数学建模非线性最小二乘问题.doc

1、非线性最小二乘问题用最小二乘法计算：sets:quantity/1.15/: x,y;endsetsmin=sum(quantity: (a+b* EXP(c*x)-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运算结果为：Local optimal solution found.Objective value: 44.78049Extended solve steps: 5Total solve iterartions: 68 Variable Value Reduced Cost A 2.430177 0.000000 B 57.33209 0.000000 C -0.4460383E-01 0.000000由此得到a的值为2.430177，b的值为57.33209，c的值为-0.04460383。线性回归方程为y=2.430177+57.33209* EXP(-0.04460383*x) 用最小一乘法计算：程序如下：sets:quantity/1.15/: x,y;endsetsmin=sum(quantity: ABS(a+b*EXP(c*x)-y);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运算结果为：Linearization components added: Constraints: 60 Variables: 60 Integers: 15Local optimal solution found.Objective value: 20.80640Extended solver steps: 2Total solver iterations: 643 Variable Value Reduced Cost A 3.398267 0.000000 B 57.11461 0.000000 C -0.4752126e-01 0.000000由上可得a的值为3.398267，b的值为57.11461，c的值为-0.04752126。线性回归方程：Y=3.398267+57.11461* EXP(-0.04752126*x)用最大偏差最小的方法计算：程序如下：sets:quantity/1.15/: x,y;endsetsmin= MAX(quantity: ABS(a+b*EXP(c*x)-y);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运行结果为：Linearization components added: Constraints: 91 Variables: 76 Integers: 30Local optimal solution found.Objective value: 3、059550Extended solver steps: 26Total solver iterations: 4843 Variable Value Reduced Cost A 3.701164 0.000000 B 54.54622 0.000000 C -0.4645980e-01 0.000000由上可得，a的值为3.701164，b的值为54.54622，c的值为-0.0464598。线性回归方程：y=3.701164+54.54622* EXP(-0.0464598*x)2.1）这个问题可以这么假设，用来生产汽油的A原油为x1桶,用来生产汽油的B原油为y1桶,用来生产民用燃料的A原油为x2桶,用来生产民用燃料的B原油为y2桶，那么就可以生产x1 +y1桶汽油，x2 +y2桶民用燃料。由题意可以知道，每桶汽油需要广告费2元，每桶民用燃料需要广告费1元。生产出来之后可以卖出的总额为250（x1 +y1）+200（x2 +y2），那么要求的目标就知道了总利润为250*(x1+y1)+200*(x2+y2)-2*(x1+y1)-(x2+y2);加上应有的约束，满足质量指数关系，编写出相应的林公程序如下:max=250*(x1+y1)+200*(x2+y2)-2*(x1+y1)-(x2+y2);(10*x1+5*y1)/(x1+y1)=8;(10*x2+5*y2)/(x2+y2)=6;(10*x2+5*y2)/(x2+y2)=8;x1+y1=5000;x2+y2=8;(1+0.6*z20.6)*(10*x2+5*y2)/(x2+y2)=6;(1+0.6*z20.6)*(10*x2+5*y2)/(x2+y2)=8;x1+y1=5000;x2+y2=10000;z1=(x1+y1)/20;z2=400;运行一下程序：max=250*(x1+y1+z1)+200*(x2+y2+z2)-2*(x1+y1+z1)-(x2+y2+z2)-100*(z1+z2); (1+z10.5)*(10*x1+5*y1)/(x1+y1)=8;(1+0.6*z20.6)*(10*x2+5*y2)/(x2+y2)=6;x1+y1=5000;x2+y2=10000;z1=(x1+y1)/20;z2=400;GIN(x1);GIN(y1);GIN(x2);GIN(y2);GIN(z1);GIN(z2); 结果; Local optimal solution found. Objective value: 3316500. Extended solver steps: 12 Total solver iterations: 176 Variable Value Reduced Cost X1 4830.000 -248.0000 Y1 170.0000 -248.0000 Z1 250.0000 -148.0000 X2 3478.000 -199.0000 Y2 6522.000 -199.0000 Z2 500.0000 -99.00000 Row Slack or Surplus Dual Price 1 3316500. 1.000000 2 157.2559 0.000000 3 169.0563 0.000000 4 0.000000 0.000000 5 0.000000 0.000000 6 0.000000 0.000000 7 0.000000 0.000000 8 350.0000 0.000000当Q又折扣的时候，结果如黄色区域内所示，此时最大利润为3316500.加分题记工地的位置为(ai，bi)，水泥日用量为di，i=1,6;料场位置为(xj，yj)，日储量为ej，j=1,2；从料场j向工地i的运送量为Xij。当用临时料场时决策变量为：Xij，当不用临时料场时决策变量为：Xij，xj，yj。 使用临时料场的情形 使用两个临时料场A(5,1)，B(2,7).求从料场j向工地i的运送量为Xij，在各工地用量必须满足和各料场运送量不超过日储量的条件下，使总的吨千米数最小，这是线性规划问题. 线性规划模型为：设X11=X1, X21= X 2, X31= X 3, X41= X 4, X51= X 5, X61= X 6 X12= X 7, X22= X 8, X32= X 9, X42= X 10, X52= X 11, X62= X 12 计算结果为： x = 3.0000 5.0000 0.0000 7.0000 0.0000 1.0000 0.0000 0.0000 4.0000 0.0000 6.0000 10.0000fval = 136.2275 改建两个新料场的情形 改建两个新料场，要同时确定料场的位置(xj,yj)和运送量Xij，在同样条件下使总吨千米数最小。这是非线性规划问题。非线性规划模型为：设 X11=X1, X21= X 2, X31= X 3, X41= X 4, X51= X 5, X61= X 6 X12= X 7, X22= X 8, X32= X 9, X42= X 10, X52= X 11, X62= X 12 x1=X13, y1=X14, x2=X15, y2=X16 计算结果为：x= 3.0000 5.0000 0.0707 7.0000 0 0.9293 0 0 3.9293 0 6.0000 10.0707 6.3875 4.3943 5.7511 7.1867fval = 105.4626exitflag = 1(4) 若修改主程序gying2.m, 取初值为上面的计算结果:x0= 3.0000 5.0000 0.0707 7.0000 0 0.9293 0 0 3.9293 0 6.0000 10.0707 6.3875 4.3943 5.7511 7.1867得结果为:x=3.0000 5.0000 0.3094 7.0000 0.0108 0.6798 0 0 3.6906 0 5.9892 10.3202 5.5369 4.9194 5.8291 7.2852fval =103.4760exitflag = 1总的吨千米数比上面结果略优. (5) 若再取刚得出的结果为初值, 却计算不出最