ECE3040 Chapter 9.pdf - Chapter 9 The lecture notes are selected from the public Powerpoint lecture documents provided by Authors Autar Kaw Sri Harsha

ECE3040 Chapter 9.pdf - Chapter 9 The lecture notes are...

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Chapter 9 The lecture notes are selected from the public Powerpoint lecture documents provided by Authors: Autar Kaw, Sri Harsha Garapati
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Golden Section Search Method
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Equal Interval Search Method Figure 1 Equal interval search method. x f(x) a b 2 2 (a+b)/2 Choose an interval [a, b] over which the optima occurs Compute and 2 2 b a f If then the interval in which the maximum occurs is otherwise it occurs in 2 2 b a f 2 2 2 2 b a f b a f b b a , 2 2 2 2 , b a a
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Golden Section Search Method The Equal Interval method is inefficient when is small. The Golden Section Search method divides the search more efficiently closing in on the optima in fewer iterations. X 2 X l X 1 X u f u f 2 f 1 f l Figure 2. Golden Section Search method
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Golden Section Search Method- Selecting the Intermediate Points a b X l X 1 X u f u f 1 f l Determining the first intermediate point a-b b X 2 a X l X 1 X u f u f 2 f 1 f l Determining the second intermediate point a b b a a b b a a b Golden Ratio=> ... 618 . 0 a b
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Golden Section Search- Determining the new search region If then the new interval is If then the new interval is All that is left to do is to determine the location of the second intermediate point. X 2 X l X 1 X u f u f 2 f 1 f l ] , , [ 1 2 x x x l ] , , [ 1 2 u x x x ) ( ) ( 1 2 x f x f ) ( ) ( 1 2 x f x f
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Example The cross-sectional area A of a gutter with equal base and edge length of 2 is given by ) cos 1 ( sin 4 A 05 . 0 . Find the angle which maximizes the cross-sectional area of the gutter. Using an initial interval of find the solution after 2 iterations. Use an initial . ] 2 / , 0 [ 2 2 2
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Solution ) cos 1 ( sin 4 ) ( f 60000 . 0 ) 5708 . 1 ( 2 1 5 5708 . 1 ) ( 2 1 5 97080 . 0 ) 5708 . 1 ( 2 1 5 0 ) ( 2 1 5 2 1 l u u l u l x x x x x x x x The function to be maximized is Iteration 1: Given the values for the boundaries of we can calculate the initial intermediate points as follows: 2 / 0 u l x and x 1654 . 5 ) 97080 . 0 ( f 1227 . 4 ) 60000 . 0 ( f X 2 X l X 1 X u f 2 f 1 X l =X 2 X 2 =X 1 X u X 1 =?
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Solution Cont 2000 . 1 ) 60000 . 0 5708 . 1 ( 2 1 5 60000 . 0 ) ( 2 1 5 1 l u l x x x x To check the stopping criteria the difference between and is calculated to be u x l x 97080 . 0 60000 . 0 5708 . 1 l u x x
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Solution Cont Iteration 2 97080 . 0 2000 . 1 5708 . 1 60000 . 0 2 1 x x x x u l 0791 . 5 ) 2000 . 1 ( f 1654 . 5 ) 97080 . 0 ( f ) ( ) ( 2 1 x f x f 82918 . 0 ) 6000 . 0 2000 . 1 ( 2 1 5 2000 . 1 ) ( 2 1 5 2 l u u x x x x X l X 2 X u X 1 97080 . 0
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