6.14 Davidon–Fletcher–Powell Method 359SOLUTION
Iteration 1 (i= 1 )
Here
∇f 1 = ∇ f(X 1 )={
1 + 4 x 1 + 2 x 2
− 1 + 2 x 1 + 2 x 2}∣∣
∣
∣
( 0 , 0 )=
{
1
− 1
}
and hence
S 1 = −[B 1 ]∇f 1 = −[
1 0
0 1
]{
1
− 1
}
=
{
− 1
1
}
To find the minimizing step lengthλ∗ 1 alongS 1 , we minimize
f(X 1 +λ 1 S 1 )=f({
0
0
}
+λ 1{
− 1
1
)}
=f (−λ 1 , λ 1 )=λ^21 − 2 λ 1with respect toλ 1. Since df/dλ 1 = at 0 λ∗ 1 = , we obtain 1
X 2 =X 1 +λ∗ 1 S 1 ={
0
0
}
+ 1
{
− 1
1
}
=
{
− 1
1
}
Since∇f 2 = ∇ f(X 2 )=
{− 1
− 1}
and||∇f 2 || = 1. 4142 >ε, we proceed to update the
matrix [Bi] by computing
g 1 = ∇f 2 − ∇f 1 ={
− 1
− 1
}
−
{
1
− 1
}
=
{
− 2
0
}
ST 1 g 1 ={
− 1 1
}
{
− 2
0
}
= 2
S 1 ST 1 =
{
− 1
1
}
{
− 1 1
}
=
[
1 − 1
−1 1
]
[B 1 ]g 1 =[
1 0
0 1
]{
− 2
0
}
=
{
− 2
0
}
([B 1 ]g 1 )T={
− 2
0
}T
=
{
− 20
}
gT 1 [B 1 ]g 1 ={
− 20
}
[
1 0
0 1
]{
− 2
0
}
=
{
−2 0
}
{
− 2
0
}
= 4
[M 1 ]=λ∗ 1S 1 ST 1
ST 1 g 1