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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.
 Fall '13

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