Engineering Optimization: Theory and Practice, Fourth Edition

6.6 Powell’s Method 327

Asdf/dλ=0 atλ∗=^12 , we haveX 2 =X 1 +λ∗S 2 =

{

0

0. 5

}

.

Next we minimizefalongS 1 =

{ 1

0

}

fromX 2 =

{ 0. 5

0. 0

}

.Since

f 2 = f(X 2 ) =f( 0. 0 , 0. 5 )= − 0. 25

f+ =f(X 2 +εS 1 ) =f( 0. 01 , 0. 50 )= − 0. 2298 >f 2

f−= f(X 2 −εS 1 ) =f(− 0. 01 , 0. 50 )= − 0. 2698

fdecreases along−S 1. As f(X 2 −λS 1 ) =f(−λ, 0. 50 )= 2 λ^2 − 2 λ− 0. 2 5,df/dλ=
0 atλ∗=^12. HenceX 3 =X 2 −λ∗S 1 =

{− 0. 5

0. 5

}

.

Now we minimizefalongS 2 =

{ 0

1

}

fromX 3 =

{− 0. 5

0. 5

}

.Asf 3 = f(X 3 ) =− 0 .75,
f+ =f(X 3 +εS 2 ) =f(− 0. 5 , 0. 51 )= − 0. 7599 < f 3 , f decreases along+S 2 direc-
tion. Since

f (X 3 +λS 2 )=f(− 0. 5 , 0. 5 +λ)=λ^2 − λ− 0. 75 ,

df

dλ

=0 at λ∗=

1

2

This gives

X 4 =X 3 +λ∗S 2 =

{

− 0. 5

1. 0

}

Cycle 2: Pattern Search

Now we generate the first pattern direction as

S(p^1 )=X 4 −X 2 =

{

−^12

1

}

−

{

0

1

2

}

=

{

− 0. 5

0. 5

}

and minimizef alongS(p^1 )fromX 4. Since

f 4 = f(X 4 ) =− 1. 0

f+= f(X 4 +εSp(^1 )) =f(− 0. 5 − 0. 005 , 1 + 0. 005 )

=f (− 0. 505 , 1. 005 )= − 1. 004975

f decreases in the positive direction ofS(p^1 ). As

f(X 4 +λSp(^1 )) =f(− 0. 5 − 0. 5 λ, 1. 0 + 0. 5 λ)

= 0. 25 λ^2 − 0. 50 λ− 1. 00 ,

df

dλ

=0 atλ∗= 1. 0 and hence

X 5 =X 4 +λ∗S(p^1 )=

{

−^12

1

}

+ 1. 0

{

−^12

1

2

}

=

{

− 1. 0

1. 5

}

The pointX 5 can be identified to be the optimum point.