Bioequivalence and Bioavailability Forum

Dear Yung-jin & Elba,

❝ ❝ I was wondering if there is any initiative in this forum to develop "public data reference" in order to evaluate the performance of BEAR (or another package).

❝ Sounds do-able. May need more opinions and time to plan it. There are plenty BE/BA guru members in this Forum from worldwide. They should be able to figure out how-to sooner or later.

I’m a little bit short in time, but I give you some ideas on what I have done back in the dark-ages validating my 60,000-lines Pascal code. A very important point in validation is to challenge the software with extreme data. Are errors trapped, are the results reasonable, etc.
I’ll split it up into two parts: NCA and BE.

NCA

► Data-limits

• Generate a dataset containing only the headers, but do data. Does the software crash or trap the error and come up with a useful (!) error-message?
  
• Dataset with only one line (i.e, time+concentration).
  
• Dataset with empty lines or – in R-terms – ‘NAs’.
  
• Dataset with negative or non-numeric values.

► C_max/t_max

• Two ore more identical values. Is the first one identified as C_max/t_max?

► λ_z-estimation

• Less than three decreasing values.
  
• C_z ≥ C_z-1.
  
• 0 or NA within the decreasing values.
  
• Dataset with known λ_z.
  Example: t(0, 1, 2), C(1, 0.5, 0.25) should give exactly λ_z=ln(2) and t_½=1.
  
• Simple dataset (use the AUC-example below) and calculate λ_z from the last three values manually.
  1. Y_i=ln(C_i)=(4.04305, 2.94444, 1.38629)
     
  2. t=(∑t_i)/3=13.33333 and Y=(∑Y_i)/3=2.79126
     
  3. λ_z=-∑(t_i-t) (Y_i-Y)/(t_i-t)²=0.13260 and
     t_½=ln(2)/λ_z=0.69315/0.13260=5.22729.
     
  4. If you want to calculate AUC_0-∞ based on the estimated Ĉ_z (instead of the observed one) you need also the regression’s intercept:
     ln(C₀)=Y-λ_zt=2.79126-0.13260×13.33333=4.55928
     All calculations in full precision; final result rounded to the software’s precision.
     A note for cross-validators: In PHX/WNL you can request this value in
     NCA > Options > Model Settings > [×] Intermediate Output.
     Results in the Core output:
     Intermediate Output
-------------------
Value for Lambda_z: 0.1326, and intercept: 4.5593

► Trapezoidal rules

• Dataset with only one concentration >0.
  
• Dataset with two identical concentrations (lin-up/log-down should switch to linear).
• Dataset which can easily be calculated manually (KISS!). Example:
  t   C inc? linear                ∑ lin-up/log-down                      ∑
 0  0  NA                    0   0                            0         0
 1 62  y  ( 1- 0)(62+ 0)/2= 31  31 apply linear              31        31
 2 70  y  ( 2- 1)(70+62)/2= 66  97 apply linear              97        97
 4 57  n  ( 4- 2)(57+70)/2=127 224 ( 4- 2)(70-57)/ln(70/57)=126.55518 223.55518
12 19  n  (12- 4)(19+57)/2=304 528 (12- 4)(57-19)/ln(57/19)=276.71272 500.26791
24  4  n  (24-12)( 4+19)/2=138 666 (24-12)(19- 4)/ln(19/ 4)=115.52201 615.78992
  Perform the manual calculation to the same precision as the software’s output.

► AUC_0-∞

• AUC_0-∞=AUC_t+C_z,obs/λ_z (based on the observed C_z) should give
  666+4/0.13260=696.16555 (linear trapezoidal; extrapolated 4.3331%) and
  615.78992+4/0.13260=645.95547 (lin-up/log-down; extrapolated 4.6699%).
  
• For the second variant we need the predicted C_z,pred=ℯ^{ln(C₀)–λ_zt_z}=ℯ^{4.55928–0.13260×24}=3.96238. Then
  666+3.96238/0.13260=695.88220 (linear trapezoidal; extrapolated 4.2941%) and
  615.78992+3.96238/0.13260=645.67212 (lin-up/log-down; extrapolated 4.6281%).

BE (coming sooner or later…)