LINGO实例

最新日志 该作者的其他文章 博主推荐 相关日志 随机阅读 首页推荐 更多>> 对“推广广告”提建议 时间运筹学与现代生活 参考文献格式详解(二） LinGo 学习（运筹学模型与实验） 数学建模 2010-05-19 09:09:51 阅读135 评论0 字号：大中小 订阅 线性规划问题（Linear Programming） max =2*x1+3*x2; 2*x1+2*x2<=12; x1+2*x2<=8; 4*x1<=16; 4*x2<=12; Solve Global optimal solution found. Objective value: 14.00000 Infeasibilities: 0.000000 Total solver iterations: 1 Variable Value Reduced Cost X1 4.000000 0.000000 X2 2.000000 0.000000 Row Slack or Surplus Dual Price 1 14.00000 1.000000 2 0.000000 0.000000 3 0.000000 1.500000 4 0.000000 0.1250000 5 4.000000 0.000000 最优解为X1=4.000000，X2=2.000000，最优值为Z=14. Generate \\Display Model MODEL: [_1] MAX= 2 * X1 + 3 * X2 ; [_2] 2 * X1 + 2 * X2 <= 12 ; [_3] X1 + 2 * X2 <= 8 ; [_4] 4 * X1 <= 16 ; [_5] 4 * X2 <= 12 ; END 程序形式（可得对偶问题的解） model: sets: stral1/1..4/:b; stral2/1..2/:x,c; matrix(stral1,stral2):A; endsets max=@sum(stral2(i):c(i)*x(i)); @for(stral1(i):@sum(stral2(j):A(i,j)*x(j))<=b(i)); data: A=2 2 1 2 4 0 0 4; b=12 8 16 12; c=2 3; enddata end Global optimal solution found. Objective value: 14.00000 Infeasibilities: 0.000000 Total solver iterations: 1 Variable Value Reduced Cost B( 1) 12.00000 0.000000 B( 2) 8.000000 0.000000 B( 3) 16.00000 0.000000 B( 4) 12.00000 0.000000 X( 1) 4.000000 0.000000 X( 2) 2.000000 0.000000 C( 1) 2.000000 0.000000 C( 2) 3.000000 0.000000 A( 1, 1) 2.000000 0.000000 A( 1, 2) 2.000000 0.000000 A( 2, 1) 1.000000 0.000000 A( 2, 2) 2.000000 0.000000 A( 3, 1) 4.000000 0.000000 A( 3, 2) 0.000000 0.000000 A( 4, 1) 0.000000 0.000000 A( 4, 2) 4.000000 0.000000 Row Slack or Surplus Dual Price 1 14.00000 1.000000 2 0.000000 0.000000 3 0.000000 1.500000 4 0.000000 0.1250000 5 4.000000 0.000000 对偶问题的最优解为：y1= 0.000000;y2=1.500000,y3=0.1250000,y4=0.000000,最优

值为14. 灵敏度分析 max=72*x1+64*x2; x1+x2<=50; 12*x1+8*x2<=480; 3*x1<=100; （1） 程序设置（激活）：运行LINGO/Options,选择General Solver Tab, 在Dual

Computations列表框中,选择Price&Ranges选项 （2） 先求解（Solve） Global optimal solution found. Objective value: 3360.000 Infeasibilities: 0.000000 Total solver iterations: 2 Variable Value Reduced Cost X1 20.00000 0.000000 X2 30.00000 0.000000 Row Slack or Surplus Dual Price 1 3360.000 1.000000 2 0.000000 48.00000 3 0.000000 2.000000 4 40.00000 0.000000 （3）再灵敏度分析（LINGO/Range） Ranges in which the basis is unchanged: Objective Coefficient Ranges Current Allowable Allowable Variable Coefficient Increase Decrease X1 72.00000 24.00000 8.000000 X2 64.00000 8.000000 16.00000 Righthand Side Ranges Row Current Allowable Allowable RHS Increase Decrease 2 50.00000 10.00000 6.666667 3 480.0000 53.33333 80.00000 4 100.0000 INFINITY 40.00000 运输问题（Transporation） !产销平衡运输问题; model: !3产地4销地的运输问题; sets: warehouses/wh1,wh2,wh3/:capacity; vendors/v1,v2,v3,v4/: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)); !数据; data: capacity=7 4 9; demand=3 6 5 6; cost=3 11 3 10 1 9 2 8 7 4 10 5; enddata end