6_10.htm

随着各行各业的发展和生产需要

    <td width="13%">0</td>
  </tr>
  <tr>
    <td width="12%">　</td>
    <td width="12%">　</td>
    <td width="12%">P<sub>0</sub></td>
    <td width="12%">P<sub>1</sub></td>
    <td width="13%">P<sub>2</sub></td>
    <td width="13%">P<sub>3</sub></td>
    <td width="13%">P<sub>4</sub></td>
    <td width="13%">P<sub>53</sub></td>
  </tr>
  <tr>
    <td width="12%">A<sub>1</sub></td>
    <td width="12%">1</td>
    <td width="12%">2</td>
    <td width="12%">1</td>
    <td width="13%">4/3</td>
    <td width="13%">-1/3</td>
    <td width="13%">0</td>
    <td width="13%">0</td>
  </tr>
  <tr>
    <td width="12%">A<sub>4</sub></td>
    <td width="12%">0</td>
    <td width="12%">-1</td>
    <td width="12%">0</td>
    <td width="13%">(-5/3)</td>
    <td width="13%">-1/3</td>
    <td width="13%">1</td>
    <td width="13%">0</td>
  </tr>
  <tr>
    <td width="12%">A<sub>5</sub></td>
    <td width="12%">0</td>
    <td width="12%">2</td>
    <td width="12%">0</td>
    <td width="13%">5/3</td>
    <td width="13%">-2/3</td>
    <td width="13%">0</td>
    <td width="13%">1</td>
  </tr>
  <tr>
    <td width="12%">　</td>
    <td width="12%">　</td>
    <td width="12%">　</td>
    <td width="12%">0</td>
    <td width="13%">2/3</td>
    <td width="13%">1/3</td>
    <td width="13%">0</td>
    <td width="13%">0</td>
  </tr>
</table>
</div>

<p>　</p>
<div align="left">

<table border="1" width="75%">
  <tr>
    <td width="12%">B</td>
    <td width="12%">C<sub>B</sub></td>
    <td width="12%">C</td>
    <td width="12%">1</td>
    <td width="13%">2</td>
    <td width="13%">0</td>
    <td width="13%">0</td>
    <td width="13%">0</td>
  </tr>
  <tr>
    <td width="12%">　</td>
    <td width="12%">　</td>
    <td width="12%">P<sub>0</sub></td>
    <td width="12%">P<sub>1</sub></td>
    <td width="13%">P<sub>2</sub></td>
    <td width="13%">P<sub>3</sub></td>
    <td width="13%">P<sub>4</sub></td>
    <td width="13%">P<sub>53</sub></td>
  </tr>
  <tr>
    <td width="12%">A<sub>1</sub></td>
    <td width="12%">1</td>
    <td width="12%">6/5</td>
    <td width="12%">1</td>
    <td width="13%">0</td>
    <td width="13%">-5/3</td>
    <td width="13%">4/5</td>
    <td width="13%">0</td>
  </tr>
  <tr>
    <td width="12%">A<sub>2</sub></td>
    <td width="12%">2</td>
    <td width="12%">3/5</td>
    <td width="12%">0</td>
    <td width="13%">1</td>
    <td width="13%">1/5</td>
    <td width="13%">-3/5</td>
    <td width="13%">0</td>
  </tr>
  <tr>
    <td width="12%">A<sub>5</sub></td>
    <td width="12%">0</td>
    <td width="12%">1</td>
    <td width="12%">0</td>
    <td width="13%">-1</td>
    <td width="13%">1</td>
    <td width="13%">1</td>
    <td width="13%">1</td>
  </tr>
  <tr>
    <td width="12%">　</td>
    <td width="12%">　</td>
    <td width="12%">12/5</td>
    <td width="12%">0</td>
    <td width="13%">0</td>
    <td width="13%">1/5</td>
    <td width="13%">2/5</td>
    <td width="13%">0</td>
  </tr>
</table>
</div>

<p>　</p>

<p>故解是 x<sub>1</sub>＝6/5，x<sub>2</sub>＝3/5，minz＝12/5<br>
</p>

<p><strong>例　</strong><br>
min z＝x<sub>1</sub> + 3x<sub>2</sub>＋2x<sub>3</sub><br>
6x<sub>1</sub>+ 3x<sub>2</sub>－4x<sub>3</sub>≤8<br>
－3x<sub>1</sub>－x<sub>2</sub>＋3x<sub>3</sub>≤－2<br>
x<sub>1</sub>，x<sub>2</sub> ,　x<sub>3</sub> ≥0<br>
<br>
<strong>解</strong><br>
引入松弛变量x<sub>4</sub>，x<sub>5</sub>，解答过程如下：<br>
</p>
<div align="left">

<table border="1" width="75%">
  <tr>
    <td width="12%">B</td>
    <td width="12%">C<sub>B</sub></td>
    <td width="12%">C</td>
    <td width="12%">1</td>
    <td width="13%">3</td>
    <td width="13%">2</td>
    <td width="13%">0</td>
    <td width="13%">0</td>
  </tr>
  <tr>
    <td width="12%">　</td>
    <td width="12%">　</td>
    <td width="12%">P<sub>0</sub></td>
    <td width="12%">P<sub>1</sub></td>
    <td width="13%">P<sub>2</sub></td>
    <td width="13%">P<sub>3</sub></td>
    <td width="13%">P<sub>4</sub></td>
    <td width="13%">P<sub>5</sub></td>
  </tr>
  <tr>
    <td width="12%">A<sub>4</sub></td>
    <td width="12%">0</td>
    <td width="12%">8</td>
    <td width="12%">6</td>
    <td width="13%">3</td>
    <td width="13%">4</td>
    <td width="13%">1</td>
    <td width="13%">0</td>
  </tr>
  <tr>
    <td width="12%">A<sub>5</sub></td>
    <td width="12%">0</td>
    <td width="12%">-2</td>
    <td width="12%">(-3)</td>
    <td width="13%">-1</td>
    <td width="13%">3</td>
    <td width="13%">0</td>
    <td width="13%">1</td>
  </tr>
  <tr>
    <td width="12%">　</td>
    <td width="12%">　</td>
    <td width="12%">　</td>
    <td width="12%">1</td>
    <td width="13%">3</td>
    <td width="13%">2</td>
    <td width="13%">0</td>
    <td width="13%">0</td>
  </tr>
</table>
</div>

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

<table border="1" width="75%">
  <tr>
    <td width="12%">B</td>
    <td width="12%">C<sub>B</sub></td>
    <td width="12%">C</td>
    <td width="12%">1</td>
    <td width="13%">3</td>
    <td width="13%">2</td>
    <td width="13%">0</td>
    <td width="13%">0</td>
  </tr>
  <tr>
    <td width="12%">　</td>
    <td width="12%">　</td>
    <td width="12%">P<sub>0</sub></td>
    <td width="12%">P<sub>1</sub></td>
    <td width="13%">P<sub>2</sub></td>
    <td width="13%">P<sub>3</sub></td>
    <td width="13%">P<sub>4</sub></td>
    <td width="13%">P<sub>5</sub></td>
  </tr>
  <tr>
    <td width="12%">A<sub>4</sub></td>
    <td width="12%">0</td>
    <td width="12%">4</td>
    <td width="12%">0</td>
    <td width="13%">1</td>
    <td width="13%">2</td>
    <td width="13%">1</td>
    <td width="13%">2</td>
  </tr>
  <tr>
    <td width="12%">A<sub>1</sub></td>
    <td width="12%">1</td>
    <td width="12%">2/3</td>
    <td width="12%">1</td>
    <td width="13%">1/3</td>
    <td width="13%">-1</td>
    <td width="13%">0</td>
    <td width="13%">-1/3</td>
  </tr>
  <tr>
    <td width="12%">　</td>
    <td width="12%">　</td>
    <td width="12%">2/3</td>
    <td width="12%">0</td>
    <td width="13%">8/3</td>
    <td width="13%">3</td>
    <td width="13%">0</td>
    <td width="13%">1/3</td>
  </tr>
</table>
</div>

<p><br>
</p>

<p>故解为x<sub>1</sub>＝2/3，x<sub>2</sub>＝x<sub>3</sub>＝0，min z＝2/3<br>
</p>
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