Engineering Optimization: Theory and Practice, Fourth Edition

686 Optimal Control and Optimality Criteria Methods

The necessary condition of optimality can be expressed as

∂f

∂zi

+λ

∂g

∂zi

= 0 , i= 1 , 2 ,... , n (12.66)

Assumingf to be linear in terms of the areas of cross section (original variables,

xi=Ai) nda gto be linear in terms ofzi, we have

∂f

∂zi

=

∂f

∂xi

∂xi

∂zi

= −

1

z^2 i

∂f

∂xi

(12.67)

andEqs. (12.66) and (12.67) yield

xi=

(

λ

∂g/∂zi

∂f/∂xi

) 1 / 2

, i= 1 , 2 ,... , n (12.68)

To findλwe first find the linear approximation ofgat a reference point (trial design)

Z 0 (orX 0 ) sa

g(Z)≈g(Z 0 )+

∑n

i= 1

∂g

∂zi

∣

∣

∣

∣

Z 0

(zi−z 0 i)≈g 0 +

∑n

i= 1

∂g

∂zi

∣

∣

∣

∣

Z 0

zi (12.69)

where

g 0 = g(Z 0 )−

∑n

i= 1

∂g

∂zi

∣

∣

∣

∣

Z 0

z 0 i= g(X 0 )+

∑n

i= 1

∂g

∂xi

∣

∣

∣

∣

X 0

x 0 i (12.70)

andz 0 iis the ith component ofZ 0 withx 0 i= 1 /z 0 i. By setting Eq. (12.69) equal to

zero and substituting Eq. (12.68) forxi, we obtain

λ=

[

1

g 0

∑n

i= 1

(

∂f

∂xi

∂g

∂zi

) 1 / 2 ]^2

(12.71)

Equations (12.71) and (12.68) can now be used iteratively to find the optimal solution

of the problem. The procedure is explained through the following example.

Example 12.5The problem of minimum weight design subject to a constraint on the

vertical displacement of nodeS(U 1 ) f the three-bar truss shown in Fig. 12.6 can beo

stated as follows:

FindX=

{x

x^1

2

}

whichminimizes

f (X)=ρ( 2

√

2 l)x 1 + ρlx 2 = 08. 0445 x 1 + 82. 3 x 2 (E 1 )

subjectto

U 1

Umax

− 1 ≤ 0

or

g(X)=

1

x 1 +

√

2 x 2

− 1 ≤ 0 (E 2 )