP code was only correct for one of the three test problems and therefore does not work properly. 4 (5.4) Given f (x) = c x + 1 x Cx, where 2 c= −4 −4 , C= 3 −1 −1 3 , the eigenvalues of C are λ = 2, with eigenvector s = (1, 1) , and λ = 4, with eigenvector s = (1, −1) . Therefore, we have S CS = D, where 1 S=√ 2 1 1 −1 1 , D= 2 0 0 4 . Solving Cx0 = −c, we have x0 = (2, 2) . Defining y by x = Sy + x0 , we have 1 2 2 f (x) = f (x0 ) + y Dy = −8 + y1 + 2y2 . 2 Therefore, the level set for f (x) = −7 is given by 2 y1 + 2 y2 =1 1/2 1 which is an ellipse with intercepts at (±1, 0) and (0, ± √2 ) . The level set for f (x) = −4 is given by 2 2 y1 y2 + =1 4 2 √ which is an ellipse with intercepts at (±2, 0) and (0, ± 2) . 5 x2 ... y 5 ... 1 y.2 ..... .... ... ....... ......... 4 ......... .... ..... ...........................
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This document was uploaded on 04/10/2014.

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