Line 124: Line 124:
 
<br>  
 
<br>  
  
 +
We need to transform the problem into the standard form. The tableau is given as follows:
  
 +
<math>
 +
\begin{matrix}
 +
  a_1 & a_2 & a_3 & a_4 &  b \\
 +
  2 & 1& 1 &0& 3 \\
 +
  1& 4& 0 &1 &4\\
 +
  -7 & -6 &0& 0& 0
 +
\end{matrix}</math>
 +
</math>
  
 
[[ QE2012 AC-3 ECE580|Back to QE2012 AC-3 ECE580]]
 
[[ QE2012 AC-3 ECE580|Back to QE2012 AC-3 ECE580]]

Revision as of 19:19, 26 January 2013

QE2012_AC-3_ECE580-4

Part 1,2,3,4,5


Solution:

        We can transfer the problem to the following standard form:
        Mnimize    x1 − 2x2
        Sbject to     − 2x1 + x2 + x3 = 2
                       x1 + x2 + x4 = 3
                       x1 + x5 = 3
                       $ x_1, x_2, x_3, x_4, x_5 \ge 0. $
       We form the corresponding tableau for the problem
     $ \begin{matrix}   a_1 & a_2 & a_3 & a_4 & a_5 & b \\   -2 & 1& 1 &0& 0& 2 \\   -1& 1& 0 &1 &0 &3\\   1 &0& 0& 0& 1& 3\\   -1& -2 &0& 0& 0& 0  \end{matrix} $
      
Since it is already in canonical form. Because r2 =  − 7, we bring a2 into the basis.
      After computing the ratios, we pivot about the (1,2) element of the tableau to obtain
     $ \begin{matrix}   a_1 & a_2 & a_3 & a_4 & a_5 & b \\   -2 & 1& 1 &0& 0& 2 \\   1& 0& -1 &1 &0 &1\\   1 &0& 0& 0& 1& 3\\   -5& 0 &2& 0& 0& 4  \end{matrix} $
      Similarly, we pivot about the (2,1) element to obtain 
     $ \begin{matrix}   a_1 & a_2 & a_3 & a_4 & a_5 & b \\   0 & 1& -1 &2& 0& 4 \\   1& 0& -1 &1 &0 &1\\   0 &0& 1& -1& 1& 2\\   0& 0 &-3& 5& 0& 9  \end{matrix} $
      Similarly, we pivot about the (3,3) element to obtain 
     $ \begin{matrix}   a_1 & a_2 & a_3 & a_4 & a_5 & b \\   0 & 1& 0 &1& 1& 6 \\   1& 0& 0 &0 &1 &3\\   0 &0& 1& -1& 1& 2\\   0& 0 &0& 2& 3& 15  \end{matrix} $
      Therefore, x1 = 3, x2 = 6, maximize the function.

Solution 2:

The standard form for this problem is

        Mnimize    x1 − 2x2
        Sbject to     − 2x1 + x2 + x3 = 2
                       x1 + x2 + x4 = 3
                       x1 + x5 = 3
                       $ x_1, x_2, x_3, x_4, x_5 \ge 0. $

$ \begin{matrix} a_1 & a_2 & a_3 & a_4 & a_5 & b \\ -2 & 1& 1 &0& 0& 2 \\ -1& 1& 0 &1 &0 &3\\ 1 &0& 0& 0& 1& 3\\ -1& -2 &0& 0& 0& 0 \end{matrix}\Rightarrow \begin{matrix} a_1 & a_2 & a_3 & a_4 & a_5 & b \\ -2 & 1& 1 &0& 0& 2 \\ 1& 0& -1 &1 &0 &1\\ 1 &0& 0& 0& 1& 3\\ -5& 0 &2& 0& 0& 4 \end{matrix} \Rightarrow \begin{matrix} a_1 & a_2 & a_3 & a_4 & a_5 & b \\ 0 & 1& -1 &2& 0& 4 \\ 1& 0& -1 &1 &0 &1\\ 0 &0& 1& -1& 1& 2\\ 0& 0 &-3& 5& 0& 9 \end{matrix} \Rightarrow \begin{matrix} a_1 & a_2 & a_3 & a_4 & a_5 & b \\ 0 & 1& 0 &1& 1& 6 \\ 1& 0& 0 &0 &1 &3\\ 0 &0& 1& -1& 1& 2\\ 0& 0 &0& 2& 3& 15 \end{matrix} $

From the last tableau, we can see that $ \begin{bmatrix} x^{1} = 3\\ x^{2} = 6 \end{bmatrix} $ maximizes the object function. The maximum value is 15

$ \color{blue} \text{ It should be noted that, } \begin{bmatrix} x^{1} = 3\\ x^{2} = 6 \end{bmatrix} \text{actually minimizes the standard form,} $ $ \color{blue} \text{which equivalently maximizes the original problem} $




$ \color{blue}\text{Related Problem: } $

Use any simplex method to solve the following linear program.

           Maximize    7x1 + 6x2
        S'ubject to    $ 2x_1+x_2 \le 3 $
                       $ x_1+4x_2 \le 4 $
                       $ x_1 \ge 0, x_2 \ge 0. $


We need to transform the problem into the standard form. The tableau is given as follows:

$ \begin{matrix} a_1 & a_2 & a_3 & a_4 & b \\ 2 & 1& 1 &0& 3 \\ 1& 4& 0 &1 &4\\ -7 & -6 &0& 0& 0 \end{matrix} $ </math>

Back to QE2012 AC-3 ECE580

Alumni Liaison

To all math majors: "Mathematics is a wonderfully rich subject."

Dr. Paul Garrett