You could use a spline to interpolate the points.
(And I'd consider increasing the number of points if possible, say to 200).
Then use a root finder, such as uniroot(), to solve for
f(i) - k
Where, k (a constant), would be 1e6, based on your example.
There are a number of variations on this approach.
My kubik package provides a solve method, and can impose some constraints.
----
library (kubik)
f <- chs (1:45, round (PCR_array),
constraints = chs.constraints (increasing=TRUE) )
plot (f)
sol <- solve (f, 1e6)
abline (v=sol, lty=2)
sol
----
Note that I had to round the values, in order to impose a
non-decreasing constraint.
Also note that I've just used the 45 points.
But re-iterating, you should increase the number of points, if possible.
On Mon, Jan 25, 2021 at 8:58 AM Luigi Marongiu <marongiu.lu...@gmail.com> wrote:
>
> Hello
> I am trying to simulate a PCR by running a logistic equation. So I set
> the function:
> ```
> PCR <- function(initCopy, dupRate, Carry) {
> ROI_T = initCopy
> A = array()
> for (i in 1:45) {
> ROI_TplusOne <- ROI_T * dupRate * (1 - ROI_T/Carry)
> A[i] <- ROI_TplusOne
> ROI_T <- ROI_TplusOne
> }
> return(A)
> }
> ```
> Which returns an array that follows the logistic shape, for instance
> ```
> d <- 2
> K <- 10^13
> A_0 <- 10000
> PCR_array <- PCR(A_0, d, K)
> plot(PCR_array)
> ```
> Given the formula `ROI_TplusOne <- ROI_T * dupRate * (1 -
> ROI_T/Carry)`, is it possible to determine at what time point `i` a
> given threshold is reached? For instance, what fractional value of i
> returns 1000 000 copies?
> Thank you
>
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