运筹学实验报告案例二.docx

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1、安阳师范学院数学与统计学院实验报告实验课程名称：运筹学 实验设计题目：配料问题 专 业：数学与应用数学 班 级：13级二班 学 生：常俊建 130800003 学 生：刘翠宇 130800004 学 生：李 燃 130800022 配料问题一、问题的描述某饲料公司生产鸡混合饲料，每千克饲料所需营养质量要求如表 C4所示。表 C4每千克饲料所需营养质量要求营养成分肉用种鸡国家标准肉用种鸡公司标准产蛋鸡标准代谢能2.72.8 Mcal/kg2.7 Mcal/kg2.65 Mcal/kg粗蛋白135145 g/kg135145 g/kg151 g/kg粗纤维=2.7;78*x1+114*x2+142

2、*x3+117*x4+402*x5+360*x6+450*x7+170*x8=135;78*x1+114*x2+142*x3+117*x4+402*x5+360*x6+450*x7+170*x8=145;16*x1+22*x2+95*x3+72*x4+49*x5+113*x6+108*x8=5.6;1.2*x1+1.7*x2+2.3*x3+2.7*x4+5.1*x5+7.1*x6+11.8*x7+2.2*x8+980*x9=2.6;0.7*x1+0.6*x2+0.3*x3+1.0*x4+3.2*x5+5.3*x6+63*x7+4.0*x8+300*x10+400*x11=30;0.3*x1+0

3、.34*x2+10.0*x3+13.0*x4+5.0*x5+8.4*x6+27*x7+4.0*x8+140*x10=5;1000*x12=3.7;X1+x2+x3+x4+x5+x6+x7+x8+x9+x10+x11+x12=1;X1=0.4;x2=0.1;x3=0.1;x4=0.15;x5=0.03;x7=0.05;x8=0.03;问题（1）Lingo程序的结果：Global optimal solution found. Objective value: 0.6553693 Infeasibilities: 0.000000 Total solver iterations: 9 Variab

4、le Value Reduced Cost X1 0.5385030 0.000000 X2 0.1000000 0.000000 X3 0.1000000 0.000000 X4 0.000000 0.1446276 X5 0.7213126E-01 0.000000 X6 0.3000000E-01 0.000000 X7 0.5000000E-01 0.000000 X8 0.3000000E-01 0.000000 X9 0.3233949E-03 0.000000 X10 0.4263719E-01 0.000000 X11 0.3270518E-01 0.000000 X12 0.

5、3700000E-02 0.000000 Row Slack or Surplus Dual Price 1 0.6553693 -1.000000 2 0.000000 -0.5218799 3 0.000000 -0.2339449E-03 4 10.00000 0.000000 5 14.51952 0.000000 6 0.3329203 0.000000 7 0.000000 -0.2461224E-01 8 0.000000 -0.5600000E-02 9 4.247413 0.000000 10 0.000000 -0.1540000E-02 11 0.000000 1.120

6、000 12 0.1385030 0.000000 13 0.000000 -0.1607394 14 0.000000 -0.3295455 15 0.1500000 0.000000 16 0.2786874E-01 0.000000 17 0.000000 -0.3059075 18 0.000000 -0.4502367 19 0.000000 -0.5434558所以按照肉用种鸡公司标准，1kg混合饲料中玉米约需要0.54kg，小麦约需要0.1kg，麦麸约需要0.1kg，米糠约需要0kg，豆饼约需要0.072kg，菜子饼约需要0.03kg，鱼粉约需要0.05kg，槐叶粉约需要0.03

7、kg，DL蛋氨酸约需要0.0003kg，骨粉约需要0.042kg，碳酸钙约需要0.032kg，食盐约需要0.0037kg，此时成本最低为约为0.655元。4.2问题（2）的模型建立与求解根据问题（2）所给数据及问题要求可列出约束条件，所以可建立混合饲料配料计划成本最低的线性规划模型如下：问题（2）的lingo程序如下：Min=0.68*x1+0.72*x2+0.23*x3+0.22*x4+0.37*x5+0.32*x6+1.54*x7+0.38*x8+23*x9+0.56*x10+1.12*x11+0.42*x12;3.35*x1+3.08*x2+1.78*x3+2.10*x4+2.40*x5

8、+1.62*x6+2.80*x7+1.61*x8=2.7;3.35*x1+3.08*x2+1.78*x3+2.10*x4+2.40*x5+1.62*x6+2.80*x7+1.61*x8=135;78*x1+114*x2+142*x3+117*x4+402*x5+360*x6+450*x7+170*x8=145;16*x1+22*x2+95*x3+72*x4+49*x5+113*x6+108*x8=5.6;1.2*x1+1.7*x2+2.3*x3+2.7*x4+5.1*x5+7.1*x6+11.8*x7+2.2*x8+980*x9=2.5;0.7*x1+0.6*x2+0.3*x3+1.0*x4+

9、3.2*x5+5.3*x6+63*x7+4.0*x8+300*x10+400*x11=23;0.7*x1+0.6*x2+0.3*x3+1.0*x4+3.2*x5+5.3*x6+63*x7+4.0*x8+300*x10+400*x11=4.6;0.3*x1+0.34*x2+10.0*x3+13.0*x4+5.0*x5+8.4*x6+27*x7+4.0*x8+140*x10=0.4;x2=0.1;x3=0.1;x4=0.15;x5=0.03;x7=0.05;x8=0.03;问题（2）Lingo程序的结果： Global optimal solution found. Objective value

10、: 0.6246589 Infeasibilities: 0.000000 Total solver iterations: 9 Variable Value Reduced Cost X1 0.4888733 0.000000 X2 0.1000000 0.000000 X3 0.1000000 0.000000 X4 0.5954138E-01 0.000000 X5 0.8930732E-01 0.000000 X6 0.3000000E-01 0.000000 X7 0.5000000E-01 0.000000 X8 0.3000000E-01 0.000000 X9 0.286969

11、4E-04 0.000000 X10 0.6262596E-02 0.000000 X11 0.4228671E-01 0.000000 X12 0.3700000E-02 0.000000 Row Slack or Surplus Dual Price1 0.6246589 -1.000000 2 0.000000 -0.3847177 3 0.1000000 0.000000 4 10.00000 0.000000 5 0.000000 0.1400426E-03 6 15.18499 0.000000 7 1.019734 0.000000 8 0.000000 -0.2411185E-

12、01 9 0.000000 -0.4374035E-02 10 17.00000 0.000000 11 0.4000000 0.000000 12 0.000000 0.8756890E-03 13 0.000000 -0.1049614E-02 14 0.000000 0.6296142 15 0.8887329E-01 0.000000 16 0.000000 -0.1373318 17 0.000000 -0.1466902 18 0.9045862E-01 0.000000 19 0.1069268E-01 0.000000 20 0.000000 -0.1897662 21 0.0

13、00000 -0.6189834 22 0.000000 -0.3469865所以按照肉用种鸡国家标准，1kg混合饲料中玉米约需要0.49kg，小麦约需要0.1kg，麦麸约需要0.1kg，米糠约需要0.06kg，豆饼约需要0.09kg，菜子饼约需要0.03kg，鱼粉约需要0.05kg，槐叶粉约需要0.03kg，DL蛋氨酸约需要0.000029kg，骨粉约需要0.0063kg，碳酸钙约需要0.042kg，食盐约需要0.0037kg，此时成本最低为约为0.625元。4.3问题（3）的模型建立与求解由题目所知，加上花生饼，得到序号原料单价元/kg代谢能Mcal/kg粗蛋白g/kg粗纤维g/kg赖氨酸

14、g/kg蛋氨酸g/kg钙g/kg有机磷g/kg食盐g/kg1玉米0.683.3578162.31.20.70.32小麦0.723.08114223.41.70.60.343麦麸0.231.78142956.02.30.3104米糠0.222.10117726.52.71.0135豆饼0.372.404024924.15.13.256菜子饼0.321.623601138.17.15.38.47鱼粉1.542.8045029.111.863278槐叶粉0.381.6117010810.62.2449DL-蛋氨酸2398010骨粉0.5630014011碳酸钙1.1240012食盐0.4210001

15、3花生饼0.62.43812000.920.150.17可建立肉用种鸡成本最低的配料方案模型如下问题（3）的lingo程序如下：Min=0.68*x1+0.72*x2+0.23*x3+0.22*x4+0.37*x5+0.32*x6+1.54*x7+0.38*x8+23*x9+0.56*x10+1.12*x11+0.42*x12+0.6*x13;3.35*x1+3.08*x2+1.78*x3+2.10*x4+2.40*x5+1.62*x6+2.80*x7+1.61*x8+2.4*x13=2.7;78*x1+114*x2+142*x3+117*x4+402*x5+360*x6+450*x7+170

16、*x8+38*x13=135;78*x1+114*x2+142*x3+117*x4+402*x5+360*x6+450*x7+170*x8+38*x13=145;16*x1+22*x2+95*x3+72*x4+49*x5+113*x6+108*x8+120*x13=5.6;1.2*x1+1.7*x2+2.3*x3+2.7*x4+5.1*x5+7.1*x6+11.8*x7+2.2*x8+980*x9+0.92*x13=2.6;0.7*x1+0.6*x2+0.3*x3+1.0*x4+3.2*x5+5.3*x6+63*x7+4.0*x8+300*x10+400*x11+0.15*x13=30;0.3

17、*x1+0.34*x2+10.0*x3+13.0*x4+5.0*x5+8.4*x6+27*x7+4.0*x8+140*x10+0.17*x13=5;1000*x12=3.7;X1+x2+x3+x4+x5+x6+x7+x8+x9+x10+x11+x12+x13=1;X1=0.4;x2=0.1;x3=0.1;x4=0.15;x5=0.03;x7=0.05;x8=0.03;问题（3）Lingo程序的结果：Global optimal solution found. Objective value: 0.6553693 Infeasibilities: 0.000000 Total solver it

18、erations: 11 Variable Value Reduced Cost X1 0.5385030 0.000000 X2 0.1000000 0.000000 X3 0.1000000 0.000000 X4 0.000000 0.1446276 X5 0.7213126E-01 0.000000 X6 0.3000000E-01 0.000000 X7 0.5000000E-01 0.000000 X8 0.3000000E-01 0.000000 X9 0.3233949E-03 0.000000 X10 0.4263719E-01 0.000000 X11 0.3270518E

19、-01 0.000000 X12 0.3700000E-02 0.000000 X13 0.000000 0.4351151 Row Slack or Surplus Dual Price 1 0.6553693 -1.000000 2 0.000000 -0.5218799 3 0.000000 -0.2339449E-03 4 10.00000 0.000000 5 14.51952 0.000000 6 0.3329203 0.000000 7 0.000000 -0.2461224E-01 8 0.000000 -0.5600000E-02 9 4.247413 0.000000 10

20、 0.000000 -0.1540000E-02 11 0.000000 1.120000 12 0.1385030 0.000000 13 0.000000 -0.1607394 14 0.000000 -0.3295455 15 0.1500000 0.000000 16 0.2786874E-01 0.000000 17 0.000000 -0.3059075 18 0.000000 -0.4502367 19 0.000000 -0.5434558所以公司采购花生饼后，肉用鸡成本最低为0.655元，其中玉米约需要0.539kg，小麦约需要0.1kg，麦麸约需要0.1kg，米糠约需要0k

21、g，豆饼约需要0.072kg，菜子饼约需要0.03kg，鱼粉约需要0.05kg，槐叶粉约需要0.03kg，DL蛋氨酸约需要0.00032kg，骨粉约需要0.0426kg，碳酸钙约需要0.0327kg，食盐约需要0.0037kg，花生饼需要0kg,4.4问题（4）的模型建立与求解根据问题（4）所给数据及问题要求可列出约束条件，所以可建立产蛋鸡的最优饲料配方方案的线性规划模型如下：问题（4）的lingo程序如下：Min=0.68*x1+0.72*x2+0.23*x3+0.22*x4+0.37*x5+0.32*x6+1.54*x7+0.38*x8+23*x9+0.56*x10+1.12*x11+0.

22、42*x12;3.35*x1+3.08*x2+1.78*x3+2.10*x4+2.40*x5+1.62*x6+2.80*x7+1.61*x8=2.65;78*x1+114*x2+142*x3+117*x4+402*x5+360*x6+450*x7+170*x8=151;16*x1+22*x2+95*x3+72*x4+49*x5+113*x6+108*x8=6.8;1.2*x1+1.7*x2+2.3*x3+2.7*x4+5.1*x5+7.1*x6+11.8*x7+2.2*x8+980*x9=6;0.7*x1+0.6*x2+0.3*x3+1.0*x4+3.2*x5+5.3*x6+63*x7+4.0*x8+300*x10+400*x11=33;0.3*x1+0.34*x2+10.0*x3+13.0*x4+5.0*x5+8.4*x6+27*x7+4.0*x8+140*x10=3;1000*x12=3;X1+x2+x3+x4+x5+x6+x7+x8+x9+x10+x11+x12=1;X1=0.4;x2=0.1;x3=0.1;x4=0.15;x5=0.03;x7=0.05;x8=0.03;问题（4）Lingo程序的结果：Global optimal solution found. Objective value: 0.8603805 Infeasibilities: