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<>model:<br>title :CUMCM2003B;<br>sets:<br>chan/1..10/:p,k,e,f;<br>xie/1..5/:q;<br>link(chan,xie):c,x,t,a,b,che;<br>endsets<br>data:<br>p=0.30 0.28 0.29 0.32 0.31 0.33 0.32 0.31 0.33 0.31;<br>q=12000 13000 13000 19000 13000;<br>c= 5.26 5.19 4.21 4.00 2.95 2.74 2.46 1.90 0.64 1.27<br> 1.90 0.99 1.90 1.13 1.27 2.25 1.48 2.04 3.09 3.51<br> 5.89 5.61 5.61 4.56 3.51 3.65 2.46 2.46 1.06 0.57<br> 0.64 1.76 1.27 1.83 2.74 2.60 4.21 3.72 5.05 6.10<br> 4.42 3.86 3.72 3.16 2.25 2.81 0.78 1.62 1.27 0.50;<br>k=9500 10500 10000 10500 11000 12500 10500 13000 13500 12500;<br>e=12500 11000 13500 10500 11500 13500 10500 11500 13500 12500;<br>enddata<br><a href="mailtmin=@sum(xie(i" target="_blank" >min=@sum(xie(i</A>):<br> @sum(chan(j):<br> f(j)*c(j,i)*x(j,i)));<br>!i到j一个周期(含装卸)所需时间;<br>t(j,i)=2*60*c(i,i)/28+5+3;<br>!i到j路线上最多能同时运行的卡车数;<br>a(j,i)=t(j,i)/5;<br>!i到j路线上一辆卡车在一个班次内最多可以运行的次数;<br>b(j,i)=60*8/t(j,i);<br>!道路能力约束;<br>@for(link(j,i):<br> x(j,i)<=a(j,i)*b(j,i)*154);<br>!装载能力约束;<br>@for(chan(j):<br> @sum(chan(i):<br> f(i)*x(j,i))<=60*8*154/5);<br>!卸载能力约束;<br>@for(xie(i):<br> @sum(chan(i):f(i)*x(j,i))<=60*8*154/3);<br>!铲位上岩石,矿石的储量约束;<br>@for(chan(i):<br> (x(i,1)+x(i,1)+x(i,5))<=k(i));<br>@for(chan(i):<br> (x(i,3)+x(i,4))<=e(i));<br>!卸点任务要求;<br>@for(xie(j)sum(chan(i):f(i)*x(i,j))>=q(j));<br>!卸点对矿石的品位要求;<br>@sum(chan (j):<br> x(1,j)*(p(j)-0.305) )<=0;<br>@sum(chan (j):<br> x(2,j)*(p(j)-0.305) )<=0;<br>@sum(chan (j):<br> x(5,j)*(p(j)-0.305) )<=0;<br>@sum(chan (j):<br> x(1,j)*(p(j)-0.285) )>=0;<br>@sum(chan (j):<br> x(2,j)*(p(j)-0.285) )>=0;<br>@sum(chan (j):<br> x(5,j)*(p(j)-0.285) )>=0; <br>!车辆具体分配;<br>@for(link(j,i):<br> che(j,i)=x(j,i)/b(j,i);<br>!卡车和;<br><a href="mailtchehe=@sum(link(j,i):che(j,i" target="_blank" >chehe=@sum(link(j,i):che(j,i</A>));<br>!车辆数目约束;<br>chehe<=20;<br>!铲车数量约束;<br>@sum(chan(j):f(i))<=7;<br>!f(i)0-1约束;<br>@for(chan(j): @bin(f(j));<br> end</P>
[此贴子已经被作者于2005-8-25 10:47:26编辑过]
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