Chapter444444444444444.docx - 4 OPTIMISATION Introduction Single Variable Unconstrained Optimisation Multivariable Unconstrained Optimisation Linear

Chapter444444444444444.docx - 4 OPTIMISATION Introduction...

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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
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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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