LINGO实验案例程序

1线性规划1model:[obj]max=2*x1+3*x2;[resources_1]x1+2*x2<=8;[resources_2]4*x1<=16;[resources_3]4*x2<=12;end2线性规划2model:sets:resources/b1b2b3/:b;var/x1x2/:x,c;tech_coe(resources,var):a;endsetsdata:b=8,16,12;c=2,3;a=124004;enddata[obj]max=@sum(var(j):c(j)*x(j));@for(resources(i):[resources_row]@sum(var(j):a(i,j)*x(j))<=b(i));end3运输问题model:!6发点8收点运输问题；sets:warehouses/wh1..wh6/:capacity;vendors/v1..v8/:demand;links(warehouses,vendors):cost,volume;endsets!目标函数；min=@sum(links:cost*volume);!需求约束；@for(vendors(J):@sum(warehouses(I):volume(I,J))=demand(J));!产量约束；@for(warehouses(I):@sum(vendors(J):volume(I,J))<=capacity(I));!这里是数据；data:capacity=605551434152;demand=3537223241324338;cost=626742954953858252197433767392712395726555228143;enddataend4一般整数规划!一般整数规划问题；Max=100*x1+150*x2;X1<=103;X2<=120;X1+2*x2<=160;@gin(x1);@gin(x2);5二进制整数规划!教材p124例6;model:sets:resources/b1b2b3b4/:b;var/x1x2x3/:x,c;tech_coe(resources,var):a;endsetsdata:b=2,4,3,6;c=3,-2,5;a=12-1141110401;enddata[obj]max=@sum(var(j):c(j)*x(j));,j)*x(j))<=b(i));@for(resources(i):[resources_row]@sum(var(j):a(,j)*x(j))<=b(i));@for(var(j):@bin(x(j)));end6指派问题model:!7个工人，7个工作的分配问题；sets:workers/w1..w7/;jobs/j1..j7/;links(workers,jobs):cost,volume;endsets!目标函数；min=@sum(links:cost*volume);!每个工人只能有一份工作；@for(workers(I):@sum(jobs(J):volume(I,J))=1);!每份工作只能有一个工人;@for(jobs(J):@sum(workers(I):volume(I,J))=1);data:cost=6267425495385852197437673927239572655228114enddata923124510;end7一维资源分配问题!教材p213例1;!在arcs集合上添加了selection和status属性后，通过决策变量selection和已知量status的联合使用，不仅轻松表达了设备总量的约束，而且为离散取值最优化的求解提供了一个有效的思路。；model:sets:!工厂；factory/123/;!设备数量；facility_amount/1..6/;!分配方案；arcs(factory,facility_amount):benefit,selection,status;endsetsdata:!盈利；benefit=0,3,7,9,12,130,5,10,11,11,110,4,6,11,12,12;!待分配设备数量；status=0,1,2,3,4,50,1,2,3,4,50,1,2,3,4,5;enddata[obj]max=@sum(arcs(i,j):benefit(i,j)*selection(i,j));!决策变量为0—1变量；@for(arcs:@bin(selection));!第1个工厂必须且只能选择一个设备分配方案；@for(factory(i):@sum(arcs(i,k):selection(i,k))=1);!3个工厂分配设备数之和等于设备总量；@sum(arcs(i,j):status(i,j)*selection(i,j))=5;end8最短路问题!最短路问题;model:data:n=10;enddatasets:cities/1..n/:F; !10个城市；roads(cities,cities)/1,21,32,52,63,53,64,85,85,96,97,108,109,10/:D,P;endsetsdata:D=65369751191875410579;enddataF(n)=0;@for(cities(i)|i#lt#n:F(i)=@min(roads(i,j):D(i,j)+F(j));