Engineering Optimization: Theory and Practice, Fourth Edition

262 Nonlinear Programming I: One-Dimensional Minimization Methods

Example 5.6 Find the minimum off=x(x− 1 .5) in the interval (0.0, 1.0) to within

10% of the exact value.

SOLUTION If the middle point of the final interval of uncertainty is taken as the

optimum point, the specified accuracy can be achieved if

1

2

Ln≤

L 0

10

or

(

1

2

)(n − 1 )/ 2

L 0 ≤

L 0

5

(E 1 )

SinceL 0 = , Eq. 1 (E 1 ) ivesg

1

2 (n^ −^1 )/^2

≤

1

5

or 2(n^ −^1 )/^2 ≥ 5 (E 2 )

Sincenhas to be odd, inequality (E 2 ) gives the minimum permissible value ofnas 7.

With this value ofn=7, the search is conducted as follows. The first three experiments

are placed at one-fourth points of the intervalL 0 = [a= 0 ,b=1] as

x 1 = 0. 25 , f 1 = 0. 25 (− 1. 25 )= − 0. 3125

x 0 = 0. 50 , f 0 = 0. 50 (− 1. 00 )= − 0. 5000

x 2 = 0. 75 , f 2 = 0. 75 (− 0. 75 )= − 0. 5625

Sincef 1 >f 0 >f 2 , we delete the interval (a, x 0 ) =(0.0, 0.5), labelx 2 andx 0 as the

newx 0 andaso thata= 0 .5,x 0 = 0 .75,andb= 1 .0. By dividing the new interval of

uncertainty,L 3 = 0.5, 1.0) into four equal parts, we obtain(

x 1 = 0. 625 , f 1 = 0. 625 (− 0. 875 )= − 0. 546875

x 0 = 0. 750 , f 0 = 0. 750 (− 0. 750 )= − 0. 562500

x 2 = 0. 875 , f 2 = 0. 875 (− 0. 625 )= − 0. 546875

Sincef 1 >f 0 andf 2 >f 0 , we delete both the intervals (a, x 1 ) nd (a x 2 , b) ,and label

x 1 ,x 0 , andx 2 as the newa,x 0 , and b,respectively. Thus the new interval of uncer-

tainty will beL 5 = (0.625,0.875). Next, this interval is divided into four equal parts

to obtain

x 1 = 0. 6875 , f 1 = 0. 6875 (− 0. 8125 )= − 0. 558594

x 0 = 0. 75 , f 0 = 0. 75 (− 0. 75 )= − 0. 5625

x 2 = 0. 8125 , f 2 = 0. 8125 (− 0. 6875 )= − 0. 558594

Again we note thatf 1 >f 0 andf 2 >f 0 and hence we delete both the intervals (a, x 1 )

and (x 2 , b) to obtain the new interval of uncertainty asL 7 = 0.6875, 0.8125). By(

taking the middle point of this interval (L 7 ) s optimum, we obtaina

xopt≈ 0. 7 5 and fopt≈ − 0. 5625

(This solution happens to be the exact solution in this case.)