Transcript mws_gen_opt_ppt_gold..

Golden Section Search Method
Major: All Engineering Majors
Authors: Autar Kaw, Ali Yalcin
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Transforming Numerical Methods Education for STEM
Undergraduates
4/7/2015
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Golden Section Search Method
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Equal Interval Search Method
•Choose an interval [a, b] over which the optima occurs
•Compute
 ab  
f
 
2
 2
and
 ab  
f
 
2
 2
f(x)
 ab  
 ab  
f
  f
 
2
2
 2
 2
•If
then the interval in
which the maximum
a  b  
occurs is  2  2 , b
otherwise it occurs in
 ab  
a, 2  2 

2
a

2
(a+b)/2
b
Figure 1 Equal interval search method.
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x
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.
f
f
2
1
fu
fl
Xl
X2
X1
Xu
Figure 2. Golden Section Search method
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Golden Section Search MethodSelecting the Intermediate Points
f2
f1
fl
fu
f1
fl
fu
a-b
b
Xl
Xu
X1
a
b
Xl
X2
a
X1
Xu
Determining the first
intermediate point
Determining the second
intermediate point
a
b

ab a
b ab

a
b
b
Golden Ratio=>
 0.618 ...
a
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Golden Section SearchDetermining the new search region
f2
f1
fl
Xl



6
fu
X2
X1
Xu
If f ( x )  f ( x ) then the new interval is [ xl , x2 , x1 ]
If f ( x )  f ( x ) then the new interval is [ x2 , x1, xu ]
All that is left to do is to determine the
location of the second intermediate point.
2
1
2
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Example
.
2
2

2

The cross-sectional area A of a gutter with equal base and edge length of 2 is given by
A  4 sin  (1  cos )
Find the angle  which maximizes the cross-sectional area of the gutter. Using an initial
interval of [0,  / 2] find the solution after 2 iterations. Use an initial   0.05 .
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Solution
The function to be maximized is f ( )  4 sin  (1  cos )
Iteration 1: Given the values for the boundaries of
xl  0 and xu   / 2 we can calculate the initial intermediate
points as follows:
5 1
5 1
( xu  xl )  0 
(1.5708)  0.97080 f (0.97080)  5.1654
2
2
5 1
5 1
x2  xu 
( xu  xl )  1.5708
(1.5708)  0.60000 f (0.60000)  4.1227
2
2
x1  xl 
f2
f1
X1=?
Xl
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X2
X1
Xl=X2
Xu
X2=X1
Xu
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Solution Cont
x1  xl 
5 1
5 1
( xu  xl )  0.60000
(1.5708 0.60000)  1.2000
2
2
To check the stopping criteria the difference between xu
and xl is calculated to be
xu  xl  1.5708 0.60000 0.97080
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Solution Cont
Iteration 2
xl  0.60000
xu  1.5708
f ( x1 )  f ( x2 )
x1  1.2000
f (1.2000)  5.0791
x2  0.97080
f (0.97080)  5.1654
xl  0.60000
xu  1.2000
x1  0.97080
x2  xu 
Xl
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X2
X 1 Xu
5 1
5 1
( xu  xl )  1.2000
(1.2000 0.6000)  0.82918
2
2
xu  xl
 1.2000  0.6000  0.9000
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Theoretical Solution and
Convergence
Iteration
1
2
3
4
5
6
7
8
9
xl
0.0000
0.6002
0.6002
0.8295
0.9712
0.9712
1.0253
1.0253
1.0253
xu
1.5714
1.5714
1.2005
1.2005
1.2005
1.1129
1.1129
1.0794
1.0588
x1
0.9712
1.2005
0.9712
1.0588
1.1129
1.0588
1.0794
1.0588
1.0460
x2
0.6002
0.9712
0.8295
0.9712
1.0588
1.0253
1.0588
1.0460
1.0381
xu  xl 1.0253  1.0588

 1.0420
2
2
f(x1)
5.1657
5.0784
5.1657
5.1955
5.1740
5.1955
5.1908
5.1955
5.1961
f(x2)
4.1238
5.1657
4.9426
5.1657
5.1955
5.1937
5.1955
5.1961
5.1957

1.5714
0.9712
0.6002
0.3710
0.2293
0.1417
0.0876
0.0541
0.0334
f (1.0420)  5.1960
The theoretically optimal solution to the problem
happens at exactly 60 degrees which is 1.0472 radians
and gives a maximum cross-sectional area of 5.1962.
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Additional Resources
For all resources on this topic such as digital audiovisual
lectures, primers, textbook chapters, multiple-choice
tests, worksheets in MATLAB, MATHEMATICA, MathCad
and MAPLE, blogs, related physical problems, please
visit
http://nm.mathforcollege.com/topics/opt_golden_section_search.html
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