Here is a reply by Bart:
Yes you're right (I should have taken off my glasses and looked closer).
However, the argument is essentially the same:
Suppose you have a solution with a,b,k,l. Then for any positive c, [a+b-bc]
+ [bc] + (bc) *exp(kl')exp(-kx) is also a solution, where l'
= l - log(c)/k
On 2011-03-01 06:38, Schatzi wrote:
Here is a reply by Bart:
Yes you're right (I should have taken off my glasses and looked closer).
However, the argument is essentially the same:
Suppose you have a solution with a,b,k,l. Then for any positive c, [a+b-bc]
+ [bc] + (bc) *exp(kl')exp(-kx) is
I am running the following nls equation. I tried it with data that excel was
fitting and got the error:
singular gradient matrix at initial parameter estimates
I thought it was due to a low number of points (6), but when I create a
dataset, I get the same problem. If I remove the parameter a,
By the way, sorry about the reposts. I subscribed but wasn't sure what was
happening. Hopefully this worked.
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Well, since no one else has replied, let me state the obvious: The
model is nonidentifiable. It is of the form:
y = a + b(1 - exp(kl)) - b exp(-kx)
Suppose you have a solution. Suppose with your solution, a + b(1
-exp(kl)) = C, say. Then a = C - b(1-exp(kl)). Hence you have
generated an
I am not sure how you simplified the model to:
y = a + b(1 - exp(kl)) - b exp(-kx)
I tried simplifying it but only got to:
y = a + b - b * exp(kl) * exp(-kx)
I agree that the model must not be identifiable. That makes sense,
especially given that removing either a or l makes the model work. Can
On 2011-02-28 14:14, Schatzi wrote:
I am not sure how you simplified the model to:
y = a + b(1 - exp(kl)) - b exp(-kx)
I tried simplifying it but only got to:
y = a + b - b * exp(kl) * exp(-kx)
I agree that the model must not be identifiable. That makes sense,
especially given that removing
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