Engineering Optimization: Theory and Practice, Fourth Edition

408 Nonlinear Programming III: Constrained Optimization Techniques

that lies outside the feasible region. Hence the following procedure is generally adopted

to find a suitable step lengthλi. Since the constraintsgj( X)are linear, we have

gj(λ)=gj(Xi+λSi)=

∑n

i= 1

aij(xi+ λsi)−bj

=

∑n

i= 1

aijxi−bj+λ

∑n

i= 1

aijsi

=gj(Xi)+λ

∑n

i= 1

aijsi, j= 1 , 2 ,... , m (7.75)

where

Xi=











x 1

x 2

..

.

xn











and Si=











s 1

s 2

..

.

sn











.

This equation shows thatgj(λ) will also be a linear function ofλ. Thus if a particular

constraint, say thekth, is not active atXi, it can be made to become active at the point

Xi+λkSiby taking a step lengthλkwhere

gk(λk)=gk(Xi)+λk

∑n

i= 1

aiksi= 0

thatis,

λk= −

gk(Xi)

∑n

i= 1 aiksi

(7.76)

Since thekth constraint is not active atXi, the value ofgk(Xi) ill be negative andw

hence the sign ofλkwill be same as that of the quantity

(∑n

i= 1 aiksi

)

.From Eqs. (7.75)

we have

dgk

dλ

(λ)=

∑n

i= 1

aiksi (7.77)

and hence the sign ofλk depends on the rate of change ofgk with respect toλ.If

this rate of change is negative, we will be moving away from thekth constraint in the

positive direction ofλ. However, if the rate of change(dgk/dλ) is positive, we will be

violating the constraintgkif we take any step lengthλlarger thanλk. Thus to avoid

violation of any constraint, we have to take the step length(λM) sa

λM= min

λk>0 and k

is any integer among

1tojmother than

1 ,j 2 ,...,jp

(λk) (7.78)

In some cases, the functionf (λ)may have its minimum along the lineSi in

betweenλ=0 and λ=λM. Such a situation can be detected by calculating the