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Unformatted text preview: E160 Operations Research I Fall 2008 Homework #5 Solution Chapter 12 Section 12.3 1. f''(x) = 6x ≥ 0, (for x ≥ 0) so f(x) is convex on S. 3. f''(x) = 2x3 >0 (for x>0). Thus f(x) is convex on S. 5. f''(x) = x2 <0, so f(x) is a concave function on S. 7. f(x 1 , x 2 ) is the sum of convex functions and is therefore a convex function. 9. 4 2 5 . 5 . 2 = H 1st Order PM's are 2, 2, 4 which are all <0. 2nd Order PM's are all >0 75 . 3 2 5 . 5 . 2 det =  8 4 2 det =  8 4 2 det =  Expanding by Row 3 we find 3rd Order PM = 4((2)(2)(.5) (.5)) = 15<0. Thus f(x 1 , x 2 , x 3 ) is concave on R 3 . 13. Since f is convex we know that for 0 ≤ k ≤ 1 (1) f(kx + (1  k)y) ≤ kf(x) + (1  k)f(y). Multiplying both sides of (1) by c ≥ 0 shows that g is also a convex function. Section 12.6 1. We wish to minimize f(x, y) = Σ {(x i x) 2 + (y i y) 2 } i = 1 i = n ∂ f/ ∂ x = 2 Σ (x i x) = 0 for x* = (x 1 + x...
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This note was uploaded on 12/18/2008 for the course IEOR 41033 taught by Professor Adler during the Spring '08 term at Berkeley.
 Spring '08
 ADLER
 Operations Research

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