Engineering Optimization: Theory and Practice, Fourth Edition

402 Nonlinear Programming III: Constrained Optimization Techniques

SOLUTION

Step 1: AtX 1 =

{ 0

0

}

:

f(X 1 ) = 8 and g 1 (X 1 ) =− 4

Iteration 1

Step 2: Sinceg 1 (X 1 ) < 0 , we take the search direction as

S 1 = −∇ f(X 1 ) =−

{

∂f/∂x 1

∂f/∂x 2

}

X 1

=

{

4

4

}

This can be normalized to obtainS 1 =

{ 1

1

}

.

Step 5:To find the new pointX 2 , we have to find a suitable step length alongS 1. For

this, we choose to minimizef (X 1 +λS 1 ) ith respect tow λ. Here

f (X 1 +λS 1 ) =f( 0 +λ, 0 +λ)= 2 λ^2 − 8 λ+ 8

df

dλ

=0 at λ= 2

Thus the new point is given byX 2 =

{ 2

2

}

andg 1 (X 2 ) = 2. As the constraint is

violated, the step size has to be corrected.

Asg 1 ′=g 1 |λ= 0 = − 4 andg′′ 1 =g 1 |λ= 2 = , linear interpolation gives the 2

new step length as

̃λ= − g

′

1

g 1 ′′−g′ 1

λ=

4

3

This givesg 1 |λ=λ ̃= and hence 0 X 2 =

{ 4

3

4

3

}

.

Step 6:f (X 2 )=^89.

Step 7:Here

∣

∣

∣

∣

f (X 1 ) −f(X 2 )

f (X 1 )

∣

∣

∣

∣=

∣

∣

∣

∣

∣

8 −^89

8

∣

∣

∣

∣

∣

=

8

9

>ε 2

||X 1 −X 2 || =[( 0 −^43 )^2 +( 0 −^43 )^2 ]^1 /^2 = 1. 887 >ε 2

and hence the convergence criteria are not satisfied.

Iteration 2

Step 2: Asg 1 = at 0 X 2 , we proceed to find a usable feasible direction.

Step 3:The direction-finding problem can be stated as [Eqs. (7.40)]:

Minimizef= −α