Engineering Optimization: Theory and Practice, Fourth Edition

5.11 Cubic Interpolation Method 283

whereε 1 andε 2 are small numbers whose values depend on the accuracy desired. The
criterion of Eq. (5.61) can be stated in nondimensional form as
∣
∣
∣
∣

ST∇f

|S||∇f|

∣

∣

∣

∣ ̃

λ∗

≤ε 2 (5.62)

If the criteria stated in Eqs. (5.60) and (5.62) are not satisfied, a new cubic equation

h′(λ)=a′+b′λ+c′λ^2 +d′λ^3

can be used to approximatef(λ). The constantsa′,b′,c′, andd′can be evaluated
by using the function and derivative values at the best two points out of the three
points currently available:A, B, and ̃λ∗. Now the general formula given by Eq. (5.54)
is to be used for finding the optimal step sizeλ ̃∗. Iff′(λ ̃∗) < 0 , the new pointsA
andBare taken asλ ̃∗and B,respectively; otherwise [iff′(λ ̃∗) ], the new points> 0
AandBare taken asAandλ ̃∗, and Eq. (5.54) is applied to find the new value of
λ ̃∗. Equations (5.60) and (5.62) are again used to test for the con vergence ofλ ̃∗. If
convergence is achieved,λ ̃∗is taken asλ∗and the procedure is stopped. Otherwise,
the entire procedure is repeated until the desired convergence is achieved.
The flowchart for implementing the cubic interpolation method is given in Fig. 5.17.

Example 5.11 Find the minimum off=λ^5 − 5 λ^3 − 02 λ+5 by the cubic interpola-
tion method.

SOLUTION Since this problem has not arisen during a multivariable optimization
process, we can skip stage 1. We takeA=0 and find that

df

dλ

(λ=A= 0 )= 5 λ^4 − 51 λ^2 − 02

∣

∣

∣

∣

λ= 0

= − 20 < 0

To findBat which df/dλis nonnegative, we start witht 0 = 0. 4 and evaluate the
derivative att 0 , 2 t 0 , 4 t 0 ,... .This gives

f′(t 0 = 0. 4 )= 5 ( 0. 4 )^4 − 51 ( 0. 4 )^2 − 02. 0 = − 22. 272

f′( 2 t 0 = 0. 8 )= 5 ( 0. 8 )^4 − 51 ( 0. 8 )^2 − 02. 0 = − 27. 552

f′( 4 t 0 = 1. 6 )= 5 ( 1. 6 )^4 − 51 ( 1. 6 )^2 − 02. 0 = − 25. 632

f′( 8 t 0 = 3. 2 )= 5 ( 3. 2 )^4 − 51 ( 3. 2 )^2 − 02. 0 = 350. 688

Thus we find that†

A= 0. 0 , fA= 5. 0 , fA′= − 20. 0

B= 3. 2 , fB= 131. 0 , fB′= 503. 688

A < λ∗< B

†Asf′has been found to be negative atλ= 1. 6 also, we can takeA= 1 .6 for faster convergence.