化归为lingo来解决的问题
<p> </p><p align="center">化归为lingo来解决的问题</p>
<p> HeLU.EEI data/ 7. 27 28 29 30</p>
<p>哪些数学建模题目可以化归为lingo来解决,它们有哪些特点?这是即将参加数模竞赛的我们关心的问题.这份总结是书本知识和对真题分析的结果。</p>
<p align="center"><b> </b></p>
<p align="center"><b>一 、 下料问题</b></p>
<p> 此类问题包括一个标准材料长约束;各种分割方法的数量约束等;目标是最有分配。</p>
<p><b> </b></p>
<p><b>例一 </b>每根长19m,A B C D 分法各为 4m 5m 6m 8m,数量为50 10 20 15根。要求不同下料方式不超过3种。求最优下料。</p>
<p><b>解: </b>设标准长L=19m,分割长为 ,标准数量为 , 为第i种切割方法所切割钢管数量。第i种下料方法得到部件j的数目为 。</p>
<p> 编写代码:</p>
<p align="left">MODEL:</p>
<p align="left"> SETS:</p>
<p align="left"> SUBLEN/1..4/:L,D;</p>
<p align="left"> CUTMETHOD/1,2,3/:X;</p>
<p align="left"> LINKS(CUTMETHOD,SUBLEN):N;</p>
<p align="left"> ENDSETS</p>
<p align="left"> DATA:</p>
<p align="left"> L= 4 5 6 8;</p>
<p align="left"> D= 50 10 20 15;</p>
<p align="left"> ENDDATA</p>
<p align="left"> MIN=@SUM(CUTMETHOD:X);</p>
<p align="left"> @FOR(SUBLEN(J):@SUM(CUTMETHOD(I):N(I,J)*X(I))>=D(J));</p>
<p align="left"> @FOR(CUTMETHOD(I):@SUM(SUBLEN(J):L(J)*N(I,J))>15);</p>
<p align="left"> @FOR(CUTMETHOD(I):@SUM(SUBLEN(J):L(J)*N(I,J))<19);</p>
<p align="left">@FOR(CUTMETHOD:@GIN(X));</p>
<p align="left">@FOR(LINKS:@GIN(N));</p>
<p align="left">END</p>
<p>结果</p>
<p align="left"> Local optimal solution found at iteration: 3010</p>
<p align="left"> Objective value: 28.00000</p>
<p align="left"> N( 1, 1) 3.000000 0.000000</p>
<p align="left"> N( 1, 2) 0.000000 0.000000</p>
<p align="left"> N( 1, 3) 1.000000 0.000000</p>
<p align="left"> N( 1, 4) 0.000000 0.000000</p>
<p align="left"> N( 2, 1) 2.000000 0.000000</p>
<p align="left"> N( 2, 2) 1.000000 0.000000</p>
<p align="left"> N( 2, 3) 1.000000 0.000000</p>
<p align="left"> N( 2, 4) 0.000000 0.000000</p>
<p align="left"> N( 3, 1) 0.000000 0.000000</p>
<p align="left"> N( 3, 2) 0.000000 0.000000</p>
<p align="left"> N( 3, 3) 0.000000 0.000000</p>
<p align="left"> N( 3, 4) 2.000000 0.000000</p>
<p align="left"> 该问题小结:此类问题可以直接化归为lingo解。</p>
<p align="left">练习: p67 11 题</p>
<p align="left"> </p>
<p><b>解:</b>标准长L=5500mm,分割长为 ,标准数量为 , 为第i种切割方法所切割钢管数量。第i种下料方法得到部件j的数目为 。</p>
<p>本题下料方法可以穷尽,但不手工穷尽。设共有m种下料方法</p>
<p align="left">得到m值的问题是背包问题,同时得到 的情况。</p>
<p align="left">若采取例题中采用的方法,编写程序:</p>
<p align="left">MODEL:</p>
<p align="left"> SETS:</p>
<p align="left"> SUBLEN/1..4/:L,D;</p>
<p align="left"> CUTMETHOD/1..11/:X;</p>
<p align="left"> LINKS(CUTMETHOD,SUBLEN):N;</p>
<p align="left"> ENDSETS</p>
<p align="left"> DATA:</p>
<p align="left"> L= 2640 1651 1770 1440;</p>
<p align="left"> D= 80 350 420 200;</p>
<p align="left"> ENDDATA</p>
<p align="left"> MIN=@SUM(CUTMETHOD:X);</p>
<p align="left"> @FOR(SUBLEN(J):@SUM(CUTMETHOD(I):N(I,J)*X(I))>=D(J));</p>
<p align="left"> @FOR(CUTMETHOD(I):@SUM(SUBLEN(J):L(J)*N(I,J))>4060);</p>
<p align="left"> @FOR(CUTMETHOD(I):@SUM(SUBLEN(J):L(J)*N(I,J))<5500);</p>
<p align="left">@FOR(CUTMETHOD:@GIN(X));</p>
<p align="left">@FOR(LINKS:@GIN(N));</p>
<p align="left">END</p>
<p align="left">运算35小时后得到可行解:</p>
<p align="left"> </p>
<p align="center"></p>
<p align="left"> </p>
<p align="left"> </p>
<p align="left"> Variable Value Reduced Cost</p>
<p align="left"> </p>
<p align="left"> N( 1, 1) 2.000000 0.000000</p>
<p align="left"> N( 1, 2) 0.000000 0.000000</p>
<p align="left"> N( 1, 3) 0.000000 0.000000</p>
<p align="left"> N( 1, 4) 0.000000 0.000000</p>
<p align="left"> N( 2, 1) 0.000000 0.000000</p>
<p align="left"> N( 2, 2) 0.000000 0.000000</p>
<p align="left"> N( 2, 3) 1.000000 0.000000</p>
<p align="left"> N( 2, 4) 2.000000 0.000000</p>
<p align="left"> N( 3, 1) 0.000000 0.000000</p>
<p align="left"> N( 3, 2) 0.000000 0.000000</p>
<p align="left"> N( 3, 3) 3.000000 0.000000</p>
<p align="left"> N( 3, 4) 0.000000 0.000000</p>
<p align="left"> N( 4, 1) 0.000000 0.000000</p>
<p align="left"> N( 4, 2) 2.000000 0.000000</p>
<p align="left"> N( 4, 3) 1.000000 0.000000</p>
<p align="left"> N( 4, 4) 0.000000 0.000000</p>
<p align="left"> N( 5, 1) 0.000000 0.000000</p>
<p align="left"> N( 5, 2) 2.000000 0.000000</p>
<p align="left"> N( 5, 3) 1.000000 0.000000</p>
<p align="left"> N( 5, 4) 0.000000 0.000000</p>
<p align="left"> N( 6, 1) 0.000000 0.000000</p>
<p align="left"> N( 6, 2) 1.000000 0.000000</p>
<p align="left"> N( 6, 3) 2.000000 0.000000</p>
<p align="left"> N( 6, 4) 0.000000 0.000000</p>
<p align="left"> N( 7, 1) 0.000000 0.000000</p>
<p align="left"> N( 7, 2) 2.000000 0.000000</p>
<p align="left"> N( 7, 3) 1.000000 0.000000</p>
<p align="left"> N( 7, 4) 0.000000 0.000000</p>
<p align="left"> N( 8, 1) 0.000000 0.000000</p>
<p align="left"> N( 8, 2) 2.000000 0.000000</p>
<p align="left"> N( 8, 3) 1.000000 0.000000</p>
<p align="left"> N( 8, 4) 0.000000 0.000000</p>
<p align="left"> N( 9, 1) 0.000000 0.000000</p>
<p align="left"> N( 9, 2) 1.000000 0.000000</p>
<p align="left"> N( 9, 3) 1.000000 0.000000</p>
<p align="left"> N( 9, 4) 1.000000 0.000000</p>
<p align="left"> N( 10, 1) 0.000000 0.000000</p>
<p align="left"> N( 10, 2) 0.000000 0.000000</p>
<p align="left"> N( 10, 3) 2.000000 0.000000</p>
<p align="left"> N( 10, 4) 1.000000 0.000000</p>
<p align="left"> N( 11, 1) 2.000000 0.000000</p>
<p align="left"> N( 11, 2) 0.000000 0.000000</p>
<p align="left"> N( 11, 3) 0.000000 0.000000</p>
<p align="left"> N( 11, 4) 0.000000 0.000000</p>
<p align="left"> </p>
<p align="left"> Row Slack or Surplus Dual Price</p>
<p align="left"> 1 364.0000 -1.000000</p>
<p align="left"> </p>
<p align="left">这个显然是不行的。在解决本类题目的时候,需要预先将所有可能出现的情况都穷举出来,然后在这个基础之上做线性规划。当然,我们也有不用穷举的方法:</p>
<p align="left">其实本题在求有多少种划钢材的方式是一个典型的找零钱问题,它的schema程序为:</p>
<p align="left"> </p>
<p align="left">(define (first-denomination kinds-of-iron)</p>
<p align="left"> (cond ((= kinds-of-iron 1)1440)</p>
<p align="left"> ((= kinds-of-iron 2)1651)</p>
<p align="left"> ((= kinds-of-iron 3)1770)</p>
<p align="left"> ((= kinds-of-iron 4)2640)</p>
<p align="left">))</p>
<p align="left">(define (cc amount kinds-of-iron)</p>
<p align="left"> (cond((< amount 1440)1)</p>
<p align="left"> ((or(< amount 0)(= kinds-of-iron 0))0)</p>
<p align="left"> (else (+ (cc amount(- kinds-of-iron 1))</p>
<p align="left"> (cc(- amount(first-denomination kinds-of-iron))</p>
<p align="left"> kinds-of-iron)))))</p>
<p align="left">(define(count-type amount)</p>
<p align="left">(cc amount 4))</p>
<p align="left"> </p>
<p align="left">结果为14.但是现在我还写不出来所有具体的情况,只能算出有多少情况(这对于非线性优化是有用的)。解决题目,还是穷举的方法:</p>
<p align="left"> </p>
<div align="center">
<table cellspacing="0" cellpadding="0" width="340" border="0">
<tbody>
<tr>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="left"> </p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">1</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">2</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">3</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">4</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">5</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">6</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">7</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">8</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">9</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">10</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">11</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">12</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">13</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">14</p></td></tr>
<tr>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="left">A</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">2</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">1</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">1</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">1</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">0</p></td></tr>
<tr>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="left">B</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">1</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">1</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">2</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">1</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">3</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">2</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">1</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">0</p></td></tr>
<tr>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="left">C</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">1</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">3</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">2</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">2</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">1</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">1</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">1</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">0</p></td></tr>
<tr>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="left">D</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">1</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">1</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="20">
<p align="right">1</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">2</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">0</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">1</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">2</p></td>
<td valign="bottom" noWrap="noWrap" width="28">
<p align="right">3</p></td></tr></tbody></table></div>
<p align="left"> </p>
<p align="left">带入lingo,解决题目</p>
<p align="left">MODEL:</p>
<p align="left"> SETS:</p>
<p align="left"> irons/1,2..14/:x;</p>
<p align="left"> ENDSETS</p>
<p align="left"> @FOR(IRONS:@GIN(X));</p>
<p align="left"> MIN=X1+X2+X3+X4+X5+X6+X7+X8+X9+X10+X11+X12+X13+X14;</p>
<p align="left"> 2*X1+X2+X3+X4>=2640;</p>
<p align="left"> X2+X6+2*X8+X9+3*X11+2*X12+X13>=1651;</p>
<p align="left"> X3+3*X5+2*X6+2*X7+X8+X9+X10>=1770;</p>
<p align="left"> X4+X7+X9+2*X10+X12+2*X13>=1440;</p>
<p align="left">END</p>
<p align="left"> </p>
<p align="left"> </p>
<p align="left">结果为:</p>
<p align="left"> Global optimal solution found at iteration: 2</p>
<p align="left"> Objective value: 2940.333</p>
<p align="left"> </p>
<p align="left"> </p>
<p align="left"> Variable Value Reduced Cost</p>
<p align="left"> X1 1320.000 0.000000</p>
<p align="left"> X2 0.000000 0.1666667</p>
<p align="left"> X3 0.000000 0.1666667</p>
<p align="left"> X4 0.000000 0.1666667</p>
<p align="left"> X5 590.0000 0.000000</p>
<p align="left"> X6 0.000000 0.000000</p>
<p align="left"> X7 0.000000 0.000000</p>
<p align="left"> X8 0.000000 0.000000</p>
<p align="left"> X9 0.000000 0.000000</p>
<p align="left"> X10 0.000000 0.000000</p>
<p align="left"> X11 310.3333 0.000000</p>
<p align="left"> X12 0.000000 0.000000</p>
<p align="left"> X13 720.0000 0.000000</p>
<p align="left"> X14 0.000000 1.000000</p>
<p align="left"> </p>
<p align="left"> </p>
<p align="center"><b>二 、 配料问题</b></p>
<p align="left"> 该问题是一个整形规划,所以具有独特的特点。</p>
<p align="left"><b>例二</b><b> </b><b>书本例</b><b>1.7.3</b><b> p53</b></p>
<p align="left"><b>解略</b><b>.</b>问题的关键就在于整形规划的特点。但是在大数据量不可手工穷尽的时候怎样解决,这是就要考虑的问题。怎样将数据链接过来?是通过数据库还是其他东西。</p>
<p align="left"> </p>
<p align="left"><b> </b><b>三</b><b> </b><b>、指派问题</b><b>&</b><b>装箱问题</b><b></b></p>
<p align="left"><b> </b></p>
<p align="left">此类问题有一个显著的特点就是都为0-1规划,比如一个人只能做一件工作;同时一个工作只能由一个人来完成。</p>
<p align="left"><b>例三</b> 分配给甲乙丙丁戊去完成A B C D E五项工作,每人完成一项,每项工作一人完成。五人完成认为所需时间如表,做出任务分配时间最少优化。</p>
<p align="left"> </p>
<div align="center">
<table cellspacing="0" cellpadding="0" width="432" border="0">
<tbody>
<tr>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="left"> </p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="left">A</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="left">B</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="left">C</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="left">D</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="left">E</p></td></tr>
<tr>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="left">甲</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">8</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">6</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">10</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">9</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">12</p></td></tr>
<tr>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="left">乙</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">9</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">12</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">7</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">11</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">9</p></td></tr>
<tr>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="left">丙</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">7</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">4</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">3</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">5</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">8</p></td></tr>
<tr>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="left">丁</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">9</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">5</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">8</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">11</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">8</p></td></tr>
<tr>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="left">戊</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">4</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">6</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">7</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">5</p></td>
<td valign="bottom" noWrap="noWrap" width="72">
<p align="right">11</p></td></tr></tbody></table></div>
<p align="left"> </p>
<p align="left">解:MODEL:</p>
<p align="left"> SETS:</p>
<p align="left"> WORKERS/W1..W5/;</p>
<p align="left"> JOBS/J1..J5/;</p>
<p align="left"> LINKS(WORKERS,JOBS):D,X;</p>
<p align="left"> </p>
<p align="left"> ENDSETS</p>
<p align="left"> DATA:</p>
<p align="left"> D=</p>
<p align="left"> </p>
<div align="center">
<table cellspacing="0" cellpadding="0" border="0">
<tbody>
<tr>
<td valign="top" width="72">
<p align="left">8</p></td>
<td valign="top" width="72">
<p align="left">6</p></td>
<td valign="top" width="72">
<p align="left">10</p></td>
<td valign="top" width="72">
<p align="left">9</p></td>
<td valign="top" width="72">
<p align="left">12</p></td></tr>
<tr>
<td valign="top" width="72">
<p align="left">9</p></td>
<td valign="top" width="72">
<p align="left">12</p></td>
<td valign="top" width="72">
<p align="left">7</p></td>
<td valign="top" width="72">
<p align="left">11</p></td>
<td valign="top" width="72">
<p align="left">9</p></td></tr>
<tr>
<td valign="top" width="72">
<p align="left">7</p></td>
<td valign="top" width="72">
<p align="left">4</p></td>
<td valign="top" width="72">
<p align="left">3</p></td>
<td valign="top" width="72">
<p align="left">5</p></td>
<td valign="top" width="72">
<p align="left">8</p></td></tr>
<tr>
<td valign="top" width="72">
<p align="left">9</p></td>
<td valign="top" width="72">
<p align="left">5</p></td>
<td valign="top" width="72">
<p align="left">8</p></td>
<td valign="top" width="72">
<p align="left">11</p></td>
<td valign="top" width="72">
<p align="left">8</p></td></tr>
<tr>
<td valign="top" width="72">
<p align="left">4</p></td>
<td valign="top" width="72">
<p align="left">6</p></td>
<td valign="top" width="72">
<p align="left">7</p></td>
<td valign="top" width="72">
<p align="left">5</p></td>
<td valign="top" width="72">
<p align="left">11</p></td></tr></tbody></table></div>
<p align="left">;</p>
<p align="left"> ENDDATA</p>
<p align="left"> MIN=@SUM(LINKS:D*X);</p>
<p align="left"> @FOR(WORKERS(I):@SUM(JOBS(J):X(I,J))=1);</p>
<p align="left"> @FOR(JOBS(J):@SUM(WORKERS(I):X(I,J))=1);</p>
<p align="left"> @FOR(LINKS:@BIN(X));</p>
<p align="left">END</p>
<p align="left">结果</p>
<p align="left"> Global optimal solution found at iteration: 0</p>
<p align="left"> Objective value: 30.00000</p>
<p align="left"> </p>
<p align="left"> </p>
<p align="left"> Variable Value Reduced Cost</p>
<p align="left"> </p>
<p align="left"> X( W1, J1) 1.000000 8.000000</p>
<p align="left"> X( W1, J2) 0.000000 6.000000</p>
<p align="left"> X( W1, J3) 0.000000 10.00000</p>
<p align="left"> X( W1, J4) 0.000000 9.000000</p>
<p align="left"> X( W1, J5) 0.000000 12.00000</p>
<p align="left"> X( W2, J1) 0.000000 9.000000</p>
<p align="left"> X( W2, J2) 0.000000 12.00000</p>
<p align="left"> X( W2, J3) 0.000000 7.000000</p>
<p align="left"> X( W2, J4) 0.000000 11.00000</p>
<p align="left"> X( W2, J5) 1.000000 9.000000</p>
<p align="left"> X( W3, J1) 0.000000 7.000000</p>
<p align="left"> X( W3, J2) 0.000000 4.000000</p>
<p align="left"> X( W3, J3) 1.000000 3.000000</p>
<p align="left"> X( W3, J4) 0.000000 5.000000</p>
<p align="left"> X( W3, J5) 0.000000 8.000000</p>
<p align="left"> X( W4, J1) 0.000000 9.000000</p>
<p align="left"> X( W4, J2) 1.000000 5.000000</p>
<p align="left"> X( W4, J3) 0.000000 8.000000</p>
<p align="left"> X( W4, J4) 0.000000 11.00000</p>
<p align="left"> X( W4, J5) 0.000000 8.000000</p>
<p align="left"> X( W5, J1) 0.000000 4.000000</p>
<p align="left"> X( W5, J2) 0.000000 6.000000</p>
<p align="left"> X( W5, J3) 0.000000 7.000000</p>
<p align="left"> X( W5, J4) 1.000000 5.000000</p>
<p align="left"> X( W5, J5) 0.000000 11.00000</p>
<p align="left"> </p>
<p align="left"> 首先可以看到,这里直接用excel作为了输入。当然也可以输入回到excel中去。</p>
<p align="left"> </p>
<p align="left"> 用lingo解决装箱子问题的时候有个技巧,不过不是很有用。</p>
<p align="left"> </p>
<p align="left"> <b> </b><b>四</b><b> </b><b>、最短路径</b><b></b></p>
<p align="left"><b> </b></p>
<p align="left"> </p>
<p align="left"><b>例五</b><b> </b>如图求各点到T的最短路</p>
<p><br/> </p>
<p>线性规划的方法解决最短路径问题真是有想法。在计算机算法中有许多为了简化计算而设计的算法,大都比较复杂;然而如果面对的问题是数学,那么重点就不一样了:数学的重点是解决问题,算法的效率是其次的(除非题目本身就是效率题),线性规划几乎就是穷举所有的情况,所以线性规划很大程度上具有通用性。</p>
<p><br/> </p>
<p>编写代码求解:</p>
<p>MODEL:</p>
<p>SETS: ! CITIES表示由1~9组成的集合,是一个基本集合;</p>
<p> CITIES /1..9/: L; !属性<i>L</i>(i)表示城市<i>i</i>到城市1的最优行驶路线的路长;</p>
<p> ROADS(CITIES, CITIES)/ ! ROADS表示网络中的弧,是由CITIES派生的集合;</p>
<p> 9,6 9,7 9,8 !由于并非所有城市间都有道路直接连接,所以将弧具体列出;</p>
<p> 6,4 6,5 7,4 7,5 8,4 8,5</p>
<p> 4,2 4,3 5,2 5,3</p>
<p> 2,1 3,1/: D; !属性D( <i>i</i>, <i>j</i>) 是城市i到j的直接距离(已知);</p>
<p>ENDSETS</p>
<p><br/> </p>
<p>DATA:</p>
<p> D = ! D赋值的顺序对应于ROADS中的弧的顺序;</p>
<p> 6 3 3</p>
<p> 6 5 8 6 7 4</p>
<p> 6 7 8 9</p>
<p> 5 6;</p>
<p>ENDDATA</p>
<p><br/> </p>
<p>L(1) = 0; !边界条件;</p>
<p>@FOR( CITIES( i)| i #GT# 1: !集合循环语句, #GT#表示逻辑关系"大于";</p>
<p> L( i) = @MIN( ROADS( i, j): D( i, j) + L( j)) !这就是动态规划基本方程;</p>
<p> );</p>
<p>END</p>
<p align="left"> </p>
<p align="left">结果为:</p>
<p align="left"> </p>
<p align="left"> Variable Value</p>
<p align="left"> L( 1) 0.000000</p>
<p align="left"> L( 2) 5.000000</p>
<p align="left"> L( 3) 6.000000</p>
<p align="left"> L( 4) 11.00000</p>
<p align="left"> L( 5) 13.00000</p>
<p align="left"> L( 6) 17.00000</p>
<p align="left"> L( 7) 19.00000</p>
<p align="left"> L( 8) 17.00000</p>
<p align="left"> L( 9) 20.00000</p>
<p align="left"> </p>
<p>可以看到,最有结果为20.当然了,这道题目比较简单,我们还需要更多的算法来验算这一个结果。</p>
[此贴子已经被作者于2008-7-31 16:31:29编辑过] 你好 啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊 你好 啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊 zanyige ,xiexie luguo luguo 最短路径那个代码只是会求出单线路程吧?怎么才可以把涉及到的点都连起来呢?
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