Theoretical Cmax value for a PK study

Phoenix Certara Forums Phoenix WNL basics
1

Could anyone know how to get the theoretical Cmax value for a PK study? Can Phoex WNL estimate this value?

Thank you very much,

2

Hi,

only for a one-compartment model (with and without lag-time) the exact Cmax can be calculated from the model parameters:

  1. The slope of tangent of the the function at any given time is positive before tmax (increasing values), zero at tmax, and negative (decreasing) afterwards. Hence, derive the function.
  2. Find the root of the derivative (i.e., its intersection with the abscissa). Trivial – this is tmax. Example (no lag-time):
    tmax = ln(k01/k10)/(k01 – k10).
  3. Plug in tmax in the function to get Cmax.

For other models the derivative cannot be obtained analytically. Therefore, you have to use a numeric method to find the maximum and its time.

Instead of paper/pencil/brain use software. In PHX/WNL both Cmax and tmax are given in the outputs as secondary parameters.

3

Thank you very much, Helmut. I will digest at your answer. Have a nice weekend!

4

Hi Helmut,

I would like to confirm that the exact Cmax in your answer is the theoretical Cmax. Am I right?

Another, how to get the secondary parameters for a group of PK data?

Appreciate your help!

5

Hi Experts,

I really need your help about how to get the theoretical Cmax value for a PK study. I am stuck here.

Could you please give me some hint about how to the theoretical value?

Appreciate your help in advance!

6

HI Li, Helmut has already suggested that you can use the models of Phoenix to generate these secondary parameters.

first choose the button to ‘set WNL model’

and then use the drop downs to make your selections and then the relevant code will be generated when you click apply.

secondary(Ke = tvCl/tvV)
secondary(Tmax = log(tvKa/Ke)/(tvKa-Ke)+tvTlag)
secondary(AUC = AaDose/tvV/Ke)
secondary(Cmax = AaDose/tvVexp(-Ke(Tmax-tvTlag)))
secondary(Ka_hl = log(2)/tvKa)
secondary(Ke_hl = log(2)/Ke)

Remember these are specific to the structural model chosen.

Simon.

set_wnl_model.jpg
set_wnl_model.jpg864×118 52.5 KB
apply.jpg
apply.jpg727×185 46 KB

7

Hi,

As I wrote already in my first post only in a one-compartment model (with and without lag-time) the theoretical Cmax can be derived by calculus/algebra. Hence, it is exact if you like.

For other models numeric methods are required. I don’t know which one is implemented is Phoenix (I’m a user with no access to its source code).

Brute force:

  • Instead of the derivative dC/dt switch to difference quotients over the entire time range of interest, i.e., calculate δC/δt, where δt is a – very – small interval.
  • Find the time point where δC/δt is sufficiently close to zero. That’s tmax. Plug it into the model to get Cmax.

More efficient are e.g., the bisection method or the uniroot algorithm (Brent 1973, 2003) implemented in R and C.

But again, if you really want the exact value, bad luck. The best you can get is a value which deviates not more than the numeric precision of the machine from the theoretical one. If you are not happy with ±3.6 ·10–16 on 64bit-machines get a number cruncher.

8

Appreciate your patience and kind help, Dr. Schutz and Dr. Davis. Got it.