数学建模课后作业第四章.doc

第四章， 最优化与存储模型实验4.2，基本实验1.非线性最小二乘问题解：由题可以得出：（1）最小二乘法，分别用x、y代替、平方和的形式时的lingo程序：sets:quantity/1.15/: x,y;endsetsmin=sum(quantity: (b*2.71828(c*x)+a-y)2);free(a); free(b);free(c);data:x=2 5 7 10 14 19 26 31 34 38 45 52 53 60 65;y=54 50 45 37 35 25 20 16 18 13 8 11 8 4 6;enddata运行程序后可以得出： Local optimal solution found. Objective value: 44.78049 Infeasibilities: 0.000000 Extended solver steps: 5 Total solver iterations: 64 Model Class: NLP Total variables: 4 Nonlinear variables: 3 Integer variables: 0 Total constraints: 2 Nonlinear constraints: 1 Total nonzeros: 4 Nonlinear nonzeros: 3 Variable Value Reduced Cost B 57.33209 0.000000 C -0.4460386E-01 0.3922619E-07 A 2.430177 0.000000 X( 1) 2.000000 0.000000 X( 2) 5.000000 0.000000 X( 3) 7.000000 0.000000 X( 4) 10.00000 0.000000 X( 5) 14.00000 0.000000 X( 6) 19.00000 0.000000 X( 7) 26.00000 0.000000 X( 8) 31.00000 0.000000 X( 9) 34.00000 0.000000 X( 10) 38.00000 0.000000 X( 11) 45.00000 0.000000 X( 12) 52.00000 0.000000 X( 13) 53.00000 0.000000 X( 14) 60.00000 0.000000 X( 15) 65.00000 0.000000 Y( 1) 54.00000 0.000000 Y( 2) 50.00000 0.000000 Y( 3) 45.00000 0.000000 Y( 4) 37.00000 0.000000 Y( 5) 35.00000 0.000000 Y( 6) 25.00000 0.000000 Y( 7) 20.00000 0.000000 Y( 8) 16.00000 0.000000 Y( 9) 18.00000 0.000000 Y( 10) 13.00000 0.000000 Y( 11) 8.000000 0.000000 Y( 12) 11.00000 0.000000 Y( 13) 8.000000 0.000000 Y( 14) 4.000000 0.000000 Y( 15) 6.000000 0.000000 Row Slack or Surplus Dual Price 1 44.78049 -1.000000即a=2.4302,b=57.3321,c=-0.0446039y=2.4302+57.3321e(-0.0446039x)=2.4302+57.3321e-0.0446039（2）最小一乘法，分别用x、y代替、的形式时的lingo程序：sets: quantity/1.15/: x,y;endsetsdata: x = 2,5,7,10,14,19,26,31,34,38,45,52,53,60,65; y = 54,50,45,37,35,25,20,16,18,13,8,11,8,4,6;enddatamin=sum(quantity: abs(a+b*exp(c*x)-y);free(a); free(b); free(c);运行程序后可得： Linearization components added: Constraints: 60 Variables: 60 Integers: 15 Local optimal solution found. Objective value: 20.80640 Objective bound: 20.80640 Infeasibilities: 0.1219750E-06 Extended solver steps: 0 Total solver iterations: 69 Model Class: MINLP Total variables: 64 Nonlinear variables: 2 Integer variables: 15 Total constraints: 62 Nonlinear constraints: 15 Total nonzeros: 196 Nonlinear nonzeros: 30 Variable Value Reduced Cost A 3.398266 0.000000 B 57.11461 0.000000 C -0.4752126E-01 0.000000 X( 1) 2.000000 0.000000 X( 2) 5.000000 0.000000 X( 3) 7.000000 0.000000 X( 4) 10.00000 0.000000 X( 5) 14.00000 0.000000 X( 6) 19.00000 0.000000 X( 7) 26.00000 0.000000 X( 8) 31.00000 0.000000 X( 9) 34.00000 0.000000 X( 10) 38.00000 0.000000 X( 11) 45.00000 0.000000 X( 12) 52.00000 0.000000 X( 13) 53.00000 0.000000 X( 14) 60.00000 0.000000 X( 15) 65.00000 0.000000 Y( 1) 54.00000 0.000000 Y( 2) 50.00000 0.000000 Y( 3) 45.00000 0.000000 Y( 4) 37.00000 0.000000 Y( 5) 35.00000 0.000000 Y( 6) 25.00000 0.000000 Y( 7) 20.00000 0.000000 Y( 8) 16.00000 0.000000 Y( 9) 18.00000 0.000000 Y( 10) 13.00000 0.000000 Y( 11) 8.000000 0.000000 Y( 12) 11.00000 0.000000 Y( 13) 8.000000 0.000000 Y( 14) 4.000000 0.000000 Y( 15) 6.000000 0.000000 Row Slack or Surplus Dual Price 1 20.80640 -1.000000a=3.3983 , b=57.1146 , c=-0.04752;=3.3983+57.1146e-0.04752 ;（3）最大偏差最小法，分别用x、y代替、的形式时的lingo程序：sets: quantity/1.15/: x,y;endsetsdata: x = 2,5,7,10,14,19,26,31,34,38,45,52,53,60,65; y = 54,50,45,37,35,25,20,16,18,13,8,11,8,4,6;enddatamin=max(quantity: (a+b*exp(c*x)-y)2)0.5);free(a); free(b); free(c);运行程序可以得出： Linearization components added: Constraints: 31 Variables: 16 Integers: 15 Local optimal solution found. Objective value: 3.287687 Objective bound: 3.287687 Infeasibilities: 0.3513839E-05 Extended solver steps: 2 Total solver iterations: 666 Model Class: MINLP Total variables: 20 Nonlinear variables: 3 Integer variables: 15 Total constraints: 33 Nonlinear constraints: 30 Total nonzeros: 152 Nonlinear nonzeros: 90 Variable Value A 4.360430 B 53.99638 C -0.4858083E-01 X( 1) 2.000000 X( 2) 5.000000 X( 3) 7.000000 X( 4) 10.00000 X( 5) 14.00000 X( 6) 19.00000 X( 7) 26.00000 X( 8) 31.00000 X( 9) 34.00000 X( 10) 38.00000 X( 11) 45.00000 X( 12) 52.00000 X( 13) 53.00000 X( 14) 60.00000 X( 15) 65.00000 Y( 1) 54.00000 Y( 2) 50.00000 Y( 3) 45.00000 Y( 4) 37.00000 Y( 5) 35.00000 Y( 6) 25.00000 Y( 7) 20.00000 Y( 8) 16.00000 Y( 9) 18.00000 Y( 10) 13.00000 Y( 11) 8.000000 Y( 12) 11.00000 Y( 13) 8.000000 Y( 14) 4.000000 Y( 15) 6.000000 Row Slack or Surplus 1 3.287687由此可知最大偏差最小时相应曲线为：=4.3604+53.9964*e(-0.04858*x) 由此对比可知三种方法时的时候相应模拟曲线为：=2.4302+57.3321e-0.0446039 （下图绿线）;=4.3604+53.9964*e(-0.04858*x) (下图蓝线)；=3.3983+57.1146e-0.04752 ;（下图红线）；相应的曲线图与散点图Matlab程序为：x=2 5 7 10 14 19 26 31 34 38 45 52 53 60 65y=54 50 45 37 35 25 20 16 18 13 11 8 8 4 6plot(x,y,.);hold on;x=2:3:65;y=2.4302+57.3321.*2.71828.(-0.0446039.*x);plot(x,y,-g);hold on;x=2:3:65;y=3.3983+57.1146.*2.71828.(-0.04752.*x);plot(x,y, -r);hold on;x=2:3:65;y=4.3604+53.9964.*2.71828.(-0.04858.*x);plot(x,y, -b);hold on;相应的曲线图为：2.非线性优化问题：解：（1）令x11、x12、x13为汽油加工所需要的A、B类原油与广告费的值，令x21、x22、x23为民用燃油所需要的A、B类原油与广告费的值。相应的目标函数为总利润Z=250*x13/10*5+200*x23/10*10-x13-x23 =124*x13+199*x23约束条件为：(10x11+5x12)/(x11+x12)8;(10x21+5x22)/(x21+x22)6;x11+x215000;x12+x2210000;5*x13/10x11+x12;10*x23/10x21+x22;则可以得出对应的lingo程序：max=124*x13+199*x23;(10*x11+5*x12)/(x11+x12)=8;(10*x21+5*x22)/(x21+x22)=6;(10*x21+5*x22)/(x21+x22)=8;x11+x21=5000;x12+x22=0.5*x13;x21+x22=x23;GIN(x11);GIN(x12);GIN(x13);GIN(x21);GIN(x22);GIN(x23);运行程序后可得： Local optimal solution found. Objective value: 3230000. Objective bound: 3230000. Infeasibilities: 0.000000 Extended solver steps: 1 Total solver iterations: 44 Model Class: PINLP Total variables: 6 Nonlinear variables: 4 Integer variables: 6 Total constraints: 8 Nonlinear constraints: 3 Total nonzeros: 18 Nonlinear nonzeros: 6 Variable Value X13 10000.00 X23 10000.00 X11 3000.000 X12 2000.000 X21 2000.000 X22 8000.000 Row Slack or Surplus 1 3230000. 2 0.000000 3 0.000000 4 2.000000 5 0.000000 6 0.000000 7 0.000000 8 0.000000由以上可得最优的生产方案即：原油类别原油A/桶原油B/桶广告费/元总利润/元汽油/桶30002000100003230000民用燃油/桶2000800010000（2）由题目中条件可令加入汽油中的SQ为x14桶，可令加入民用燃油中的SQ为x24桶，则可得新的目标函数为：Z=250*x13/10*5+200*x23/10*10-x13-x23-200x14-200x24;约束条件为：x11+x215000;x12+x2210000;(10*x11+5*x12)/(x11+x12+x14)+(x14/(x11+x12+x14)0.5=8;(10x21+5x22)/(x21+x22+x24)+0.6*(x24/(x21+x22+x24)0.6=6;x13/10*5x11+x12+x14;x23/10*10x21+x22+x24;x14(x11+x12)*5%;x24(x21+x22)*5%;则可以得到新的lingo程序：max=124*x13+199*x23-200*x14-200*x24;(1+(x14/(x11+x12+x14)0.5)*(10*x11+5*x12)/(x11+x12)=8;(1+0.6*(x24/(x21+x22+x24)0.6)*(10*x21+5*x22)/(x21+x22)=6;x11+x21=5000;x12+x22=0.5*x13;x21+x22+x24=x23;0.05*x11+0.05*x12=x14;0.05*x21+0.05*x22=x24;GIN(x11);GIN(x12);GIN(x13);GIN(x14);GIN(x21);GIN(x22);GIN(x23);GIN(x24);运行程序后可得： Local optimal solution found. Objective value: 3755856. Objective bound: 3755856. Infeasibilities: 0.000000 Extended solver steps: 2 Total solver iterations: 902 Model Class: PINLP Total variables: 8 Nonlinear variables: 6 Integer variables: 8 Total constraints: 9 Nonlinear constraints: 2 Total nonzeros: 28 Nonlinear nonzeros: 6 Variable Value X13 31494.00 X23 0.000000 X14 747.0000 X24 0.000000 X11 5000.000 X12 10000.00 X21 0.000000 X22 0.000000 Row Slack or Surplus 1 3755856. 2 0.1186784 3 0.1000000E+31 4 0.000000 5 0.000000 6 0.000000 7 0.000000 8 3.000000则可以得到新的生产方案：原油类别原油A/桶原油B/桶广告费/元SQ/桶总利润/元汽油/桶500010000314947473755856民用燃油/桶0000(3)当x14