数学建模非线性最小二乘问题.docx

1.非线性最小二乘问题用最小二乘法计算：sets:quantity/l・・15/: x,y;en dsetsmin=@sum(quantity: (a+b* @EXP(c*x)-y)A2); @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:68VariableValueReduced CostA2.4301770.000000B57.332090.000000C-0.4460383E-010.000000由此得到a的值为2.430177, b的值为57.33209, c的值为-0.04460383o 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/l.・15/: x,y;endsetsmin=@sum(quantity: @ABS(a+b*@EXP(c*x)-y));@free(a); @free(b);@free(c);data: x=2,5,7,10,14,19,26,31,34,38,45,5乙 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: 60Variables: 60Integers:15Local optimal solution found.Objective value:20.80640Extended solver steps:Total solver iterations:643VariableValueReduced CostA3.3982670.000000B57.114610.000000c-0.4752126e-010.000000由上可得a的值为3.398267, b的值为57.11461, c的值为-0.04752126。

线性冋归方程：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/l・・15/: x,y;endsetsmin= @MAX(quantity: @ABS(a+b*@EXP(c*x)-y)); @free(a); @free(b);@free(c);data:x=2,5,7,10,14,19,26,31,34,3&45,5 乙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:Con straints:91Variables:76Integers:Local optimal solution found.Objective value:Extended solver steps:Total solver iterations:303、 059550264843Variable Value Reduced CostA 3.701164 0.000000B 54.54622 0.000000C -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xl桶，用来生产汽油的B原油为yl桶，用來生产民用燃料的A原油为X2桶，用來生产民用燃料的B原油为y2桶，那么就可以生产xl +yl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 (xl +yl) +200 (x2 +y2),那么要求的目标就知道了总利润为 250*(xl+yl)+200*(x2+y2)-2*(xl+yl)-(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*(xl+yl)+200*(x2+y2)-2*(xl+yl)-(x2+y2);(10*xl+5*yl)/(xl+yl)>=8;(10*x2+5*y2)/(x2+y2)>=6;(10*x2+5*y2)/(x2+y2)<=8;xl+yl<=5000;x2+y2<=10000;@GIN(xl);@GIN(yl);@GIN(x2);@GIN(y2);计算结果是：Local optimal solution found at iteration: 211Objective value: 3230000.RowSlack or SurplusDual PriceVariableValueReduced CostXl3000.0000.000000Yl2000.0000.000000X25000.0000.000000Y25000.0000.00000013230000.1.00000020.0000000.00000031.5000000.00000040.50000000.00000050.000000248.000060.000000199.0000由此可知:用3000桶A类原油和2000桶B类原油化合成汽油，用2000桶A类和 8000桶B类花和成民用燃料油，同时花费10000元用作民用燃料油广告，汽油 销售量能竺到5000桶，民用燃料油的销售量能达到100000桶，获得最大利润为 3230000 元。

2)设在汽油中加入的Q为zl桶，在民用燃料中加入的为Z2桶，那么相应的质量指数就 是：汽油(l+zlA0.5)*(10*xl+5*yl)/(xl+yl)民用燃料(l+0.6*z2A0.6)*(10*x2+5*y2)/(x2+y2)这次的成本中就要减去Q的成本，再次列出程序如下：max=250*(xl+yl+zl)+200*(x2+y2+z2)-2*(xl+yl+zl)-(x2+y2+z2)-200*(zl+z2);(l+zlA0.5)*(10*xl+5*yl)/(xl+yl)>=8;(l+0.6*z2A0.6)*(10*x2+5*y2)/(x2+y2)>=6;(l+0.6*z2A0.6)*(10*x2+5*y2)/(x2+y2)<=8;xl+yl<=5000;x2+y2<=10000;zl<=(xl+yl)/2O;z2<=(x2+y2)/2O;@GIN(xl);@GIN(yl);@GIN(x2);@GIN(y2);@GIN(zl);@GIN(z2);得到的一种结果如下283242000.Local optimal solution found at iteration:Objective value:VariableValueReduced CostXI2303.0000.000000Y12697.0000.000000Z1250.00000.000000X25000.0000.000000Y25000.0000.000000Z20.0000001.000000RowSlack or SurplusDual Price13242000.1.0000002114.77360.00000031.5000000.00000040.50000000.00000050.000000250.400060.000000199.000070.00000048.000008500.00000.000000这时的战略方法应该是黄色区域的方法，最大的利润是3242000(3)购货大于400时,再进行约束zl+z2>=400;运行一下程序：max=250*(xl+yl+zl)+200*(x2+y2+z2)-2*(xl+yl+zl)-(x2+y2+z2)-100*(zl+z2);(l+zlA0.5)*(10*xl+5*yl)/(xl+yl)>=8;(l+0.6*z2A0.6)*(10*x2+5*y2)/(x2+y2)>=6; xl+yl<=5000;x2+y2<=10000;zl<=(xl+yl)/20;z2<=(x2+y2)/20;zl+z2>=400;@GIN(xl);@GIN(yl);@GIN(x2);@GIN(y2);@GIN(zl);@GIN(z2);结果；3316500.12176Local optimal solution found.Objective value:Extended solver steps: Total solver iterations:VariableValueReduced CostXI4830.000-248.0000Y1170.0000-248.0000Z1250.0000-148.0000X23478.000・199.0000Y26522.000・ 199.0000Z2500.0000-99.00000RowSlack or SurplusDual Price13316500.1.0000002157.25590.0000003169.05630.00000040.0000000.00000050.0000000.00000060.0000000.00000070.0000000.0000008350.00000.000000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=l,...,6;料场位置为(xj, yj),日储量为ej,。