Engineering Optimization: Theory and Practice, Fourth Edition

278 Nonlinear Programming I: One-Dimensional Minimization Methods

Table 5.5 Refitting Scheme

New points for refitting

Case Characteristics New Old

1 λ ̃∗> B A B

f < f ̃ B B λ ̃∗

C C

Neglect oldA

2 λ ̃∗> B A A

f > f ̃ B B B

C λ ̃∗

Neglect oldC

3 λ ̃∗< B A A

f < f ̃ B B λ ̃∗

C B

Neglect oldC

4 λ ̃∗< B A λ ̃∗

f > f ̃ B B B

C C

Neglect oldA

Example 5.10 Find the minimum off=λ^5 − 5 λ^3 − 02 λ+5.

SOLUTION Since this is not a multivariable optimization problem, we can proceed

directly to stage 2. Let the initial step size be taken ast 0 = 0. 5 andA=0.

Iteration 1

fA= f(λ= 0 )= 5

f 1 =f(λ=t 0 )= 0. 03125 − 5 ( 0. 125 )− 20 ( 0. 5 )+ 5 = − 5. 59375

Sincef 1 < fA, we setfB=f 1 = − 5 .59375,and find that

f 2 = f(λ= 2 t 0 = 1. 0 )=− 19. 0

Asf 2 < f 1 , we set newt 0 = and 1 f 1 = − 19 .0. Again we find thatf 1 < fAand hence

setfB=f 1 = − 19 .0, and find thatf 2 = f(λ= 2 t 0 = ) 2 = −43. Sincef 2 < f 1 , we

again sett 0 = 2 andf 1 = − 4 3. As thisf 1 < fA, setfB=f 1 = − 4 3 and evaluate

f 2 =f(λ= 2 t 0 = 4 )= 6 29. This timef 2 >f 1 and hence we setfC=f 2 = 629 and

computeλ ̃∗from Eq. (5.40) as

λ ̃∗=^4 (−^43 )−^3 (^5 )−^629

4 (− 43 )− 2 ( 629 )− 2 ( 5 )

( 2 )=

1632

1440

= 1. 135

Convergence test: SinceA=0,fA= , 5 B=2,fB= − 4 3,C=4, andfC= 29, 6

the values ofa,b, andccan be found to be

a= 5 , b= − 204 , c= 90