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是什么原因导致在SAS中采用“Jacobian矩阵”。
原标题:What causes "Jacobian matrix" to be singular in SAS?

我有一个简单的SAS(第9.2章)方案,如下:

proc model;
cdf( normal ,log(V/100)+1)=0.5;
bounds V>0;
solve V/solveprint;
run;

它 exception倒了一种例外,即说亚科比亚矩阵是单体,

The Newton method Jacobian matrix of partial derivatives of the
equations with respect to the variables to be solved is singular.

这一错误的可能原因是什么?

最新情况:我把这个问题简化了一步。 如果改为“cdf(正常,X)=0.5”,则无一例外。

更新2:约束更新至V>0;但例外情况仍然存在

最佳回答

什么投入数据集? 例如,该法典一贯运作:

data a;
 v=100;
run;

proc model data=a;
  cdf( normal ,log(V/100)+1) = 0.5;
  bounds V>0;
  solve V / solveprint;
run;
quit;

并提供<代码>V=36.78794的解决办法。

但是,对投入数据作一些改动(见下文),将始终造成Jacobian矩阵的错误。

data a;
 v=0.00001;
run;

proc model data=a;
  cdf( normal ,log(V/100)+1) = 0.5;
  bounds V>0;
  solve V / solveprint;
run;
quit;
问题回答

你们要求《联合国系统会计准则》解决一项没有解决办法的职能。 你们要求的是<代码>V>1000,使这一方程式成为现实。 但是,没有这种数值,因为<代码>log(1000/100+1)大约为3.3,而CDF的正常随机变量为0,标准偏离1,按3.3评价为0.9995。 <代码>的任何较大价值 V只会使这一功能更接近1,而不是转移到0.5,因此对你的问题没有答案。

通过告诉你,部分衍生物的矩阵是单体的,SAS正使用正数字表示“功能没有解决办法”。 (从它所说的话来看,“我把你的问题变成了同等的最大限度的问题,这个问题没有最大的问题,因此我可以帮助你”。)





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