Engineering Optimization: Theory and Practice, Fourth Edition

(Martin Jones) #1
5.10 Quadratic Interpolation Method 273

SOLUTION The new design pointXcan be expressed as

X=

{

x 1
x 2

}

=X 1 +λS=

{

− 2 +λ
− 2 + 0. 25 λ

}

By substitutingx 1 = − 2 +λandx 2 = − 2 + 0. 25 λin Eq. (E 1 ) we obtain, f as a
function ofλas

f (λ)=f

(

− 2 +λ
− 2 + 0. 25 λ

)

=[(− 2 +λ)^2 −(− 2 + 0. 25 λ)]^2

+ 1 [ −(− 2 +λ)]^2 =λ^4 − 8. 5 λ^3 + 13. 0625 λ^2 − 75. 0 λ+ 45. 0

The value ofλat whichf (λ)attains a minimum givesλ∗.

In the following sections, we discuss three different interpolation methods with
reference to one-dimensional minimization problems that arise during multivariable
optimization problems.

5.10 Quadratic Interpolation Method


The quadratic interpolation method uses the function values only; hence it is useful
to find the minimizing step (λ∗) of functionsf(X)for which the partial derivatives
with respect to the variablesxi are not available or difficult to compute [5.2, 5.5].
This method finds the minimizing step lengthλ∗in three stages. In the first stage the
S-vectoris normalized so that a step length ofλ=1 is acceptable. In the second stage
the functionf (λ)is approximated by a quadratic functionh(λ)and the minimum, ̃λ∗,
ofh(λ)is found. Ifλ ̃∗is not sufficiently close to the true minimumλ∗, a third stage is
used. In this stage a new quadratic function (refit)h′( )λ =a′+b′λ+c′λ^2 is used to
approximatef (λ), and a new value of ̃λ∗is found. This procedure is continued until
aλ ̃∗that is sufficiently close toλ∗is found.

Stage 1. In this stage,†the Svector is normalized as follows: Find =max
i

|si|,
wheresiis the ith component ofSand divide each component ofSby. Another
method of normalization is to find =(s 12 +s^22 + · · · +sn^2 )^1 /^2 and divide each com-
ponent ofSby.

Stage 2. Let

h(λ)=a+bλ+cλ^2 (5.29)

be the quadratic function used for approximating the functionf (λ). It is worth noting
at this point that a quadratic is the lowest-order polynomial for which a finite minimum
can exist. The necessary condition for the minimum ofh(λ)is that
dh

=b+ 2 cλ= 0

†This stage is not required if the one-dimensional minimization problem has not arisen within a multivariable
minimization problem.
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