418 lines
16 KiB
TeX
418 lines
16 KiB
TeX
% vim: set ts=2 sw=2 et tw=80:
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\documentclass[unicode,11pt,a4paper,oneside,numbers=endperiod,openany]{scrartcl}
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\usepackage{pgfplots}
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\pgfplotsset{compat=1.17}
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\usepackage{mathrsfs}
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\usepackage{hyperref}
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\usetikzlibrary{arrows}
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\input{assignment.sty}
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\definecolor{zzttqq}{rgb}{0.6,0.2,0}
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\definecolor{uuuuuu}{rgb}{0.26666666666666666,0.26666666666666666,0.26666666666666666}
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\definecolor{ccqqqq}{rgb}{0.8,0,0}
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\definecolor{qqqqff}{rgb}{0,0,1}
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\begin{document}
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\setassignment
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\setduedate{Friday, December 18, 2020, 11:59 PM}
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\serieheader{Numerical Computing}{2020}{Student: Claudio Maggioni}{Discussed
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with: Gianmarco De Vita (Exercise 2)}{Solution for Project 6}{}
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\newline
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\assignmentpolicy
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The purpose of this mini-project is to implement the Simplex Method to find the
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solution of linear programs, involving both the minimisation and the maximisation
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of the objective function.
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\section{Graphical Solution of Linear Programming Problems [25 points]}
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Please consider the following two problems:
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\begin{enumerate}
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\item[(1)] \begin{equation*}
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\begin{aligned}
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& \text{min} & z=4x&+y\\
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& \text{ s.t.} & x+2y &\leq 40 \\
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& & x+y &\geq 30\\
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& & 2x+3y &\geq 72 \\
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& & x,\,y &\geq0 \\
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\end{aligned}
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\end{equation*}
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\item[(2)] A tailor plans to sell two types of trousers, with production costs
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of 25 CHF and 40 CHF, respectively. The former type can be sold for 85 CHF,
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while the latter for 110 CHF. The tailor estimates a total monthly demand
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of 265 trousers. Find the number of units of each type of trousers that
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should be produced in order to maximise the net profit of the tailor, if we
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assume that the he cannot spend more than 7000 CHF in raw materials.
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\end{enumerate}
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Start by writing problem (2) as a linear programming problem. Then complete the
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following tasks:
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\begin{itemize}
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\item Solve the system of inequalities.
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\item Plot the feasible region identified by the constraints.
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\item Find the optimal solution and the value of the objective function in
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that point.
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\end{itemize}
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\subsection{Solving problem 1}
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We first solve the system of inequalities for y:
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\[\begin{cases}
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y \leq 20 - \frac{1}{2} x \\
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y \geq 30 - x \\
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y \geq 24 - \frac{2}{3} x \\
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x, y \geq 0 \\
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\end{cases}
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\]
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Here is the feasability region plot, which was derived geometrically by plotting
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the inequalities above.
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\begin{center}
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\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1cm,y=1cm]
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\begin{axis}[
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x=0.2cm,y=0.2cm,
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xmin=-1, ymin=-1, xmax=40,ymax=32,
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axis lines=middle,
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ymajorgrids=true,xmajorgrids=true]
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\draw [line width=2pt,color=qqqqff,domain=-11.510887860744262:56.37536226673365] plot(\x,{(--20-0.5*\x)/1});
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\draw [line width=2pt,color=ccqqqq,domain=-11.510887860744262:56.37536226673365] plot(\x,{(--30-1*\x)/1});
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\draw [line width=2pt,color=ccqqqq,domain=-11.510887860744262:56.37536226673365] plot(\x,{(--24-0.6666666666666666*\x)/1});
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\draw [line width=2pt,domain=-11.510887860744262:56.37536226673365] plot(\x,{(-0-0*\x)/1});
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\draw [line width=2pt] (0,-15.596475658682538) -- (0,33.531731670413244);
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\draw [line width=2pt,color=zzttqq] (24,8)-- (36,0);
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\draw [line width=2pt,color=zzttqq] (36,0)-- (40,0);
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\draw [line width=2pt,color=zzttqq] (40,0)-- (24,8);
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\fill[line width=2pt,color=zzttqq,fill=zzttqq,fill opacity=0.10000000149011612] (24,8) -- (36,0) -- (40,0) -- cycle;
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\end{axis}
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\end{tikzpicture}
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\end{center}
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Red lines represent feasibility constraints keeping the upper part of the graph,
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while blue lines keep the low part. The resulting region is the area in brown.
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We therefore identify the verticies $(24,8)$, $(36,0)$ and $(40,0)$ as candidate
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solutions, and therefore we evaluate the minimization function in these points:
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\[ m(x, y) = 4x + y \]
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\[\begin{aligned}
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m(24,8) = 104 \\
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m(36,0) = 144 \\
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m(40,0) = 160 \\
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\end{aligned}\]
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And therefore we have as a solution $(x, y) = (24, 8)$, with $z = 104$, as this
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is a minimization problem.
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\subsection{Problem 2 as linear programming problem}
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The following is the linear programming formulation of problem 2:
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\begin{equation*}
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\begin{aligned}
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\text{max} &\; (85 - 25) x + (110 - 40) y \\
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\text{s.t.} & \begin{cases}
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25 x + 40 y \leq 7000 \\
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x + y \leq 265 \\
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x, y \geq 0 \\
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\end{cases}
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\end{aligned}
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\end{equation*}
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\subsection{Solving problem 2}
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We first solve the system of inequalities for y:
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\[
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\begin{cases}
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25 x + 40 y \leq 7000 \\
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x + y \leq 265 \\
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x, y \geq 0 \\
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\end{cases}
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=
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\begin{cases}
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y < 175 - \frac{5}{8}x \\
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x \leq 265 - x \\
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x, y \geq 0
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\end{cases}
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\]
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From this solution it is clear that the region of satisfactory $x$s and $y$s is
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bounded by $x = 0$, $y = 0$, $y = 265 - x$, and $y = 175 - \frac58 x$.
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Here is a plot of the feasability region:
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\begin{center}
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\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45]
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\begin{axis}[
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x=0.015cm,y=0.015cm,
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axis lines=middle,
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xmin=-10,ymin=-10,xmax=300,ymax=300]
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\fill[line width=2pt,color=zzttqq,fill=zzttqq,fill opacity=0.10000000149011612]
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(265,0) -- (240,25) -- (0,175) -- (0,0) -- cycle;
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\draw [line width=2pt] (0,175)-- (280,0);
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\draw [line width=2pt] (265,0)-- (0,265);
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\end{axis}
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\end{tikzpicture}
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\end{center}
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We then proceed to evaluate the maximization function at all verticies of the
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feasability region, i.e. $\{ (0, 0), (265, 0), (0, 175), (240,25) \}$.
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\[ m(x, y) = 60x + 70y \]
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\[\begin{aligned}
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m(0,0) = 0 \\
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m(0,175) = 12250 \\
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m(240,25) = 16150 \\
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m(265,0) = 15900 \\
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\end{aligned}\]
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We then conclude the tailor should produce 240 trousers of the first type and
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25 trousers of the second type, for
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a revenue of 16,150.-- Fr.
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\section{Implementation of the Simplex Method [35 points]}
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The implementation of the simplex method can be found in the
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\texttt{Project\_6\_Claudio\_Maggioni/} folder.
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\section{Applications to Real-Life Example: Cargo Aircraft [25 points]}
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In this second part of the assignment, you are required to use the simplex
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method implementation to solve a real-life problem taken from economics
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(constrained profit maximisation).
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A cargo aircraft has 4 compartments (indicated simply as $S_1,\dots,S_4$) used
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to store the goods to be transported. Details about the weight capacity and
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storage capacity of the different compartments can be inferred from the data
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reported in the following table:
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\begin{center}
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\begin{tabular}{||c | c | c ||}
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\hline
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Compartment & Weight Capacity ($t$) & Storage Capacity ($m^3$) \\ [0.5ex]
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\hline\hline
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$S_1$ & 18 & 11930\\
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\hline
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$S_2$ & 32 & 22552\\
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\hline
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$S_3$ & 25 & 11209\\
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\hline
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$S_4$ & 17 & 5870\\
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\hline
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\end{tabular}
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\end{center}
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The following four cargos are available for shipment during the next flight:
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\begin{center}
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\begin{tabular}{|| c | c | c | c ||}
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\hline
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Cargo & Weight ($t$) & Volume ($m^3/t$) & Profit ($\text{CHF}/t$) \\ [0.5ex]
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\hline\hline
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$C_1$ & 16 & 320 & 135 \\
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\hline
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$C_2$ & 32 & 510 & 200 \\
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\hline
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$C_3$ & 40 & 630 & 410 \\
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\hline
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$C_4$ & 28 & 125 & 520 \\
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\hline
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\end{tabular}
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\end{center}
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Any proportion of the four cargos can be accepted, and the profit obtained for
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each cargo is increased by $10\%$ if it is put in $S_2$, by $20\%$ if it is put
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in $S_3$ and by $30\%$ if it is put in $S_4$, due to the better storage conditions
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. The objective of this problem is to determine which amount of the different
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cargos will be transported and how to allocate it among the different compartments,
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while maximising the profit of the owner of the cargo plane.
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\subsection{Formulate the problem above as a linear program: what is the
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objective function? What are the constraints? Write down all equations, with
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comments explaining what you are doing.}
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From the problem description we can determine that this problem has 16 variables
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$x_{i,j}$, where $1 \leq i,j \leq 4 \; i,j \in N$. Each variable represents a
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weight in tonnes, with $i$ determining the type of cargo used and $j$ the
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compartment used. Therefore the variables of this problem to optimize represent
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how much cargo to put where and in which quantity.
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From this initial discussion we can easily derive the objective function to
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maximize (as the objective function will output revenue, and we want to be
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greedy as most Swiss people):
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\[\text{max}\; z =
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\text{grandsum}\left[\left(\begin{bmatrix}135\\200\\410\\520\end{bmatrix}\cdot\begin{bmatrix}1&1.1&1.2&1.3
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\end{bmatrix}\right) \odot \begin{bmatrix}
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x_{1,1}&x_{1,2}&x_{1,3}&x_{1,4}\\
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x_{2,1}&x_{2,2}&x_{2,3}&x_{2,4}\\
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x_{3,1}&x_{3,2}&x_{3,3}&x_{3,4}\\
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x_{4,1}&x_{4,2}&x_{4,3}&x_{4,4}\\\end{bmatrix}\right]\]
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where $\odot$ is an element-wise or Hadamard product and $\text{grandsum}(A)$
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simply sums all elements of $A$. The column vector represents the value per
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ton of each good while the row vector represents the revenue multipliers
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associated with each cargo hold.
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We can move to cargo hold constraints on weight, which check if the cargo holds are
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overloaded:
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\[\begin{aligned}
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\sum_{i = 1}^4 x_{i,1} \leq 18 & \hspace{1cm} & \sum_{i = 1}^4 x_{i,2} \leq 32
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& \hspace{1cm} &
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\sum_{i = 1}^4 x_{i,3} \leq 25 & \hspace{1cm} & \sum_{i = 1}^4 x_{i,4} \leq 17
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\end{aligned}\]
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Then we consider compartment constraints on space, which check if the cargos
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will fit:
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\[\begin{aligned}
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320 x_{1,1} + 510 x_{2,1} + 630 x_{3,1} + 125 x_{4,1} \leq &\; 11930 \\
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320 x_{1,2} + 510 x_{2,2} + 630 x_{3,2} + 125 x_{4,2} \leq &\; 22552 \\
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320 x_{1,3} + 510 x_{2,3} + 630 x_{3,3} + 125 x_{4,3} \leq &\; 11209 \\
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320 x_{1,4} + 510 x_{2,4} + 630 x_{3,4} + 125 x_{4,4} \leq &\; 5870 \\
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\end{aligned}\]
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Then we have a final set of constraints to check if we are loading more cargo
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than we own:
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\[\begin{aligned}
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\sum_{i=1}^4 x_{1,i} \leq 16 & \hspace{1cm} &
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\sum_{i=1}^4 x_{1,i} \leq 32 & \hspace{1cm} &
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\sum_{i=1}^4 x_{1,i} \leq 40 & \hspace{1cm} &
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\sum_{i=1}^4 x_{1,i} \leq 28
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\end{aligned}\]
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Therefore, we have a final minimization problem in standard form with 16
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variables and 12 feasability constraints.
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\subsection{Create a script \emph{exercise3.m} which uses the simplex method
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implemented in the previous exercise to solve the problem. What is the optimal
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solution? Visualise it graphically and briefly comment the results obtained
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(are you surprised of this outcome on the basis of your data?)}
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The file that solves this problem using our MATLAB's implementation of the
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simplex method can be found in the \emph{exercise3.m} file.
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The optimal solution according to the simplex method is the following:
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\[\begin{bmatrix}
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x_{1,1}&x_{1,2}&x_{1,3}&x_{1,4}\\
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x_{2,1}&x_{2,2}&x_{2,3}&x_{2,4}\\
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x_{3,1}&x_{3,2}&x_{3,3}&x_{3,4}\\
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x_{4,1}&x_{4,2}&x_{4,3}&x_{4,4}\\\end{bmatrix} =
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\begin{bmatrix}0&0&0&0\\18&6&0&0\\0&26&14&0\\0&0&11&17\end{bmatrix}\]
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The revenue for this load is of 41890.- Fr.
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Figure~\ref{fig:hist} shows a graphical representation of this solution.
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The solution seems quite intuitive: the simplex method seems to have found quite
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a greedy solution, placing the most valuable cargo in the ``best'' cargo hold,
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proceeding with less valuable cargos and ``worse'' cargo holds once either the
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filling constraints of the hold are reached or we run out of our most valuable
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cargo left.
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It is notable that weight limits are reached before space limits, but in
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hindsight it is a quite likely scenario for a cargo plane. This problem has
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quite realistic data, as the total weight limit (82 tonnes) matches almost
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exactly the designed load capacity of a cargo variant Boeing 747
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(81.6 tonnes\footnote{Source: Wikipedia at
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\url{https://en.wikipedia.org/wiki/Boeing_747\#Background}}).
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\begin{figure}[h]
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\centering
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\begin{tikzpicture}
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\begin{axis}[ybar interval,
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ymax=40,
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ymin=0,
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minor y tick num = 3,
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xticklabels={{$S_1$},{$S_2$},{$S_3$},{$S_4$}}]
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\addplot coordinates { (1,0) (2,0) (3,0) (4,0) (5,0) };
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\addplot coordinates { (1,18) (2,6) (3,0) (4,0) (5,0) };
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\addplot coordinates { (1,0) (2,26) (3,14) (4,0) (5,0) };
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\addplot coordinates { (1,0) (2,0) (3,11) (4,17) (5,0) };
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\legend {$C_1$, $C_2$, $C_3$, $C_4$}
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\end{axis}
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\end{tikzpicture}
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\caption{Solution to the Cargo Airplane as an histogram of tonnes loaded per
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cargo hold per cargo type. Y-axis is the amount tonnes loaded, X-axis is the cargo hold
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where the cargo should be loaded, bar color is the cargo type loaded.}\label{fig:hist}
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\end{figure}
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\section{Cycling and Degeneracy [15 points]}
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Consider now the following simple problem:
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\begin{alignat*}{2}
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&\text{max}\;\, && z = 3x_1+4x_2,\\
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&\text{s.t.} && 4x_1+3x_2\leq 12\\
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& && 4x_1+x_2\leq 8\\
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& && 4x_1+2x_2\leq 8\\
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& && x_1, x_2 \geq 0.
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\end{alignat*}
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\subsection{Create a script \emph{exercise4.m} which uses the simplex method
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implemented above to solve this problem. Do you achieve convergence within the
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maximum number of iterations (given by the maximum number of possible basic
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solutions)? Do you notice any strange behaviour? (\emph{hint:} check, e.g., the
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indices of the entering and departing variables)}
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The MATLAB implementation for this problem can be found under file
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\emph{exercise4.m} in the resources folder (i.e.
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\texttt{Project\_6\_Claudio\_Maggioni/}).
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The simplex method implementation for this problem fails with an error in the
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auxiliary problem initialization. The error message is ``The original LP problem does not admit a feasible
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solution''. This error message appears when the simplex method does not
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converge in the expected maximum number of iterations. The large number of
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iterations (793 before forced termination) is caused by an ``oscillation'' or
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vicious cycle of swap operations between matrix $B$ and matrix $D$. This
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oscillation is cause by the fact that entering and departing variable indexes do
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not change, and stay fixed at values 6 and 4 from the second iteration onwards.
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This causes an unnecessary continuous oscillation between two solutions that are
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equally not optimal ($z = 28$) according to the auxiliary problem definition,
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terminating the auxiliary algorithm abnormally after too many iterations.
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\subsection{Look at the number of constraints and at the number of unknowns:
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what can you notice about the underlying system of equations? Represent them
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graphically and try to use this information to explain the behaviour of your
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solver in the previous point.}
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In figure~\ref{fig:strange} lies a graphical representation of this linear
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programming problem.
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The anomaly in the feasability region of this problem is that its perimeter has
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triangular shape, and it is made up only of one constraint other than the
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trivial positivity constraints at the axes (i.e. the constraints different from
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$4x +2y \leq 8$ do not make a difference in the feasability region).
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\begin{figure}[h]
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\centering
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\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45]
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\begin{axis}[
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x=1cm,y=1cm,
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axis lines=middle,
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ymajorgrids=true,
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xmajorgrids=true,
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xmin=-0.5,
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ymin=-0.5,
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xmax=6,
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ymax=6]
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\clip(-4.052471790978185,-1.8483084480511438) rectangle (6.897018211059206,6.075664579739015);
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\fill[line width=2pt,color=zzttqq,fill=zzttqq,fill opacity=0.10000000149011612] (0,4) -- (0,0) -- (2,0) -- cycle;
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\draw [line width=2pt,domain=0:6] plot(\x,{(-12--4*\x)/-3});
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\draw [line width=2pt,domain=0:6] plot(\x,{(-8--4*\x)/-2});
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\draw [line width=2pt,domain=0:6] plot(\x,{(--8-4*\x)/1});
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\draw [line width=2pt] (0,-1.8483084480511438) -- (0,6.075664579739015);
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\draw [line width=2pt,domain=0:6] plot(\x,{(-0-0*\x)/1});
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\draw [line width=2pt,color=zzttqq] (0,4)-- (0,0);
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\draw [line width=2pt,color=zzttqq] (0,0)-- (2,0);
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\draw [line width=2pt,color=zzttqq] (2,0)-- (0,4);
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\end{axis}
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\end{tikzpicture}
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\caption{Feasability region plot of maximization problem for exercise 4. Note
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the fact that the region perimeter is a triangle.}\label{fig:strange}
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\end{figure}
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\end{document}
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