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# Numerical Methods Golden Section Search Method - Theory nm.mathforcollege - PowerPoint PPT Presentation

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Numerical MethodsGolden Section Search Method - Theoryhttp://nm.mathforcollege.com

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Equal Interval Search Method

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

• Compute and

• If

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

Figure 1 Equal interval search method.

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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.

Figure 2. Golden Section Search method

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Golden Section Search Method-Selecting the Intermediate Points

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Determining the first intermediate point

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Determining the second intermediate point

Let ,hence

Golden Ratio=>

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

=Initial Iteration

Second Iteration

Only 1 new inserted location need to be completed!

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Golden Section Search-Determining the new search region

• Case1:

If then the new interval is

• Case2:

If then the new interval is

Case 2

Case 1

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Golden Section Search-Determining the new search region

• 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.

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

0.382(αU -αl)

Min.g(α)=Max.[-g(α)]

1st

jth

j

j - 2

αL = Σδ(1.618)v

αU = Σδ(1.618)v

V = 0

j-2th

V = 0

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αb

αU

αL

αa

= αL

j-1th

Figure 2.5 Golden section partition.

Figure 2.4 Bracketing the minimum point.

α

2nd

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δ

2.618δ

5.232δ

9.468δ

1.6182δ

δ

j - 2

j - 2

1.618δ

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

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

3rd

V = 0

V = 0

j - 1

4th

V = 0

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

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

• Ifas shown in Figure 2.5, Then the minimum will bebetween&and .

Notice that:

And

Thusαb (wrtαU & αL ) plays same role as αa(wrtαU & αL ) !!

_

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

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 areand.

Step 2: Compute g(αb) ,whereαa= 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),thenαL ≤ αi ≤ αb. By the choice of αaandαb, the new pointsandhave . Compute , whereand go to Step 7.

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

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 = αaand 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.

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