Occupancy, Half Value Duration, Plateau Time in Phoenix/WinNonlin 8 [🇷 for BE/BA]

posted by Helmut Homepage – Vienna, Austria, 2017-10-30 14:18 (2426 d 11:24 ago) – Posting: # 17951
Views: 23,954

Hi Astea,

❝ […] an easy way of calculation T50% early and late, T75%, T90% and similar parameters via Phoenix?

I don’t know any method to obtain the intersections in PHX/WNL. As discussed before, starting with release 8.0 the intersection in the decreasing part(s) depends on the selected trapezoidal method (was always lin/log in previous releases). Simple example: \(C(t)=20(e^{-log(2)/4\cdot t}-e^{-log(2)\cdot t})\)
 t    C
 0   0.000
 0.5 4.198
 1   6.818
 2   9.142
 3   9.392
 4   8.750
 6   6.759
 9   4.165
12   2.495
16   1.250
24   0.312

Calculation of the Occupancy Time (interval where concentrations are above a fixed value) is trivial.
More demanding are the Half Value Duration (HVD, t50%, POT-50), Plateau Time (t75%, POT-25), and their relatives because they depend on the subject’s Cmax. Hence, we need two steps (I used the data of above for subject 1 and ½C for subject 2); example for HVD:
  1. Send the data to NCA, map as usual (sort by subject).
    User Defined Parameters > Additional NCA Parameters Add
    Parameter: Cmax_2
    Definition: Cmax/2

    Include with Final Parameters
    Parameter NamesUse Internal Worksheet > Include in Workbook:
    Set to No for all except Cmax
    Rename the NCA object to Preliminary and execute.
  2. Send the data to NCA, map as usual (sort by subject).
    Therapeutic Response, link to Preliminary.Final Parameters Pivoted
    Sort by subject; map Cmax_2 to Lower and Cmax to Upper, execute.
TimeBetween is the HVD. IMHO, TimeLow is only interesting for a fixed limit (Occupancy Time) and if t is the intended τ.
For the linear trapezoidal you should get

subject Cmax  Cmax_2 TimeLow TimeBetween
1       9.392 4.696  16.2089 7.7911
2       4.696 2.348  16.2089 7.7911

and for the lin-up/log-down

subject Cmax  Cmax_2 TimeLow TimeBetween
1       9.392 4.696  16.3383 7.6617
2       4.696 2.348  16.3383 7.6617

Linear interpolation of the intersections will always give a larger interval than the lin/log interpolation.

0.595 to 8.386 = 7.7911

0.595 to 8.257 = 7.6617

I posted an example project at the Certara Forum.

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