Average Concentration Over A Defined Window

Why Define the Time Window?

A concentration changes over time, so an average concentration is meaningful only when the time interval is specified. Dividing the area under the concentration-time curve (AUC) by the duration of that interval gives a time-averaged concentration. The result has the same units as concentration, but it summarizes the whole window rather than describing the concentration at one instant.

The window should match the exposure question. For example, an average over 0 to 12 hours summarizes exposure during those 12 hours; an average over 0 to 24 hours summarizes a different period. Even when both calculations use the same concentration profile, their results need not be equal. The window endpoints are therefore part of the definition of the metric, not a minor reporting detail.

Definition: Let \(C(t)\) be concentration at time \(t\), and let \([t_1,t_2]\) be a defined time window with \(t_2>t_1\). The average concentration over that window is

C_{\mathrm{avg},[t_1,t_2]}=\frac{\mathrm{AUC}_{t_1,t_2}}{t_2-t_1} =\frac{1}{t_2-t_1}\int_{t_1}^{t_2} C(t)\,dt.

Here, \(\mathrm{AUC}_{t_1,t_2}\) is the area under the concentration-time curve from \(t_1\) through \(t_2\). The formula applies when that area represents the concentration profile over the entire window.

If concentration is measured in \(\mathrm{mg/L}\) and time in hours, AUC is measured in \(\mathrm{mg\cdot h/L}\). Dividing by hours gives \(C_{\mathrm{avg}}\) in \(\mathrm{mg/L}\). This unit check is a useful first diagnostic: an answer still in \(\mathrm{mg\cdot h/L}\) is an AUC, not an average concentration.

The definition is a time-weighted mean of concentration across the window. It is not necessarily the arithmetic mean of the measured concentration values. A simple arithmetic mean gives equal weight to each recorded value, whereas the time average reflects how long the curve spends at different concentrations. With observations at unevenly spaced times, those weights are particularly unlikely to be equal.

Calculating the Average

When AUC over the window is already available, the calculation is direct: divide by the elapsed time between the two endpoints. The endpoints and duration must refer to the same window as the AUC. If AUC is estimated from concentration samples, the calculation also depends on how the curve between samples is represented. A common numerical approach is the linear trapezoidal rule, which estimates the area between consecutive observations as the area of a trapezoid.

Worked Example: Average Concentration From AUC

Suppose a hypothetical concentration profile has an estimated AUC of \(48.0\ \mathrm{mg\cdot h/L}\) from hour 0 to hour 12. Calculate the average concentration over that window.

The window duration is \(12-0=12\) hours. Apply the definition:

C_{\mathrm{avg},[0,12]} =\frac{\mathrm{AUC}_{0,12}}{12-0} =\frac{48.0\ \mathrm{mg\cdot h/L}}{12\ \mathrm{h}} =4.00\ \mathrm{mg/L}.

The result summarizes the area over hours 0 through 12 as a constant concentration of \(4.00\ \mathrm{mg/L}\) lasting 12 hours: \(4.00\ \mathrm{mg/L}\times12\ \mathrm{h}=48.0\ \mathrm{mg\cdot h/L}\). It does not mean concentration was \(4.00\ \mathrm{mg/L}\) at every instant, or that the concentration at hour 12 was \(4.00\ \mathrm{mg/L}\).

The formula uses elapsed time, not the number of samples. If the interval is \([t_1,t_2]\), its duration is \(t_2-t_1\), regardless of how many observations were collected. When the window begins after time zero, use the AUC within that window; do not divide by the time from zero unless zero is the actual starting endpoint.

Worked Example: Estimating Averages Over Different Windows

Consider a hypothetical profile measured at the following times. Assume concentration changes linearly between successive observations. Calculate the average concentration over hours 0 to 4, 0 to 8, and 2 to 8.

Time (h)	Concentration (mg/L)
0	4
2	12
4	8
8	4

For a segment of duration \(\Delta t\) with endpoint concentrations \(C_{\mathrm{start}}\) and \(C_{\mathrm{end}}\), the linear trapezoidal area is \(\Delta t(C_{\mathrm{start}}+C_{\mathrm{end}})/2\).

From hour 0 to hour 2, the area is \(2(4+12)/2=16\ \mathrm{mg\cdot h/L}\). From hour 2 to hour 4, it is \(2(12+8)/2=20\ \mathrm{mg\cdot h/L}\). Thus the AUC from 0 to 4 hours is \(16+20=36\ \mathrm{mg\cdot h/L}\), and

C_{\mathrm{avg},[0,4]}=\frac{36}{4-0}=9.00\ \mathrm{mg/L}.

From hour 4 to hour 8, the area is \(4(8+4)/2=24\ \mathrm{mg\cdot h/L}\). The AUC from 0 to 8 hours is therefore \(36+24=60\ \mathrm{mg\cdot h/L}\), giving

C_{\mathrm{avg},[0,8]}=\frac{60}{8-0}=7.50\ \mathrm{mg/L}.

For the window from hour 2 to hour 8, use only the areas from 2 to 4 and 4 to 8. Their sum is \(20+24=44\ \mathrm{mg\cdot h/L}\), and the duration is \(8-2=6\) hours:

C_{\mathrm{avg},[2,8]}=\frac{44}{8-2}=7.33\ \mathrm{mg/L}.

The three values differ because they summarize different parts of the profile. Including the relatively lower concentrations from hours 4 to 8 reduces the average for the 0-to-8-hour window compared with the 0-to-4-hour window. Moving the starting boundary to hour 2 also changes which part of the profile contributes to the calculation. These are not conflicting estimates of one fixed-window quantity; each is an average for its stated interval.

Why Combined Averages Are Duration-Weighted

A longer interval contributes more time to a combined window than a shorter interval. Consequently, when separate intervals are combined, their averages must be weighted by their durations. This follows directly from adding their AUCs and dividing by the total duration.

Result: If a window is split into non-overlapping intervals with durations \(\Delta t_i\) and interval averages \(C_{\mathrm{avg},i}\), then

C_{\mathrm{avg,combined}} =\frac{\sum_i \Delta t_i C_{\mathrm{avg},i}}{\sum_i \Delta t_i}.

The combined average lies between the smallest and largest interval averages. If the intervals have equal duration, it is the arithmetic mean of their averages. In general, it is weighted toward the average from the longer-duration interval.

To justify the result, for each interval \(i\), the definition gives \(\mathrm{AUC}_i=\Delta t_i C_{\mathrm{avg},i}\). The AUC over the combined window is the sum of the interval AUCs, and the combined duration is the sum of the interval durations. Dividing the summed AUC by the summed duration gives the stated formula. The bounds follow because the result is a weighted mean with nonnegative weights that sum to one.

Worked Example: Combining Intervals With Unequal Durations

In a hypothetical profile, the average concentration is \(2.00\ \mathrm{mg/L}\) during a 10-hour interval and \(10.0\ \mathrm{mg/L}\) during a separate 1-hour interval. Calculate the average over both intervals together.

The first interval contributes an AUC of \(10\ \mathrm{h}\times2.00\ \mathrm{mg/L}=20.0\ \mathrm{mg\cdot h/L}\). The second contributes \(1\ \mathrm{h}\times10.0\ \mathrm{mg/L}=10.0\ \mathrm{mg\cdot h/L}\). Total AUC is \(30.0\ \mathrm{mg\cdot h/L}\), and total duration is \(10+1=11\) hours. Therefore,

C_{\mathrm{avg,combined}} =\frac{20.0+10.0}{11} =\frac{30.0}{11} =2.73\ \mathrm{mg/L}.

The combined average is closer to \(2.00\ \mathrm{mg/L}\) than to \(10.0\ \mathrm{mg/L}\), because the first interval lasts much longer. It is between the two interval averages, as the weighted-mean result requires. In contrast, the unweighted arithmetic mean \((2.00+10.0)/2=6.00\ \mathrm{mg/L}\) would give the 1-hour interval the same influence as the 10-hour interval and would not represent the time average.

Assumptions, Diagnostics, and Interpretation

The definition itself is straightforward, but a calculated value is only as appropriate as the AUC and window used. Check the following before interpreting an average concentration:

Window alignment: Confirm that the AUC covers exactly the stated start and end times. An AUC that includes additional time cannot be used with a shorter duration without first restricting the area to the intended window.
Complete coverage: The curve must be represented over the full interval. If a boundary concentration is missing, or a substantial time gap is poorly characterized, the estimated AUC may be unreliable.
Area estimation: The trapezoidal rule assumes a specified shape between observations—in this example, a straight line. Sparse measurements or a rapidly changing profile can make that approximation inadequate. More sampling or a justified curve model may be needed.
Units and arithmetic: AUC units divided by time units must yield concentration units. Check that the denominator is \(t_2-t_1\), not the number of samples or a dosing interval that differs from the stated window.
Interpretation: Report the interval with the value and its units. An average is a summary of exposure across time, not a concentration measured at a particular time.

A useful diagnostic is to compare the computed average with the concentrations observed within the window. For a nonnegative concentration profile, the time average cannot be negative; for a continuous profile it must also lie between the minimum and maximum concentrations attained over the interval. A value outside those bounds suggests a calculation, unit, or window error. If the profile is incompletely observed, however, an apparent discrepancy may instead indicate that the estimated area does not adequately represent the full window.

Average concentration is complementary to AUC, \(C_{\max}\), \(C_{\min}\), and trough exposure. AUC reports cumulative area and therefore depends on the length of the window; average concentration rescales that area by duration. \(C_{\max}\), \(C_{\min}\), and trough describe particular features or times in the profile rather than its time average. None of these summaries should be substituted for another without considering the question being asked. In particular, averaging does not preserve the timing or shape of concentration changes, so it cannot by itself describe every feature of a concentration-time profile.

Common Mistakes

Using the wrong duration. Dividing an AUC from hour 2 to hour 8 by 8 hours is incorrect; its duration is \(8-2=6\) hours.
Calling AUC an average concentration. AUC has concentration-times-time units. Divide by the window duration to obtain concentration units.
Taking an unweighted mean of interval averages. This is only appropriate when the intervals have equal durations. Otherwise, weight each interval average by its duration.
Treating the average as an observed concentration. The average represents an equivalent constant concentration over the window; the actual profile may vary substantially around it.
Leaving the window implicit. The value \(C_{\mathrm{avg}}\) is incomplete without the time interval, because changing the window can change the result and what it summarizes.

The main practical safeguard is to report the window endpoints, duration, AUC, and average concentration together. That makes the calculation reproducible and helps a reader judge whether the time period matches the exposure question.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

A hypothetical AUC from hour 4 to hour 10 is \(30.0\ \mathrm{mg\cdot h/L}\). What is the average concentration over this window?
AUC from hour 0 to hour 8 is \(56\ \mathrm{mg\cdot h/L}\). What is the average concentration, and what does it mean?
Two consecutive intervals have durations of 2 and 6 hours, with average concentrations of 8 and \(3\ \mathrm{mg/L}\), respectively. Calculate their combined average.
Why is the arithmetic mean of concentration samples generally not the same as the time average when sample times are unevenly spaced?
A reported average concentration is greater than every concentration in the window, although all values are nonnegative. Give a likely diagnostic conclusion.

Solutions. (1) The duration is \(10-4=6\) hours, so \(30.0/6=5.00\ \mathrm{mg/L}\). (2) The duration is 8 hours, so \(56/8=7.00\ \mathrm{mg/L}\). This is the equivalent constant concentration whose area over hours 0 to 8 is \(56\ \mathrm{mg\cdot h/L}\); it does not mean the concentration stayed at \(7.00\ \mathrm{mg/L}\). (3) The interval AUCs are \(2\times8=16\) and \(6\times3=18\ \mathrm{mg\cdot h/L}\). The combined average is \((16+18)/(2+6)=34/8=4.25\ \mathrm{mg/L}\). (4) The arithmetic mean gives equal weight to each sample, while a time average weights concentration according to the time represented between observations. (5) For a nonnegative profile, an average above the maximum concentration is inconsistent with the stated window; check the AUC, duration, units, and window boundaries.