Engineering Optimization: Theory and Practice, Fourth Edition

6.14 Davidon–Fletcher–Powell Method 357

Cubic Interpolation Method (First Fitting)

Stage 1: As the search directionS 1 is normalized already, we go to stage 2.
Stage 2: To establish lower and upper bounds on the optimal step sizeλ∗ 1 , we have to
find two pointsAandBat which the slopedf/dλ 1 has different signs. We
takeA=0 and choose an initial step size oft 0 = 0. 2 5 to findB.
Atλ 1 = A= 0 :
fA= f(λ 1 =A= 0 )= 3609

fA′=

df

dλ 1

∣

∣

∣

∣

λ 1 =A= 0

= − 4956. 64

Atλ 1 =t 0 = 0. 2 5:

f= 2535. 62

df

dλ 1

= − 3680. 82

Asdf/dλ 1 is negative, we accelerate the search by takingλ 1 = 4 t 0 = 1. 0 0.

Atλ 1 = 1. 0 0:

f= 795. 98

df

dλ 1

= − 1269. 18

Sincedf/dλ 1 is still negative, we takeλ 1 = 2. 0 0.

Atλ 1 = 2. 0 0:

f= 227. 32

df

dλ 1

= − 113. 953

Althoughdf/dλ 1 is still negative, it appears to have come close to zero and

hence we take the next value ofλ 1 as 2.50.

Atλ 1 = 2. 5 0:

f= 241. 51

df

dλ 1

= 741. 684 =positive

Sincedf/dλ 1 is negative atλ 1 = 2. 0 and positive atλ 1 = 2 .5,we takeA=

2 .0 (instead of zero for faster convergence) andB= 2 .5. Therefore,

A= 2. 0 , fA= 272. 32 , fA′= − 113. 95

B= 2. 5 , fB= 412. 51 , fB′= 741. 68

Stage 3: To find the optimal step lengthλ ̃∗ 1 using Eq. (5.54), we compute

Z=

3 ( 227. 32 − 241. 51 )

2. 5 − 2. 0

− 113. 95 + 174. 68 = − 24. 41

Q=[( 24. 41 )^2 + 13 ( 1. 95 )( 174. 68 )]^1 /^2 = 431. 2