7.9 Generalized Reduced Gradient Method 419we find, atX 1 ,∇Yf=
∂f
∂x 1
∂f
∂x 2
X 1=
{
2 (− 2. 6 − 2 )
− 2 (− 2. 6 − 2 )+ 4 ( 2 − 2 )^3
}
=
{
− 9. 2
9. 2
}
∇Zf={
∂f
∂x 3}
X 1={− 4 (x 2 −x 3 )^3 }X 1 = 0[C]=
[
∂g 1
∂x 1∂g 1
∂x 2]
X 1=[ 5 − 10. 4 ]
[D]=
[
∂g 1
∂x 3]
X 1=[32]
D−^1 =[ 321 ], [D]−^1 [C]= 321 [5 − 10 .4]=[0. 15625 − 0 .325]
GR= ∇Yf −[[D]−^1 [ ]C]T∇Zf=
{
− 9. 2
9. 2
}
−
{
0. 15625
− 0. 325
}
( 0 )=
{
− 9. 2
9. 2
}
Step 3: Since the components ofGRare not zero, the pointX 1 is not optimum, and
hence we go to step 4.
Step 4: We use the steepest descent method and take the search direction as
S= −GR=
{
9. 2
− 9. 2
}
Step 5: We find the optimal step length alongS.
(a) Considering the design variables, we use Eq. (7.111) to obtain Fory 1 =x 1 :λ=3 −(− 2. 6 )
9. 2
= 0. 6087
Fory 2 =x 2 :
λ=− 3 −( 2 )
− 9. 2
= 0. 5435
Thus the smaller value givesλ 1 = 0. 5 435. Equation (7.113) givesT= −([D]−^1 [ C])S=−( 0. 15625 − 0. 325 )
{
9. 2
− 9. 2
}
= − 4. 4275
and hence Eq. (7.114) leads toForz 1 =x 3 :λ=− 3 −( 2 )
− 4. 4275
= 1. 1293
Thusλ 2 = 1. 1 293.