|
|
< >各位lingo 高手帮忙看一下这个程序，</P>
<>这是我下载的程序，但是呢不能被执行，我自己也改不好！</P>
<>model:
title CUMCM-2003B-01;
sets:
cai / 1..10 /:crate,cnum,cy,ck,flag;
xie / 1 .. 5 /:xsubject,xnum;
link( xie,cai ):distance,lsubject,number,che,b;
endsets
data:
crate=30 28 29 32 31 33 32 31 33 31;
xsubject= 1.2 1.3 1.3 1.9 1.3;
distance= 5.26 5.19 4.21 4.00 2.95 2.74 2.46 1.90 0.64 1.27
1.90 0.99 1.90 1.13 1.27 2.25 1.48 2.04 3.09 3.51
5.89 5.61 5.61 4.56 3.51 3.65 2.46 2.46 1.06 0.57
0.64 1.76 1.27 1.83 2.74 2.60 4.21 3.72 5.05 6.10
4.42 3.86 3.72 3.16 2.25 2.81 0.78 1.62 1.27 0.50;
cy = 1.25 1.10 1.35 1.05 1.15 1.35 1.05 1.15 1.35 1.25;
ck = 0.95 1.05 1.00 1.05 1.10 1.25 1.05 1.30 1.35 1.25;
enddata
!目标函数;
<a href="http://www.shumo.com/bbs/mailtmin=@sum" target="_blank" >min=@sum</A>( cai (i):
@sum ( xie (j):
number (j,i)*154*distance (j,i)));
!max <a href="http://www.shumo.com/bbs/mailt=@sum(link(i,j):number(i,j" target="_blank" >=@sum(link(i,j):number(i,j</A>));
!max=xnum (3)+xnum (4)+xnum (1)+xnum (2)+xnum(5);
<a href="http://www.shumo.com/bbs/mailt!min=@sum" target="_blank" >!min=@sum</A>( cai (i):
! @sum ( xie (j):
! number (j,i)*154*distance (j,i)));
!xnum (1)+xnum (2)+xnum(5)=340;
!xnum (1)+xnum (2)+xnum(5)=341;
!xnum (3)=160;
!xnum (4)=160;
!卡车每一条路线上最多可以运行的次数;
@for (link (i,j):
b(i,j)=@floor((8*60-(@floor((distance(i,j)/28*60*2+3+5)/5)-1)*5)/(distance(i,j)/28*60*2+3+5)));</P>
<P>!b(i,j)=@floor(8*60/(distance(i,j)/28*60*2+3+5)));
!t(i,j)=@floor((distance(i,j)/28*60*2+3+5)/5);
!b(i,j)=@floor((8*60-(@floor((distance(i,j)/28*60*2+3+5)/5))*5)/(distance(i,j)/28*60*2+3+5)));
!每一条路线上的最大总车次的计算;
@for( link (i,j):
lsubject(i,j)=(@floor((distance(i,j)/28*60*2+3+5)/5))*b(i,j));
!计算各个铲位的总产量;
@for (cai(j):
cnum(j)=@sum(xie(i):number(i,j)));
!计算各个卸点的总产量;
@for (xie(i):
xnum(i)=@sum(cai(j):number(i,j)));
!道路能力约束;
@for (link (i,j):
number(i,j)<=lsubject(i,j));
!电铲能力约束;
@for (cai (j) :
cnum(j) <= flag(j)*8*60/5 );
!电铲数量约束 ---- added by Xie Jinxing, 2003-09-07;
@sum(cai(j): flag(j) ) <=7;
!卸点能力约束;
@for (xie (i):
xnum (i)<=8*20);
!铲位产量约束;
@for (cai (i):number(1,i)+number(2,i)+number(5,i)<=ck(i)*10000/154);
@for (cai (i):number(3,i)+number(4,i)<=cy(i)*10000/154);
!产量任务约束;
@for (xie (i):
xnum (i)>= xsubject (i)*10000/154);
!铁含量约束;
@sum(cai (j):
number(1,j)*(crate(j)-30.5) )<=0;
@sum(cai (j):
number(2,j)*(crate(j)-30.5) )<=0;
@sum(cai (j):
number(5,j)*(crate(j)-30.5) )<=0;
@sum(cai (j):
number(1,j)*(crate(j)-28.5) )>=0;
@sum(cai (j):
number(2,j)*(crate(j)-28.5) )>=0;
@sum(cai (j):
number(5,j)*(crate(j)-28.5) )>=0;
!关于车辆的具体分配;
@for (link (i,j):
che (i,j)=number (i,j)/b(i,j));
!各个路线所需卡车数简单加和;
<a href="http://www.shumo.com/bbs/mailthehe=@sum" target="_blank" >hehe=@sum</A> (link (i,j): che (i,j));
!整数约束;
@for (link (i,j): @gin(number (i,j)));
@for (cai (j): @bin(flag (j)));
!车辆能力约束;
hehe<=20;
<a href="http://www.shumo.com/bbs/mailtccnum=@sum(cai(j" target="_blank" >ccnum=@sum(cai(j</A>): cnum(j) );
end
</P> |
|
发表于 2004-9-11 07:57:31