Interpolating Concentrations at Partial-AUC Boundaries

Why Boundary Interpolation Matters

A partial-AUC window may begin or end at a specified time that was not an actual sampling time. To calculate the exposure within that window, the concentration at the boundary must be estimated from the samples around it. That estimate also determines the area between the boundary and the nearest sample, so the interpolation method affects both the boundary concentration and the boundary area.

The introductory material on partial-AUC window boundaries establishes that the chosen start and end times define the exposure being summarized. Here the focus is local: estimating concentration at a boundary between two observations, then calculating the area over the boundary segment. This does not make the interpolated concentration an observed measurement. It is an estimate under an assumed shape for the concentration-time profile between samples.

Definition: Suppose samples at times $t_1$ and $t_2$, with $t_1 < t_2$, have concentrations $C_1$ and $C_2$. A boundary time $t^*$ is bracketed by these observations when $t_1 \le t^* \le t_2$. Interpolation estimates $C(t^*)$ from the two observations under a specified model for the profile between them.

Two Interpolation Assumptions

Both methods below assume the boundary lies within the interval between the selected observations, and that the sampling times and concentrations are correctly recorded. The methods differ in the assumed shape of the concentration-time profile over that interval. They do not determine which shape is biologically correct; the method should follow the analysis plan or the convention being applied.

Linear interpolation: Concentration changes linearly with time between the two observations. Define $u=(t^*-t_1)/(t_2-t_1)$, so $0 \le u \le 1$. Then $$ C(t^*)=(1-u)C_1+uC_2 $$

The estimate is a weighted average of the endpoint concentrations. It equals $C_1$ at $t_1$, equals $C_2$ at $t_2$, and remains between them for a boundary inside the interval. This method permits zero concentrations and does not require concentrations to be strictly positive.

Log-linear interpolation: When both endpoint concentrations are positive, assume that the natural logarithm of concentration changes linearly with time. Then $$ \ln C(t^*)=(1-u)\ln C_1+u\ln C_2 $$ and equivalently $$ C(t^*)=C_1^{1-u}C_2^u $$

Log-linear interpolation also gives an estimate between the endpoint concentrations, but it assumes a straight line on the log-concentration scale, corresponding to an exponential profile on the concentration scale. It requires $C_1>0$ and $C_2>0$; a zero or negative endpoint cannot be used in the logarithm.

From an Interpolated Concentration to Boundary Area

Once a boundary concentration has been estimated, the area from the adjacent observation to the boundary follows the same local profile assumption. Let $t_a$ and $t_b$ be the endpoints of the boundary segment, with $t_a<t_b$, and let their concentrations be $C_a$ and $C_b$. Write $h=t_b-t_a$. Under a linear profile, the area is the ordinary trapezoid:

A_{\mathrm{lin}}=h\frac{C_a+C_b}{2}

Under a log-linear profile, concentration changes exponentially between the endpoints. Integrating that exponential gives the log-linear segment area:

A_{\mathrm{log}}=h\frac{C_b-C_a}{\ln(C_b/C_a)}

The log-linear expression requires positive endpoint concentrations. If $C_a=C_b$, its displayed form has a zero denominator, but the profile is constant and the area is $hC_a$. This is also the limit of the expression as $C_b$ approaches $C_a$.

Why the Log-Linear Area Formula Works

For a log-linear segment, let $x$ be elapsed time from $t_a$, so $0\le x\le h$. Straight-line change in log concentration means $\ln C(x)=\ln C_a+kx$, where $k=\ln(C_b/C_a)/h$. Exponentiating gives $C(x)=C_a e^{kx}$. Integrating this profile over the segment yields:

\begin{aligned} A_{\mathrm{log}} &=\int_0^h C_a e^{kx}\,dx\\ &=C_a\frac{e^{kh}-1}{k}\\ &=\frac{C_b-C_a}{\ln(C_b/C_a)/h}\\ &=h\frac{C_b-C_a}{\ln(C_b/C_a)} \end{aligned}

The same expression applies whether concentration rises or declines: numerator and denominator change sign together, leaving a positive area. For a constant concentration, the integral is $hC_a$. For a linear concentration profile, integrating the straight line gives its average endpoint concentration times the duration, which is the linear trapezoid formula.

A Practical Boundary Workflow

Locate the bracketing samples.
Find consecutive observation times $t_1$ and $t_2$ such that $t_1 \le t^* \le t_2$. If the boundary equals an observed time, use that measured concentration rather than interpolating.

Choose the interpolation assumption.
Use the specified linear or log-linear convention. Check that the boundary is within the observed interval and, for log-linear interpolation, both bracketing concentrations are positive.

Estimate the boundary concentration.
Calculate $u$, then substitute it and the observed endpoint concentrations into the chosen interpolation formula.

Calculate only the in-window boundary segment.
For a window start inside an interval, the relevant segment runs from the start to the next sample. For a window end inside an interval, it runs from the previous sample to the end. Apply the area formula matching the chosen profile assumption.

If both window boundaries fall inside the same sampling interval, calculate the area directly between the two boundary concentrations using the profile assumed over that interval. Do not add areas from each boundary to the outside sample times: those pieces extend beyond the window and can double-count or include exposure that is not part of it.

Worked Example: Linear Interpolation at a Window End

Suppose hypothetical samples are $8.0$ mg/L at 2 h and $5.0$ mg/L at 5 h. A partial-AUC window ends at 4 h, and the analysis uses linear interpolation. Estimate the boundary concentration and the area from 2 h to the boundary.

The 4 h boundary is between the observations, and the interval duration is $5-2=3$ h. Its relative position is $u=(4-2)/(5-2)=2/3$.

The interpolated concentration is $C(4)=(1-2/3)(8.0)+(2/3)(5.0)=8.0/3+10.0/3=6.0$ mg/L.

The in-window boundary segment runs from 2 h to 4 h, so its duration is 2 h. Its linear area is $A_{\mathrm{lin}}=2(8.0+6.0)/2=14.0$ mg·h/L.

The result estimates a concentration of 6.0 mg/L at 4 h under a straight-line assumption. The segment area is 14.0 mg·h/L; it is not an observed exposure measurement independent of that assumption.

Worked Example: Log-Linear Interpolation at a Window End

Suppose hypothetical concentrations are $12.0$ mg/L at 2 h and $3.0$ mg/L at 6 h. A window ends at 4 h, and log-linear interpolation is specified. Calculate the boundary concentration and area from 2 h to 4 h.

Both concentrations are positive, so log-linear interpolation is available. The relative position is $u=(4-2)/(6-2)=0.5$. Therefore, $C(4)=12.0^{0.5}3.0^{0.5}=\sqrt{36}=6.0$ mg/L.

For the segment from 2 h to 4 h, $h=2$ h, $C_a=12.0$ mg/L, and $C_b=6.0$ mg/L. Thus,

A_{\mathrm{log}} =2\frac{6.0-12.0}{\ln(6.0/12.0)} =2\frac{-6.0}{\ln(0.5)} =17.31\ \mathrm{mg\cdot h/L}

The denominator is negative because concentration declines, as is the numerator; their ratio and the resulting area are positive. The interpolated concentration is 6.0 mg/L, and the area over the two-hour boundary segment is 17.31 mg·h/L.

Worked Example: Boundary Areas at Both Ends of a Window

Consider hypothetical samples at 1, 2, 4, and 6 h with concentrations 4.0, 10.0, 5.0, and 2.5 mg/L, respectively. A window runs from 1.5 h to 4.5 h, and the selected method is linear throughout. Find each boundary concentration and the area of each boundary segment.

The start at 1.5 h lies halfway between 1 and 2 h, so $u=(1.5-1)/(2-1)=0.5$. The estimate is $C(1.5)=(1-0.5)(4.0)+0.5(10.0)=7.0$ mg/L. The in-window start segment is 1.5 to 2 h, with duration 0.5 h. Its area is $0.5(7.0+10.0)/2=4.25$ mg·h/L.

The end at 4.5 h lies one-quarter of the way from 4 to 6 h, so $u=(4.5-4)/(6-4)=0.25$. Its estimate is $C(4.5)=(1-0.25)(5.0)+0.25(2.5)=4.375$ mg/L. The in-window end segment is 4 to 4.5 h, also 0.5 h. Its area is $0.5(5.0+4.375)/2=2.34375$, or 2.344 mg·h/L.

These are the two boundary-segment areas only. The samples at 2 and 4 h delimit an interior segment, which is not needed to establish either boundary estimate or boundary area.

Choosing a Method and Checking Its Limits

Linear interpolation assumes a constant absolute rate of concentration change over time. Log-linear interpolation assumes a constant proportional rate of change, so equal time increments correspond to equal multiplicative changes in concentration. These assumptions can produce different estimates between the same two observations. For example, a declining profile with positive endpoints generally has a log-linear curve between the endpoints rather than the straight line used by linear interpolation.

Method	Assumed profile between samples	Key requirement	Boundary area
Linear	Straight line in concentration	Bracketing observations and boundary within their time interval	Duration times arithmetic mean of endpoint concentrations
Log-linear	Straight line in log concentration; exponential in concentration	Same bracketing condition; both concentrations positive	Log-mean area, or duration times the common concentration if endpoints are equal

Several checks help detect mistakes. Confirm that the boundary lies between consecutive samples, that $u$ is between 0 and 1, and that an interpolated concentration lies between the endpoint concentrations. Check that the segment duration is positive and that the area has units of concentration multiplied by time. A boundary outside the observed time range requires extrapolation, not interpolation, and the formulas here do not justify such an estimate.

Log-linear interpolation is unavailable when either endpoint is zero or negative. Do not silently replace a zero with an arbitrary small positive value to make the logarithm possible; that changes the data and the calculation. Use a method allowed by the analysis convention or obtain suitable data. Conversely, linear interpolation can be calculated with a zero endpoint, though whether the resulting profile is scientifically appropriate remains a separate question.

Common Mistakes and Limitations

Using the wrong time interval. A boundary must be bracketed by the adjacent observations around it, not by a more distant pair selected for convenience.
Reversing the boundary segment. A window start uses the segment from the start to the next sample; a window end uses the segment from the previous sample to the end.
Mixing interpolation and area assumptions. If the local profile is treated as linear, use the linear segment area. If it is treated as log-linear, use the log-linear area, subject to positive concentrations.
Including out-of-window exposure. The area from a sample outside the window to an interior boundary is not all part of the window. Restrict the boundary segment to the portion inside the window.
Treating an estimate as observed data. An interpolated concentration depends on the assumed between-sample profile. Sparse sampling, rapid changes, or an unobserved local feature can make that estimate inaccurate.

Interpolation is a local numerical convention, not a substitute for denser sampling or a validated model of the concentration-time profile. Linear and log-linear methods are practical because they use nearby observations and have explicit area formulas, but neither can recover unobserved features between samples. Consistent application and transparent reporting of the chosen method are therefore important when a boundary falls between observations.

Check Your Understanding

Work through each question, then compare your reasoning with the solutions.

Samples are 6.0 mg/L at 3 h and 10.0 mg/L at 7 h. Under linear interpolation, estimate concentration at 5 h.
Samples are 8.0 mg/L at 2 h and 2.0 mg/L at 6 h. Under log-linear interpolation, estimate concentration at 4 h and calculate the area from 2 h to 4 h.
A boundary equals an observed sampling time. Is interpolation required? What concentration should be used?
Can log-linear interpolation be applied when one bracketing concentration is zero? Explain.

Solutions. 1. $u=(5-3)/(7-3)=0.5$, so $C(5)=0.5(6.0)+0.5(10.0)=8.0$ mg/L. 2. $u=0.5$, so $C(4)=\sqrt{8.0\times2.0}=4.0$ mg/L. The two-hour area is $2(4.0-8.0)/\ln(4.0/8.0)=11.54$ mg·h/L. 3. No. Use the measured concentration at that time; there is no interval to interpolate across. 4. No. The logarithm of zero is undefined, so this log-linear formula cannot be used with that endpoint.