Question

数学建模 复合肥生产

我们老师给的期末考试作业，交一篇数学建模论文，网上找了发现是中北大学今天的数学建模题目，还有几天就要交了可是我做不来，还忘哪位大哥大姐给我点资料些什么的。题目如下，高分追加啊。

复合肥料生产
某复合肥料由几种基本肥料组合而成，基础肥料有5种，其中氮肥3种：N1,N2,N3，磷肥2种P1,P2,各种基础肥料由其它化工厂购进，未来半年中各种基础肥料的价格如下：
月份\肥料 P1 P2 N1 N2 N3
一 1650 1800 1950 1650 1725
二 1950 1950 1650 1350 1725
三 1650 2100 1950 1500 1425
四 1800 1650 1800 1800 1875
五 1500 1800 2250 1650 1575
六 1350 1500 2100 1200 2025
对几种基础肥料加工，然后混合为复合肥。复合肥售价为2250元/吨。氮肥和磷肥在不同生产线加工，每个月最多可以加工磷肥200吨，氮肥250吨。加工过程没有重量损失，费用不考虑。每种基础肥料最多可以存储1000吨备用，存储费用为每吨每月75元。成品复合肥和加工过的基础肥料不能存储。对复合肥的杂质指标限制在3－6%个单位之间，假设杂质是线性混合的。各种基础肥料的杂质含量见下表
基础肥料 P1 P2 N1 N2 N3
杂质（%） 8.8 6.1 2.0 4.2 5.0

为使公司获得最大利润，应采取什么样的采购和加工方案。现存有5种基础肥料每种500吨，要求在6月底仍然有这样多存货。
研究总利润和采购与加工方案适应不同的未来市场价格应如何变化。考虑如下的价格变化方式：2月份基础磷肥价格上升x%，基础氮肥价格上升2x%；3月份基础磷肥上升2x%，基础氮肥上升4x%；其余月份保持这种线性的上升势头。对不同的值x（直到20），就方案的必要的变化及对利润的影响，作出全面计划。

Answer 1

　　model:
　　sets:
　　shijian/1..6/;
　　feiliao/1..5/:t;
　　link(shijian,feiliao):y,x,m;
　　endsets
　　data:
　　m=1650 1800 1950 1650 1725
　　1950 1950 1650 1350 1725
　　1650 2100 1950 1500 1425
　　1800 1650 1800 1800 1875
　　1500 1800 2250 1650 1575
　　1350 1500 2100 1200 2025;
　　t=0.088 0.061 0.020 0.042 0.050;
　　enddata
　　!目标函数;
　　max=@sum(link:2250*y-m*x)-75*(30*500+
　　@sum(feiliao(j):x(1,j)-y(1,j))+
　　@sum(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#2:x(i,j)-y(i,j)))+
　　@sum(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#3:x(i,j)-y(i,j)))+
　　@sum(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#4:x(i,j)-y(i,j)))+
　　@sum(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#5:x(i,j)-y(i,j))));
　　!加工能力约束;
　　@for(shijian(i):
　　@sum(feiliao(j)|j#ge#1#and#j#le#2:y(i,j))<=200);
　　@for(shijian(i):
　　@sum(feiliao(j)|j#ge#3#and#j#le#5:y(i,j))<=250);
　　!储存量约束;
　　@for(feiliao(j):x(1,j)-y(1,j)<=500);
　　@for(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#2:x(i,j)-y(i,j))<=500);
　　@for(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#3:x(i,j)-y(i,j))<=500);
　　@for(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#4:x(i,j)-y(i,j))<=500);
　　@for(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#5:x(i,j)-y(i,j))<=500);

　　@for(feiliao(j):x(1,j)-y(1,j)+500>=0);
　　@for(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#2:x(i,j)-y(i,j))>=-500);
　　@for(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#3:x(i,j)-y(i,j))>=-500);
　　@for(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#4:x(i,j)-y(i,j))>=-500);
　　@for(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#5:x(i,j)-y(i,j))>=-500);

　　!杂质含量约束;
　　@sum(feiliao(j):y(1,j)*t(j))>0.03*@sum(feiliao(j):y(1,j));
　　@sum(feiliao(j):y(2,j)*t(j))>0.03*@sum(feiliao(j):y(2,j));
　　@sum(feiliao(j):y(3,j)*t(j))>0.03*@sum(feiliao(j):y(3,j));
　　@sum(feiliao(j):y(4,j)*t(j))>0.03*@sum(feiliao(j):y(4,j));
　　@sum(feiliao(j):y(5,j)*t(j))>0.03*@sum(feiliao(j):y(5,j));
　　@sum(feiliao(j):y(6,j)*t(j))>0.03*@sum(feiliao(j):y(6,j));
　　@sum(feiliao(j):y(1,j)*t(j))<0.06*@sum(feiliao(j):y(1,j));
　　@sum(feiliao(j):y(2,j)*t(j))<0.06*@sum(feiliao(j):y(2,j));
　　@sum(feiliao(j):y(3,j)*t(j))<0.06*@sum(feiliao(j):y(3,j));
　　@sum(feiliao(j):y(4,j)*t(j))<0.06*@sum(feiliao(j):y(4,j));
　　@sum(feiliao(j):y(5,j)*t(j))<0.06*@sum(feiliao(j):y(5,j));
　　@sum(feiliao(j):y(6,j)*t(j))<0.06*@sum(feiliao(j):y(6,j));

　　!月末余量约束;
　　@sum(shijian(i):x(i,1)-y(i,1))=0;
　　@sum(shijian(i):x(i,2)-y(i,2))=0;
　　@sum(shijian(i):x(i,3)-y(i,3))=0;
　　@sum(shijian(i):x(i,4)-y(i,4))=0;
　　@sum(shijian(i):x(i,5)-y(i,5))=0;
　　!复合肥基本条件;
　　@for(shijian(i):
　　@sum(feiliao(j)|j#ge#1#and#j#le#2:y(i,j))>0.0005);
　　@for(shijian(i):
　　@sum(feiliao(j)|j#ge#3#and#j#le#5:y(i,j))>0.0005);

　　end
　　第二问：
　　model:
　　sets:
　　shijian/1..6/;
　　feiliao/1..5/:t,m1;
　　link(shijian,feiliao):y,x;
　　endsets
　　data:
　　m1=1650 1800 1950 1650 1725;
　　t=0.088,0.061,0.020,0.042,0.050;
　　r=?;
　　enddata
　　!目标函数;
　　max=@sum(link:2250*y)-@sum(feiliao(j):m1(j)*x(1,j))-
　　(1950*(1+r/100)*x(2,1)+1950*(1+r/100)*x(2,2)+1650*(1+2*r/100)*x(2,3)+1350*(1+2*r/100)*x(2,4)+1725*(1+2*r/100)*x(2,5)+
　　1650*(1+2*r/100)*x(3,1)+2100*(1+2*r/100)*x(3,2)+1950*(1+4*r/100)*x(3,3)+1500*(1+4*r/100)*x(3,4)+1425*(1+4*r/100)*x(3,5)+
　　1800*(1+4*r/100)*x(4,1)+1650*(1+4*r/100)*x(4,2)+1800*(1+8*r/100)*x(4,3)+1800*(1+8*r/100)*x(4,4)+1875*(1+8*r/100)*x(4,5)+
　　1500*(1+8*r/100)*x(5,1)+1800*(1+8*r/100)*x(5,2)+2250*(1+16*r/100)*x(5,3)+1650*(1+16*r/100)*x(5,4)+1575*(1+16*r/100)*x(5,5)+
　　1350*(1+16*r/100)*x(6,1)+1500*(1+16*r/100)*x(6,2)+2100*(1+32*r/100)*x(6,3)+1200*(1+32*r/100)*x(6,4)+2025*(1+32*r/100)*x(6,5)
　　)-75*(30*500+
　　@sum(feiliao(j):x(1,j)-y(1,j))+
　　@sum(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#2:x(i,j)-y(i,j)))+
　　@sum(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#3:x(i,j)-y(i,j)))+
　　@sum(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#4:x(i,j)-y(i,j)))+
　　@sum(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#5:x(i,j)-y(i,j))));
　　!加工能力约束;
　　@for(shijian(i):
　　@sum(feiliao(j)|j#ge#1#and#j#le#2:y(i,j))<=200);
　　@for(shijian(i):
　　@sum(feiliao(j)|j#ge#3#and#j#le#5:y(i,j))<=250);
　　!储存量约束;
　　@for(feiliao(j):x(1,j)-y(1,j)<=500);
　　@for(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#2:x(i,j)-y(i,j))<=500);
　　@for(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#3:x(i,j)-y(i,j))<=500);
　　@for(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#4:x(i,j)-y(i,j))<=500);
　　@for(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#5:x(i,j)-y(i,j))<=500);

　　@for(feiliao(j):x(1,j)-y(1,j)+500>=0);
　　@for(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#2:x(i,j)-y(i,j))>=-500);
　　@for(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#3:x(i,j)-y(i,j))>=-500);
　　@for(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#4:x(i,j)-y(i,j))>=-500);
　　@for(feiliao(j):@sum(shijian(i)|i#ge#1#and#i#le#5:x(i,j)-y(i,j))>=-500);

　　!杂质含量约束;
　　@sum(feiliao(j):y(1,j)*t(j))>0.03*@sum(feiliao(j):y(1,j));
　　@sum(feiliao(j):y(2,j)*t(j))>0.03*@sum(feiliao(j):y(2,j));
　　@sum(feiliao(j):y(3,j)*t(j))>0.03*@sum(feiliao(j):y(3,j));
　　@sum(feiliao(j):y(4,j)*t(j))>0.03*@sum(feiliao(j):y(4,j));
　　@sum(feiliao(j):y(5,j)*t(j))>0.03*@sum(feiliao(j):y(5,j));
　　@sum(feiliao(j):y(6,j)*t(j))>0.03*@sum(feiliao(j):y(6,j));
　　@sum(feiliao(j):y(1,j)*t(j))<0.06*@sum(feiliao(j):y(1,j));
　　@sum(feiliao(j):y(2,j)*t(j))<0.06*@sum(feiliao(j):y(2,j));
　　@sum(feiliao(j):y(3,j)*t(j))<0.06*@sum(feiliao(j):y(3,j));
　　@sum(feiliao(j):y(4,j)*t(j))<0.06*@sum(feiliao(j):y(4,j));
　　@sum(feiliao(j):y(5,j)*t(j))<0.06*@sum(feiliao(j):y(5,j));
　　@sum(feiliao(j):y(6,j)*t(j))<0.06*@sum(feiliao(j):y(6,j));

　　!月末余量约束;
　　@sum(shijian(i):x(i,1)-y(i,1))=0;
　　@sum(shijian(i):x(i,2)-y(i,2))=0;
　　@sum(shijian(i):x(i,3)-y(i,3))=0;
　　@sum(shijian(i):x(i,4)-y(i,4))=0;
　　@sum(shijian(i):x(i,5)-y(i,5))=0;
　　!复合肥基本条件;
　　@for(shijian(i):
　　@sum(feiliao(j)|j#ge#1#and#j#le#2:y(i,j))>0.0005);
　　@for(shijian(i):
　　@sum(feiliao(j)|j#ge#3#and#j#le#5:y(i,j))>0.0005);

　　end

Answer 2

还是自己做吧！