356 Nonlinear Programming II: Unconstrained Optimization Techniques
The quantitySTi+ 1 [A]Sican be written as
STi+ 1 [A]Si= −([Bi+ 1 ]∇fi+ 1 )T[A]Si
= −∇fiT+ 1 [Bi+ 1 ][A]Si= −∇fiT+ 1 Si= 0 (E 11 )
sinceλ∗i is the minimizing step in the directionSi. Equation (E 11 ) proves that the
successive directions generated in the DFP method are [A]-conjugate and hence the
method is a conjugate gradient method.
Example 6.14 Minimizef (x 1 , x 2 )= 100 (x^21 −x 2 )^2 +( 1 −x 1 )^2 takingX 1 =
{− 2
− 2
}
as
the starting point. Use cubic interpolation method for one-dimensional minimization.
SOLUTION Since this method requires the gradient off, we find that
∇f=
{
∂f /∂x 1
∂f /∂x 2
}
=
{
400 x 1 (x 12 −x 2 )− 2 ( 1 −x 1 )
− 200 (x 12 −x 2 )
}
Iteration 1
We take
[B 1 ]=
[
10
0 1
]
AtX 1 =
{− 2
− 2
}
,∇f 1 = ∇ f(X 1 )=
{− 8064
− 1200
}
andf 1 = 609. Therefore, 3
S 1 = −[B 1 ]∇f 1 =
{
4806
1200
}
By normalizing, we obtain
S 1 =
1
[( 4806 )^2 + 200 ( 1 )^2 ]^1 /^2
{
4806
1200
}
=
{
0. 970
0. 244
}
To findλ∗i, we minimize
f(X 1 +λ 1 S 1 ) =f(− 2 + 0. 970 λ 1 , − 2 + 0. 244 λ 1 )
= 100 ( 6 − 4. 124 λ 1 + 0. 938 λ^21 )^2 + ( 3 − 0. 97 λ 1 )^2 (E 1 )
with respect toλ 1. Equation (E 1 ) gives
df
dλ 1
= 002 ( 6 − 4. 124 λ 1 + 0. 938 λ^21 )( 1. 876 λ 1 − 4. 124 )− 1. 94 ( 3 − 0. 97 λ 1 )
Since the solution of the equationdf/dλ 1 = cannot be obtained in a simple manner, 0
we use the cubic interpolation method for findingλ∗i.