How to solve a system of equations involving distribution?

2011 12 26
2 个回答
回答已采纳
4 次查看（30 天）

 采纳的回答

Amin Zargar
Amin Zargar 2011-12-26

0 个投票

this worked:
F = [gamcdf(6.63,x(1),x(2))-0.4;
gamcdf(8.89,x(1),x(2))-0.6];
x0 = [2; 0.8];
options=optimset('Display','iter');
[x,fval] = fsolve(@SPI05,x0,options)
spasiba Andrei !

1 个评论
显示 -1更早的评论 隐藏 -1更早的评论

Walter Roberson
Walter Roberson 2011-12-26
Okay, but what do the 6.63 and 8.89 have to do with the requirements to use 3.01 and 4.63 ? And what is SPI05 ?
Also, please cross-check the solutions. When I was testing here, I found that expressing the solution in that form (as a difference), allowed an infinite number of solutions in which the gamcdf() values equations were both wrong by the same amount.

请先登录，再进行评论。

更多回答（1 个）

Andrei Bobrov
Andrei Bobrov 2011-12-26

0 个投票

EDIT [20:23MSD 26.12.2011]:
fh1 = @(x)[gamcdf(3.01,x(1),x(2));gamcdf(4.63,x(1),x(2))] - [.4;.6];
out = fsolve(fh1,[1;1])
Equation solved.
fsolve completed because the vector of function values is near zero
as measured by the default value of the function tolerance, and
the problem appears regular as measured by the gradient.
<stopping criteria details>
out =
1.6466
2.8322

2 个评论
显示 无 隐藏 无

Amin Zargar
Amin Zargar 2011-12-26
Hi Andrei, I get this answer ..
Warning: Matrix is singular, close to singular or badly scaled.
Results may be inaccurate. RCOND = NaN.
> In optim\private\dogleg at 72
In optim\private\trustnleqn at 186
In fsolve at 377
No solution found.
fsolve stopped because the problem appears to be locally singular.
<stopping criteria details>
Walter Roberson
Walter Roberson 2011-12-26
Amin, which MATLAB version are you using?

请先登录，再进行评论。

类别

在 帮助中心 和 File Exchange 中查找有关 Loops and Conditional Statements 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by