Engineering Optimization: Theory and Practice, Fourth Edition

7.13 Interior Penalty Function Method 441

By using inequalities (7.178) and (7.186), inequality (7.185) becomes

f (X∗) ≤φ(X∗k, rk) < f (X∗)+

ε

2

+

ε

2

=f (X∗)+ε

or
φ(X∗k, rk) −f(X∗) < ε (7.187)

Given anyε>0 (however small it may be), it is possible to choose a value ofkso as
to satisfy the inequality (7.187). Hence ask→ ∞(rk→ 0 ),we have

lim

rk→ 0

φ(X∗k, rk) =f(X∗)

This completes the proof of the theorem.

Additional Results. From the proof above, it follows that asrk→ 0 ,

lim

k →∞

f (X∗k) =f(X∗) (7.188)

lim

k →∞

rk



−

∑m

j= 1

1

gj(X∗k)



= 0 (7.189)

It can also be shown that ifr 1 , r 2 ,... is a strictly decreasing sequence of positive values,
the sequencef (X∗ 1 ), f (X∗ 2 ),... will also be strictly decreasing. For this, consider two
consecutive parameters, say,rkandrk+ 1 , with

0 <rk+ 1 < rk (7.190)

Then we have

f (X∗k+ 1 )−rk+ 1

∑m

j= 1

1

gj(X∗k+ 1 )

<f (X∗k)−rk+ 1

∑m

j= 1

1

gj(X∗k)

(7.191)

sinceX∗k+ 1 alone minimizesφ(X, rk+ 1 ) Similarly,.

f (X∗k)−rk

∑m

j= 1

1

gj(X∗k)

<f (X∗k+ 1 )−rk

∑m

j= 1

1

gj(X∗k+ 1 )

(7.192)

Divide Eq. (7.191) byrk+ 1 , Eq. (7.192) byrk, and add the resulting inequalities to
obtain

1

rk+ 1

f (X∗k+ 1 )−

∑m

j= 1

1

gj(X∗k+ 1 )

+

1

rk

f (X∗k)−

∑m

j= 1

1

gj(X∗k)

<

1

rk+ 1

f (X∗k)−

∑m

j= 1

1

gj(X∗k)

+

1

rk

f (X∗k+ 1 )−

∑m

j= 1

1

gj(X∗k+ 1 )

(7.193)