非线性拟合的1stOpt与Lingo实现

shamohu

非线性拟合也是一优化问题，即残差平方和最小，Lingo因而可用于曲线拟合。Lingo也不需要赋初值。
拟合数据：
X = [0.25,0.5,0.75,1,1.5,2,2.5,3,3.5,4,4.5,5,6,7,8,9,10,11,12,13,14,15,16],
Y = [30,68,75,82,82,77,68,68,58,51,50,41,38,35,28,25,18,15,12,10,7,7,4];
拟合公式：
y = A1*(Exp(-A2*X)-Exp(-A3*X))

Lingo代码：
Model:
   Sets:
       BAC/R1..R23/:X,Y;
   EndSets
   Data:
       X = 0.25,0.5,0.75,1,1.5,2,2.5,3,3.5,4,4.5,5,6,7,8,9,10,11,12,13,14,15,16;
       Y = 30,68,75,82,82,77,68,68,58,51,50,41,38,35,28,25,18,15,12,10,7,7,4;
   EndData
   Min = @Sum(BAC: A1*(@Exp(-A2*X)-@Exp(-A3*X))-y)^2);
End

1stOpt代码：
Constant X = [0.25,0.5,0.75,1,1.5,2,2.5,3,3.5,4,4.5,5,6,7,8,9,10,11,12,13,14,15,16],
             Y = [30,68,75,82,82,77,68,68,58,51,50,41,38,35,28,25,18,15,12,10,7,7,4];
MinFunction Sum(i=1:23,X,Y)((A1*(Exp(-A2*X)-Exp(-A3*X))-y)^2);

Lingo不需初值，成功率不是很高（有兴趣可自己试试），还有如果想赋初值，在Lingo里不知如何设定。1stOpt也不需初值，本题成功率为100％。有兴趣者也可用其它数学工具练练手，对比一下。

FreddyMusic

This is Mathematica.

What is your result with 1stOpt ?

wanglei5201118

优化算法: 麦夸特法(Levenberg-Marquardt) + 通用全局优化法
函数表达式: (((a1*(exp(-a2*(0.25))-exp(-a3*(0.25)))-(30))^2)+((a1*(exp(-a2*(0.5))-exp(-a3*(0.5)))-(68))^2)+((a1
          *(exp(-a2*(0.75))-exp(-a3*(0.75)))-(75))^2)+((a1*(exp(-a2*(1))-exp(-a3*(1)))-(82))^2)+((a1*(exp(-a2
          *(1.5))-exp(-a3*(1.5)))-(82))^2)+((a1*(exp(-a2*(2))-exp(-a3*(2)))-(77))^2)+((a1*(exp(-a2*(2.5))-exp(
          -a3*(2.5)))-(68))^2)+((a1*(exp(-a2*(3))-exp(-a3*(3)))-(68))^2)+((a1*(exp(-a2*(3.5))-exp(-a3*(3.5)))
          -(58))^2)+((a1*(exp(-a2*(4))-exp(-a3*(4)))-(51))^2)+((a1*(exp(-a2*(4.5))-exp(-a3*(4.5)))-(50))^2)+((a1
          *(exp(-a2*(5))-exp(-a3*(5)))-(41))^2)+((a1*(exp(-a2*(6))-exp(-a3*(6)))-(38))^2)+((a1*(exp(-a2*(7))-exp(
          -a3*(7)))-(35))^2)+((a1*(exp(-a2*(8))-exp(-a3*(8)))-(28))^2)+((a1*(exp(-a2*(9))-exp(-a3*(9)))-(25))^2)
          +((a1*(exp(-a2*(10))-exp(-a3*(10)))-(18))^2)+((a1*(exp(-a2*(11))-exp(-a3*(11)))-(15))^2)+((a1*(exp(
          -a2*(12))-exp(-a3*(12)))-(12))^2)+((a1*(exp(-a2*(13))-exp(-a3*(13)))-(10))^2)+((a1*(exp(-a2*(14))-exp(
          -a3*(14)))-(7))^2)+((a1*(exp(-a2*(15))-exp(-a3*(15)))-(7))^2)+((a1*(exp(-a2*(16))-exp(-a3*(16)))-(4))^2)
          )
目标函数值(最小): 225.341698421196
a1: 114.43252478145
a2: 0.185501959293161
a3: 2.00793803739701

mathcd

以后应该有一些机会将各种科学软件的各方面的能力加以比较，这样可以学到很大东西，眼界也会更开阔。

wanglei5201118

说的对阿，1stopt有很强大的功能，希望各位网友共同讨论阿