Engineering Optimization: Theory and Practice, Fourth Edition

274 Nonlinear Programming I: One-Dimensional Minimization Methods

that is,

λ ̃∗= −b

2 c

(5.30)

The sufficiency condition for the minimum ofh(λ)is that

d^2 h

dλ^2

∣

∣

∣

∣ ̃

λ∗

> 0

thatis,

c> 0 (5.31)

To evaluate the constantsa, b, andcin Eq. (5.29), we need to evaluate the function

f (λ)at three points. Letλ=A, λ=B, andλ=Cbe the points at which the function

f (λ)is evaluated and letfA, fB, andfCbe the corresponding function values, that is,

fA= a+bA+cA^2

fB= a+bB+cB^2

fC= a+bC+cC^2 (5.32)

The solution of Eqs. (5.32) gives

a=

fA BC(C−B)+fBCA(A −C)+fCAB(B −A)

(A−B)(B−C)(C−A)

(5.33)

b=

fA(B^2 −C^2 )+fB(C^2 −A^2 )+fC(A^2 −B^2 )

(A−B)(B−C)(C−A)

(5.34)

c= −

fA (B−C)+fB(C −A)+fC(A −B)

(A−B)(B−C)(C−A)

(5.35)

From Eqs. (5.30), (5.34), and (5.35), the minimum ofh(λ)can be obtained as

λ ̃∗=−b

2 c

=

fA(B^2 −C^2 )+fB(C^2 −A^2 )+fC(A^2 −B^2 )

2[fA(B −C)+fB(C −A)+fC(A −B)]

(5.36)

provided thatc, as given by Eq. (5.35), is positive.

To start with, for simplicity, the pointsA, B, andCcan be chosen as 0,t, and 2t,

respectively, wheretis a preselected trial step length. By this procedure, we can save

one function evaluation sincefA= f(λ=0) is generally known from the previous

iteration (of a multivariable search). For this case, Eqs. (5.33) to (5.36) reduce to

a=fA (5.37)

b=

4 fB− 3 fA−fC

2 t

(5.38)

c=

fC+fA− 2 fB

2 t^2

(5.39)

λ ̃∗=^4 fB−^3 fA−fC

4 fB− 2 fC− 2 fA

t (5.40)