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1、P94，例3.4选址问题目录题目1第一步，旧址基础上只求运量的ＬＰ程序1第二步，旧址基础上选择新址的NLP程序2题目6个工地的地址(坐标表示，距离单位KM)及水泥用量(单位：吨)如下表，而在P(5,1)及Q(2,7)处有两个临时料场，日储量各有20t，如何安排运输，可使总的吨公里数最小?新料场应选何处？能节约多少吨公里数？123456A1.258.750.55.7537.25B1.250.754.7556.57.75D3547611第一步，旧址基础上只求运量的ＬＰ程序MODEL:TitleLocationProblem;
2、sets:demand/1..6/:a,b,d;supply/1..2/:x,y,e;link(demand,supply):c;endsetsdata:!locationsforthedemand(需求点的位置);a=1.25,8.75,0.5,5.75,3,7.25;b=1.25,0.75,4.75,5,6.5,7.75;!quantitiesofthedemandandsupply（供需量）;d=3,5,4,7,6,11;e=20,20;x,y=5,1,2,7;enddatainit:!initiallocatio
3、nsforthesupply（初始点）;endinit!Objectivefunction（目标）;[OBJ]min=@sum(link(i,j):c(i,j)*((x(j)-a(i))^2+(y(j)-b(i))^2)^(1/2));!demandconstraints（需求约束）;@for(demand(i):[DEMAND_CON]@sum(supply(j):c(i,j))=d(i););!supplyconstraints（供应约束）;@for(supply(i):[SUPPLY_CON]@sum(demand
4、(j):c(j,i))<=e(i););!@for(supply:@free(x);!@free(Y);!);@for(supply:@bnd(0.5,X,8.75);@bnd(0.75,Y,7.75););END运行可得到全局最优解Globaloptimalsolutionfound.Objectivevalue:136.2275Totalsolveriterations:1ModelTitle:LocationProblemVariableValueReducedCostX(1)5.0000000.000000X(2
5、)2.0000000.000000Y(1)1.0000000.000000Y(2)7.0000000.000000E(1)20.000000.000000E(2)20.000000.000000第二步，旧址基础上选择新址的NLP程序!选新址的ＮＬＰ程序;MODEL:TitleLocationProblem;sets:demand/1..6/:a,b,d;supply/1..2/:x,y,e;link(demand,supply):c;endsetsdata:!locationsforthedemand(需求点的位置);a
6、=1.25,8.75,0.5,5.75,3,7.25;b=1.25,0.75,4.75,5,6.5,7.75;!quantitiesofthedemandandsupply（供需量）;d=3,5,4,7,6,11;e=20,20;enddatainit:!initiallocationsforthesupply（初始点）;!x,y=5,1,2,7;endinit!Objectivefunction（目标）;[OBJ]min=@sum(link(i,j):c(i,j)*((x(j)-a(i))^2+(y(j)-b(i))^
7、2)^(1/2));!demandconstraints（需求约束）;@for(demand(i):[DEMAND_CON]@sum(supply(j):c(i,j))=d(i););!supplyconstraints（供应约束）;@for(supply(i):[SUPPLY_CON]@sum(demand(j):c(j,i))<=e(i););!@for(supply:@free(x);!@free(Y);!);@for(supply:@bnd(0.5,X,8.75);@bnd(0.75,Y,7.75););END求
8、解结果只得到局部最优解Localoptimalsolutionfound.Objectivevalue:89.88347Totalsolveriterations:67ModelTitle:LocationProblemVariableValueReducedCostX(1)5.6959660.000000X(2)7
