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群组: LINGO 群组: 环境-数学-建模群 |
我们暑期建模培训,老师布置了一个作业,是2003年b题露天矿生产的车辆安排第一问。我们的思路是先用“贪心算法”选出六个最优点(铲位),然后再分别选取剩下的四个铲位作为第七个点,对其分别进行lingo编程,由于小弟初学,lingo的一些基础都没掌握,希望各位朋友老师多多指教(第一次在咱们论坛发帖。。呵呵)。话不多说,看代码(附2003b题)——- model:
- title 2003B露天矿生产的车辆安排;
- sets:
- chan /1,2,3,4,5,6,7/:ks,ys,chandi,grade;
- xie /1,2,3,4,5/:need,xiedian;
- link(chan,xie):d,x,k,n,t;
- endsets
- data:
- ks= 0.95 1.05 1.00 1.05 1.10 1.35 1.25;
- ys= 1.25 1.10 1.35 1.05 1.15 1.35 1.25;
- grade=30 28 29 32 31 33 31;
- d= 5.26 5.19 4.21 4.00 2.95 0.64 1.27
- 1.90 0.99 1.90 1.13 1.27 3.09 3.51
- 5.89 5.61 5.61 4.56 3.51 1.06 0.57
- 0.64 1.76 1.27 1.83 2.74 5.05 6.10
- 4.42 3.86 3.72 3.16 2.25 1.27 0.50;
- need=1.2 1.3 1.3 1.9 1.3 ;
- enddata
- !目标函数;
- min=@sum( chan (i):
- @sum ( xie (j):
- x(i,j)*154*d(i,j)));
- !卸点的产量要求;
- @for (xie(j):
- xiedian(j)>=@sum(chan(i):154*x(i,j)));
- !卸点的品位要求;
- @sum(chan (j):
- x(1,j)*(grade(j)-30.5) )<=0;
- @sum(chan (j):
- x(2,j)*(grade(j)-30.5) )<=0;
- @sum(chan (j):
- x(5,j)*(grade(j)-30.5) )<=0;
- @sum(chan (j):
- x(1,j)*(grade(j)-28.5) )>=0;
- @sum(chan (j):
- x(2,j)*(grade(j)-28.5) )>=0;
- @sum(chan (j):
- x(5,j)*(grade(j)-28.5) )>=0;
- !铲位允许最大开采量;
- !矿石;
- @for (chan (i): x(1,i)+x(2,i)+x(5,i)<=ks(i)*10000/154);
- !岩石;
- @for (chan (i): x(3,i)+x(4,i)<=ks(i)*10000/154);
- !一个班次最多装载次数小于等于480除以5;
- @sum(xie(i):
- x(1,i))<=96;
- @sum(xie(i):
- x(2,i))<=96;
- @sum(xie(i):
- x(3,i))<=96;
- @sum(xie(i):
- x(4,i))<=96;
- @sum(xie(i):
- x(5,i))<=96;
- @sum(xie(i):
- x(6,i))<=96;
- @sum(xie(i):
- x(7,i))<=96;
- !一个班次最多缷载次数小于等于480除以3;
- @sum(chan(j):
- x(j,1))<=160;
- @sum(chan(j):
- x(j,2))<=160;
- @sum(chan(j):
- x(j,3))<=160;
- @sum(chan(j):
- x(j,4))<=160;
- @sum(chan(j):
- x(j,5))<=160;
- !实际用去卡车数;
- (@sum(link (i,j):(3+5+120*d(i,j)/28)/480))<=20;
- !对实际作去卡车数取整;
- !一辆卡车一班次往返次数;
- @for (link(i,j):k(i,j)=480/(3+5+120*d(i,j)));
- !每条线路上卡车数;
- @for (link(i,j):n(i,j)=@floor(link(i,j):x(i,j)/link(i,j):k(i,j)));
- !从i到j的周期;
- @for (link(i,j):t(i,j)=(3+5+120*d(i,j)/28));
- !同一路线不发生冲突;
- @for (link(i,j):n(i,j)<=@floor(link(i,j):t(i,j)/5));
- end
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