Engineering Optimization: Theory and Practice, Fourth Edition

7.8 Rosen’s Gradient Projection Method 411

as

∇f (X 1 )=

{

2 x 1 − 2

2 x 2 − 4

}

X 1

=

{

0

− 2

}

The normalized search direction can be obtained as

S 1 =

1

[(− 0. 4707 )^2 + ( 0. 1177 )^2 ]^1 /^2

{

− 0. 4707

0. 1177

}

=

{

− 0. 9701

0. 2425

}

Step 5: SinceS 1 = 0 ,we go step 6.
Step 6: To find the step lengthλM, we set

X=

{

x 1

x 2

}

=X 1 +λS

=

{

1. 0 − 0. 9701 λ

1. 0 + 0. 2425 λ

}

Forj=2:

g 2 (X)=( 2. 0 − 1. 9402 λ)+( 3. 0 + 0. 7275 λ)− 6. 0 =0 at λ=λ 2

= − 0. 8245

Forj=3:

g 3 ( X)=−( 1. 0 − 0. 9701 λ)=0 at λ=λ 3 = 1. 03

Forj=4:

g 4 ( X)=−( 1. 0 + 0. 2425 λ)=0 at λ=λ 4 = − 4. 124

Therefore,

λM=λ 3 = 1. 03

Also,

f (X)=f (λ)=( 1. 0 − 0. 9701 λ)^2 +( 1. 0 + 0. 2425 λ)^2

− 2 ( 1. 0 − 0. 9701 λ)− 4 ( 1. 0 + 0. 2425 λ)

= 0. 9998 λ^2 − 0. 4850 λ− 4. 0

df

dλ

= 1. 9996 λ− 0. 4850

df

dλ

(λM )= 1. 9996 ( 1. 03 )− 0. 4850 = 1. 5746

Asdf/dλ(λM) , we compute the minimizing step length> 0 λ∗ 1 by setting

df/dλ=0. This gives

λ 1 =λ∗ 1 =

0. 4850

1. 9996

= 0. 2425