铁路平板车装货问题

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1、铁路平板车装货问题令xi1为第i种的货物装到第一辆车的数量,xi2为第i种的货物装到第二辆车的数量;ti为第i种货物的厚度；wi为第i种货物的重量；ni为第i种货物的件数。建立数学模型:Maxz=i=17ti(xi1+xi2)s.t.i=17xi1+xi2≤nii=17xij*wi≤40000(j=1,2)i=17xij*ti≤1020(j=1,2)i=57xij≤302.7(j=1,2)解出可行解,有如下lingo程序:model:sets:type/1..7/:t,w,c;car/1,2/;links(type,car):x

2、;endsetsdata:c=8796648;t=48.753.061.372.048.752.064.0;w=200030001000500400020001000;enddatamax=@sum(type(i):t(i)*(x(i,1)+x(i,2)));@for(type(i):@sum(type(i):x(i,1)+x(i,2))

3、i#

4、GT#4:(x(i,1)+x(i,2))*t(i))<302.7;@for(links:@gin(x)));end得到以下结果：Feasiblesolutionfound.Objectivevalue:2039.800Objectivebound:2040.000Infeasibilities:0.000000Extendedsolversteps:10314Totalsolveriterations:32716VariableValueReducedCostT(1)48.700000.000000T(2)53.000000.0

5、00000T(3)61.300000.000000T(4)72.000000.000000T(5)48.700000.000000T(6)52.000000.000000T(7)64.000000.000000W(1)2000.0000.000000W(2)3000.0000.000000W(3)1000.0000.000000W(4)500.00000.000000W(5)4000.0000.000000W(6)2000.0000.000000W(7)1000.0000.000000C(1)8.0000000.000000C(

6、2)7.0000000.000000C(3)9.0000000.000000C(4)6.0000000.000000C(5)6.0000000.000000C(6)4.0000000.000000C(7)8.0000000.000000X(1,1)2.000000-48.70000X(1,2)6.000000-48.70000X(2,1)2.000000-53.00000X(2,2)5.000000-53.00000X(3,1)7.000000-61.30000X(3,2)2.000000-61.30000X(4,1)2.000

7、000-72.00000X(4,2)4.000000-72.00000X(5,1)5.000000-48.70000X(5,2)0.000000-48.70000X(6,1)0.000000-52.00000X(6,2)1.000000-52.00000X(7,1)0.000000-64.00000X(7,2)0.000000-64.00000RowSlackorSurplusDualPrice12039.8001.00000020.0000000.00000030.0000000.00000040.0000000.000000

8、50.0000000.00000061.0000000.00000073.0000000.00000088.0000000.00000092000.0000.000000100.0000000.000000117.2000000.000000127000.0000.000000130.20000000.000000147.2000000.000000最优解为：在第一辆平板车上分别装c1，c2，c3，c4，c5，c6，c7类的包装箱的件数为2，2，7，2，5，0，0；第二辆平板车上分别装c1，c2，c3，c4，c5，c6，c7类的

9、包装箱的件数为6，5，2，4，0，1，0；剩余空间为：1020*2—2039.8=0.2展厅监控问题令变量xi为是否在通道i上装摄像机(i=1,...,13)，aij为通道i处装摄像机是否能监控到展厅j（j=1,...,8）。xi=1,(在通道i处安装摄像头)0