Golden Section Search Method - Numerical Methods - Lecture Slides, Slides of Mathematical Methods for Numerical Analysis and Optimization

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Golden Section Search Method

Equal Interval Search Method

Figure 1 Equal interval search method.

x f(x) a (^) b

(a+b)/

• Choose an interval [a, b] over which the optima occurs
• Compute  and     

2 2 a b ε f

• If then the interval in which the maximum occurs is otherwise it occurs in       −

2 2 a b ε f       −

>     

2 2 2 2 ε a b ε f a b f

b

a b

ε

a b ε

a

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 a b a =

• (^) b a b a b − = Golden Ratio=> (^) = 0. 618 ... a b

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 (^) l X^2 X 1 X (^) u f (^) u f (^2) f (^1) f (^) l [ , , ] 2 1 x x x l [ , , ] 2 1 u x x x f ( x 2 )> f ( x 1 ) f ( x 2 )< f ( x 1 )

Solution

f (θ ) = 4 sinθ( 1 +cos θ) ( 1. 5708 ) 0. 60000 2 5 1 ( ) 1. 5708 2 5 1 ( 1. 5708 ) 0. 97080 2 5 1 ( ) 0 2 5 1 2 1 = − − = − − = − = − − = + − = + u u l 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:

xl = 0 and xu = π / 2 f ( 0. 97080 )= 5.^1654 f ( 0. 60000 )= 4. 1227 X X^2 l X (^1) X u f (^2) f (^1) X (^) l =X 2 X^2 =X^1 X (^) u X 1 =?

Solution Cont

( 1. 5708 0. 60000 ) 1. 2000 2 5 1 ( ) 0. 60000 2 5 1 1 − = − − = + − x = xl + xu xl

To check the stopping criteria the difference between

and is calculated to be

u

x

l

x

u l

x x

Theoretical Solution and

Convergence

Iteration xl x (^) u x 1 x 2 f(x 1 ) f(x 2 ) ε 1 0.0000 1.5714 0.9712 0.6002 5.1657 4.1238 1. 2 0.6002 1.5714 1.2005 0.9712 5.0784 5.1657 0. 3 0.6002 1.2005 0.9712 0.8295 5.1657 4.9426 0. 4 0.8295 1.2005 1.0588 0.9712 5.1955 5.1657 0. 5 0.9712 1.2005 1.1129 1.0588 5.1740 5.1955 0. 6 0.9712 1.1129 1.0588 1.0253 5.1955 5.1937 0. 7 1.0253 1.1129 1.0794 1.0588 5.1908 5.1955 0. 8 1.0253 1.0794 1.0588 1.0460 5.1955 5.1961 0. 9 1.0253 1.0588 1.0460 1.0381 5.1961 5.1957 0.

1. 0420 2
2. 0253 1. 0588 2 =

= xu + xl 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.