O
PTIMISATION
Introduction
Single Variable Unconstrained Optimisation
Multivariable Unconstrained Optimisation
Linear Programming
4

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4.1
Introduction
In an engineering analysis, sometimes
extremities
, either
minimum
or
maximum
value, has to be obtained.
f
x
FIGURE 4.1
Extremities for a single variable function
Extremity value can be obtained via
optimisation
, which is divided into:
1.
Unconstrained optimisation —
f
x
0
.
2.
Constrained optimisation — linear/non-linear programming.
Global maximum
Local maximum
x
Local minimum
Global minimum
Chapter 4
Optimisation / 2

4.2
Single Variable Unconstrained Optimisation
Extremities, if any, can be evaluated using either the quadratic interpolation
method or the Newton method using the condition of
f
x
0
.
For
the
quadratic
interpolation
method
,
consider
a
second
order
Lagrange interpolation equation as followed:
f
x
x
x
1
x
x
2
f
x
x
x
0
x
x
2
f
x
0
1
0
2
1
0
1
2
x
x
0
x
x
1
f
x
(4.1)
x
2
x
0
x
2
x
1
FIGURE 4.2
Evaluation of extremities using a quadratic function
Eq. (4.1) is differentiated to yield:
f
x
0
2
x
x
1
x
2
f
x
2
x
x
0
x
2
f
x
x
0
x
1
x
0
x
2
x
1
x
0
x
1
x
2
2
x
x
0
x
1
x
2
x
0
x
2
x
1
f
x
2
Thus, it can be rearranged to get an optimised value of
x
x
3
:
f
x
x
2
x
2
f
x
x
2
x
2
f
x
x
2
x
2
x
3
0
1
2
1
2
0
2
0
1
(4.2)
2
f
x
0
x
1
x
2
2
f
x
1
x
2
x
0
2
f
x
2
x
0
x
1
Eq. (4.2) can be repeated until converged.
x
x
x
0
x
x
1
2
f
x
Approximated
maximum
Actual
maximum
x
0
x
1
x
3
x
2
x
0
1

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Example 4.1
Use the quadratic interpolation method to obtain a maximum value of the
following function accurate to four decimal places:
f
x
sin
x
0.2
x
2
using initial values of
x
0
0,
x
1
1 dan
x
2
2.
Solution
From the given function:
f
0
sin
0
0.2
0
2
0
f
1
sin
1
0.2
1
2
0.6415
f
2
sin
2
0.2
2
2
0.1093
Using Eq. (4.2), the value of
x
3
can be estimated as followed:
0
1
2
2
2
0.6415
2
2
0
2
0.1093
0
2
1
2
x
3
2
0
1
2
2
0.6415
2
0
2
0.1093
0
1
f
x
sin
1.0466
0.2
1.0466
2
0.6466
1.0466
The overal process is as followed:
i
x
0
f
(
x
0
)
x
1
f
(
x
1
)
x
2
f
(
x
2
)
x
3
f
(
x
3
)
1
0
0
1
0.64
15
2
0.10
93
1.04
66
0.64
66
2
1
0.64
15
1.04
66
0.64
66
2
0.10
93
1.10
57
0.64
93
3
1
0.64
15
1.04
66
0.64
66
1.10
57
0.64
93
1.11
10
0.64
93
4
1.04
66
0.64
66
1.10
57
0.64
93
1.11
10
0.64
93
1.11
05
0.64
93
5
1.10
57
0.64
93
1.11
05
0.64
93
1.11
10
0.64
93
1.11
05
0.64
93
Hence, the maximum value is
f
(
x
)
0.6493 at
x
1.1105.

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