Engineering Optimization: Theory and Practice, Fourth Edition

344 Nonlinear Programming II: Unconstrained Optimization Techniques

SOLUTION

Iteration 1

∇f =

{

∂f/∂x 1

∂f/∂x 2

}

=

{

1 + 4 x 1 + 2 x 2

− 1 + 2 x 1 + 2 x 2

}

∇f 1 = ∇ f(X 1 )=

{

1

− 1

}

The search direction is taken asS 1 = −∇f 1 =

{− 1

1

}

.To find the optimal step length

λ∗ 1 alongS 1 , we minimizef(X 1 +λ 1 S 1 ) ith respect tow λ 1. Here

f(X 1 +λ 1 S 1 ) =f(−λ 1 ,+λ 1 )=λ^21 − 2 λ 1

df

dλ 1

= 0 at λ∗ 1 = 1

Therefore,

X 2 =X 1 +λ∗ 1 S 1 =

{

0

0

}

+ 1

{

− 1

1

}

=

{

− 1

1

}

Iteration 2

Since∇f 2 = ∇ f(X 2 )=

{− 1

− 1

}

,Eq. (6.81) gives the next search direction as

S 2 = −∇f 2 +

|∇f 2 |^2

|∇f 1 |^2

S 1

where

|∇f 1 |^2 = 2 and |∇f 2 |^2 = 2

Therefore,

S 2 = −

{

− 1

− 1

}

+

(

2

2

){

− 1

1

}

=

{

0

+ 2

}

To findλ∗ 2 , we minimize

f(X 2 +λ 2 S 2 ) =f(− 1 , 1 + 2 λ 2 )

=− 1 −( 1 + 2 λ 2 )+ 2 − 2 ( 1 + 2 λ 2 )+( 1 + 2 λ 2 )^2

= 4 λ^22 − 2 λ 2 − 1

with respect toλ 2. As df/dλ 2 = 8 λ 2 − 2 = 0 atλ∗ 2 =^14 , we obtain

X 3 =X 2 +λ∗ 2 S 2 =

{

− 1

1

}

+

1

4

{

0

2

}

=

{

− 1

1. 5

}