80分~生产任务分配问题~需要方程式和
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80分~生产任务分配问题~需要方程式和
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2450件和3600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,公司拥有水泥与钢材的数量为25000吨和6500吨,各构件厂所需材料由公司供应,公司到各构件厂的距离为20,35,60,40公里,水泥与钢材的每吨公里运输单价分别为0.2与0.3元.订货企业与各构件厂的距离见表13,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,建立综合考虑生产费用和运输费用,按公司利润最大建立并求解模型.
![](http://img.wesiedu.com/upload/0/28/028c69c11966fae7f7bc3ff933252acc.jpg)
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2450件和3600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,公司拥有水泥与钢材的数量为25000吨和6500吨,各构件厂所需材料由公司供应,公司到各构件厂的距离为20,35,60,40公里,水泥与钢材的每吨公里运输单价分别为0.2与0.3元.订货企业与各构件厂的距离见表13,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,建立综合考虑生产费用和运输费用,按公司利润最大建立并求解模型.
![](http://img.wesiedu.com/upload/0/28/028c69c11966fae7f7bc3ff933252acc.jpg)
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)))
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)))