# hw-answers - | f ( x k + k d k ) T d k | c 2 | f ( x k ) T...

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IE 417 – Nonlinear Programming Selected Answers Problem 4.10 Since B is symmetric, it can be “diagonalized” into B = QΛQ T where Q is an orthogonal matrix of eigenvectors and Λ = diag ( λ 1 , λ 2 , . . . , λ n } with λ 1 λ 2 . . . λ n is a diagonal matrix of eigenvalues. If λ 1 0 we are done, since the matrix itself is positive deﬁnite. Since Q is orthogonal, we know B + λI = Q ( Λ + λI ) Q T . That is, the eigenvectors of B + λI are unchanged and the eigenvalues are shifted by λ . So if we choose λ - λ 1 , the matrix B + λI is positive semideﬁnite. Problem 8.1 First, note that if f is strongly convex, then d T 2 f ( x ) d > 0, d R n , x R n . Taylor’s Theorem tells us that f ( x k + α k d k ) - f ( x k ) = Z 1 0 2 f ( x k + k d k ) T ( α k d k ) dt for some t ( 0, 1 ) . The left-hand side of the above equation is y k , and s k = α k d k . So s T k y k = ( α k d k ) T Z 1 0 2 f ( x k + k d k ) T ( α k d k ) dt = α 2 k Z 1 0 d T k 2 f ( x k + k d k ) T d k dt. The proof is complete by noting that surely Z 1 0 d T k 2 f ( x k + k d k ) T d k dt > 0 since d T k 2 f ( x k + k d k ) T d k > 0 d k R n and x k + k d k R n .

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IE417 Homework #1 – Selected Answers Prof Jeﬀ Linderoth Problem 8.2 Show that the second of the strong Wolfe Conditions
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Unformatted text preview: | f ( x k + k d k ) T d k | c 2 | f ( x k ) T d k | implies the curvature condition s T k y k , where s k = x k + 1-x k and y k = f ( x k + 1 ) - f ( x k ) . The second strong Wolfe condition implies that f ( x k + k d k ) T d k -c 2 | f ( x k ) T d k | = c 2 f ( x k ) T d k , since d k is a descent direction. (Note that this is just the regular second Wolfe condition.) Therefore, by subtracting the same term to both sides, we see that f ( x k + k d k ) T d k- f ( x k ) T d k ( c 2-1 ) f ( x k ) T d k &gt; 0. (1) Recall that k d k = x k + 1-x k , multiply both sides of (1) by k and substitute to get f ( x k + 1 ) T ( x k + 1-x k ) - f ( x k ) T ( x k + 1-x k ) &gt; 0 which is equivalent to y T k s k &gt; 0 . Problem 0 Page 2...
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## hw-answers - | f ( x k + k d k ) T d k | c 2 | f ( x k ) T...

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