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数学中国»论坛 › 数学中国论坛 › 数学建模 › 谁能帮我看看这个lingo程序哪有问题?
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谁能帮我看看这个lingo程序哪有问题?

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angellys
发表于 2007-11-11 02:01 | 显示全部楼层 |阅读模式
sets:
   shangxian/1..8/:l;
   xiangmu/1..8/:a;
   nianshu/1..5/:b,w,p1,p2,p3,p4,p5,p6,p7,p8,q3,q4,q5,q6,o5,o6,o8;
   niantouzishangxian/1..5/:m;
   niantouzizhongshu/1..5/:n;
   link(nianshu,xiangmu):x,c;
endsets
data:
     p1=0.1474,0.1482,0.1461,0.1457,0.1491;
   p2=0.1718,0.169,0.1659,0.1689,0.1704;
   p3=0.3618,0.3776,0.3365,0.3639,0.3475;
   p4=0.3757,0.3894,0.4005,0.3919,0.3814;
    p5=0.7267,0.9288,0.8719,0.8035,0.8416;
    p6=1.2738,1.255,1.2313,1.8816,1.162;
    p7=3.9696,10.7582,7.3639,9.0611,8.2125;
    p8=1.3197,1.3729,1.1813,1.53,1.8892;
    q3=0.5027,0.482,0.4745,0.5047,0.5146;
    q4=0.4628,0.4686,0.4619,0.4323,0.4366;
    q5=1.1307,1.1495,0.9213,0.9993,1.1135;
    q6=0.8444,1.0044,1.1981,1.1119,1.5098;
    o5=1.0299,1.0592,1.2352,1.2926,0.9166;
    o6=0.7095,0.7385,0.6545,0.6357,0.6456;
    o8=0.7301,0.32,0.6532,0.4996,0.5507;
   
   l=60000,60000,35000,30000,30000,40000,30000,30000;
   b=1,2,3,4,5;
   a=1,2,3,4,5,6,7,8;
enddata
[obj]max=@sum(link(i,j):x(i,j)*c(i,j));
@for(link(i,j):x(i,1)<=l(1));
@for(link(i,j):x(i,2)<=l(2));
@for(link(i,j)|i&#35;gt&#35;1:x(i,3)+x(i-1,3)<=l(3));
@for(link(i,j)|i&#35;gt&#35;1:x(i,4)+x(i-1,4)<=l(4));
@for(link(i,j)|i&#35;gt&#35;2:x(i,5)+x(i-1,5)+x(i-2,5)<=l(5));
@for(link(i,j)|i&#35;gt&#35;2:x(i,6)+x(i-1,6)+x(i-2,6)<=l(6));
@for(nianshu(i):c(i,1)=p1(i);c(i,2)=p2(i);c(i,7)=p7(i););
@for(nianshu(i):c(i,3)=@if(x(i,3)*x(i,4)&#35;eq&#35;0,p3(i),q3(i)));
@for(nianshu(i):c(i,4)=@if(x(i,3)*x(i,4)&#35;eq&#35;0,p4(i),q4(i)));
@for(nianshu(i):c(i,5)=@if(x(i,5)*x(i,6)&#35;eq&#35;0 ,p5(i),@if(x(i,8)&#35;ne&#35;0,o5(i),q5(i))));
@for(nianshu(i):c(i,6)=@if(x(i,5)*x(i,6)&#35;eq&#35;0 ,p6(i),@if(x(i,8)&#35;ne&#35;0,o6(i),q6(i))));
@for(nianshu(i):c(i,8)=@if(x(i,5)*x(i,6)*x(i,8)&#35;eq&#35;0,p8(i),o8));
x(2,7)<=l(7);
x(3,8)<=l(8);
x(5,3)=0;x(5,4)=0;x(4,5)=0;x(5,5)=0;x(4,6)=0;x(5,6)=0;
x(1,7)=0;x(3,7)=0;x(4,7)=0;x(5,7)=0;x(1,8)=0;x(2,8)=0;x(4,8)=0;x(5,8)=0;
x(1,5)=500*k1;
x(2,5)=500*k2;
x(3,5)=500*k3;
@for(nianshu(i)sum(xiangmu(j):x(i,j))=n(i););
@for(niantouzizhongshu(i):n(i)<=m(i));
@for(nianshu(i):w(i)=@if(x(i,1)&#35;gt&#35;20000,0.01*x(i,1),0););
m(1)=200000+w(1);
m(2)=m(1)-n(1)+x(1,1)*(1+c(1,1))+x(1,2)*(1+c(1,2))+w(2);
m(3)=m(2)-n(2)+x(2,1)*(1+c(2,1))+x(2,2)*(1+c(2,2))+x(1,3)*(1+c(1,3))+x(1,4)*(1+c(1,4))+w(3);
m(4)=m(3)-n(3)+x(3,1)*(1+c(3,1))+x(3,2)*(1+c(3,2))+x(2,3)*(1+c(2,3))+x(2,4)*(1+c(2,4))+x(1,5)*(1+c(1,5))+x(1,6)*(1+c(1,6))+w(4);
m(5)=m(4)-n(4)+x(4,1)*(1+c(4,1))+x(4,2)*(1+c(4,2))+x(3,3)*(1+c(3,3))+x(3,4)*(1+c(3,4))+x(2,5)*(1+c(2,5))+x(2,6)*(1+c(2,6))+w(5);
@gin(k1);
@gin(k2);
@gin(k3);
end
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发表于 2007-12-8 18:36 | 显示全部楼层

谁能帮我看看这个lingo程序哪有问题?

这个我不怎么懂
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发表于 2007-12-8 18:37 | 显示全部楼层

谁能帮我看看这个lingo程序哪有问题?

不错的帖子支持一下楼主
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发表于 2007-12-9 15:44 | 显示全部楼层

谁能帮我看看这个lingo程序哪有问题?

没看出来有问题啊
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\frac{\square}{\square}\sqrt{\square}\square_{\baguet}^{\baguet}\overarc{\square}\ \dot{\baguet}\left(\square\right)\binom{\square}{\square}\begin{cases}\square\\\square\end{cases}\ \begin{bmatrix}\square&\square\\\square&\square\end{bmatrix}\to\Rightarrow\mapsto\alpha\ \theta\ \pi\times\div\pm\because\angle\ \infty
\frac{\square}{\square}\sqrt{\square}\sqrt[\baguet]{\square}\square_{\baguet}\square^{\baguet}\square_{\baguet}^{\baguet}\sum_{\baguet}^{\baguet}\prod_{\baguet}^{\baguet}\coprod_{\baguet}^{\baguet}\int_{\baguet}^{\baguet}\lim_{\baguet}\lim_{\baguet}^{\baguet}\bigcup_{\baguet}^{\baguet}\bigcap_{\baguet}^{\baguet}\bigwedge_{\baguet}^{\baguet}\bigvee_{\baguet}^{\baguet}
\underline{\square}\overline{\square}\overrightarrow{\square}\overleftarrow{\square}\overleftrightarrow{\square}\underrightarrow{\square}\underleftarrow{\square}\underleftrightarrow{\square}\dot{\baguet}\hat{\baguet}\vec{\baguet}\tilde{\baguet}
\left(\square\right)\left[\square\right]\left\{\square\right\}\left|\square\right|\left\langle\square\right\rangle\left\lVert\square\right\rVert\left\lfloor\square\right\rfloor\left\lceil\square\right\rceil\binom{\square}{\square}\boxed{\square}
\begin{cases}\square\\\square\end{cases}\begin{matrix}\square&\square\\\square&\square\end{matrix}\begin{pmatrix}\square&\square\\\square&\square\end{pmatrix}\begin{bmatrix}\square&\square\\\square&\square\end{bmatrix}\begin{Bmatrix}\square&\square\\\square&\square\end{Bmatrix}\begin{vmatrix}\square&\square\\\square&\square\end{vmatrix}\begin{Vmatrix}\square&\square\\\square&\square\end{Vmatrix}\begin{array}{l|l}\square&\square\\\hline\square&\square\end{array}
\to\gets\leftrightarrow\nearrow\searrow\downarrow\uparrow\updownarrow\swarrow\nwarrow\Leftarrow\Rightarrow\Leftrightarrow\rightharpoonup\rightharpoondown\impliedby\implies\Longleftrightarrow\leftharpoonup\leftharpoondown\longleftarrow\longrightarrow\longleftrightarrow\Uparrow\Downarrow\Updownarrow\hookleftarrow\hookrightarrow\mapsto
\alpha\beta\gamma\Gamma\delta\Delta\epsilon\varepsilon\zeta\eta\theta\Theta\iota\kappa\varkappa\lambda\Lambda\mu\nu\xi\Xi\pi\Pi\varpi\rho\varrho\sigma\Sigma\tau\upsilon\Upsilon\phi\Phi\varphi\chi\psi\Psi\omega\Omega\digamma\vartheta\varsigma\mathbb{C}\mathbb{H}\mathbb{N}\mathbb{P}\mathbb{Q}\mathbb{R}\mathbb{Z}\Re\Im\aleph\partial\nabla
\times\cdot\ast\div\pm\mp\circ\backslash\oplus\ominus\otimes\odot\bullet\varnothing\neq\equiv\not\equiv\sim\approx\simeq\cong\geq\leq\ll\gg\succ\prec\in\ni\cup\cap\subset\supset\not\subset\not\supset\notin\not\ni\subseteq\supseteq\nsubseteq\nsupseteq\sqsubset\sqsupset\sqsubseteq\sqsupseteq\sqcap\sqcup\wedge\vee\neg\forall\exists\nexists\uplus\bigsqcup\bigodot\bigotimes\bigoplus\biguplus\bigcap\bigcup\bigvee\bigwedge
\because\therefore\angle\parallel\perp\top\nparallel\measuredangle\sphericalangle\diamond\diamondsuit\doteq\propto\infty\bowtie\square\smile\frown\bigtriangledown\triangle\triangleleft\triangleright\bigcirc \wr\amalg\models\preceq\mid\nmid\vdash\dashv\nless\ngtr\ldots\cdots\vdots\ddots\surd\ell\flat\sharp\natural\wp\clubsuit\heartsuit\spadesuit\oint\lfloor\rfloor\lceil\rceil\lbrace\rbrace\lbrack\rbrack\vert\hbar\aleph\dagger\ddagger

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