11_hw4_sol

11_hw4_sol - MS&E 211 Linear and Nonlinear Programming...

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Homework #4 Solution Linear and Nonlinear Programming Nov 3, 2011 HOMEWORK ASSIGNMENT 4 SOLUTION Problem 1 1. Let x = 1 y , we have: lim x 0 + f ( x ) = lim y + f ( 1 y ) = lim y + - ln y y = 0 so f ( x ) is continuous. x > 0 f 00 ( x ) = (1 + ln x ) 0 = 1 x > 0 0 α 1 f ((1 - α )0 + αx ) = αx ln( αx ) = αx (ln x + ln α ) αx ln x = αf ( x ) so f ( x ) is convex. 0 < x < 1 f ( x ) < 0 0 is not a global minimizer. lim x →∞ f ( x ) = ∞ ⇒ f ( x ) doesn’t have a global maximizer. f 0 ( x 0 ) = 0 (1 + ln x 0 ) = 0 x 0 = 1 e , so 1 e is a local (and global) minimizer. 2. We know that f ( x ) is differentiable on (0 , + ). We get its KKT conditions at some inner point ¯ x = ( ¯ x 1 , ¯ x 2 ,..., ¯ x n ) > 0, such that:
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(1 + ln x i ) - μa i = λ i i λ i 0 i λ i x i = 0 i N X i =1 a i x i = 1 x i 0 i Here ¯ x = ( ¯ x 1 , ¯ x 2 ,..., ¯ x n ) must be an inner point in R + N because any boundary point on R + N
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11_hw4_sol - MS&amp;E 211 Linear and Nonlinear Programming...

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