6_6.htm

随着各行各业的发展和生产需要

    <td width="10%">-1</td>
    <td width="10%">12/5</td>
    <td width="10%">-3/5</td>
    <td width="10%">←λ</td>
  </tr>
</table>
</div>

<p>故x<sub>1</sub>＝2/5, x<sub>2</sub>＝1/5，x<sub>3</sub>＝0时, maxz＝12/5</p>

<p>比较本例在上节中的解答，可以看出，只有在n较大时，用改善的单纯形表格才有优势．</p>

<p><strong>例２</strong>　<br>
maxz＝－x<sub>1</sub>－x<sub>2</sub>＋4x<sub>3</sub>，<br>
x<sub>1</sub>＋x<sub>2</sub>＋x<sub>3</sub>≤9<br>
x<sub>1</sub>＋x<sub>2</sub>－x<sub>3</sub>≤2<br>
－x<sub>1</sub>＋x<sub>2</sub>＋x<sub>3</sub>≤4<br>
x<sub>1</sub>,x<sub>2</sub>,x<sub>3</sub>≥0<br>
<strong>解</strong>　引进松弛变量x<sub>4</sub>,x<sub>5</sub>,x<sub>6</sub>，得<br>
maxz＝－x<sub>1</sub>－x<sub>2</sub>＋4x<sub>3</sub>＋0x<sub>4</sub>＋0x<sub>5</sub>＋0x<sub>6</sub><br>
x<sub>1</sub>＋x<sub>2</sub>＋x<sub>3</sub>＋x<sub>4</sub>＝9<br>
x<sub>1</sub>＋x<sub>2</sub>－x<sub>3</sub>＋x<sub>5</sub>＝2<br>
－x<sub>1</sub>＋x<sub>2</sub>＋x<sub>3</sub>＋x<sub>6</sub>＝4<br>
x<sub>1</sub>,x<sub>2</sub>,x<sub>3</sub>, x<sub>4</sub> , x<sub>5</sub>, x<sub>6</sub>≥0<br>
结果如下：<br>
</p>
<div align="left">

<table border="1" width="85%">
  <tr>
    <td width="9%">B</td>
    <td width="9%">C<sub>B</sub></td>
    <td width="9%">C<sub>B</sub>B<sup>-1</sup></td>
    <td width="9%">C</td>
    <td width="9%">-1</td>
    <td width="9%">-1</td>
    <td width="9%">4</td>
    <td width="9%">0</td>
    <td width="9%">0</td>
    <td width="9%">0</td>
    <td width="10%">β</td>
  </tr>
  <tr>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="9%">P<sub>0</sub></td>
    <td width="9%">P<sub>1</sub></td>
    <td width="9%">P<sub>2</sub></td>
    <td width="9%">P<sub>3</sub></td>
    <td width="9%">P<sub>4</sub></td>
    <td width="9%">P<sub>5</sub></td>
    <td width="9%">P<sub>6</sub></td>
    <td width="10%">　</td>
  </tr>
  <tr>
    <td width="9%">A<sub>4</sub></td>
    <td width="9%">0</td>
    <td width="9%">0</td>
    <td width="9%">　</td>
    <td width="9%">9</td>
    <td width="9%">1</td>
    <td width="9%">1</td>
    <td width="9%">2</td>
    <td width="9%">1</td>
    <td width="9%">0</td>
    <td width="10%">0</td>
  </tr>
  <tr>
    <td width="9%">A<sub>5</sub></td>
    <td width="9%">0</td>
    <td width="9%">0</td>
    <td width="9%">2</td>
    <td width="9%">1</td>
    <td width="9%">1</td>
    <td width="9%">-1</td>
    <td width="9%">0</td>
    <td width="9%">1</td>
    <td width="9%">0</td>
    <td width="10%">　</td>
  </tr>
  <tr>
    <td width="9%">A<sub>6</sub></td>
    <td width="9%">0</td>
    <td width="9%">0</td>
    <td width="9%">4</td>
    <td width="9%">-1</td>
    <td width="9%">1</td>
    <td width="9%">(1)</td>
    <td width="9%">0</td>
    <td width="9%">0</td>
    <td width="9%">1</td>
    <td width="10%">4</td>
  </tr>
  <tr>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="9%">z=0</td>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="9%">4</td>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="10%">← λ</td>
  </tr>
</table>
</div>

<p><br>
</p>
<div align="left">

<table border="1" width="85%">
  <tr>
    <td width="9%">B</td>
    <td width="9%">C<sub>B</sub></td>
    <td width="9%">C<sub>B</sub>B<sup>-1</sup></td>
    <td width="9%">C</td>
    <td width="9%">-1</td>
    <td width="9%">-1</td>
    <td width="9%">4</td>
    <td width="9%">0</td>
    <td width="9%">0</td>
    <td width="9%">0</td>
    <td width="10%">β</td>
  </tr>
  <tr>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="9%">P<sub>0</sub></td>
    <td width="9%">P<sub>1</sub></td>
    <td width="9%">P<sub>2</sub></td>
    <td width="9%">P<sub>3</sub></td>
    <td width="9%">P<sub>4</sub></td>
    <td width="9%">P<sub>5</sub></td>
    <td width="9%">P<sub>6</sub></td>
    <td width="10%">　</td>
  </tr>
  <tr>
    <td width="9%">A<sub>4</sub></td>
    <td width="9%">0</td>
    <td width="9%">0</td>
    <td width="9%">1</td>
    <td width="9%">(3)</td>
    <td width="9%">　</td>
    <td width="9%">0</td>
    <td width="9%">1</td>
    <td width="9%">0</td>
    <td width="9%">-2</td>
    <td width="10%">1/3</td>
  </tr>
  <tr>
    <td width="9%">A<sub>5</sub></td>
    <td width="9%">0</td>
    <td width="9%">0</td>
    <td width="9%">6</td>
    <td width="9%">0</td>
    <td width="9%">　</td>
    <td width="9%">0</td>
    <td width="9%">0</td>
    <td width="9%">1</td>
    <td width="9%">1</td>
    <td width="10%">　</td>
  </tr>
  <tr>
    <td width="9%">A<sub>6</sub></td>
    <td width="9%">4</td>
    <td width="9%">4</td>
    <td width="9%">4</td>
    <td width="9%">-1</td>
    <td width="9%">　</td>
    <td width="9%">1</td>
    <td width="9%">0</td>
    <td width="9%">0</td>
    <td width="9%">1</td>
    <td width="10%">　</td>
  </tr>
  <tr>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="9%">z=16</td>
    <td width="9%">　</td>
    <td width="9%">3</td>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="10%">← λ</td>
  </tr>
</table>
</div>

<p><br>
</p>
<div align="left">

<table border="1" width="85%">
  <tr>
    <td width="9%">B</td>
    <td width="9%">C<sub>B</sub></td>
    <td width="9%">C<sub>B</sub>B<sup>-1</sup></td>
    <td width="9%">C</td>
    <td width="9%">-1</td>
    <td width="9%">-1</td>
    <td width="9%">4</td>
    <td width="9%">0</td>
    <td width="9%">0</td>
    <td width="9%">0</td>
    <td width="10%">β</td>
  </tr>
  <tr>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="9%">P<sub>0</sub></td>
    <td width="9%">P<sub>1</sub></td>
    <td width="9%">P<sub>2</sub></td>
    <td width="9%">P<sub>3</sub></td>
    <td width="9%">P<sub>4</sub></td>
    <td width="9%">P<sub>5</sub></td>
    <td width="9%">P<sub>6</sub></td>
    <td width="10%">　</td>
  </tr>
  <tr>
    <td width="9%">A<sub>1</sub></td>
    <td width="9%">-1</td>
    <td width="9%">1</td>
    <td width="9%">1/3</td>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="9%">1/3</td>
    <td width="9%">0</td>
    <td width="9%">-2/3</td>
    <td width="10%">　</td>
  </tr>
  <tr>
    <td width="9%">A<sub>5</sub></td>
    <td width="9%">0</td>
    <td width="9%">0</td>
    <td width="9%">6</td>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="9%">0</td>
    <td width="9%">1</td>
    <td width="9%">1</td>
    <td width="10%">　</td>
  </tr>
  <tr>
    <td width="9%">A<sub>3</sub></td>
    <td width="9%">4</td>
    <td width="9%">2</td>
    <td width="9%">13/3</td>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="9%">1/3</td>
    <td width="9%">0</td>
    <td width="9%">1/3</td>
    <td width="10%">　</td>
  </tr>
  <tr>
    <td width="9%">　</td>
    <td width="9%">　</td>
    <td width="9%">z=17</td>
    <td width="9%">　</td>
    <td width="9%">0</td>
    <td width="9%">-4</td>
    <td width="9%">0</td>
    <td width="9%">-1</td>
    <td width="9%">0</td>
    <td width="9%">-2</td>
    <td width="10%">← λ</td>
  </tr>
</table>
</div>

<p><br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 在上面的单纯形法中，当出现c<sub>3</sub>－z<sub>3</sub>＞0，便可立即判定可取P<sub>3</sub>作为进入基，<br>
相应地确定P<sub>6</sub>退出．只要对P<sub>3</sub>进行列消元，使之出现(0 
0 1)<sup>T</sup>,<br>
则相应可得新的P<sub>0</sub>,P<sub>4</sub>,P<sub>5</sub>,P<sub>6</sub>便是所求的P<sub>0</sub>及新的逆矩阵．即以P<sub>4</sub>,P<sub>5</sub>,P<sub>6</sub>为<br>
基的逆矩阵便是<br>
<img src="6_6_3.gif"> </p>
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