From Concentration Samples to Bioequivalence Endpoints
A bioequivalence analysis does not begin with an AUC or Cmax value. It begins with a concentration-time profile for each subject and treatment period. The endpoint calculation turns that profile into a compact summary while retaining the information needed for a valid comparison. In particular, sampling times, concentration units, the integration rule, and the definition of the endpoint must be handled consistently across profiles.
This tutorial focuses on AUC from time zero to the last included sample, denoted AUC0-t, and the observed maximum concentration, Cmax. AUC summarizes exposure over the sampled interval; Cmax summarizes the highest concentration actually measured. Neither is a direct observation independent of the sampling plan: AUC is calculated from the observations, and the observed Cmax can miss the true peak if sampling is too sparse.
The trapezoidal rule requires ordered sample times and concentrations recorded on a consistent scale. In the examples, time is measured in hours and concentration in mg/L, so AUC has units mg·h/L, while Cmax has units mg/L. The first time is time zero. The last included sample should be defined by the analysis protocol and applied consistently; calculations should not silently switch between scheduled and observed endpoints.
Linear Trapezoidal AUC
Suppose adjacent observations occur at \(t_i\) and \(t_{i+1}\), with concentrations \(C_i\) and \(C_{i+1}\). The linear trapezoidal rule treats concentration as changing linearly between those points. The estimated interval area is the width of the interval multiplied by the average of its endpoint concentrations. Summing those areas gives the profile AUC:
The interval widths need not be equal. Each interval uses its own \(t_{i+1}-t_i\), so using a single common width when the sampling schedule is uneven gives the wrong result. If concentration is represented by a straight line within each interval, integrating that line gives exactly the trapezoid formula. For observed data, the straight-line segment is an approximation to the unknown concentration curve, not a claim that the biological concentration-time course is truly piecewise linear.
Worked Example: Linear Trapezoidal AUC and Observed Cmax
Consider a hypothetical profile with concentrations in mg/L at the following times in hours: \(0,1,2,4,8,12\), with concentrations \(0,2,5,4,2,1\), respectively. Calculate AUC0-12 and observed Cmax.
The interval widths are \(1,1,2,4,4\) hours. The five trapezoid areas are:
From 0 to 1 hour: \((0+2)(1)/2=1\) mg·h/L. From 1 to 2 hours: \((2+5)(1)/2=3.5\) mg·h/L. From 2 to 4 hours: \((5+4)(2)/2=9\) mg·h/L. From 4 to 8 hours: \((4+2)(4)/2=12\) mg·h/L. From 8 to 12 hours: \((2+1)(4)/2=6\) mg·h/L.
Therefore, \(\mathrm{AUC}_{0-12}=1+3.5+9+12+6=31.5\) mg·h/L. The largest measured concentration is \(5\) mg/L, observed at 2 hours, so Cmax is \(5\) mg/L and the observed \(T_{\max}\) is 2 hours. Cmax is selected from the measured values; this calculation does not estimate an unobserved concentration between samples.
Choosing and Documenting the Integration Rule
A common alternative to linear trapezoids is the linear-up/log-down method. It uses linear trapezoids when concentration rises and logarithmic trapezoids when concentration falls between two positive concentrations. The rationale is that a declining pharmacokinetic profile may be closer to exponential than linear over an interval. The rule must be specified in advance and applied consistently; changing methods after seeing the treatment results can introduce avoidable bias.
For a declining interval with positive endpoint concentrations, exponential interpolation assumes that the logarithm of concentration changes linearly with time. The interpolated concentration is \(C(t)=C_i\exp[-k(t-t_i)]\), where \(k=\ln(C_i/C_{i+1})/(t_{i+1}-t_i)\). Integrating this curve over the interval gives the logarithmic trapezoid:
This expression applies when both concentrations are positive and unequal. If the concentrations are equal, its limit is \(C_i(t_{i+1}-t_i)\). An interval that rises is handled linearly under the linear-up/log-down rule. If an endpoint concentration is zero, a logarithmic interval is not defined, so the linear rule is used for that interval. The analysis specification should state the convention for such cases.
Worked Example: Comparing Linear and Linear-Up/Log-Down Areas
A hypothetical profile has times \(0,0.5,1.5,3\) hours and concentrations \(0,4,8,4\) mg/L. Calculate AUC0-3 using both the all-linear trapezoidal rule and the linear-up/log-down rule.
Under the all-linear rule, the interval areas are \((0+4)(0.5)/2=1\), \((4+8)(1)/2=6\), and \((8+4)(1.5)/2=9\) mg·h/L. Their sum is \(16\) mg·h/L.
Under linear-up/log-down, the first two intervals are increasing and retain areas of 1 and 6 mg·h/L. The final interval declines from 8 to 4 mg/L over 1.5 hours, so its logarithmic area is \(1.5(8-4)/\ln(8/4)=6/\ln(2)=8.656\) mg·h/L. The total is \(1+6+8.656=15.66\) mg·h/L. Both methods give observed Cmax \(=8\) mg/L at 1.5 hours. The difference in AUC results reflects different interpolation assumptions, not an arithmetic inconsistency. The prespecified method determines which value is carried into the analysis.
Preparing Subject-Level Endpoint Values
For bioequivalence, calculate endpoints separately for each subject and each treatment period. Do not first average concentrations across subjects and then calculate one AUC per treatment: in general, the AUC of an average profile is not the same as the average of individual AUCs, and that procedure discards the subject-level pairing used by the crossover analysis. Preserve identifiers for subject, treatment, period, and sequence alongside each derived endpoint. Also retain the raw time-concentration records and calculation method so that an endpoint can be audited.
For the usual log-scale analysis, AUC and Cmax are transformed after endpoint calculation. The logarithm is taken for each subject-period endpoint, not for a pooled treatment mean. Logs are understood as logs of the numerical values expressed in fixed, common units; changing those common units adds the same constant to all log values and does not alter a treatment contrast. The earlier material on Complete 2×2 Crossover Analysis for AUC and Cmax covers the subsequent crossover comparison; the task here is to create accurate input values for that analysis.
Worked Example: Two Period Profiles for One Hypothetical Subject
A hypothetical subject receives reference and test treatments in separate periods. The sample times are \(0,1,2,4,8,12\) hours in both periods. The reference concentrations are \(0,2,5,4,2,1\) mg/L; the test concentrations are \(0,2.2,5.4,4.2,2.1,1.0\) mg/L. Use the linear trapezoidal rule and derive the endpoint values to store.
For reference, the interval areas are \(1,3.5,9,12,6\) mg·h/L, giving AUC0-12 \(=31.5\) mg·h/L. Reference Cmax is \(5.0\) mg/L. For test, the interval areas are \((0+2.2)(1)/2=1.1\), \((2.2+5.4)(1)/2=3.8\), \((5.4+4.2)(2)/2=9.6\), \((4.2+2.1)(4)/2=12.6\), and \((2.1+1.0)(4)/2=6.2\) mg·h/L. Thus test AUC0-12 \(=1.1+3.8+9.6+12.6+6.2=33.3\) mg·h/L. Test Cmax is \(5.4\) mg/L.
The corresponding natural-log endpoints, using the displayed numerical values in the stated units, are \(\ln(31.5)=3.450\), \(\ln(33.3)=3.506\), \(\ln(5.0)=1.609\), and \(\ln(5.4)=1.686\). A tidy subject-period record would contain the subject identifier, treatment, period, AUC0-12, Cmax, and their log values. These are the two paired endpoint records for this subject, not a treatment-level conclusion. No bioequivalence decision follows from the endpoint arithmetic for one subject.
Checks, Failure Modes, and Limitations
A reliable derivation can be checked without fitting a pharmacokinetic model. Confirm that times are in increasing order, each concentration is matched to its correct time and subject-period, and units remain consistent. Recompute interval widths and areas, then verify that their sum equals the reported AUC. Cmax must equal one of the observed concentrations, and its recorded time must correspond to an observation attaining that maximum. If several observations tie for the maximum, retain the tie information or use the protocol-defined \(T_{\max}\) convention.
Common warning signs include a negative interval width, an AUC with concentration units but no time unit, a Cmax that is not in the raw profile, or identical derived endpoints copied across different subject-period records. A concentration-time plot can reveal swapped samples, unexpected discontinuities, or a missed peak between widely spaced samples. Such checks identify possible data or derivation problems; they do not justify changing recorded concentrations without a documented, authorized correction.
The all-linear trapezoidal rule is simple and transparent, but it can overestimate or underestimate area when the true curve is appreciably curved between samples. Linear-up/log-down can better represent some declining segments, but it also rests on an interpolation assumption and can yield a different AUC. Model-based integration is another approach, but requires a suitable model and introduces model assumptions beyond this endpoint calculation. The observed Cmax is particularly sensitive to sampling frequency and timing: a real peak may occur between scheduled samples. AUC0-t also covers only the stated observed interval; extending it beyond the last sample is a distinct endpoint calculation and should not be implied by the notation or the reported value.
Check Your Understanding
Work each question, then compare with the solutions below.
- A profile has times 0, 2, and 5 hours and concentrations 0, 3, and 3 mg/L. Find linear trapezoidal AUC0-5 and Cmax.
- For times 0, 1, and 3 hours, concentrations are 1, 5, and 2 mg/L. Calculate all-linear AUC0-3, showing the interval areas.
- Why should endpoint values be calculated separately for each subject and treatment period rather than from an average concentration profile?
- For a positive declining interval, what interpolation assumption leads to the logarithmic trapezoid, and why is that method not interchangeable with the linear rule?
Solutions. (1) The areas are \((0+3)(2)/2=3\) and \((3+3)(3)/2=9\) mg·h/L, so AUC0-5 is \(12\) mg·h/L; Cmax is \(3\) mg/L. (2) The areas are \((1+5)(1)/2=3\) and \((5+2)(2)/2=7\) mg·h/L, giving AUC0-3 \(=10\) mg·h/L. (3) Subject-level derivation preserves each profile and the within-subject pairing required for crossover analysis; an average profile generally does not retain those individual endpoints. (4) The method assumes log concentration changes linearly with time, equivalent to exponential interpolation within the interval. Linear interpolation makes a different assumption, so the resulting area may differ and the prespecified rule must be used consistently.