R help - Jul 2020 - Error in chol.default

Hello,
I am trying to fit a Richards model to some cumulative incidence data
of infection. I got this example:
```
rich = function(p, x) {
  a = p["curvature"]
  k = p["finalPop"]
  r = p["growthRate"]
  y = r * x * (1-(x/k)^a)
  return(y)
}
ricky = function(p, x, y) p$r * x * (1-(x/p$k)^p$a) -y
# values
Y <- c(41,   41,   41,   41,   41,   41,   45,   62,  121,  198,  275,
 288,  859, 1118)
X <- 1:14
a = 1
k = 83347
r = 40
fit = rich(c(curvature=a, finalPop=k, growthRate=r), X)
plot(Y ~ X,
     col = "red", lwd = 2,
     main = "Richards model",
     xlab = expression(bold("Days")),
     ylab = expression(bold("Cases")))
points(X, fit, type = "l", lty = 2, lwd = 2)
library("minpack.lm")
o <- nls.lm(par = list(a=a, k=k, r=r), fn = ricky, x = X, y = Y)
summary(o)
```
When I run the optimization (given that I can't find parameters that
fit the data by eyeball), I get the error:
```
Error in chol.default(object$hessian) :
  the leading minor of order 1 is not positive definite
```
What is this about?
Thank you




-- 
Best regards,
Luigi

The error msg says it all if you know how to read it.
> When I run the optimization (given that I can't find parameters that
> fit the data by eyeball), I get the error:
> ```
> Error in chol.default(object$hessian) :
>   the leading minor of order 1 is not positive definite

Your Jacobian (derivatives of the residual function w.r.t. the parameters)
is singular -- rather spectacularly so.

Try the short addition to your code to use analytic derivatives:

rich = function(p, x) {
  a = p["curvature"]
  k = p["finalPop"]
  r = p["growthRate"]
  y = r * x * (1-(x/k)^a)
  return(y)
}
ricky = function(p, x, y) p$r * x * (1-(x/p$k)^p$a) -y
# values
Y <- c(41,   41,   41,   41,   41,   41,   45,   62,  121,  198,  275,
       288,  859, 1118)
X <- 1:14
a = 1
k = 83347
r = 40
fit = rich(c(curvature=a, finalPop=k, growthRate=r), X)
plot(Y ~ X,
     col = "red", lwd = 2,
     main = "Richards model",
     xlab = expression(bold("Days")),
     ylab = expression(bold("Cases")))
points(X, fit, type = "l", lty = 2, lwd = 2)
library("minpack.lm")
o <- nls.lm(par = list(a=a, k=k, r=r), fn = ricky, x = X, y = Y)
summary(o)
print(o1)
library(nlsr)
xy=data.frame(x=X, y=Y)
rcky2 = y ~ r * x * (1-(x/k)^a) -y
o1 <- nlxb(rcky2, start = c(a=a, k=k, r=r), data=xy, trace=TRUE)


You should find
> o1
nlsr object: x
residual sumsquares =  3161769  on  14 observations
    after  8    Jacobian and  13 function evaluations
  name            coeff          SE       tstat      pval      gradient   
JSingval
a                294.113            NA         NA         NA           0      
31.86
k                  83347            NA         NA         NA           0        
0
r                74.9576            NA         NA         NA  -6.064e-05        
0
>
Note the singular values of the Jacobian. Actual zeros!

Even so, nlsr had managed to make some progress.

nls() and nls.lm() use approximate derivatives. Often that's fine (and it is
more flexible
in handling awkward functions), but a lot of the time it is NOT OK.

JN