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Find a basic feasible solution to: $$ \begin{aligned} &2 \mathrm{x}_{1}+\mathrm{x}_{2}+2 \mathrm{x}_{3}=4 \\ &3 \mathrm{x}_{1}+3 \mathrm{x}_{2}+\mathrm{x}_{3}=3 \\ &\mathrm{x}_{1} \geq 0, \mathrm{x}_{2} \geq 0, \mathrm{x}_{2} \geq 0 \end{aligned} $$

Short Answer

Expert verified
The basic feasible solution to the given system is \((x_1, x_2, x_3, a_1, a_2) = (0, 0, 0, 4, 3)\), where \(a_1\) and \(a_2\) are artificial variables introduced to create an identity matrix in the augmented coefficient matrix. Note that this solution is not optimal and may need further refining using the Simplex Method.

Step by step solution

01

Write the given system in its standard form

The standard form of a linear system includes slack variables for each inequality (equal to zero, in our case) that represent the difference between the left and right sides of the inequality. However, all our equations are given in the equality form, so we don't need to introduce slack variables. The given system can be considered already in its standard form.
02

Identify the basic variables

Basic variables are the ones that correspond to the identity matrix in the coefficient matrix of the system. In this case, none of the variables x鈧, x鈧, and x鈧 form an identity matrix (notice the coefficients 2, 3, and 1 in the first column). Therefore, we don't have a basic feasible solution for the system in its current state.
03

Introduce artificial variables

To find a basic feasible solution, we will introduce artificial variables to create an identity matrix in the augmented coefficient matrix. Add the variables a鈧 and a鈧 so that the matrix forms an identity matrix. The new system becomes: $$ \begin{aligned} & 2x_1 + x_2 + 2x_3 + a_1 = 4 \\ & 3x_1 + 3x_2 + x_3 + a_2 = 3 \\ & x_1 \geq 0, x_2 \geq 0, x_3 \geq 0, a_1 \geq 0, a_2 \geq 0 \end{aligned} $$ Now the basic variables are a鈧 and a鈧.
04

Calculate the basic feasible solution

The values of the basic variables a鈧 and a鈧 are equal to the corresponding right-hand-side values. So we have: \(a_1 = 4\) \(a_2 = 3\) And the remaining variables \(x_1, x_2\), and \(x_3\) are equal to 0 at this stage. The basic feasible solution is therefore: \((x_1, x_2, x_3, a_1, a_2) = (0, 0, 0, 4, 3)\) Note: This basic feasible solution does not describe an optimal solution to the given system, since artificial variables are used only to facilitate finding the feasible starting point. By further refining this solution iteratively through the Simplex Method, we could potentially find an optimal solution without involving artificial variables.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Basic Feasible Solution
In linear programming, a basic feasible solution is a foundational concept that provides an initial solution to a system of linear equations or inequalities. It's where some variables, called basic variables, take non-zero values while others, the non-basic variables, are zero.
To find a basic feasible solution, we usually aim to set up a system where the number of equations equals the number of variables that are allowed to be non-zero. These non-zero variables form what is called the 'basis', and they correspond with the columns forming an identity matrix within our system's augmented matrix.
This solution is feasible if it satisfies all the constraints of the problem, including the non-negativity restrictions. However, it's important to note that not every basic feasible solution is optimal.
When we apply this to our exercise, the original set of equations didn't provide such a perfect match because of their coefficients. However, by introducing artificial variables, we were able to manipulate this system to find a basic feasible solution.
Artificial Variables
Artificial variables are a technique used in linear programming to assist in finding a basic feasible solution. When your system of equations doesn't naturally lend itself to forming a basic feasible solution, you might introduce artificial variables to create an identity matrix in the augmented system.
These variables are temporarily added to convert the set of equations into a form that allows you to solve for an initial basic feasible solution easily. However, they do not have any actual physical or practical significance in the problem itself.
In the given exercise, artificial variables \(a_1\) and \(a_2\) were added to the equations to help achieve a basic feasible solution. This means that our system of equations could be manipulated to make a starting point from which an actual solution could be derived, although these artificial variables are not part of the final or true solution as they represent auxiliary components introduced solely for the purpose of solving.
Simplex Method
The Simplex Method is a well-known iterative process used in linear programming to find the optimal solution of an objective function, given a set of linear constraints. It's particularly handy when dealing with multi-variable systems with many constraints.
The method takes the basic feasible solution as a starting point and systematically improves upon it. In essence, it finds a path through vertices of the feasible region, defined by the constraints, towards the optimal solution.
While the initial step often involves introducing artificial variables to obtain a starting solution, the Simplex Method iteratively refines this result by pivoting through basic feasible solutions. The aim is to drive these artificial variables out of the solution and find the real-valued solutions that satisfy the original constraints.
In our context, after identifying an initial basic feasible solution using artificial variables, the Simplex Method would be the tool used to iteratively refine this solution, ultimately seeking an optimal configuration where all the artificial variables are zero and only the realistic variables remain as determined by the original problem statement.

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Most popular questions from this chapter

Apply the Dual Simplex algorithm to the following dual tableau to obtain a terminal tableau: $$ \begin{array}{|c|c|c|c|c|c|} \hline & { }^{\mathrm{s}}_{1} & \mathrm{~s}_{2} & 1 & & \\ \hline \mathrm{q}_{1} & -(2 / 3) & -(1 / 3)^{*} & 12 & =-\mathrm{r}_{1} & \\ \hline \mathrm{q}_{2} & 4 / 3 & 1 / 3 & -60 & =-\mathrm{r}_{3} & \\ \hline-1 & -(20 / 3) & -(5 / 3) & 200 & =\mathrm{u} & \\ \hline \multicolumn{3}{|c|} {=\mathrm{p}_{1}} & \mathrm{~T} . \mathrm{K}) \\ \hline \end{array} $$

Maximize $$ z=7 x_{1}+10 x_{2} $$ subject to: $$ \begin{aligned} 5 \mathrm{x}_{1}+4 \mathrm{x}_{2} & \leq 24 \\ 2 \mathrm{x}_{1}+5 \mathrm{x}_{2} & \leq 13 \\ \mathrm{x}_{1}, \mathrm{x}_{2} & \geq 0 \end{aligned} $$ Find an initial feasible solution.

Minimize \(\quad z=10 \mathrm{x}_{1}+5 \mathrm{x}_{2}+4 \mathrm{x}_{3}\) subject to $$ \begin{aligned} &3 \mathrm{x}_{1}+2 \mathrm{x}_{2}-3 \mathrm{x}_{3} \geq 3 \\ &4 \mathrm{x}_{1}+2 \mathrm{x}_{3} \geq 10 \\ &\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3} \geq 0 \end{aligned} $$ Solve the primal problem by applying the Dual Simplex algorithm. Also, solve the dual problem by the Simplex method.

Maximize \(\mathrm{f}\left(\mathrm{x}_{1}, \mathrm{x}_{2}\right)=3 \mathrm{x}_{1}+13 \mathrm{x}_{2}\) subject to: $$ \begin{aligned} &2 \mathrm{x}_{1}+9 \mathrm{x}_{2} \leq 40 \\ &11 \mathrm{x}_{1}-8 \mathrm{x}_{2} \leq 82 \end{aligned} $$ \(\mathrm{x}_{1}, \mathrm{x}_{2}>0\) and integer. Solve by the simplex method first, ignoring the integer requirements. Try to approximate an integer solution by using the result of the simplex algorithm. Also solve by graphical means, and compare with the continuous solution obtained from the simplex tableaux.

Put into standard form: Maximize \(\quad 3 \mathrm{x}+\mathrm{y}\) subject to: \(\quad x \geq 0\) $$ \begin{aligned} y \geq 0 & \\ 2 x-y & \leq-10 \\ x+2 y & \leq 14 \\ x & \leq 12 \end{aligned} $$

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