Time Above A Concentration Threshold

Why Time Above Threshold Is a Distinct Exposure Measure

A concentration threshold can represent a minimum level associated with a pharmacologic objective, or a level at which a safety concern becomes relevant. In either case, knowing whether a measured concentration is above the threshold at one sampling time does not tell us how long it was above it between samples. Estimating that duration requires a rule for describing concentration over the intervals between observations.

This tutorial uses piecewise-linear interpolation: concentration is assumed to change linearly between each pair of consecutive observations. Under that assumption, a crossing between the observed concentration and the threshold can be located by solving a linear equation. The resulting time above threshold is an estimate for the specified observation window, not a direct measurement of every instant in that window.

Definition: For a concentration curve \(C(t)\), a threshold \(C_{\mathrm{thr}}\), and a defined interval \([t_0,t_1]\), the time above threshold is the total duration within that interval for which \(C(t)>C_{\mathrm{thr}}\). The fraction of the interval above threshold is that duration divided by \(t_1-t_0\). If the intended criterion is “at or above” the threshold, use \(C(t)\geq C_{\mathrm{thr}}\) instead and state that convention.

Time above threshold differs from AUC and average concentration. AUC summarizes concentration accumulated over time, while time above threshold records how long concentration satisfies a specified level criterion. The earlier material on average concentration over a defined window establishes how to summarize concentration over an interval; here the focus is locating threshold crossings and adding the durations that meet the criterion.

Linear Interpolation at a Threshold Crossing

Suppose two consecutive observations are \((t_i,C_i)\) and \((t_{i+1},C_{i+1})\), with \(t_{i+1}>t_i\). Piecewise-linear interpolation represents concentration between them as a straight line. If the threshold lies between the two observed concentrations, solve for the time \(t^*\) at which the interpolated concentration equals the threshold:

t^*=t_i+\frac{C_{\mathrm{thr}}-C_i}{C_{i+1}-C_i}(t_{i+1}-t_i)

The ratio in this expression is the fraction of the concentration change from \(C_i\) to \(C_{i+1}\) needed to reach the threshold. Multiplying that fraction by the elapsed time gives the elapsed time from \(t_i\) to the crossing. This calculation works for either an upward or downward crossing: the numerator and denominator change direction together, leaving a crossing time between the observations when the threshold is between their concentrations.

Assumptions and conditions: Observation times are ordered, concentrations and the threshold use the same units, and concentration is treated as linear between consecutive observations. Calculate only within the defined observation window; do not extrapolate beyond its first or last time unless a separate model explicitly justifies that extrapolation. If both endpoint concentrations are on the same side of the threshold, a linear segment has no crossing in its interior. If both equal the threshold, the entire segment is exactly at threshold.

For each adjacent pair, determine whether the segment is above threshold throughout, below it throughout, or crosses the threshold. For a crossing segment, count only the part on the above-threshold side. Add the durations from all segments. A segment whose endpoints are both above contributes its full elapsed time; one whose endpoints are both below contributes zero. If an endpoint equals the threshold, the crossing occurs at that endpoint. For a strict \(>\) definition, a single instant exactly at threshold has zero duration, but a whole plateau exactly at threshold contributes no time above. A plateau does contribute time under a \(\geq\) definition.

The fraction above threshold is calculated using the full length of the specified window as its denominator, not just the time between samples that happen to be above threshold. With time expressed in hours, the duration is in hours and the fraction is dimensionless; multiply the fraction by 100 to report a percentage.

Worked Calculations

Worked Example: One Upward and One Downward Crossing

Suppose hypothetical concentrations over a 6-hour interval are recorded as follows. Concentration is in mg/L and time is in hours.

Time (h)	Concentration (mg/L)
0	0
2	10
5	2
6	0

Estimate the time and fraction above \(C_{\mathrm{thr}}=5\) mg/L, using the strict \(>\) criterion. The threshold is crossed upward between 0 and 2 hours:

t_{\mathrm{up}}=0+\frac{5-0}{10-0}(2-0)=1.00\ \mathrm{h}

The downward crossing occurs between 2 and 5 hours:

t_{\mathrm{down}}=2+\frac{5-10}{2-10}(5-2) =2+\frac{5}{8}(3) =3.875\ \mathrm{h}

Under the piecewise-linear assumption, concentration is above 5 mg/L from 1.00 to 3.875 hours. Thus,

T_{>} = 3.875-1.00=2.875\ \mathrm{h}, \qquad \text{fraction above}=\frac{2.875\ \mathrm{h}}{6\ \mathrm{h}}=0.4792

The estimate is 2.875 hours, or about 47.92% of the interval. The units cancel in the fraction. The calculation does not claim that the concentration was directly observed at either crossing; both crossing times are interpolated estimates.

Worked Example: Adding Time From Two Separate Peaks

Consider hypothetical samples at 0, 1, 3, 4, and 7 hours with concentrations of 2, 6, 3, 7, and 1 mg/L, respectively. Estimate the total duration and fraction above 4 mg/L over the 7-hour window.

There are four crossing segments. The first, from 2 to 6 mg/L over 0–1 hour, crosses upward at

t_1=0+\frac{4-2}{6-2}(1-0)=0.5000\ \mathrm{h}.

The next segment declines from 6 to 3 mg/L over 1–3 hours, so its downward crossing is

t_2=1+\frac{4-6}{3-6}(3-1) =1+\frac{2}{3}(2) =2.333\ \mathrm{h}.

The concentration is therefore above threshold from 0.5000 to 2.333 hours, for \(2.333-0.5000=1.833\) hours. On the next rising segment, from 3 to 7 mg/L over 3–4 hours, the upward crossing is

t_3=3+\frac{4-3}{7-3}(4-3)=3.250\ \mathrm{h}.

On the final segment, which declines from 7 to 1 mg/L over 4–7 hours, the downward crossing is

t_4=4+\frac{4-7}{1-7}(7-4) =4+\frac{1}{2}(3) =5.500\ \mathrm{h}.

The second above-threshold period lasts \(5.500-3.250=2.250\) hours. Adding the separate periods gives

T_{>}=1.833+2.250=4.083\ \mathrm{h}, \qquad \text{fraction above}=\frac{4.083}{7}=0.5833.

The estimated concentration is above threshold for 4.083 hours, or 58.33% of the window. Adding the durations of both periods is essential: using only the first peak would omit time above threshold after the second rise.

Worked Example: A Segment Exactly at the Threshold

Suppose hypothetical concentrations at 0, 2, 5, and 8 hours are 5, 3, 3, and 6 mg/L. Let the threshold be 3 mg/L over the full 8-hour interval. First use the strict \(>\) convention.

The first segment falls from 5 to 3 mg/L, reaching the threshold at 2 hours. It is above threshold for 2 hours. The segment from 2 to 5 hours stays exactly at 3 mg/L, so it contributes zero time under the strict convention. On the final segment, concentration rises from 3 to 6 mg/L. It is above threshold after 5 hours, contributing \(8-5=3\) hours. Therefore,

T_{>}=2+0+3=5\ \mathrm{h}, \qquad \text{fraction above}=\frac{5}{8}=0.625.

If the intended criterion were instead \(C(t)\geq 3\) mg/L, every point in all three segments would meet the criterion, including the 3-hour plateau exactly at threshold. The duration would be 8 hours and the fraction would be 1.000. This example shows why the threshold convention should be specified, particularly when the curve may remain exactly at the threshold for a nonzero period.

A Practical Calculation and Its Checks

Define the window and criterion.
Record the start and end times, the threshold value and units, and whether the rule is strictly above or at-or-above.

Inspect each adjacent pair.
Compare both concentrations with the threshold. Identify full segments above, full segments below, and segments that cross it.

Interpolate each crossing.
Use the linear crossing formula for every segment with the threshold between its endpoint concentrations. Treat an endpoint equal to the threshold as a crossing at that endpoint.

Sum durations and divide by window length.
Add all portions above threshold, then divide by \(t_1-t_0\) to obtain the fraction. Check that total duration is between zero and the window length.

A short derivation explains why the interpolation formula locates the crossing. On a segment, the assumed line has slope \((C_{i+1}-C_i)/(t_{i+1}-t_i)\), so its concentration at time \(t\) is \(C_i+\frac{C_{i+1}-C_i}{t_{i+1}-t_i}(t-t_i)\). Setting this expression equal to \(C_{\mathrm{thr}}\) and solving for \(t\) gives the stated crossing formula. If the threshold lies between the endpoint concentrations, the ratio \((C_{\mathrm{thr}}-C_i)/(C_{i+1}-C_i)\) lies between zero and one; consequently, \(t^*\) lies between \(t_i\) and \(t_{i+1}\). This provides a direct arithmetic check on a computed crossing.

Diagnostics, Alternatives, and Limitations

Several checks help identify calculation errors. Every interpolated crossing should fall within its own pair of observation times. A computed duration must be nonnegative and cannot exceed the window length; the fraction must be between 0 and 1. When the threshold lies between two endpoint concentrations, the segment should contribute time on one side of the crossing, not both. Confirm that the time units match throughout and that the denominator is the full defined interval.

The estimate also depends on the quality and spacing of the concentration observations. Long gaps can conceal rapid changes, such as a sharp peak or multiple crossings, that a straight line between endpoints cannot represent. If both endpoints are below threshold, piecewise-linear interpolation implies no crossing within that segment, but the true concentration could have risen above threshold and returned between samples. Sparse observations therefore do not guarantee that the estimated duration captures all actual above-threshold time.

Alternatives answer different questions or impose different assumptions. Counting only sample times above threshold is simple, but it does not estimate the crossing times and can change substantially with the sampling schedule. AUC or average concentration cannot replace a threshold-duration calculation: they summarize concentration magnitude over time rather than the duration meeting a cutoff. A concentration-time curve predicted by a pharmacokinetic model can be used to locate crossings on that model-defined curve, but the result then depends on that model and its parameters. Piecewise-linear interpolation is transparent and needs only observed concentrations, while a fitted curve may represent nonlinear changes but requires additional modeling assumptions.

The estimate is specific to the chosen threshold, interval, observed concentration data, and interpolation rule. It does not establish that the threshold is biologically valid or that time above it causes a particular effect. Measurement error near the threshold can also change whether observations are classified above or below, and hence change interpolated crossing times. Report the threshold and units, time window, criterion (\(>\) or \(\geq\)), interpolation method, and estimated duration or fraction so that the result can be interpreted and reproduced.

Common Mistake

Common pitfall: “Two of four sample times are above threshold, so the fraction of the interval above threshold is one-half.” This treats observations as if they each represent equal-duration blocks and ignores when crossings occur. Instead, locate each crossing using the stated interpolation assumption, add the actual above-threshold durations, and divide by the full interval length. Sample-count fractions are not generally time fractions.

Check Your Understanding

Use piecewise-linear interpolation unless a question specifies otherwise. Assume strict exceedance for questions that say “above.”

Concentrations are 1 mg/L at 0 hours, 7 mg/L at 3 hours, and 1 mg/L at 6 hours. For a threshold of 4 mg/L, find the total duration and fraction above threshold.
Concentrations are 6 mg/L at 0 hours, 2 mg/L at 2 hours, and 5 mg/L at 5 hours. For a threshold of 4 mg/L, find the total duration and fraction above threshold over the 5-hour window.
Both endpoint concentrations on a linearly interpolated segment are below a threshold. Can that segment contribute time above threshold under the piecewise-linear assumption? Explain.
A concentration segment stays exactly at the threshold for 2 hours. How does that segment contribute under \(>\) and under \(\geq\)?

Solutions. (1) The upward crossing is \(0+\frac{4-1}{7-1}(3)=1.5\) hours. The downward crossing is \(3+\frac{4-7}{1-7}(3)=4.5\) hours. Duration is \(4.5-1.5=3.00\) hours; fraction is \(3/6=0.500\). (2) The downward crossing is \(0+\frac{4-6}{2-6}(2)=1\) hour. The upward crossing is \(2+\frac{4-2}{5-2}(3)=4\) hours. The above-threshold portions are 0–1 and 4–5 hours, totaling 2.00 hours; the fraction is \(2/5=0.400\). (3) No. A straight line between two concentrations below the threshold remains below it, so the segment contributes zero time above threshold. (4) It contributes zero time under \(>\), because concentration is never greater than the threshold on that plateau. It contributes the full 2 hours under \(\geq\).