6.12 Marquardt Method 349whereλ∗i is found using any of the one-dimensional search methods described in
Chapter 5.
Example 6.12 Minimizef (x 1 , x 2 )=x 1 −x 2 + 2 x 12 + 2 x 1 x 2 +x^22 from the starting
point X 1 =
{ 0
0}
using Marquardt method with α 1 = 014 , c 1 =^14 , c 2 = , and 2
ε= 10 −^2.
SOLUTION
Iteration 1 (i= 1 )
Heref 1 = f(X 1 ) = 0. 0 and
∇f 1 =
∂f
∂x 1
∂f
∂x 2
( 0 , 0 )=
{
1 + 4 x 1 + 2 x 2
− 1 + 2 x 1 + 2 x 2}
( 0 , 0 )=
{
1
− 1
}
Since||∇f 1 || = 1. 4142 >ε, we compute
[J 1 ]=
∂^2 f
∂x^21∂^2 f
∂x 1 x 2∂^2
∂x 1 x 2∂^2 f
∂x^22
( 0 , 0 )=
[
42
2 2
]
X 2 =X 1 − [[J 1 ]+α 1 [ ]I]−^1 ∇f 1=
{
0
0
}
−
[
4 + 104 2
2 2 + 104
]− (^1) {
1
− 1
}
=
{
− 0. 9998
1. 0000
}
10 −^4
Asf 2 = f(X 2 ) =− 1. 9997 × 10 −^4 < f 1 , we setα 2 =c 1 α 1 = 500, 2 i=2, and proceed
to the next iteration.
Iteration 2 (i= 2 )
The gradient vector corresponding to X 2 is given by ∇f 2 =
{ 0 998. 9
− 1. 0000}
,||∇f 2 || =
- 4141 >ε, and hence we compute
X 3 =X 2 − [[J 2 ]+α 2 [ ]I]−^1 ∇f 2=
{
− 0. 9998 × 10 −^4
1 000. 0 × 10 −^4
}
−
[
2504 2
2 2502
]− 1 {
0. 9998
− 1. 0000
}
=
{
− 4. 9958 × 10 −^4
5 000. 0 × 10 −^4
}
Sincef 3 = f(X 3 ) =− 0. 9993 × 10 −^3 < f 2 , we setα 3 =c 1 α 2 = 256 , i=3, and pro-
ceed to the next iteration. The iterative process is to be continued until the convergence
criterion,||∇fi|| <ε, is satisfied.