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# 数学代写|运筹学代写Operations Research代考|MTH360 ALGEBRAIC FORM OF THE SIMPLEX ALGORITHM

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## 数学代写|运筹学代写Operations Research代考|ALGEBRAIC FORM OF THE SIMPLEX ALGORITHM

Let us consider the same problem which was discussed earlier to illustrate this algorithm.
Maximize $Z=6 X_1+5 X_2+0 X_3+0 X_4$

Subject to
\begin{aligned} X_1+X_2+X_3 &=5 \ 3 X_1+2 X_2+X_4 &=12 \ X_1, X_2 & \geq 0 \end{aligned}
Iteration 1
We start with a basic feasible solution with $X_3$ and $X_4$ as basic variables. We write the basic variables in terms of the non-basic variables as:
\begin{aligned} X_3 &=5-X_1-X_2 \ X_4 &=12-3 X_1-2 X_2 \ Z &=0+6 X_1+5 X_2 \end{aligned}
The present solution has $Z=0$, since $X_1$ and $X_2$ are presently non-basic with value zero. We want to increase $Z$ and this is possible by increasing $X_1$ or $X_2$. We choose to increase $X_1$ by bringing it to the basis because it has the highest rate of increase.

Considering the equation $X_3=5-X_1-X_2, X_1$ can be increased to 5 beyond which $X_3$ will become negative and infeasible. Considering the equation $X_4=12-3 X_1-2 X_2, X_1$ can be increased to 4 beyond which $X_4$ will become negative. The limiting value of $X_1$ (or the allowable upper limit on $X_1$ ) is 4 from the second equation.

## 数学代写|运筹学代写Operations Research代考|ABULAR FORM OF THE SIMPLEX ALGORITHM

The simplex method can be represented in a tabular form, where only the numbers are written in a certain easily understandable form. Several forms are available but we present only one version of the simplex method in tabular form. Table $2.1$ represents the simplex tableau for the first iteration.

The second row has all the variables listed. Above them (in the first row) are the objective function coefficients of the variables. Under the basic variables, variables $X_3$ and $X_4$ are shown. To their left are the objective function values of the basic variables. To the right of the basic variables are the constraints written in the form of the equations along with the right hand side values of the constraints.

The $C_j-Z_j$ for a variable is $C_j$ minus the dot product of the $C_B$ and the column corresponding to the variable $j$. For example, $C_1-Z_1=6-(0 \times 1+0 \times 3)=6$. The variable with maximum positive value of $C_j-Z_j$ enters. In our example, it is variable $X_1$ shown with an arrow. The $\theta$ values are the ratios between the RHS value and the coefficient under the entering variable column. In our example, these are $5 / 1=5$ and $12 / 3=4$, respectively. The minimum $\theta$ is 4 and variable $X_4$ is the leaving variable. Now, variable $X_1$ replaces $X_4$ as the basic variable in the next iteration.

In the previous iteration, we were solving for variables $X_3$ and $X_4$. They had an identity matrix as their coefficients (or $X_3$ and $X_4$ appeared in one equation only with $+1$ coefficient), so that we can directly solve them. In the next iteration, we need to rewrite the constraints (rows) such that $X_3$ and $X_1$ have the identity matrix as coefficients. We call the row corresponding to the leaving variable as the pivot row and the corresponding element in the entering column as the pivot element. The pivot element is shown in bold in Table 2.1. Table $2.2$ shows the first two iterations of the simplex algorithm.

## 数学代写|运筹学代写运筹学代考|THE graphics METHOD

. THE . THE 数学代写|运筹学代写

## 数学代写|运筹学代写运筹学代考|THE algeaic METHOD

Subject to
\begin{aligned} X_1+X_2 & \leq 5 \ 3 X_1+2 X_2 & \leq 12 \end{aligned}

Maximize $Z=6 X_1+5 X_2+0 X_3+0 X_4$
Subject to
$$\begin{gathered} X_1+X_2+X_3=5 \ 3 X_1+2 X_2+X_4=12 \ X_1, X_2, X_3, X_4 \geq 0 \end{gathered}$$

1. 变量 $X_1$ 和 $X_2$ 是非基本的，并设置为零。代入，我们得到 $X_3=5, X_4=12$ 和目标函数的值 $Z=0$.
2. 变量 $X_1$ 和 $X_3$ 是非基本的，并设置为零。代入，我们求解 $X_2=5$ 和 $2 X_2+X_4=12$ 然后得到 $X_2=5, X_4=2$ 和目标函数的值 $Z=25$.
3. 变量 $X_1$ 和 $X_4$ 是非基本的，并设置为零。代入，我们求解 $X_2+X_3=5$ 和 $2 X_2=12$ 然后得到 $X_2=6, X_3=-1$
4. 变量 $X_2$ 和 $X_3$ 是非基本的，并设置为零。代入，我们求解 $X_1=5$ 和 $3 X_1+X_4=12$ 然后得到 $X_1=5, X_4=-3$.
5. 变量 $X_2$ 和 $X_4$ 是非基本的，并设置为零。代入，我们求解 $X_1+X_3=5$ 和 $3 X_1=12$ 然后得到 $X_1=4, X_3=1$ 和目标函数的值 $Z=24$.
6. 变量 $X_3$ 和 $X_4$ 是非基本的，并设置为零。代入，我们求解 $X_1+X_2=5$ 和 $3 X_1+2 X_2=12$ 然后得到 $X_1=2, X_2=3$ 和目标函数的值 $Z=27$.

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