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