

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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On 24/01/2021 2:57 p.m., Luigi Marongiu 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?
There are two answers:
The brute force answer is just to try it and count how far you need to
go. This is really simple, but really inefficient.
The faster and more elegant way is to solve the recursive relation for
an explicit solution. You've got a quadratic recurrence relation;
there's no general solution to those, but there are solutions in special
cases. See https://math.stackexchange.com/q/3179834 and links therein
for some hints.
Duncan Murdoch
______________________________________________
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Thanks, I'll check it out. I ran the simulation and I got:
```
t = 1, N = 20,000
t = 2, N = 40,000
t = 3, N = 80,000
t = 4, N = 160,000
t = 5, N = 320,000
t = 6, N = 640,000
t = 7, N = 1,280,000
```
Hence the answer is t=6.{...} but the problem is to get that
fractional value. Would be possible to use some kind of interpolation?
I have the known Xs (the t values), the known Ys (the Nt), I need to
get x when y is 10⁶
On Sun, Jan 24, 2021 at 9:40 PM Duncan Murdoch < [hidden email]> wrote:
>
> On 24/01/2021 2:57 p.m., Luigi Marongiu 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?
>
> There are two answers:
>
> The brute force answer is just to try it and count how far you need to
> go. This is really simple, but really inefficient.
>
> The faster and more elegant way is to solve the recursive relation for
> an explicit solution. You've got a quadratic recurrence relation;
> there's no general solution to those, but there are solutions in special
> cases. See https://math.stackexchange.com/q/3179834 and links therein
> for some hints.
>
> Duncan Murdoch

Best regards,
Luigi
______________________________________________
[hidden email] mailing list  To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


If I run this:
```
Y = c(10000, 20000, 40000, 80000, 160000, 320000, 640000, 1280000)
X = 0:7
plot(Y~X, log='y')
model < lm(log10(Y) ~ X)
abline(model)
predict(model, data.frame(Y=log10(1000000)))
```
I get a funny answer:
```
1 2 3 4 5 6 7 8
4.00000 4.30103 4.60206 4.90309 5.20412 5.50515 5.80618 6.10721
Warning message:
'newdata' had 1 row but variables found have 8 rows
```
but:
```
> data.frame(Y=log10(1000000))
Y
1 6
```
What is the correct use?
Thank you
On Mon, Jan 25, 2021 at 3:20 PM Luigi Marongiu < [hidden email]> wrote:
>
> Thanks, I'll check it out. I ran the simulation and I got:
> ```
>
> t = 1, N = 20,000
> t = 2, N = 40,000
> t = 3, N = 80,000
> t = 4, N = 160,000
> t = 5, N = 320,000
> t = 6, N = 640,000
> t = 7, N = 1,280,000
> ```
> Hence the answer is t=6.{...} but the problem is to get that
> fractional value. Would be possible to use some kind of interpolation?
> I have the known Xs (the t values), the known Ys (the Nt), I need to
> get x when y is 10⁶
>
> On Sun, Jan 24, 2021 at 9:40 PM Duncan Murdoch < [hidden email]> wrote:
> >
> > On 24/01/2021 2:57 p.m., Luigi Marongiu 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?
> >
> > There are two answers:
> >
> > The brute force answer is just to try it and count how far you need to
> > go. This is really simple, but really inefficient.
> >
> > The faster and more elegant way is to solve the recursive relation for
> > an explicit solution. You've got a quadratic recurrence relation;
> > there's no general solution to those, but there are solutions in special
> > cases. See https://math.stackexchange.com/q/3179834 and links therein
> > for some hints.
> >
> > Duncan Murdoch
>
>
>
> 
> Best regards,
> Luigi

Best regards,
Luigi
______________________________________________
[hidden email] mailing list  To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


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
nondecreasing constraint.
Also note that I've just used the 45 points.
But reiterating, you should increase the number of points, if possible.
On Mon, Jan 25, 2021 at 8:58 AM Luigi Marongiu < [hidden email]> 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
>
> ______________________________________________
> [hidden email] mailing list  To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/rhelp> PLEASE do read the posting guide http://www.Rproject.org/postingguide.html> and provide commented, minimal, selfcontained, reproducible code.
______________________________________________
[hidden email] mailing list  To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.

