# HW4_Sol - E160 Operations Research I Fall 2008 Homework #4...

This preview shows pages 1–2. Sign up to view the full content.

E160 Operations Research I Fall 2008 Homework #4 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).

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 12/18/2008 for the course IEOR 41033 taught by Professor Adler during the Spring '08 term at University of California, Berkeley.

### Page1 / 2

HW4_Sol - E160 Operations Research I Fall 2008 Homework #4...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online