numerical on dichotomous search

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OPTIMIZATION ALGORITHMS NUMERICALS ON DICHOTOMOUS SEARCH BY Sumita Das

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Page 1: Numerical on dichotomous search

OPTIMIZATION ALGORITHMS

NUMERICALS ON DICHOTOMOUS SEARCH

BY Sumita Das

Page 2: Numerical on dichotomous search

Dichotomous Search

• It is a Search Based Method

• Requirements for Dichotomous Search:– Interval of uncertainty which contains

minimum of function must be bounded [a b]

– Function must be unimodal.

PowerPoint Presentation by Sumita Das, GHRCE

Page 3: Numerical on dichotomous search

Algorithm

• Input: Level of uncertainty [a b]Initialization: k=0ak =abk =bϵ > 0l : Permitted length of interval

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Page 4: Numerical on dichotomous search

While(bk -ak )>lλk= (ak +bk )/2 - ϵμk =(ak +bk )/2 + ϵ

if f(λk)>=f(μk ) ak+1 = λk

bk+1 = bk

Else

bk+1 = μk

ak+1 = ak

end if

k=k+1

end whilePowerPoint Presentation by Sumita Das, GHRCE

If (b-a) is greater than permitted length of interval. Usually taken as 0.1

Find λFind μ

λ is now a.b remains same

μ is now b. a remains same

Page 5: Numerical on dichotomous search

In Simple words• We are using 4 point information i.e. a, b, λ and μ to find the

minima/maxima

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Midpointλ μ

ϵϵ

a b

How to calculate λ

λ= (Midpoint of a and b) - ϵλ=(a+b)/2 - ϵ

How to calculate μ

μ= (Midpoint of a and b) + ϵμ= (a+b)/2 + ϵ

Place λ and μ symmetrically, each at a distance of ‘ϵ’ from the midpoint of a and b

Page 6: Numerical on dichotomous search

If f(λ) >=f(μ), so the solution is on the right side . Hence solution is between λ and b.

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In Simple words: Case A

Midpointλ μ

ϵ

a b

ϵ

λ is now a. Again Perform Procedure for new a and b.

a bis now aMidpoint

μ

ϵ

λ

ϵ

Page 7: Numerical on dichotomous search

If ff(λ)<f(μ), so the solution is on the left side . Hence solution is between a and μ.

PowerPoint Presentation by Sumita Das, GHRCE

In Simple words: Case B

Midpointλ μ

ϵ

a b

ϵ

Midpointa bis now b

μ is now b. Again Perform Procedure for a and new b

λ

ϵ ϵ

μ

Page 8: Numerical on dichotomous search

ExampleQue: Find Minima f(x)=(X-1)2 +3 [-3 6]Solution: Let l=1, ϵ =0.5

k ak bk (bk -ak ) λk μk f(λk) f(μk )

0 -3 6 9 (-3+6)/2 -0.5=1

(-3+6)/2 +0.5=2

(1-1)2 +3 =3

(2-1)2 +3 =4

1 -3 2 5 -1 0 7 4

2 -1 2 3 0 1 4 7

3 -1 1 2 -0.5 0.5 5.25 3.25

PowerPoint Presentation by Sumita Das, GHRCEContinue until (bk -ak ) becomes <= l

Page 9: Numerical on dichotomous search

AssignmentFind the minimum of the function f(x)=x(x-1.5) in the interval

[0 1] using dichotomous search.

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k ak bk (bk -ak ) λk μk f(λk) f(μk )

0 0 1 1 0.25 0.75 -0.3125 -0.5625

1 0.25 1 0.75 0.4375 0.8125 -0.464 -0.5585

2 0.4375 1 0.5625 0.578 0.859 -0.5329 -.5509

Page 10: Numerical on dichotomous search

References

PowerPoint Presentation by Sumita Das, GHRCE

[1] Singiresu S. Rao, “Engineering Optimization, Chapter 5: Nonlinear Programming I: One-Dimensional Minimization Methods”, 4th Edition