1、线性规划模型运筹学实验实验一 线性规划模型一、 实验目的掌握数学软件Lingo编程求解线性规划模型。二、 实验内容1 安装并启动Lingo软件，了解Lingo软件子菜单内容及其功能，掌握操作命令。2 输入模型，求解模型，结果的简单分析，用Lingo软件求解书P47练习1.1。3 用Lingo软件完成下列问题(1) 写出对偶线性规划；(2) 求原问题和对偶问题的最优解；(3) 分别写出价值系数和右端常数的最大允许变化范围；(4) 目标函数改为C=(5,3,6),同时常数改为b=(120,140,100),求最优解；(5) 增加一个设备约束和一个变量,系数为()=(7,5,4,1,2),求最优解。
2、4 思考题 书P52案例1.2。三、 实验指导参考PDF文档。四、 实验程序和结果（学生填）2.题目1.1（a）输入程序：min=2*x1+3*x2;4*x1+6*x2=6;4*x1+2*x2=4;x1=0;x2=0;运行 Global optimal solution found at iteration: 0 Objective value: 3.000000 Variable Value Reduced Cost X1 0.7500000 0.000000 X2 0.5000000 0.000000 Row Slack or Surplus Dual Price 1 3.000000 -
3、1.000000 2 0.000000 -0.5000000 3 0.000000 0.000000 4 0.7500000 0.000000 5 0.5000000 0.000000即x1=0.75,x2=0.5,min=3(b) max=3*x1+2*x2;2*x1+x2=12;x1=0;x2=0; Variable Value Reduced Cost X1 0.000000 0.7500000E+10 X2 2.000000 0.000000 Row Slack or Surplus Dual Price 1 4.000000 1.000000 2 0.000000 0.6000000
4、E+10 3 -4.000000 -0.1500000E+10 4 0.000000 0.000000 5 2.000000 0.000000无可行解(c)max=x1+x2;6*x1+10*x2=5;x1=3;x2=2;-2*x1+3*x2=0;x2=0; Variable Value Reduced Cost X1 2.000000 0.000000 X2 2.000000 0.000000 Row Slack or Surplus Dual Price 1 22.00000 1.000000 2 0.000000 6.750000 3 0.000000 4.250000 4 2.0000
5、00 0.000000 5 2.000000 0.000000即无界解2.(1)对偶问题为min=100*y1+100*y2+120*y3;2*y1+3*y2+3*y3=42*y1+y2+y3=24*y1+6*y2+2*y3=3y1=0y2=0y3=0（2）求原问题最优解，输入：max=4*x1+2*x2+3*x3;2*x1+2*x2+4*x3=100;3*x1+x2+6*x3=100;3*x1+x2+2*x3=0;x2=0;x3=0; Global optimal solution found at iteration: 2 Objective value: 150.0000 Variabl
6、e Value Reduced Cost X1 25.00000 0.000000 X2 25.00000 0.000000 X3 0.000000 5.000000 Row Slack or Surplus Dual Price 1 150.0000 1.000000 2 0.000000 0.5000000 3 0.000000 1.000000 4 20.00000 0.000000 5 25.00000 0.000000 6 25.00000 0.000000 7 0.000000 0.000000即下，x1=25,x2=25,x3=0,max=150求对偶问题最优解，输入min=10
7、0*y1+100*y2+120*y3;2*y1+3*y2+3*y3=4;2*y1+y2+y3=2;4*y1+6*y2+2*y3=3;y1=0;y2=0;y3=0; Global optimal solution found at iteration: 0 Objective value: 150.0000 Variable Value Reduced Cost Y1 0.5000000 0.000000 Y2 1.000000 0.000000 Y3 0.000000 20.00000 Row Slack or Surplus Dual Price 1 150.0000 -1.000000 2
8、 0.000000 -25.00000 3 0.000000 -25.00000 4 5.000000 0.000000 5 0.5000000 0.000000 6 1.000000 0.000000 7 0.000000 0.000000即，y1=0.5,y2=1,y3=0,min=150.(3) 即（4）输入max=5*x1+3*x2+6*x3;2*x1+2*x2+4*x3=120;3*x1+x2+6*x3=140;3*x1+x2+2*x3=120;Global optimal solution found at iteration: 2 Objective value: 240.000
9、0 Variable Value Reduced Cost X1 30.00000 0.000000 X2 30.00000 0.000000 X3 0.000000 0.000000 Row Slack or Surplus Dual Price 1 240.0000 1.000000 2 0.000000 1.000000 3 20.00000 0.000000 4 0.000000 1.000000即，x1=30,x2=30,x3=0,max=240(5)输入max=4*x1+2*x2+3*x3+7*x4;2*x1+2*x2+4*x3+5*x4=100;3*x1+x2+6*x3+4*x4
10、=100;3*x1+x2+2*x3+x4=120;6*x1+5*x2+x3+2*x4=0;x2=0;x3=0;x4=0; Global optimal solution found at iteration: 0 Objective value: 157.1429 Variable Value Reduced Cost X1 14.28571 0.000000 X2 0.000000 0.2857143 X3 0.000000 5.000000 X4 14.28571 0.000000 Row Slack or Surplus Dual Price 1 157.1429 1.000000 2 0.000000 0.7142857 3 0.000000 0.8571429 4 62.85714 0.000000 5 85.71429 0.000000 6 14.28571 0.000000 7 0.000000 0.000000 8 0.000000 0.000000 9 14.28571 0.000000即x1=14.25771,x2=0,x3=0,x4=14.28571,max=157.1429
