运筹学lingo软件解生产任务分配

某构件公司有四个构件厂，现接受五个企业预应力梁和预制桩的订货，订货量分别为2260件和3270件，单价分别是1万元和1.2万元。各构件厂生产能力、单位成本、材料单耗等资料见表12，各公司拥有的材料见表13，订货企业与各构件厂的距离见表14，预应力梁单件重5吨，预制桩单件重3吨，每吨公里运费1元，按公司利润最大建立并求解模型。
表12 各构件厂生产能力、单位成本、材料单耗资料
项目

企业 生产能力（件） 单位成本（元） 材料单耗（㎏）
预应力梁 预制桩 预应力梁 预制桩 水泥 钢材
预应力梁 预制桩 预应力梁 预制桩
1 800 1000 3800 6400 4000 2000 1000 600
2 600 700 4000 6380 4050 2050 1050 510
3 450 800 4100 6500 4050 2060 1030 510
4 550 1000 3950 6600 4000 1990 990 515
合计 2400 3500 — — — — — —
表13 各构件厂拥有的材料数量
企业
材料 1 2 3 4 合计
水泥 12000 3000 4000 6000 25000
钢材 6000 1400 1500 3000 11900
表14 构件厂厂与订货企业之间的距离（公里）
订货企业
构件厂 1 2 3 4 5
1 12 15 25 18 9
2 12 10 15 18 17
3 17 18 14 11 15
4 16 12 18 13 20
预制桩订货量 630 720 700 420 800
预制梁订货量 520 420 400 620 300

model:
sets:
supply/1..4/:;
demand/1..5/:;
goods/1..2/:tnum,weight,price;
material/1..2/:;
link1(supply,demand):distance;
link2(supply,goods):mproduce,cost;
link3(goods,demand):booking;
link4(supply,demand,goods):num;
link5(supply,material,goods):expend;
link6(material,supply):tmaterial;
endsets
data:
tnum=2260 3270;
price=10000 12000;
weight=5 3;
transportcost=1;
mproduce=
800 1000
600 700
450 800
550 1000;
cost=
3800 6400
4000 6380
4100 6500
3950 6600;
expend=
4000 2000 1000 600
4050 2050 1050 510
4050 2060 1030 510
4000 1990 990 515;
tmaterial=
12000 3000 4000 6000
6000 1400 1500 3000;
distance=
12 15 25 18 9
12 10 15 18 17
17 18 14 11 15
16 12 18 13 20;
booking=
520 420 400 620 300
630 720 700 420 800;
enddata
max=@sum(goods:tnum*price)-@sum(link2(i,j):cost(i,j)*@sum(demand(k):num(i,k,j)))
-transportcost*@sum(link1(i,k):distance(i,k)*@sum(goods(j):weight(j)*num(i,k,j)));
@for(link6(l,i):@sum(goods(j):@sum(demand(k):num(i,k,j))*expend(i,l,j))<=tmaterial(l,i)*1000);
@for(link2(i,j):@sum(demand(k):num(i,k,j))<=mproduce(i,j));
@for(link3(j,k):@sum(supply(i):num(i,k,j))=booking(j,k));
@for(link4:@gin(num));
end

model: sets: supply/1..4/:distance1; demand/1..5/:; goods/1..2/:tnum,weight,price; link1(supply,demand):distance2; link2(supply,goods):mproduce,cost,cementcost,steelcost; link3(goods,demand):booking; link4(supply,demand,goods):num; endsets data: tnum=2450 3600; price=10000 12000; distance1=20 35 60 40; weight=5 3; mproduce= 1000 1000 800 700 500 800 450 1200; cost= 6000 8600 6500 8500 6400 8550 5950 8700; cementcost= 4000 2000 4050 2050 4050 2060 4000 1990; steelcost= 1000 600 1050 510 1030 510 990 515; distance2= 15 12 19 25 9 12 18 15 18 17 17 10 14 11 15 16 9 18 13 20; booking= 450 520 600 500 380 900 600 1000 500 600; enddata max=@sum(goods:tnum*price)-@sum(link2(i,j):cost(i,j)*@sum(demand(k):num(i,k,j))) -0.2*@sum(supply(i):distance1(i)*@sum(goods(j):@sum(demand(k):num(i,k,j))*cementcost(i,j)/1000)) -0.3*@sum(supply(i):distance1(i)*@sum(goods(j):@sum(demand(k):num(i,k,j))*steelcost(i,j)/1000)) -@sum(link1(i,k):distance2(i,k)*@sum(goods(j):weight(j)*num(i,k,j))); @sum(supply(i):@sum(goods(j):@sum(demand(k):num(i,k,j))*cementcost(i,j)))&lt;=23000000; @sum(supply(i):@sum(goods(j):@sum(demand(k):num(i,k,j))*steelcost(i,j)))&lt;=9000000; @for(link2(i,j):@sum(demand(k):num(i,k,j))&lt;=mproduce(i,j)); @for(link3(j,k):@sum(supply(i):num(i,k,j))=booking(j,k)); @for(link4:@gin(num)); end

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