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

0
Utf
6/5/2010 1:44:21 AM
excel.worksheet.functions 4936 articles. 2 followers. Follow

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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
Mike@DecisionToolworks.com


"Barbo" <Barbo@discussions.microsoft.com> wrote in message 
news:CFCE454F-D077-4526-BFE5-66902FFE0A12@microsoft.com...
> 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
> 
0
Mike
6/5/2010 3:36:07 AM
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
> Mike@DecisionToolworks.com
> 
> 
> "Barbo" <Barbo@discussions.microsoft.com> wrote in message 
> news:CFCE454F-D077-4526-BFE5-66902FFE0A12@microsoft.com...
> > 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
> > 
> .
> 
0
Utf
6/5/2010 1:51:03 PM
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
Mike@DecisionToolworks.com



"Barbo" <Barbo@discussions.microsoft.com> wrote in message 
news:BD2F239D-B24A-49BD-9274-B09576D1F741@microsoft.com...
> 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
>> Mike@DecisionToolworks.com
>>
>>
>> "Barbo" <Barbo@discussions.microsoft.com> wrote in message
>> news:CFCE454F-D077-4526-BFE5-66902FFE0A12@microsoft.com...
>> > 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
>> >
>> .
>> 
0
Mike
6/5/2010 11:19:02 PM
Reply:

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