260 Nonlinear Programming I: One-Dimensional Minimization Methods
The final set of experiments will be conducted atx 5 =(
0. 74925 +
1. 0 − 0. 74925
2
)
− 0. 0005 = 0. 874125
x 6 =(
0. 74925 +
1. 0 − 0. 74925
2
)
+ 0. 0005 = 0. 875125
The corresponding function values are
f 5 =f(x 5 )= 0. 874125 (− 0. 625875 )= − 0. 5470929844
f 6 =f(x 6 )= 0. 875125 (− 0. 624875 )= − 0. 5468437342
Sincef 5 < f 6 , the new interval of uncertainty is given by (x 3 , x 6 )= (0.74925,
0 .875125). The middle point of this interval can be taken as optimum, and hence
xopt≃ 0. 8 121875 and fopt≃ − 0. 55863271485.6 Interval Halving Method
In theinterval halving method, exactly one-half of the current interval of uncertainty
is deleted in every stage. It requires three experiments in the first stage and two exper-
iments in each subsequent stage. The procedure can be described by the following
steps:
1.Divide the initial interval of uncertaintyL 0 = [a, b] into four equal parts and
label the middle pointx 0 and the quarter-interval pointsx 1 andx 2.
2 .Evaluate the functionf (x)at the three interior points to obtainf 1 = f(x 1 ),
f 0 = f(x 0 ) and, f 2 = f(x 2 ).
3 .(a) Iff 2 >f 0 >f 1 as shown in Fig. 5.8a,delete the interval (x 0 , b)^ ,labelx 1
andx 0 as the newx 0 and b,respectively, and go to step 4.
(b) Iff 2 < f 0 < f 1 as shown in Fig. 5.8b,delete the interval (a, x 0 ) label, x 2
andx 0 as the newx 0 and a,respectively, and go to step 4.
(c) If f 1 >f 0 andf 2 >f 0 as shown in Fig. 5.8c,delete both the intervals
(a, x 1 ) nd (a x 2 , b) ,labelx 1 andx 2 as the newaandb, respectively, and
go to step 4.
4.Test whether the new interval of uncertainty,L=b−a, satisfies the conver-
gence criterionL≤ε, whereεis a small quantity. If the convergence criterion
is satisfied, stop the procedure. Otherwise, set the newL 0 = Land go to step 1.
Remarks:
1.In this method, the function value at the middle point of the interval of uncer-
tainty,f 0 , will be available in all the stages except the first stage.
2 .The interval of uncertainty remaining at the end ofnexperiments (n≥3 and
odd) is given byLn=(
1
2
)(n − 1 )/ 2
L 0 (5.4)