Bioequivalence and Bioavailability Forum

Dear Yu!

❝ Thank you for your detailed explanation.

Welcome!

❝ Could you clarify "observed" and "predicted" concentration C_last? Doesn't "predicted" conc. mean extrapolated C_last(24 h) of test drug?

The last observed/measured concentrations are 0.82 at 24h (reference) and 3.22 at 16.92h (test). The 24h value of the test (0.78) is just below the LLOQ and is given in the table for demonstrational purposes only.
If we estimate k_el from the last three values above the LLOQ for each formulation (linear regression of t vs. ln C), we obtain (interval of reference 11.90h – 24.00h, and test 8.37h – 16.92h):
┌───────────┬──────────┬──────────┐
│ parameter │    ref   │   test   │
├───────────┼──────────┼──────────┤
│ slope     │ -0.20027 │ -0.19992 │
│ intercept │  4.60838 │  4.55224 │
└───────────┴──────────┴──────────┘
Whereas k_el = |slope|, and C₀ = ℯ^intercept.
Therefore we can estimate the concentration (C_pred) at t_last by calculating
C_pred = ℯ^{(intercept+k_el·t_last)}.

❝ I wonder why the AUC values of the test are same in method 4 and 5.

Because in my simple example there was no noise in the data; only due to rounding to two decimal places small differences are observed. The observed and predicted concentrations show a bias of 0.04% (reference) and 0.02% (test) only. This will not be the case with ‘real world’ data. Since we already have agreed in using a couple of data points (≥3) in the estimation of k_el, it’s justified to use the estimated concentration – instead of the observed one – in extrapolating beyond t_last.

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Edit: If you use WinNonlin, check "Output intermediate calculations" in the Model Options. You will get the Intercept in "NCA Text", but not in the result worksheet. Example:
Intermediate Output
-------------------
  Value for Lambda_z:      0.2003, and intercept:     4.6084
  Value for Lambda_z:      0.1999, and intercept:     4.5522

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Edit: Gabrielsson and Weiner (see here) give a formula without requiring the intercept:
C₂₄ = C_obs · ℯ^{-lambda_z·(24-t_obs)} or C₂₄ = 3.22 · ℯ^{-0.19992·(24-16.92)} = 0.78…