400 Nonlinear Programming III: Constrained Optimization Techniques
the constraintgjis active ifgj(Xi+ 1 ) = 10 −^2 , 10−^3 , 10−^8 , and so on? We immediately
noticethat a small marginεhas to be specified to detect an active constraint. Thus we
can accept a pointXto be lying on the constraint boundary if|gj( X)|≤εwhereεis
a prescribed small number. If pointXi+ 1 lies in the infeasible region, the step length
has to be reduced (corrected) so that the resulting point lies in the feasible region only.
It is to be noted that an initial trial step size(ε 1 ) hasto be specified to initiate the
one-dimensional minimization process.
Method 2. Even if we do not want to find the optimal step length, some sort of
a trial-and-error method has to be adopted to find the step lengthλiso as to satisfy
the relations (7.42). One possible method is to choose an arbitrary step lengthεand
compute the values of
f ̃=f (Xi+εSi) nda g ̃j=gj(Xi+εSi)
Dependingon the values off ̃andg ̃j, we may need to adjust the value ofεuntil we
improve the objective function value without violating the constraints.
Initial Trial Step Length. It can be seen that in whatever way we want to find the
step sizeλi, we need to specify an initial trial step lengthε.The value ofεcan be
chosen in several ways. Some of the possibilities are given below.
1.The average of the final step lengthsλiobtained in the last few iterations can
be used as the initial trial step lengthεfor the next step. Although this method
is often satisfactory, it has a number of disadvantages:
(a) This method cannot be adopted for the first iteration.
(b) This method cannot take care of the rate of variation off (X)in different
directions.
(c) This method is quite insensitive to rapid changes in the step length that take
place generally as the optimum point is approached.
2.At each stage, an initial step lengthεis calculated so as to reduce the objective
function value by a given percentage. For doing this, we can approximate the
behavior of the functionf (λ)to be linear inλ. Thus if
f (Xi) =f(λ= 0 )=f 1 (7.43)
df
dλ
(Xi)=
df
dλ
(Xi+λSi)
∣
∣
∣
∣
λ= 0
=ST∇fi=f 1 ′ (7.44)
are known to us, the linear approximation off (λ)is given by
f (λ)≃f 1 +f 1 ′λ
To obtain a reduction ofδ% in the objective function value compared to|f 1 |,
the step lengthλ=εis given by
f 1 +f 1 ′ε=f 1 −
δ
100
|f 1 |