420 Nonlinear Programming III: Constrained Optimization Techniques
(b) The upper bound onλis given by the smaller ofλ 1 andλ 2 , which is equal
to 0.5435. By expressingX=
{
Y+λS
Z+λT}
we obtainX=
x 1
x 2
x 3
=
− 2. 6
2
2
+λ
9. 2
− 9. 2
− 4. 4275
=
− 2. 6 + 9. 2 λ
2 − 9. 2 λ
2 − 4. 4275 λ
and hencef (λ)=f (X)=(− 2. 6 + 9. 2 λ− 2 + 9. 2 λ)^2+( 2 − 9. 2 λ− 2 + 4. 4275 λ)^4= 185. 7806 λ^4 + 383. 56 λ^2 − 691. 28 λ+ 21. 16df/dλ=0 gives2075. 1225 λ^3 + 776. 12 λ− 169. 28 = 0from which we find the root asλ∗≈ 0. 2 2. Sinceλ∗is less than the upper
bound value 0.5435, we useλ∗.
(c) The new vectorXnewis given byXnew={
Yold+dY
Zold+dZ}
=
{
Yold+λ∗S
Zold+λ∗T}
=
− 2. 6 + 0. 22 ( 9. 2 )
2 + 0. 22 (− 9. 2 )
2 + 0. 22 (− 4. 4275 )
=
− 0. 576
− 0. 024
1. 02595
with
dY={
2. 024
− 2. 024
}
, dZ= {− 0. 97405 }Now, we need to check whether this vector is feasible. Sinceg 1 (Xnew )=(− 0. 576 )[1+(− 0. 024 )^2 ] +( 1. 02595 )^4 − 3 =− 2. 4684 = 0the vectorXnewis infeasible. Hence we holdYnewconstant and modify
Znewusing Newton’s method [Eq. (7.108)] asdZ=[D]−^1 [−g(X)−[C]dY]