HW3Sol - Fall 2009 IEOR 160 Industrial Engineering...

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Fall 2009 IEOR 160 September 23, 2009 Page 1 of 1 HOMEWORK 3 SOLUTIONS Chapter 12.5 1. a = - 3 and b = 5. We need k iterations where 8( . 618) k < 0 . 8. This requires k = 5. x 1 = 5 - . 618(8) = . 056, x 2 = - 3 + . 618(8) = 1 . 944 f ( x 1 ) = - . 115 > f ( x 2) = - 7 . 67. Thus I 1 = [ - 3 , 1 . 944]. x 3 = 1 . 944 - . 618(4 . 944) = - 1 . 111, x 4 = x 1 = . 056 f ( x 3 ) = . 988 > f ( x 4 ) = - . 115. Thus I 2 = [ - 3 , . 056]. x 5 = . 056 - . 618(3 . 056) = - 1 . 833, x 6 = x 3 = - 1 . 111 f ( x 5 ) = - 306 < f ( x 6) = . 988. Thus I 3 = [ - 1 . 833 , . 056]. x 7 = x 6 = - 1 . 111, x 8 = - 1 . 833 + . 618(1 . 89) = - . 666 f ( x 7 ) = . 988 > f ( x 8 ) = 0 . 888. Thus I 4 = [ - 1 . 833 , - . 666]. x 9 = - . 666 - . 618(1 . 167) = - 1 . 387, x 10 = x 7 f ( x 9 ) = . 850 < f ( x 10 ) = . 988. Thus I 5 = [ - 1 . 387 , - . 666]. This interval has length . 72 < . 8. Actual maximum occurs for x = - 1. 2.
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