47 lines
1.4 KiB
Markdown
47 lines
1.4 KiB
Markdown
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<!-- vim: set ts=2 sw=2 et tw=80: -->
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---
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title: Homework 5 -- Optimization Methods
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author: Claudio Maggioni
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header-includes:
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- \usepackage{amsmath}
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- \usepackage{hyperref}
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- \usepackage[utf8]{inputenc}
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- \usepackage[margin=2.5cm]{geometry}
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- \usepackage[ruled,vlined]{algorithm2e}
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- \usepackage{float}
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- \floatplacement{figure}{H}
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- \hypersetup{colorlinks=true,linkcolor=blue}
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---
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\maketitle
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# Exercise 2
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## Exercise 2.1
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The resulting MATLAB plot of each constraint and of the feasible region is shown
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below:
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## Exercise 2.3
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We then compute the objective function value for each basic feasible point
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found, The smallest objective value will correspond with the constrained
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minimizer problem solution.
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$$
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x_1 = \begin{bmatrix}0\\0\end{bmatrix} \;\;\; f(x_1) = 4 \cdot 0 + 3 \cdot 0 =
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0$$$$
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x_2 = \frac12 \cdot \begin{bmatrix}0\\3\end{bmatrix} \;\;\;
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f(x_2) = 4 \cdot 0 + 3 \cdot \frac32 = \frac92$$$$
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x_3 = \frac{1}{13} \cdot \begin{bmatrix}3\\24\end{bmatrix} \;\;\; f(x_3) = 4
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\cdot \frac{3}{13} + 3 \cdot \frac{24}{13} = \frac{84}{13}$$$$
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x_4 = \frac12 \cdot \begin{bmatrix}3\\2\end{bmatrix} \;\;\; f(x_4) = 4 \cdot
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frac32 + 3 \cdot 1 = 9$$$$
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x_5 = \begin{bmatrix}2\\0\end{bmatrix} \;\;\; 4 \cdot 2 + 1 \cdot 0 = 8$$
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Therefore, $x^* = x_1$ is the global constrained minimizer with $\lambda^* = \lambda_1 = NaN$ as
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the slack variable value.
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