Skip to content

How to find when a value is reached given a function?

5 messages · Duncan Murdoch, Luigi Marongiu, Abby Spurdle

Original
Luigi Marongiu
# 
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
Duncan Murdoch
#
On 24/01/2021 2:57 p.m., Luigi Marongiu wrote:
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
Luigi Marongiu
# 
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 10?
On Sun, Jan 24, 2021 at 9:40 PM Duncan Murdoch <murdoch.duncan at gmail.com> wrote:

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      Luigi Marongiu
    

    

      
      #
    

  


  

      
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:
```

            

      
      
      
    

        
Y
1 6
```
What is the correct use?
Thank you

        
On Mon, Jan 25, 2021 at 3:20 PM Luigi Marongiu <marongiu.luigi at gmail.com> wrote:

            

      
      
      
    

        

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      Abby Spurdle
    

    

      
      #
    

  


  

      
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.luigi at gmail.com> wrote: