1平板车装货问题与展厅安保问题lingo解法.doc

第一题基本模型：决策变量：引入变量xij,是第j辆车上第i种包装箱的数量（i=1,2,3.7。j=1,2）；ni是第i种包箱所装件数， ti是第i种包箱的厚度, wi第i中包厢的重量，clj第j节车的长度,cwj第j车的载重量。S为特殊限制（s=302.7）目标函数：约束条件：两节车的装箱数不能超过需要装的件数（i=1,2,3,4,5,6,7）每节车可装的长度不能超过车能提供的长度（j=1,2）每节车能装的重量不能超过车能承受的重量（j=1,2）对于C5,C6,C7三类包装箱的总数特别限制整数线性规划模型： （i=1,2,3,4,5,6,7） （j=1,2） （j=1,2） Model:!铁路平板车装货问题;sets: cars/1.2/:l,wet; boxes/1.7/: t,w,n; link(boxes,cars): x;endsets!这里是数据;data: m=4; l=1020,1020; wet=40000,40000; t=48.7,53.0,61.3,72.0,48.7,52.0,64.0; w=2000,3000,1000,500,4000,200,100; n=8,7,9,6,6,4,8; enddata!目标函数; max=sum(boxes(i):t(i)*sum(cars(j):x(i,j); for(boxes(i):sum(cars(j): x(i,j)n(i); for(cars(j):sum(boxes(i): t(i)*x(i,j)l(j); for(cars(j):sum(boxes(i): w(i)*x(i,j)wet(j); for(cars(j):sum(boxes(i)|i#GT#m: t(i)*x(i,j)1;x1+x2+x31;x4+x3+x5+x71;x6+x8+x121;x9+x7+x8+x101;x2+x5+x9+x111;x10+x12+x131;x11+x131;gin(x1);gin(x2);gin(x3);gin(x4);gin(x5);gin(x6);gin(x7);gin(x8);gin(x9);gin(x10);gin(x11);gin(x12);gin(x13);End Global optimal solution found. Objective value: 4.000000 Objective bound: 4.000000 Infeasibilities: 0.000000 Extended solver steps: 0 Total solver iterations: 0 Variable Value Reduced Cost X1 1.000000 1.000000 X2 0.000000 1.000000 X3 0.000000 1.000000 X4 0.000000 1.000000 X5 1.000000 1.000000 X6 0.000000 1.000000 X7 0.000000 1.000000 X8 1.000000 1.000000 X9 0.000000 1.000000 X10 0.000000 1.000000 X11 0.000000 1.000000 X12 0.000000 1.000000 X13 1.000000 1.000000 Row Slack or Surplus Dual Price 1 4.000000 -1.000000 2 0.000000 0.000000 3 0.000000 0.000000 4 0.000000 0.000000 5 0.000000 0.000000 6 0.000000 0.000000 7 0.000000 0.000000 8 0.000000 0.000000 9 0.000000 0.000000matlabf=1111111111111;A=-100-10-10000000;00000-10-1000-10;-1-1-10000000000;00-1-1-10-1000000;000000-1-1-1-1000;000000000-10-1-1;0-1