Combining Peak and Trough Measurements

Why Pair a Peak With a Trough?

A concentration measured at one time can be consistent with many combinations of clearance and distribution volume. A suitably timed peak and trough add different information: their ratio describes how quickly concentration falls during the interval, while their scale relative to the dose informs the volume. Under a specified model, the pair can therefore identify both an individual elimination rate and an individual volume, from which clearance follows.

This is a model-based calculation, not a rule that any two concentrations automatically determine reliable patient parameters. The dose history, sampling times, steady-state condition, and structural model must all be appropriate. Here, the derivation uses repeated intravenous bolus doses in a one-compartment model with linear elimination. A peak is the concentration immediately after a bolus, and a trough is the concentration immediately before the next dose.

Definition: Let \(D\) be the dose, \(V\) the apparent distribution volume, \(CL\) clearance, \(k=CL/V\) the first-order elimination rate constant, and \(\tau\) the dosing interval. At steady state, write \(C_{\text{peak}}\) for the post-bolus concentration and \(C_{\text{trough}}\) for the pre-dose concentration. The model assumes \(D\), \(V\), and \(CL\) are constant across intervals, elimination is linear and first order, and the measured concentrations correspond to the stated times. With concentration in mg/L, dose in mg, and time in hours, \(V\) is in L, \(CL\) in L/h, and \(k\) in h\(^{-1}\).

How the Pair Identifies Elimination and Volume

For a repeated bolus regimen, each earlier dose contributes a residual concentration at the time of the next peak. Define \(r=e^{-k\tau}\), the fraction remaining after one interval. At steady state, the contributions to the peak form a geometric series: \(D/V\) from the current dose, \(rD/V\) from the preceding dose, and so on. Since \(k>0\), \(0<r<1\), and the series converges.

C_{\text{peak}}=\frac{D}{V}(1+r+r^2+\cdots) =\frac{D}{V(1-r)}

During the interval, concentration declines exponentially. At the trough, one full interval has elapsed since the peak, so \(C_{\text{trough}}=C_{\text{peak}}r\). Dividing the trough by the peak isolates \(r\); taking its logarithm gives \(k\). The peak equation then gives \(V\), and \(CL=kV\).

r=\frac{C_{\text{trough}}}{C_{\text{peak}}},\qquad k=-\frac{\ln(r)}{\tau} =\frac{\ln(C_{\text{peak}}/C_{\text{trough}})}{\tau}

V=\frac{D}{C_{\text{peak}}(1-r)},\qquad CL=kV

These equations require \(0<C_{\text{trough}}<C_{\text{peak}}\). That condition is necessary under the stated model, but it is not sufficient to establish that the model or sampling history is correct. The algebra also presumes that the two values describe the same dose and interval, with no intervening dose change or missed dose.

Worked Example: Deriving Clearance and Volume From a Paired Sample

Suppose, hypothetically, a patient receives 1,000 mg as a repeated intravenous bolus every 12 hours. At steady state, the concentration immediately after a dose is 20.0 mg/L, and the concentration just before the next dose is 4.00 mg/L. Assume the measurements and dose times are accurate and the one-compartment linear model is suitable.

First calculate the fraction remaining over the 12-hour interval:

r=\frac{4.00\ \text{mg/L}}{20.0\ \text{mg/L}}=0.200

Then calculate the elimination rate:

k=\frac{\ln(20.0/4.00)}{12\ \text{h}} =\frac{\ln(5)}{12\ \text{h}} =\frac{1.60944}{12\ \text{h}} =0.1341\ \text{h}^{-1}

The volume follows from the peak equation:

V=\frac{1{,}000\ \text{mg}} {(20.0\ \text{mg/L})(1-0.200)} =\frac{1{,}000\ \text{mg}}{16.0\ \text{mg/L}} =62.50\ \text{L}

Finally, calculate clearance:

CL=\left(\frac{\ln(5)}{12\ \text{h}}\right)(62.50\ \text{L}) =8.382\ \text{L/h}

The pair therefore implies \(k=0.1341\ \text{h}^{-1}\), \(V=62.50\ \text{L}\), and \(CL=8.382\ \text{L/h}\) under the stated assumptions. The calculation does not show that these parameters are known without error; it shows the values that exactly reproduce the two concentrations in this model.

Why One Concentration Is Usually Not Enough

A single peak supplies one equation involving both \(V\) and \(k\). For any admissible choice of \(k\), the volume can be chosen to reproduce the same peak. A single trough also supplies one equation in two unknowns. Thus, without an additional constraint—such as fixing one parameter from reliable external information—a single concentration does not generally identify both parameters.

Worked Example: Many Parameter Pairs Fit One Peak

Use a hypothetical 1,000 mg bolus every 12 hours and a steady-state peak of 20.0 mg/L. Consider two possible elimination rates. For \(k=0.100\ \text{h}^{-1}\), \(r=e^{-1.2}\approx0.3012\), so:

V=\frac{1{,}000}{20.0(1-e^{-1.2})} \approx71.55\ \text{L},\qquad CL=(0.100)(71.55)=7.155\ \text{L/h}

For \(k=0.200\ \text{h}^{-1}\), \(r=e^{-2.4}=0.09072\), giving:

V=\frac{1{,}000}{20.0(1-e^{-2.4})} \approx54.99\ \text{L},\qquad CL=(0.200)(54.99)=11.00\ \text{L/h}

Both parameter pairs reproduce the same 20.0 mg/L peak, yet the implied clearance values differ. Their predicted troughs differ as well: \(20.0(0.3012)=6.024\ \text{mg/L}\) for the first rate and \(20.0(0.09072)=1.814\ \text{mg/L}\) for the second. A trough can distinguish between such possibilities, provided it is measured in the defined interval and the model assumptions hold.

Comparing Forecasts Based on One Sample or Both

A one-sample estimate can be made if an additional parameter is fixed, but the resulting forecast inherits that constraint. The following illustration holds \(V=60.0\ \text{L}\) fixed for each single-sample calculation. It compares those calculations with the paired estimate from the earlier example, which allowed \(V\) to be estimated.

Worked Example: Forecasting a New Trough Under Different Fits

Use the hypothetical 1,000 mg dose, 12-hour interval, peak of 20.0 mg/L, and trough of 4.00 mg/L from the paired example. For a peak-only fit with \(V=60.0\ \text{L}\), the peak equation implies \(r=1-D/(VC_{\text{peak}})=1-1{,}000/(60.0\times20.0)=1/6\). Thus \(k=\ln(6)/12=0.1493\ \text{h}^{-1}\), and \(CL=\left(\frac{\ln(6)}{12\ \text{h}}\right)(60.0\ \text{L})=8.959\ \text{L/h}\).

For a trough-only fit with that same fixed volume, use \(C_{\text{trough}}=(D/V)r/(1-r)\). Rearranging gives \(r=C_{\text{trough}}V/(D+C_{\text{trough}}V)\):

r=\frac{(4.00)(60.0)}{1{,}000+(4.00)(60.0)} =\frac{6}{31},\qquad k=\frac{\ln(31/6)}{12}=0.1369\ \text{h}^{-1}

That trough-only fit gives \(CL=\left(\frac{\ln(31/6)}{12\ \text{h}}\right)(60.0\ \text{L})=8.211\ \text{L/h}\). These estimates differ because each one-sample fit is constrained to use the stipulated volume and each sample carries different information.

Now forecast the steady-state trough for the same 1,000 mg dose given every 8 hours. For any fitted \(V\) and \(k\), the new trough is \((D/V)e^{-8k}/(1-e^{-8k})\). The peak-only fit predicts approximately 7.240 mg/L; the trough-only fit predicts approximately 8.381 mg/L. The paired fit has \(V=62.50\ \text{L}\) and \(k=\ln(5)/12\ \text{h}^{-1}\), and predicts:

C_{\text{trough,new}} =\frac{1{,}000}{62.50} \frac{e^{-8\ln(5)/12}}{1-e^{-8\ln(5)/12}} \approx 8.316\ \text{mg/L}

The paired forecast uses information from both measured concentrations and does not impose \(V=60.0\ \text{L}\). The difference among these predictions is not evidence that one sample is inherently wrong; it reflects the assumptions used to fill in the parameter that one sample cannot determine by itself.

Measurement Error, Diagnostics, and Alternatives

Direct substitution is useful for understanding what the two concentrations imply, but a real pair is measured with error. The ratio \(C_{\text{trough}}/C_{\text{peak}}\) then also contains measurement error, and the resulting \(k\), \(V\), and \(CL\) are estimates rather than exact patient properties. With a specified residual-error model, both concentrations can instead be used together in a likelihood or individual Bayesian fit, as covered in the earlier material on constructing the individual concentration likelihood. Such a fit can account for uncertainty and population information; simply taking the ratio does not.

Worked Example: Checking a Noisy Pair Before Interpreting It

Suppose hypothetical measured concentrations are 19.6 mg/L after a 1,000 mg bolus and 4.20 mg/L immediately before the next dose, 12 hours later. They have the expected order, and the time and dose history are assumed correct. Direct substitution gives:

r=\frac{4.20}{19.6}=0.2143,\qquad k=\frac{\ln(19.6/4.20)}{12} =\frac{\ln(14/3)}{12} =0.1284\ \text{h}^{-1}

The volume and clearance are:

V=\frac{1{,}000}{19.6(1-0.2143)} =\frac{1{,}000}{15.40} =64.94\ \text{L},\qquad CL=\left(\frac{\ln(14/3)}{12\ \text{h}}\right)(64.94\ \text{L})=8.336\ \text{L/h}

These values are an exact algebraic fit to the rounded pair, not a measure of precision. Because two parameters are fitted to two concentrations, this pair alone leaves no residual degree of freedom for checking whether the one-compartment model describes the concentration-time profile. More samples at known times, or other independent information, are needed for a stronger model check.

Before interpreting any paired estimate, check the conditions that make its algebra meaningful. Confirm the actual dose and interval, whether steady state was reached, and whether the samples were taken at the specified post-bolus and pre-dose times. Confirm that concentrations are positive and the peak exceeds the trough. Then assess whether linear one-compartment disposition is plausible for the drug and interval. A model with an infusion, absorption phase, multiple compartments, nonlinear elimination, or an inaccurate dose history needs its corresponding concentration-time equation; the bolus formulas above should not be applied unchanged.

A particularly tempting mistake is to treat two measured values as sufficient proof that both individual parameters are precisely determined. The algebra yields a unique pair under the exact model, but does not test that model, account for assay error, or establish that the patient was at steady state. The correction is to distinguish identification within an assumed model from precision and model adequacy. When the assumptions are uncertain, use an appropriate error model and individual fitting approach, and interpret the estimates with the available information about parameter uncertainty.

The main alternatives serve different purposes. A single concentration can update an individual only in combination with a constraint or population information; it cannot freely determine both \(CL\) and \(V\). A paired algebraic calculation is transparent and useful under the narrow assumptions given here, but it does not accommodate measurement error or departures from the model. A likelihood-based or Bayesian fit can combine multiple timed samples with an explicit error model and population distribution, but its conclusions depend on those inputs and cannot repair an incorrect dose history or unsuitable structural model.

Key takeaway: For a steady-state one-compartment IV bolus regimen, the peak-to-trough ratio over a known interval identifies \(k\); the peak scale and known dose then identify \(V\), and \(CL=kV\). Treat these as model-dependent estimates, and verify timing, dosing history, and structural assumptions before using them to forecast.

Check Your Understanding

Answer each question, then compare with the concise solutions below.

What conditions are required for the peak-to-trough ratio formula to estimate \(k\) in this tutorial?
For a 10-hour interval, \(C_{\text{peak}}=15.0\ \text{mg/L}\), and \(C_{\text{trough}}=3.00\ \text{mg/L}\), calculate \(r\) and \(k\).
Why can one peak not generally identify both \(V\) and \(CL\)?
Under the stated model, what does a measured trough higher than its corresponding peak suggest checking?

Solutions. (1) The concentrations must be from the same steady-state interval with a known interval, correctly timed post-bolus peak and pre-dose trough, and the stated one-compartment linear-elimination model must be appropriate. (2) \(r=3.00/15.0=0.200\), and \(k=\ln(15.0/3.00)/(10\ \text{h})=\ln(5)/10=0.1609\ \text{h}^{-1}\). (3) The peak equation contains both \(V\) and \(k=CL/V\), so different rates can be paired with different volumes to reproduce that same peak. (4) Check whether sample labels or times were reversed, whether dose history or interval was recorded correctly, and whether the model assumptions are plausible; the stated model requires \(C_{\text{trough}}<C_{\text{peak}}\).