数学模型实验三-线性规划问题

《数学模型实验三-线性规划问题》由会员上传分享，免费在线阅读，更多相关内容在教育资源-天天文库。

1、3线性规划实验3.1基本实验3.1.1生产计划安排（1）确定获利最大的生产方案；max=3*x1+x2+4*x3;6*x1+3*x2+5*x3<45;3*x1+4*x2+5*x3<30;@gin(x1);@gin(x2);@gin(x3);求解得到输出：Globaloptimalsolutionfound.Objectivevalue:27.00000Objectivebound:27.00000Infeasibilities:0.000000Extendedsolversteps:0Totalsolveriterations:0VariableValueReducedCostX15.00

2、0000-3.000000X20.000000-1.000000X33.000000-4.000000RowSlackorSurplusDualPrice127.000001.00000020.0000000.00000030.0000000.000000结论：最佳的生产方案为A产品5件，不生产B产品，C产品3件，获利27元。（2）A、B、C产品的利润变化范围；Rangesinwhichthebasisisunchanged:ObjectiveCoefficientRangesCurrentAllowableAllowableVariableCoefficientIncreaseDecre

3、aseX13.0000001.8000000.6000000X21.0000002.000000INFINITYX34.0000001.0000001.500000RighthandSideRangesRowCurrentAllowableAllowableRHSIncreaseDecrease245.0000015.0000015.00000330.0000015.000007.500000结论：A产品例润变化范围为（3-0.6,3+1.8）；B产品例润变化范围为（0,1+2）；C产品例润变化范围为（4-1.5,4+1）。（3）购买原料多少为宜；max=3*x1+x2+4*x3-0.4*

4、z;6*x1+3*x2+5*x3<45;3*x1+4*x2+5*x3-z<30;@gin(x1);@gin(x2);@gin(x3);@gin(z);求解得到输出：Globaloptimalsolutionfound.Objectivevalue:30.00000Objectivebound:30.00000Infeasibilities:0.000000Extendedsolversteps:0Totalsolveriterations:3VariableValueReducedCostX10.000000-3.000000X20.000000-1.000000X39.000000-4.

5、000000Z15.000000.4000000RowSlackorSurplusDualPrice130.000001.00000020.0000000.00000030.0000000.000000结论：购买15份材料为宜。（4）D产品是否值得生产；max=3*x1+x2+4*x3+3*x4;6*x1+3*x2+5*x3+8*x4<45;3*x1+4*x2+5*x3+2*x4<30;@gin(x1);@gin(x2);@gin(x3);@gin(x4);求解得到输出：Globaloptimalsolutionfound.Objectivevalue:27.00000Objectiveb

6、ound:27.00000Infeasibilities:0.000000Extendedsolversteps:0Totalsolveriterations:0VariableValueReducedCostX15.000000-3.000000X20.000000-1.000000X33.000000-4.000000X40.000000-3.000000RowSlackorSurplusDualPrice127.000001.00000020.0000000.00000030.0000000.000000结论：D产品不值得生产。3.1.3工程进度安排max=z1+z2+z3+z4;a

7、1+a2+a3=1;b1+b2+b3+b4=1;c1+c2+c3+c4+c5=1;d1+d2=1;5*a1+15*c1<3;5*a2+8*b1+15*c2<6;5*a3+8*b2+15*c3+1.2*d1<7;8*b3+15*c4+1.2*d2<7;8*b4+15*c5<7;z1=50*a1+50*(a1+a2)+50*2;z2=70*b1+70*(b1+b2)+70*(b1+b2+b3);z3=150*c1+150*(