sing levar opt

7/27/2019 Sing Levar Opt

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STEPS IN FORMULISATION OF ANOPTIMISATION PROBLEM

NEED IDENTIFICATION

CHOOSE DESIGN VARIABLES

FORMULATE CONSTRAINTS

FORMULATE OBJECTIVE FUNCTION

SET UP VARIABLE BOUNDS

CHOOSE OPTIMISATION ALGORITHM

OBTAIN SOLUTION(S)


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CLASSICAL OPTIMIZATIONTECHNIQUES

SINGLE-VARIABLE OPTIMIZATION

MULTI-VARIABLE OPTIMIZATION

- WITH NO CONSTRAINTS

- WITH EQUALITY CONSTRAINTS

- WITH INEQUALITY CONSTRAINTS


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FUNCTION IS SAID TO HAVE

at if *xx =- RELATIVE OR LOCALMINIMUM

)*(*)( hxfxf +

- RELATIVE OR LOCALMAXIMUM

at if *xx = )*(*)( hxfxf +For all sufficiently small positive and negative values of h


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FUNCTION IS SAID TO HAVE

at if *x- ABSOLUTE OR GLOBAL MINIMUM

)(*)( xfxf

- ABSOLUTE OR GLOBAL MAXIMUM

at if *x )(*)( xfxf for all x in the domain defined for )(xf


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Single variable optimization Bracketing (exhaustive search)

Region elimination (internal halving)


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Step 3 Is ? If yes, go

to Step 2;Else no minimum

exists in (a,b) or a boundarypoint (a or b) is the minimum

point.

Exhaustive search algorithm (given f(x), a & b)

Step 1 set = a, x = (b-a)/n (n is the number ofintermediate points), and .

Step 2 If , the minimum point lies in ,Terminate;

Else , and go to Step 3.

1x

,12 xxx += xxx += 23

( ) ( ) ( )321 xfxfxf ( )31,xx

xxxxxxx +=== 233221 ,,

bx 3

a 1x 2x 3x ..... b


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Step 1 Choose a lower bound a and an upper bound b. Choosealso a small number. Let ,Compute

Step 2 Set Compute and

Step 3 If set go to Step 5;

Else go to Step 4.

Interval Halving Method

( ) 2baxm += .abLLo ==).(

mxf

,4/1 Lax += .4/2 Lbx = ).( 2xf)( 1xf

)()( 1 mxfxf < 1; xxxb mm ==

Step 4 If f(x2) < f(xm) {a = xm; xm = x2; go to step 5}Else {a = x1; b = x2; go to step 5}.

Step 5 Calculate L = b-a If |L| < Terminate;Else go to Step 2.

a1

x2

xm

x b

L


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EXAMPLE Minimize in the interval (0,5).

Step 1: ; a=0; b=5; ; ;

Step 2:

Step 3: IS NO. Step 4: IS NO.

hence [1.25 - 3.75] i.e a=1.25; b=3.75.

Step 5: L=2.5; a=1.25;b=3.75;

xxxf 54)( 2 +=

310= 50 =L 5.2=mx .85.27)( =mxf

75.3;25.1 21 == xx

4.28)(;7.44)( 21 == xfxf

?)()( 1 mxfxf

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