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@ -23,24 +23,127 @@ header-includes:
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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 2.2
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<!--The Simplex method is used to solve linear programs which are defined as
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follows:
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$$\min c^Tx, \text{ subject to } Ax = b, x > 0$$
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And when we have inequalities constraints such as:
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$$\min c^Tx,\text{ subject to }Ax \leq b$$
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We can introduce slack variable to convert the inequalities into equalities:
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$$\min c^Tx,\text{ subject to }Ax + z = 0, z>0$$
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Each iterate generated by the simplex method is a basic feasible point which
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is a-->
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According to Nocedal, a vector $x$ is a basic feasible point if it is in the
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feasible region and if there exists a subset $\beta$ of the
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index set $1, 2, \ldots, n$ such that:
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- $\beta$ contains exactly $m$ indices, where $m$ is the number of rows of $A$;
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- For any $i \notin \beta$, $x_i = 0$, meaning the bound $x_i \geq 0$ can be
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inactive only if $i \in \beta$;
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- The $m$ x $m$ matrix $B$ defined by $B = [A_i]_{i \in \beta}$ (where $A_i$ is
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the i-th column of A) is non-singular, i.e. all columns corresponding to the
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indices in $\beta$ are linearly independent from each other.
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The geometric interpretation of basic feasible points is that all of them
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are verticies of the polytope that bounds the feasible region. We will use this
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proven property to manually solve the constrained minimization problem presented
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in this section by aiding us with the graphical plot of the feasible region in
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figure \ref{fig:a}.
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<!--
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And as already been
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said, the basic feasible point is the basic feasible solution for the problem.
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Regarding the section 13.3, which outline how the steps are done, we know that
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most steps consist of a move from one vertex to an adjacent one for which the
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basis $\beta$ differs exactly one component.The step is an edge along which the
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objective function is reduced.
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Then, one major issue at every simplex iteration
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is to decide which index must be removed from the basis. As the book specify,
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unless the step is a direction of unboundedness, a single index must be removed
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by replacing it with another from outside $\beta$.
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From a geometrical point of view,
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a move, is along an edge of the feasible polytope that decreases $c^Tx$ and we
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continue moving along that edge until we find a new vertex. One we find a vertex
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we defined the new constraint $x_p > 0$ which is one of the component $x_p,p \in \beta$
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decreased to zero. Afterwards we can remove the index $p$ from the basis $\beta$ and
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replace it with $q$. A more detailed visualisation can be see in the following
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Figure 1 taken from the book.-->
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## Exercise 2.3
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Since the geometrical interpretation of the definition of basic feasible point
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states that these point are non-other than the vertices of the feasible region,
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we first look at the plot above and to these points (i.e. the verticies of the
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bright green non-trasparent region). Then, we look which constraint boundaries cross these
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edges, and we formulate an algebraic expression to find these points. In
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clockwise order, we have:
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- The lower-left point at the origin, given by the boundaries of the constraints
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$x_1 \geq 0$ and $x_2 \geq 0$:
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$$x^*_1 = \begin{bmatrix}0\\0\end{bmatrix}$$
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- The top-left point, at the intersection of constraint boundaries $x_1 \geq 0$ and
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$-3x_1 + 2x_2 \leq 3$:
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$$x_1 = 0 \;\;\; 2x_2 = 3 \Leftrightarrow x_2 = \frac32 \;\;\; x^*_2 =
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\frac12 \cdot \begin{bmatrix}0\\3\end{bmatrix}$$
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- The top-center-left point at the intersection of constraint boundaries $-3x_1 +
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2x_2 \leq 3$ and $2x_1 + 3x_2 \leq 6$:
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$$-3x_1 + 2x_2 + 3x_1 + \frac92 x_2 = \frac{13}{2} x_2 = 3 + 9 = 12
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\Leftrightarrow x_2 = 12 \cdot \frac2{13} = \frac{24}{13}$$$$
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-3x_1 + 2 \cdot \frac{24}{13} = 3 \Leftrightarrow x_1 = \frac{39 - 48}{13}
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\cdot \frac{1}{-3} = \frac{3}{13} \;\;\; x^*_3 = \frac{1}{13} \cdot
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\begin{bmatrix}3\\24\end{bmatrix}$$
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- The top-center-right point at the intersection of constraint boundaries $2x_1
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+ 3x_2 \leq 6$ and $2x_1 + x_2 \leq 4$:
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$$2x_1 + 3x_2 - 2x_1 - x_2 = 2x_2 = 6 - 4 = 2 \Leftrightarrow x_2 = 1 \;\;\;
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2x_1 + 1 = 4 \Leftrightarrow x_1 = \frac32 \;\;\; x^*_4 = \frac12 \cdot
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\begin{bmatrix}3\\2\end{bmatrix}$$
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- The right point at the intersection of $2x_1 + x_2 \leq 4$ and $x_2 \geq
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0$:
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$$x_2 = 0 \;\;\; 2x_1 + 0 = 4 \Leftrightarrow x_1 = 2 \;\;\; x^*_5 =
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\begin{bmatrix}2\\0\end{bmatrix}$$
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Therefore, $x^*_1$ to $x^*_5$ are all of the basic feasible points for this
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constrained minimization problem.
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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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$$
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x_1 = \begin{bmatrix}0\\0\end{bmatrix} \;\;\; f(x_1) = 4 \cdot 0 + 3 \cdot 0 =
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x^*_1 = \begin{bmatrix}0\\0\end{bmatrix} \;\;\; f(x^*_1) = 4 \cdot 0 + 3 \cdot 0 =
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0$$$$
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x_2 = \frac12 \cdot \begin{bmatrix}0\\3\end{bmatrix} \;\;\;
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f(x_2) = 4 \cdot 0 + 3 \cdot \frac32 = \frac92$$$$
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x_3 = \frac{1}{13} \cdot \begin{bmatrix}3\\24\end{bmatrix} \;\;\; f(x_3) = 4
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x^*_2 = \frac12 \cdot \begin{bmatrix}0\\3\end{bmatrix} \;\;\;
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f(x^*_2) = 4 \cdot 0 + 3 \cdot \frac{3}{2} = \frac92$$$$
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x^*_3 = \frac{1}{13} \cdot \begin{bmatrix}3\\24\end{bmatrix} \;\;\; f(x^*_3) = 4
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\cdot \frac{3}{13} + 3 \cdot \frac{24}{13} = \frac{84}{13}$$$$
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x_4 = \frac12 \cdot \begin{bmatrix}3\\2\end{bmatrix} \;\;\; f(x_4) = 4 \cdot
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frac32 + 3 \cdot 1 = 9$$$$
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x_5 = \begin{bmatrix}2\\0\end{bmatrix} \;\;\; 4 \cdot 2 + 1 \cdot 0 = 8$$
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x^*_4 = \frac12 \cdot \begin{bmatrix}3\\2\end{bmatrix} \;\;\; f(x^*_4) = 4 \cdot
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\frac32 + 3 \cdot 1 = 9$$$$
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x^*_5 = \begin{bmatrix}2\\0\end{bmatrix} \;\;\; f(x^*_5) = 4 \cdot 2 + 1 \cdot 0 = 8$$
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Therefore, $x^* = x^*_1 = \begin{bmatrix}0 & 0\end{bmatrix}^T$ is the global
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constrained minimizer.
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# Exercise 3
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## Exercise 3.1
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## Exercise 3.2
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The MATLAB code used to find the solution can be found under section 3.2 of the
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`main.m` script. The solution is:
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$$x=\begin{bmatrix}0.7692\\-2.2308\\2.2308\end{bmatrix} \;\;\; \lambda=
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\begin{bmatrix}-10.3846\\2.1538\end{bmatrix}$$
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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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@ -1,79 +0,0 @@
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clc
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clear
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close all
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%% Exercise 2.1
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syms x1 x2
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c1 = 2 * x1 + 3 * x2 - 6;
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c2 = -3 * x1 + 2 * x2 - 3;
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c3 = 2 * x2 - 5;
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c4 = 2 * x1 + x2 - 4;
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c1 = solve(c1 == 0, x2, 'Real', true);
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c2 = solve(c2 == 0, x2, 'Real', true);
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c3 = solve(c3 == 0, x2, 'Real', true);
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c4 = solve(c4 == 0, x2, 'Real', true);
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i1 = solve(c1 == c2, x1, 'Real', true);
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i2 = solve(c1 == c4, x1, 'Real', true);
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i3 = solve(c4 == 0, x1, 'Real', true);
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px = double([0 0 i1 i2 i3]);
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pysym = [0 subs(c2, x1, 0) subs(c1, x1, i1) subs(c4, x1, i2) subs(c4, x1, i3)];
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py = double(pysym);
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xl = -0.05;
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xh = 2.05;
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orange = [232/255 128/255 18/255];
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grey = [0.5 0.5 0.5];
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colors = [orange; 0 0 1; 1 0 0; 0 1 0];
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dirs = [30 30 30 30];
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axis([-0.05 2.05 -0.15 5.15])
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i = 1;
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hold on
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for c = [c1 c2 c3 c4]
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pl = patch([xl, xl, xh, xh], ...
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double([-0.15, subs(c, x1, xl), subs(c, x1, xh), -0.15]), ...
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colors(i, :));
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pl.EdgeColor = colors(i, :);
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pl.FaceAlpha = .2;
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pl.EdgeAlpha = .2;
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i = i + 1;
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end
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pl = patch([0, 0, xh, xh], ...
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double([0, 5.15, 5.15, 0]), ...
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grey);
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pl.EdgeColor = grey;
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pl.FaceAlpha = .2;
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pl.EdgeAlpha = .2;
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pl = patch(px, py, 'green');
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pl.EdgeColor = 'green';
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legend('2x1 + 3x2 <= 6', '-3x1 + 2x2 <= 3', '2x2 <= 5', ...
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'2x1 + x2 <= 4', 'x1 > 0 and x2 > 0', 'feasible region');
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hold off
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%% gsppn
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for i=1:5
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obj = 4 * px(i) + 3 * py(i);
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fprintf("x1=%g x2=%g y=%g\n", px(i), py(i), obj);
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end
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%% Exercise 3.2
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G = [6 2 1; 2 5 2; 1 2 4];
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c = [-8; -3; -3];
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A = [1 0 1; 0 1 1];
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b = [3; 0];
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[x, lambda] = uzawa(G, c, A, b, [0;0;0], [0;0], 1e-8, 100);
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display(x);
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display(lambda);
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