numerical methods golden section search method - theory nm.mathforcollege

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Equal Interval Search Method

Figure 1 Equal interval search method.

x

f(x)

a b

2

2

Choose an interval [a, b] over which the optima occurs.

Compute and

22baf

If then the interval in which the maximum occurs is otherwise it occurs in

22baf

2222 bafbaf

bba ,

22

22

, baa

.Opt

2ba

2ba

6

Golden Section Search Method

The Equal Interval method is inefficient when is small. Also, we need to compute 2 interior points !

The Golden Section Search method divides the search more efficiently closing in on the optima in fewer iterations.

X2XL X1 Xu

fu

f2 f1

fL

Figure 2. Golden Section Search method

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7

Golden Section Search Method-Selecting the Intermediate Points

a bXL X1 Xu

fu

f1

fL

Determining the first intermediate point

a-b

b

X2

a

XL X1 Xu

fu

f2f1

fL

Determining the second intermediate point

)(*382.0),(*618.0

?);(618.0)(

lulu

lu

XXbandXXa

hencewhyab

XXbaa

bXaXX lu 2

Golden Ratio=> ...618.0ab

)(382.0 Lu xxb

ab

aba

ba

1

Let ,hence abR

)(618.0 Lu xxa

b

61803.02

)15(0111 2

RRRRRR

abXaXX ul 1


Golden Section Search Method

8

3

2

1X2X

)(f

0]1cos2[4cos4

0)2cos(4cos4)()2sin(2sin4)(

)cos1(sin4)(

2

fff

Hence, after solving quadratic equation, with initial guess = (0, 1.5708 rad)

3. Opt

=Initial Iteration Second Iteration st1

3

Only 1 new inserted location need to be completed!


Golden Section Search-Determining the new search region

Case1:If then the new interval is Case2:If then the new interval is

9

X2XL X1 Xu

fu

f2f1

fL

],,[ 12 xxxL

],,[ 12 uxxx

)()( 12 xfxf

)()( 12 xfxf

X2XL X1 Xu

fu

f2f1

fL

Case 2 Case 1


At each new interval ,one needs to determine only 1(not 2) new inserted location (either compute the new ,or new )

Max. Min. It is desirable to have automated

procedure to compute and initially.

10

Golden Section Search-Determining the new search region

1x 2x )cos1(sin4)( f )cos1(sin4)( f

Lx ux


11

0

δ

2.618δ 5.232δ 9.468δδ

1.618δ

1.6182δ

Min.g(α)=Max.[-g(α)]

α

jth

j-1thj-2th

αL = Σδ(1.618)vV = 0

j - 2

αU = Σδ(1.618)vV = 0

j • •

• •

αL αa αb αU

0.382(αU - αl)

1st

2nd

αa = αL + 0.382(αU – αl) = Σδ(1.618)v + 0.382δ(1.618)j-1(1+1.618)V = 0

j - 2

3rd

αa = Σδ(1.618)v + 1δ(1.618)j-1 = Σδ(1.618)v = already known ! V = 0

j - 2

V = 0

j - 1

4th

Figure 2.4 Bracketing the minimum point.

Figure 2.5 Golden section partition.= αL

_

Golden Section Search-(1-D) Line Search Method


12

If , Then the minimum will be between αa & αb.

If as shown in Figure 2.5, Then the minimum will be between & and .

Notice that: And

Thus αb (wrt αU & αL ) plays same role as αa(wrt αU &

αL ) !!

_ _


jaULU )618.1(

)()( ba gg

)()( ba gg

a aLU UU

)(382.0]618.1[()382.0(618.1618.1)]618.1[(618.0])618.11[]618.1[)(236.0(

)]618.1[]618.1[)(236.0())(382.021(

11

1

LUj

Lb

jj

jjLUabLb


13

Step 1 : For a chosen small step size δ in α,say ,let j be the smallest integer such that.

The upper and lower bound on αi are and .

Step 2: Compute g(αb) ,where αa= αL+ 0.382(αU- αL) ,and αb = αL+ 0.618(αU- αL).

Note that , so g(αa) is already known.

Step 3: Compare g(αa) and g(αb) and go to Step 4,5, or 6.

Step 4: If g(αa)<g(αb),then αL ≤ αi ≤ αb. By the choice of αa and αb, the new points and have .

Compute , where and go to Step 7.


12 1010

j

V

j

VVV gg

0

1

0))618.1(())618.1((

J

V

VU

0

)618.1(

2

0

)618.1(j

V

VL

1

0

)618.1(j

V

Va

LL bu ab )( ag )(382.0 LuLa


Step 5: If g(αa) > g(αb), then αa ≤ αi ≤ αU. Similar to the procedure in Step 4, put and .

Compute ,where and go to Step 7.

Step 6: If g(αa) = g(αb) put αL = αa and αu = αb and return to Step 2.

Step 7: If is suitably small, put and stop. Otherwise, delete the bar symbols on ,and and

return to Step 3.

14


aL )(618.0 LuLb

uu )( bg

Lu )(21

Lui

baL ,, u


THE END



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This material is based upon work supported by the National Science Foundation under Grant # 0717624. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.

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Numerical Methods Golden Section Search Method - Examplehttp://nm.mathforcollege.com


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Go to http://nm.mathforcollege.com Click on Keyword Click on Golden Section Search Method



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display and perform the work to Remix – to make derivative works

Under the following conditions Attribution — You must attribute the

work in the manner specified by the author or licensor (but not in any way that suggests that they endorse you or your use of the work).

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23

Example

The cross-sectional area A of a gutter with equal base and edge length of 2 is given by (trapezoidal area):

)2sin(2sin4)cos1(sin4)(. AfMax

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.Convergence achieved if “ interval length ” is within

]2

,0[

2

2

2


24

Solution)cos1(sin4)( f

60000.0)5708.1(2

155708.1)(2

15

97080.0)5708.1(2

150)(2

15

2

1

Luu

LuL

xxxx

xxxx

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 uL xandx

1654.5)97080.0( f

1227.4)60000.0( f

X2XLX1 Xu

f2f1

XL=X2X2=X1 Xu

X1=?


25

Solution Cont2000.1)60000.05708.1(

21560000.0)(

215

1

LuL xxxx

To check the stopping criteria the difference between and is calculated to be

uxLx

97080.060000.05708.1 Lu xx


26

Solution ContIteration 2

97080.02000.15708.160000.0

2

1

xxxx

u

L

0791.5)2000.1( f

1654.5)97080.0( f

)()( 21 xfxf

82918.0)6000.02000.1(2

152000.1)(2

152

Luu xxxx

XLX2 XuX1

97080.02000.160000.0

1

xxx

u

L

9000.06000.02000.12

Lu xx


27

Theoretical Solution and Convergence

Iteration xl xu x1 x2 f(x1) f(x2) 1 0.0000 1.5714 0.9712 0.6002 5.1657 4.1238 1.57142 0.6002 1.5714 1.2005 0.9712 5.0784 5.1657 0.97123 0.6002 1.2005 0.9712 0.8295 5.1657 4.9426 0.60024 0.8295 1.2005 1.0588 0.9712 5.1955 5.1657 0.37105 0.9712 1.2005 1.1129 1.0588 5.1740 5.1955 0.22936 0.9712 1.1129 1.0588 1.0253 5.1955 5.1937 0.14177 1.0253 1.1129 1.0794 1.0588 5.1908 5.1955 0.08768 1.0253 1.0794 1.0588 1.0460 5.1955 5.1961 0.05419 1.0253 1.0588 1.0460 1.0381 5.1961 5.1957 0.0334

0420.12

0588.10253.12

Lu xx 1960.5)0420.1( f

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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This material is based upon work supported by the National Science Foundation under Grant # 0717624. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.

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numerical methods golden section search method - theory nm.mathforcollege

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