340 Nonlinear Programming II: Unconstrained Optimization Techniques
SOLUTIONIteration 1
The gradient offis given by∇f={
∂f/∂x 1
∂f/∂x 2}
=
{
1 + 4 x 1 + 2 x 2
− 1 + 2 x 1 + 2 x 2}
∇f 1 = ∇ f(X 1 )={
1
− 1
}
Therefore,
S 1 = −∇f 1 ={
− 1
1
}
To findX 2 , we need to find the optimal step lengthλ∗ 1. For this, we minimizef(X 1 +
λ 1 S 1 ) =f(−λ 1 , λ 1 )=λ^21 − 2 λ 1 with respect toλ 1. Since df/dλ 1 = at 0 λ∗ 1 = , we 1
obtainX 2 =X 1 +λ∗ 1 S 1 ={
0
0
}
+ 1
{
− 1
1
}
=
{
− 1
1
}
As∇f 2 = ∇ f(X 2 )={
− 1
− 1
}
=
{
0
0
}
,X 2 is not optimum.Iteration 2S 2 = −∇f 2 ={
1
1
}
To minimize
f (X 2 +λ 2 S 2 ) =f(− 1 +λ 2 , 1 +λ 2 )= 5 λ^22 − 2 λ 2 − 1we setdf/dλ 2 =. This gives 0 λ∗ 2 =^15 , and henceX 3 =X 2 +λ∗ 2 S 2 ={
− 1
1
}
+
1
5
{
1
1
}
=
{
− 0. 8
1. 2
}
Since the components of the gradient atX 3 ,∇f 3 ={
0. 2
− 0. 2
}
,are not zero, we proceed
to the next iteration.Iteration 3S 3 = −∇f 3 =