HW3_Sol

# HW3_Sol - E160 Operations Research I Spring 2009 Homework#3...

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E160 Operations Research I Spring 2009 Homework #3 Solution Chapter 12 Section 12.5 1. a = -3 b = 5, so b - a = 8. x 1 = 5 - .618(8) = .056 x 2 = -3 + .618(8) = 1.944. f(x 1 ) = .115, f(x 2 ) = 7.67 f(x 2 )>f(x 1 ) so interval of uncertainty is now (.056,5]. Then x 3 = x 2 , x 4 = .056 + .618(4.944) = 3.11. f(x 4 ) = 15.89>f(x 3 ) = f(x 2 ) = 7.67. Thus new interval of uncertainty is (1.944,5]. Now x 5 = x 4 = 3.11 and x 6 = 1.944 + .618(3.056) = 3.83. f(x 5 ) = 15.89, f(x 6 ) = 22.33. Since f(x 6 )>f(x 5 ) new interval of uncertainty is (3.11,5]. x 7 = x 6 = 3.83 and x 8 = 3.11 + .618(1.89) = 4.28. f(x 8 ) = 26.88 and f(x 8 )>f(x 7 ). Thus new interval of uncertainty is (3.83,5]. x 9 = x 8 = 4.28 and x 10 = 3.83 + . 618(1.17) = 4.55. f(x 10 ) = 29.8>f(x 9 ) so new interval of uncertainty is (4.28,5]. This interval has length .72<.8. Thus we know that maximum occurs for some value of x on interval (4.28,5]. (maximum actually occurs for x = 5).

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## This note was uploaded on 09/06/2009 for the course IEOR 160 taught by Professor Hochbaum during the Spring '07 term at Berkeley.

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HW3_Sol - E160 Operations Research I Spring 2009 Homework#3...

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