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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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 Spring '07
 HOCHBAUM
 Operations Research

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