2024年数学实验实验报告

《数學试验》试验汇报（

年

4

月

8

曰）班级：数學系0902班學号：姓名：馬骁一、试验問題

1．（指派問題）考虑指定n個人完毕n项任务（每人單独承担一项任务），使所需的總完毕時间（成本）尽量短.已知某指派問題的有关数据（每人完毕各任务所需的時间）如下表所示，试建模并求解该指派問題。2．（二次指派問題）某企业指派n個员工到n個都市工作（每個都市單独一人），但愿使所花费的總電话费用尽量少。n個员工两两之间每月通话的時间表达在下面的矩阵的上三角部分（由于通话的時间矩阵是對称的，没有必要写出下三角部分），n個都市两两之间通话费率表达在下面的矩阵的下三角部分（同样道理，由于通话的费率矩阵是對称的，没有必要写出上三角部分）.试求解该二次指派問題。3、謝金星第四章課後习題第1或3題任选一題。二、問題的分析（波及的理论知识、数學建模与求解的措施等）

1)根据实际問題，建立数學优化模型2)根据优化模型，运用LINGO来求解模型。三、计算過程、結论和成果分析1.模型：model:sets:m/1..4/;n/1..4/;link(m,n):a,x;endsetsmin=@sum(link(i,j):x(i,j)*a(i,j));@for(m(i):@sum(n(j):a(i,j))=1);@for(n(j):@sum(m(i):a(i,j))=1);data:x=15182124192322182618161919212317;enddataend成果：Globaloptimalsolutionfound.Objectivevalue:70.00000Infeasibilities:0.000000Totalsolveriterations:7VariableValueReducedCostA(1,1)0.0000000.000000A(1,2)1.0000000.000000A(1,3)0.0000005.000000A(1,4)0.00000010.00000A(2,1)1.0000000.000000A(2,2)0.0000001.000000A(2,3)0.0000002.000000A(2,4)0.0000000.000000A(3,1)0.00000011.00000A(3,2)0.0000000.000000A(3,3)1.0000000.000000A(3,4)0.0000005.000000A(4,1)0.0000001.000000A(4,2)0.0000000.000000A(4,3)0.0000004.000000A(4,4)1.0000000.000000X(1,1)15.000000.000000X(1,2)18.000000.000000X(1,3)21.000000.000000X(1,4)24.000000.000000X(2,1)19.000000.000000X(2,2)23.000000.000000X(2,3)22.000000.000000X(2,4)18.000000.000000X(3,1)26.000000.000000X(3,2)18.000000.000000X(3,3)16.000000.000000X(3,4)19.000000.000000X(4,1)19.000000.000000X(4,2)21.000000.000000X(4,3)23.000000.000000X(4,4)17.000000.000000RowSlackorSurplusDualPrice170.00000-1.00000020.000000-14.0000030.000000-18.0000040.000000-14.0000050.000000-17.0000060.000000-1.00000070.000000-4.00000080.000000-2.00000090.0000000.000000第1個人完毕第2项，第2人完毕第1项，第3人完毕第3项，第4人完毕第4项。最短時间為70.2.模型：model:sets:pe/1,2,3,4,5/;ci/1,2,3,4,5/;lin(pe,ci):x;lin1(pe,pe):d;lin2(ci,ci):c;endsetsmin=@sum(lin1(i,j):d(i,j)*@sum(lin2(a,b):c(a,b)*x(a,i)*x(b,j)));@for(pe(m):@sum(ci(n):x(m,n))=1);@for(ci(n):@sum(pe(m):x(m,n))=1);@for(lin:@bin(x));data:d=0112310212120122110132210;c=0524150302230004000512050;enddataend成果：Localoptimalsolutionfound.Objectivevalue:50.00000Objectivebound:50.00000Infeasibilities:0.000000Extendedsolversteps:18Totalsolveriterations:802VariableValueReducedCostX(1,1)0.00000014.00000X(1,2)0.0000000.000000X(1,3)1.0000005.999996X(1,4)0.0000000.000000X(1,5)0.0000006.000000X(2,1)0.00000016.00000X(2,2)0.0000009.999992X(2,3)0.0000000.000000X(2,4)1.0000001.999992X(2,5)0.0000000.000000X(3,1)0.00000014.00000X(3,2)0.0000006.000000X(3,3)0.0000000.8523365E-08X(3,4)0.0000000.000000X(3,5)1.0000000.000000X(4,1)1.0000007.999996X(4,2)0.0000000.000000X(4,3)0.0000005.999996X(4,4)0.0000006.000000X(4,5)0.00000014.00000X(5,1)0.0000000.000000X(5,2)1.0000001.999981X(5,3)0.0000000.000000X(5,4)0.0000009.999981X(5,5)0.00000015.99998D(1,1)0.0000000.000000D(1,2)1.0000000.000000D(1,3)1.0000000.000000D(1,4)2.0000000.000000D(1,5)3.0000000.000000D(2,1)1.0000000.000000D(2,2)0.0000000.000000D(2,3)2.0000000.000000D(2,4)1.0000000.000000D(2,5)2.0000000.000000D(3,1)1.0000000.000000D(3,2)2.0000000.000000D(3,3)0.0000000.000000D(3,4)1.0000000.000000D(3,5)2.0000000.000000D(4,1)2.0000000.000000D(4,2)1.0000000.000000D(4,3)1.0000000.000000D(4,4)0.0000000.000000D(4,5)1.0000000.000000D(5,1)3.0000000.000000D(5,2)2.0000000.000000D(5,3)2.0000000.000000D(5,4)1.0000000.000000D(5,5)0.0000000.000000C(1,1)0.0000000.000000C(1,2)5.0000000.000000C(1,3)2.0000000.000000C(1,4)4.0000000.000000C(1,5)1.0000000.000000C(2,1)5.0000000.000000C(2,2)0.0000000.000000C(2,3)3.0000000.000000C(2,4)0.0000000.000000C(2,5)2.0000000.000000C(3,1)2.0000000.000000C(3,2)3.0000000.000000C(3,3)0.0000000.000000C(3,4)0.0000000.000000C(3,5)0.0000000.000000C(4,1)4.0000000.000000C(4,2)0.0000000.000000C(4,3)0.0000000.000000C(4,4)0.0000000.000000C(4,5)5.0000000.000000C(5,1)1.0000000.000000C(5,2)2.0000000.000000C(5,3)0.0000000.000000C(5,4)5.0000000.000000C(5,5)0.0000000.000000RowSlackorSurplusDualPrice150.00000-1.00000020.000000-10.0000030.000000-6.00000040.0000008.00000050.0000000.00000060.000000-0.8674480E-0570.000000-10.0000080.000000-16.0000090.000000-14.00000100.000000-12.00000110.000000-22.00000第一种人去第三個都市，第二個人去第四個都市，第三個人去第五個都市，第四個人去第一种都市，第五個人去第二個都市，最低费用50。3.模型：model:min=100*x1+100*x2+40*y1+40*y2+40*y3+40*y4+40*y5;x1+x2+y1>4;x1+x2+y1+y2>3;x1+x2+y1+y2+y3>4;x2+y1+y2+y3+y4>6;x1+y2+y3+y4+y5>5;x1+x2+y3+y4+y5>6;x1+x2+y4+y5>8;x1+x2+y5>8;y1+y2+y3+y4+y5<3;@gin(x1);@gin(x2);@gin(y1);@gin(y2);@gin(y3);@gin(y4);@gin(y5);endGlobaloptimalsolutionfound.Objectivevalue:820.0000Objectivebound:820.0000Infeasibilities:0.000000Extendedsolversteps:0Totalsolveriterations:46VariableValueReducedCostX13.000000100.0000X24.000000100.0000Y10.00000040.00000Y22.00000040.00000Y30.00000040.00000Y40.00000040.00000Y51.00000040.00000RowSlackorSurplusDualPrice1820.0000-1.00000023.0000000.00000036.0000000.00000045.0000000.00000050.0000000.00000061.0000000.00000072.0000000.00000080.0000000.00000090.0000000.000000100.0000000.000000阐明：该储蓄雇用7名全時服务员，其中3名12：00~13：00吃午餐，4名13：00~14：00吃午餐；并在10：00~14：00雇用2名、13：00~17：00雇用1名半時服务员，符合題目规定，且花费至少，為820元。第二問y1+y2+y3+y4+y5<0;Globaloptimalsolutionfound.Objectivevalue:1100.000Objectivebound:1100.000Infeasibilities:0.000000Extendedsolversteps:0Totalsolveriterations:0VariableValueReducedCostX15.000000100.0000X26.000000100.0000Y10.00000040.00000Y20.00000040.00000Y30.00000040.00000Y40.00000040.00000Y50.00000040.00000RowSlackorSurplusDualPrice11100.000-1.00000027.0000000.00000038.0000000.00000047.0000000.00000050.0000000.00000060.0000000.00000075.0000000.00000083.0000000.00000093.0000000.000000100.0000000.000000阐明：不能雇用半時服务员，则需要雇用5名12：00~13：00吃午餐、6名13：00~14：00吃午餐的全時服务员，總花费為1100元，则每天至少增長280元經费。第三問y1+y2+y3+y4+y5>0;Globaloptimalsolutionfound.Objectivevalue:560.0000Objective