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Weibull function in Excel



 
 
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  #1  
Old June 5th, 2010, 02:44 AM posted to microsoft.public.excel.worksheet.functions
Barbo
external usenet poster
 
Posts: 3
Default Weibull function in Excel

I have a survival dataset (see below). I want to fit to Weibull function
S(t)=EXP(-alpha*Time^gamma). If I use Excel Solver, alpha=0.00367 and
gamma=2.32. If I use a method tranforming probability to Ln(Ln(1/S(t)) which
will be a linear function of Ln (t). But I got a different alpha and gamma.
When I compared two alphas adn gammas, the Solver results had a better
goodness of fit. What is the problem? THanks

Month Probability of survival
0 1
1 1
2 0.92
3 0.90
4 0.88
5 0.83
6 0.76
7 0.74
8 0.69
9 0.57
10 0.48
11 0.39
12 0.27
13 0.27
14 0.13

  #2  
Old June 5th, 2010, 04:36 AM posted to microsoft.public.excel.worksheet.functions
Mike Middleton[_2_]
external usenet poster
 
Posts: 90
Default Weibull function in Excel

Barbo -

What is the problem?


There is no problem. You should expect a difference.

Your Solver method (which I prefer) minimizes sum of squared deviations
between actual S and fitted S.

The other method uses transformed values, so it does not yield a better fit.

(Excel's trendline features use transformations to fit the logarithmic,
power, and exponential functions. The approach appears to be a computational
convenience. Better fits are obtained using Solver.)

- Mike
http://www.MikeMiddleton.com



"Barbo" wrote in message
...
I have a survival dataset (see below). I want to fit to Weibull function
S(t)=EXP(-alpha*Time^gamma). If I use Excel Solver, alpha=0.00367 and
gamma=2.32. If I use a method tranforming probability to Ln(Ln(1/S(t))
which
will be a linear function of Ln (t). But I got a different alpha and
gamma.
When I compared two alphas adn gammas, the Solver results had a better
goodness of fit. What is the problem? THanks

Month Probability of survival
0 1
1 1
2 0.92
3 0.90
4 0.88
5 0.83
6 0.76
7 0.74
8 0.69
9 0.57
10 0.48
11 0.39
12 0.27
13 0.27
14 0.13

  #3  
Old June 5th, 2010, 02:51 PM posted to microsoft.public.excel.worksheet.functions
Barbo
external usenet poster
 
Posts: 3
Default Weibull function in Excel

Hi Mike, thanks for the reply. So if I want to find a function with best fit,
adding a trendline is not an proper method.

I got the weibull function estimates from someone else and then try to
replicate the results in Excel. When I used Solver, most of the time I can
get the parameter estimates with the best fit. I was told there is no
feasible solution. How to properly use Solver? Thanks

Barbo


"Mike Middleton" wrote:

Barbo -

What is the problem?


There is no problem. You should expect a difference.

Your Solver method (which I prefer) minimizes sum of squared deviations
between actual S and fitted S.

The other method uses transformed values, so it does not yield a better fit.

(Excel's trendline features use transformations to fit the logarithmic,
power, and exponential functions. The approach appears to be a computational
convenience. Better fits are obtained using Solver.)

- Mike
http://www.MikeMiddleton.com



"Barbo" wrote in message
...
I have a survival dataset (see below). I want to fit to Weibull function
S(t)=EXP(-alpha*Time^gamma). If I use Excel Solver, alpha=0.00367 and
gamma=2.32. If I use a method tranforming probability to Ln(Ln(1/S(t))
which
will be a linear function of Ln (t). But I got a different alpha and
gamma.
When I compared two alphas adn gammas, the Solver results had a better
goodness of fit. What is the problem? THanks

Month Probability of survival
0 1
1 1
2 0.92
3 0.90
4 0.88
5 0.83
6 0.76
7 0.74
8 0.69
9 0.57
10 0.48
11 0.39
12 0.27
13 0.27
14 0.13

.

  #4  
Old June 6th, 2010, 12:19 AM posted to microsoft.public.excel.worksheet.functions
Mike Middleton[_2_]
external usenet poster
 
Posts: 90
Default Weibull function in Excel

Barbo -

How to properly use Solver?


Solver's success with nonlinear functions may depend on the initial values
for the changing cells.

For simple functions, like the Weibull, you may have some idea of reasonable
initial values.

- Mike
http://www.MikeMiddleton.com




"Barbo" wrote in message
...
Hi Mike, thanks for the reply. So if I want to find a function with best
fit,
adding a trendline is not an proper method.

I got the weibull function estimates from someone else and then try to
replicate the results in Excel. When I used Solver, most of the time I can
get the parameter estimates with the best fit. I was told there is no
feasible solution. How to properly use Solver? Thanks

Barbo


"Mike Middleton" wrote:

Barbo -

What is the problem?


There is no problem. You should expect a difference.

Your Solver method (which I prefer) minimizes sum of squared deviations
between actual S and fitted S.

The other method uses transformed values, so it does not yield a better
fit.

(Excel's trendline features use transformations to fit the logarithmic,
power, and exponential functions. The approach appears to be a
computational
convenience. Better fits are obtained using Solver.)

- Mike
http://www.MikeMiddleton.com



"Barbo" wrote in message
...
I have a survival dataset (see below). I want to fit to Weibull
function
S(t)=EXP(-alpha*Time^gamma). If I use Excel Solver, alpha=0.00367 and
gamma=2.32. If I use a method tranforming probability to Ln(Ln(1/S(t))
which
will be a linear function of Ln (t). But I got a different alpha and
gamma.
When I compared two alphas adn gammas, the Solver results had a better
goodness of fit. What is the problem? THanks

Month Probability of survival
0 1
1 1
2 0.92
3 0.90
4 0.88
5 0.83
6 0.76
7 0.74
8 0.69
9 0.57
10 0.48
11 0.39
12 0.27
13 0.27
14 0.13

.

 




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