难难难！烦烦烦！
做完这个都有一篇论文了！！
lingo常用于线性优化问题，结合运筹学中主要的优化问题类型，本系列将使用lingo求解常见的运筹学优化模型，主要包含：
一般来说，一个优化模型将由以下三部分组成： 1. 目标函数（Objective Function） ：要达到的目标。 2. 决策变量（Decision variables） ：每组决策变量的值代表一种方案。在优化模型中需要确定决策变量的最优值，优化的目标就是找到决策变量的最优值使得目标函数取得最优。 3. 约束条件（Constraints） ：对于决策变量的一些约束，它限定决策变量可以取的值。 在写数学模型时，一般第一行是目标函数，接下来是约束条件，再接着是一些非负限制等。
model:
sets:
warehouse/1..2/:ownership2;
factory/1..2/:ttime;
item/1..5/:demand3;
material/1..3/:;
production/1..2/:;
link1(material,warehouse):ownership1;
link2(production,material):expend;
link3(factory,production):time,cost,produce,ability;
link4(item,production):demand1;
link5(item,material):demand2;
link6(factory,item):transportcost1;
link7(factory,item,production):transport1;
link8(warehouse,factory):transportcost2;
link9(warehouse,factory,material):transport2;
link10(warehouse,item):transportcost3,transport3;
link11(warehouse,item,material):transport4;
endsets
data:
ownership1=
5000 7000
5500 6500
10000 6000;
ownership2=380 450;
expend=
9 13 23
11 15 20;
time=
30 40
40 35;
ttime=20000 10000;
ability=
120 80
100 120;
cost=
320 300
280 360;
demand1=
80 0
100 90
100 60
60 50
30 120;
demand2=
50 30 40
0 120 100
70 120 160
40 90 30
60 90 40;
demand3=140 120 100 220 70;
transportcost1=
20 40 100 50 40
40 50 120 70 60;
transportcost2=
90 50
60 40;
transportcost3=
10 10 20 30 20
10 15 15 5 30;
enddata
min=@sum(link11(i,k,m):transport4(i,k,m)*transportcost3(i,k))
+@sum(link10(i,k):transport3(i,k)*transportcost3(i,k))
+@sum(link9(i,j,m):transport2(i,j,m)*transportcost2(i,j))
+@sum(link7(j,k,n):transport1(j,k,n)*transportcost1(j,k))
+@sum(link3(j,n):produce(j,n)*cost(j,n)+@if(produce(j,n)#gt#ability(j,n),200*(produce(j,n)-ability(j,n)),0));
@for(factory(j):@for(material(m):@sum(warehouse(i):transport2(i,j,m))=@sum(production(n):produce(j,n)*expend(n,m))));
@for(production(n):@for(factory(j):@sum(item(k):transport1(j,k,n))=produce(j,n)));
@for(warehouse(i):@sum(item(k):transport3(i,k))<=ownership2(i));
@for(warehouse(i):@for(material(m):@sum(item(k):transport4(i,k,m))+@sum(factory(j):transport2(i,j,m))<=ownership1(m,i)));
@for(factory(i):@sum(production(n):produce(i,n)*time(i,n))<=ttime(i));
@for(item(k):@for(production(n):@sum(factory(j):transport1(j,k,n))=demand1(k,n)));
@for(item(k):@for(material(m):@sum(warehouse(i):transport4(i,k,m))=demand2(k,m)));
@for(item(k):@sum(warehouse(i):transport3(i,k))=demand3(k));
end
现在和将来的角度，结合你所学
我可以写，比较多
num表示要求的某工厂供应某企业的某产品的件数
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=2500 3500;
price=10000 8000;
weight=5 3;
transportcost=1;
mproduce=
1000 1000
900 1000
500 800
450 1200;
cost=
8600 5800
8700 6300
8750 6200
8750 5750;
expend=
600
510
510
4000 1990 990 515;
tmaterial=
6000
1500;
distance=
15 12 19 25 9
12 18 15 18 17
17 10 14 11 15
16 9 18 13 20;
booking=
500 500 700 400 400
400 400;
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
