Engineering Optimization: Theory and Practice, Fourth Edition

440 Nonlinear Programming III: Constrained Optimization Techniques

Proof: IfX∗is the optimum solution of the constrained problem, we have toprove

that

lim

rk→ 0

[min φ(X, rk )]=φ(X∗k, rk) =f(X∗) (7.177)

Sincef (X)is continous andf (X∗) ≤f(X)for all feasible pointsX, we can choose

feasible point

̃

Xsuch that

f (

̃

X) < f (X∗)+

ε

2

(7.178)

for any value ofε>0. Next select a suitable value ofk, sayK, such that

rk≤

{

ε

2 m

/

min

j

[

−

1

gj(

̃

X)

]}

(7.179)

From the definition of theφfunction, we have

f (X∗) ≤minφ(X, rk) =φ(X∗k, rk) (7.180)

whereX∗kis the unconstrained minimum ofφ(X, rk) Further,.

φ(X∗k, rk) ≤φ(X∗K, rk) (7.181)

sinceX∗kminimizesφ(X,rk) nd anya Xother thanX∗kleads to a value ofφgreater

than or equal toφ(X∗k, rk) Further, by choosing. rk< rK, we obtain

φ(X∗K, rK) =f(X∗K)−rK

∑m

j= 1

1

gj(X∗K)

>f (X∗K)−rk

∑m

j= 1

1

gj(X∗K)

>φ(X∗k, rk) (7.182)

asX∗kis the unconstrained minimum ofφ(X, rk) Thus.

f (X∗) ≤φ(X∗k, rk) ≤φ(X∗K, rk) < φ(X∗K, rK) (7.183)

But

φ(X∗K, rK) ≤φ(

̃

X, rK) =f(

̃

X)−rK

∑m

j= 1

1

gj(

̃

X)

(7.184)

Combining the inequalities (7.183) and (7.184), we have

f (X∗)≤ φ(X∗k, rk) ≤f(

̃

X)−rK

∑m

j= 1

1

gj(

̃

X)

(7.185)

Inequality (7.179) gives

−rK

∑m

j= 1

1

gj(

̃

X)

<

ε

2

(7.186)