Time-Specific Sensitivities in Individual PK Models

Why Sensitivity Depends on Sampling Time

A concentration measurement can respond strongly to a small change in one pharmacokinetic parameter and hardly respond to a change in another. That response also depends on when the sample is taken. Early concentrations may be especially responsive to volume, while concentrations at other times can respond more strongly to clearance. A sensitivity profile makes this time dependence explicit.

The purpose here is to calculate and interpret these profiles for an individual PK model. A sensitivity is a local property: it describes how the model-predicted concentration changes near specified parameter values. It is not, by itself, a guarantee that a parameter can be estimated precisely. Measurement error, parameter scaling, and the sensitivities to other parameters also matter.

Definition: For a model-predicted concentration \(C(t;\boldsymbol{\theta})\), the sensitivity to parameter \(\theta_j\) at time \(t\) is the partial derivative \(\partial C(t;\boldsymbol{\theta})/\partial \theta_j\), evaluated at a specified parameter vector. It gives the local change in predicted concentration per unit change in that parameter, holding the other parameters fixed.

Model, Parameters, and Assumptions

Consider a hypothetical one-compartment model after a single IV bolus dose \(D\). The individual parameters are clearance \(CL\) and volume of distribution \(V\), both positive. Concentration at time \(t\geq 0\) is

C(t;CL,V)=\frac{D}{V}\exp\left(-\frac{CL}{V}t\right).

This model assumes linear elimination, instantaneous bolus input, a well-mixed compartment, and no absorption phase. The dose and sample time are treated as known. The calculations describe model-predicted concentration for one individual at specified values of \(CL\) and \(V\); they do not account for uncertainty in those values.

It is useful to write \(k=CL/V\), the elimination rate constant, and \(x=kt\), a dimensionless time. Then \(C=(D/V)e^{-x}\). Because \(CL\) and \(V\) are positive, sensitivities with respect to their logarithms are often easier to compare than derivatives per unit of the original parameters. A unit change in \(\log(CL)\) or \(\log(V)\) corresponds locally to a proportional change in that parameter.

Definition: The log-parameter sensitivity is \(\partial C/\partial\log(\theta_j)=\theta_j\,\partial C/\partial\theta_j\). It has the same units as concentration. The relative sensitivity, or elasticity, is \(\partial\log(C)/\partial\log(\theta_j)\); it is dimensionless and describes the local proportional change in concentration for a proportional change in a parameter.

Deriving the Individual Sensitivity Profiles

For clarity, take derivatives on the log-parameter scale while holding the other original parameter fixed. Since \(\log C=\log D-\log V-(CL/V)t\), differentiation with respect to \(\log(CL)\) gives

\frac{\partial C}{\partial\log(CL)} =C\frac{\partial\log C}{\partial\log(CL)} =-xC.

Likewise, differentiating with respect to \(\log(V)\) while holding \(CL\) fixed gives

\frac{\partial C}{\partial\log(V)} =C\frac{\partial\log C}{\partial\log(V)} =(x-1)C.

The corresponding relative sensitivities are \(-x\) for \(CL\) and \(x-1\) for \(V\). These signs have a direct interpretation. Increasing clearance while holding volume fixed lowers concentration at every positive time. Increasing volume while holding clearance fixed lowers the initial concentration, but also reduces \(CL/V\), slowing the subsequent decline. Thus the volume effect can change sign over time.

At \(t=0\), \(x=0\): concentration is \(D/V\), so it has no sensitivity to clearance and has log-volume sensitivity \(-C\). At \(x=1\), concentration is locally insensitive to volume on the log scale, since the decrease in initial concentration and the slower decline balance at that time. At later times, the volume sensitivity can be positive: increasing \(V\), with \(CL\) fixed, can increase the predicted concentration relative to its value at the original parameter values.

Worked Example: Calculating Sensitivities at a Specific Time

Suppose a hypothetical individual receives \(D=100\) mg, with \(V=20\) L and \(CL=4\) L/h. At \(t=2\) h, calculate the predicted concentration and the log-parameter sensitivities.

First, \(k=CL/V=4/20=0.2000\) h\(^{-1}\), so \(x=kt=(0.2000)(2)=0.4000\). The concentration is

C(2)=\frac{100\ \mathrm{mg}}{20\ \mathrm{L}}e^{-0.4000} =5.000\ \mathrm{mg/L}\times 0.6703 =3.352\ \mathrm{mg/L}.

The log-clearance sensitivity is

\frac{\partial C}{\partial\log(CL)} =-xC=-(0.4000)(3.352) =-1.341\ \mathrm{mg/L}.

The log-volume sensitivity is

\frac{\partial C}{\partial\log(V)} =(x-1)C=(-0.6000)(3.352) =-2.011\ \mathrm{mg/L}.

Locally, a 1% increase in clearance, with volume fixed, corresponds to about a 1% decrease in concentration at this time, because the clearance elasticity is \(-0.4000\): the approximate concentration change is \(-0.4000\%\). A 1% increase in volume corresponds to an approximate \(-0.6000\%\) concentration change. These are local approximations, not exact predictions for finite parameter changes.

When Do the Sensitivity Profiles Peak?

The timing of a peak depends on what “sensitivity” means. For log-clearance, the absolute concentration sensitivity is \(xC=(D/V)xe^{-x}\). Differentiating its time-shape with respect to \(x\) gives

\frac{d}{dx}(xe^{-x})=e^{-x}(1-x).

This derivative is positive for \(0\leq x<1\) and negative for \(x>1\), so the absolute log-clearance sensitivity is largest at \(x=1\), or \(t=1/k=V/CL\). This is a peak in the magnitude of the local concentration response, not a universal optimal sampling time.

For log-volume, the absolute sensitivity is \((D/V)|x-1|e^{-x}\). On \(0\leq x<1\), the shape \((1-x)e^{-x}\) decreases from 1. On \(x>1\), the shape \((x-1)e^{-x}\) reaches a local maximum at \(x=2\), as its derivative is \((2-x)e^{-x}\). But that later maximum has height \(e^{-2}\), which is smaller than the value 1 at \(x=0\). Thus, if a sample at time zero is possible, the largest absolute log-volume sensitivity occurs there. Practical sampling constraints or the measurement error model can change which times are useful.

Worked Example: Finding the Peak for Clearance Sensitivity

For the hypothetical individual above, \(k=0.2000\) h\(^{-1}\). The absolute log-clearance sensitivity peaks when \(x=1\), so

t_{\mathrm{peak}}=\frac{1}{k}=\frac{1}{0.2000\ \mathrm{h}^{-1}}=5.000\ \mathrm{h}.

At 5 h, \(C=5e^{-1}=1.839\) mg/L and the log-clearance sensitivity is \(-xC=-1.839\) mg/L. Its magnitude is \(1.839\) mg/L per unit change in \(\log(CL)\). At the same time, \(x=1\), so the log-volume sensitivity is \((1-1)C=0\) mg/L. The sample is locally responsive to clearance but not to volume on this scale, holding clearance fixed for the volume derivative.

This does not mean that a 5 h sample alone can determine clearance. A sensitivity peak addresses response to one parameter in isolation; it does not assess measurement noise or whether changes in other parameters could produce a similar concentration response.

Residual Error Changes the Practical Comparison

A raw sensitivity has concentration units, but a concentration change should be judged relative to measurement variability. For a measurement with additive error standard deviation \(\sigma\), a simple local standardized sensitivity is \(|\partial C/\partial\log(\theta_j)|/\sigma\). Under a proportional-error model with coefficient of variation \(c\), the standard deviation is approximately \(cC\), so the corresponding standardized sensitivity is \(|\partial C/\partial\log(\theta_j)|/(cC)\). These quantities compare a local response with expected noise; they are not complete measures of parameter precision.

Worked Example: Additive Versus Proportional Error

Using the same hypothetical model, compare samples at 2, 5, and 10 h. Suppose first that the additive residual standard deviation is \(0.2000\) mg/L. The values of \(x\) are 0.4, 1, and 2, and the concentrations are 3.352, 1.839, and 0.6767 mg/L, respectively.

At 2 h, the absolute log-clearance and log-volume sensitivities are 1.341 and 2.011 mg/L. Dividing by \(0.2000\) mg/L gives standardized magnitudes of 6.703 and 10.06. At 5 h the magnitudes are 1.839 and 0 mg/L, giving 9.197 and 0. At 10 h they are 1.353 and 0.6767 mg/L, giving 6.767 and 3.383. Under this additive-error assumption, the 5 h sample has the largest standardized log-clearance sensitivity among these three times.

Now suppose instead the residual error is proportional with \(c=0.1000\), or approximately 10% of concentration. Dividing each absolute sensitivity by \(0.1000C\) is equivalent to dividing the absolute elasticity by 0.1000. For clearance, the standardized magnitudes at 2, 5, and 10 h are \(0.4/0.1=4\), \(1/0.1=10\), and \(2/0.1=20\). For volume they are \(0.6/0.1=6\), \(0\), and \(1/0.1=10\). Under this error assumption, the 10 h sample has the largest standardized clearance sensitivity among the three.

The comparison shows why a raw sensitivity peak is not enough to choose a sampling time. The error model changes the response relative to noise. The calculations also concern one sample and one parameter at a time; they do not account for whether a collection of observations can distinguish clearance from volume.

Diagnostics, Alternatives, and Limitations

Sensitivity calculations should be checked for sign, scale, and model behavior. For this model, clearance sensitivity must be zero at \(t=0\) and negative at positive times. Volume sensitivity is negative at \(x<1\), zero at \(x=1\), and positive at \(x>1\). Violating these checks often points to an algebra or parameter-holding error. Units provide another check: derivatives with respect to \(CL\) have units of concentration per (L/h), whereas derivatives with respect to \(\log(CL)\) have concentration units.

A numerical finite-difference check is useful when a model has no convenient analytic derivative. Perturb a parameter slightly upward and downward, calculate the corresponding concentrations, and approximate the derivative by the central difference. The perturbation should be small enough to reflect local behavior but not so small that numerical rounding dominates. For positive parameters, perturbing on the log scale helps preserve positivity. A disagreement between analytic and numerical derivatives can reveal a coding error, a parameterization mismatch, or an excessively large finite-difference step.

The main alternative is to use derivatives with respect to the original parameters. For example, \(\partial C/\partial CL=(-t/V)C\) and \(\partial C/\partial V=((x-1)/V)C\). These are correct, but their magnitudes depend on the units used to express the parameters. Log-parameter sensitivities are useful for comparing proportional changes, while original-scale derivatives are appropriate when changes in absolute parameter units are the scientific question. Relative sensitivities additionally remove concentration units, but can be unhelpful when concentration is near zero or when only absolute measurement error is plausible.

A further limitation is that sensitivity to one parameter alone does not establish joint identifiability. At a given time, the concentration is one model output; several parameters may affect it in similar ways. For example, at early times both clearance and volume sensitivities in this model are negative. A schedule that samples across times where the profiles differ in magnitude or sign may help separate their effects, but the full assessment requires the observation model and the combined information from all selected samples.

Common pitfall: “The parameter with the largest raw sensitivity at a time is the parameter that will be estimated most precisely.” This is not generally true. Raw sensitivity depends on parameter scale, and precision also depends on residual error, the number and timing of samples, and overlap with sensitivities to other parameters. Compare sensitivities on a meaningful parameter scale and relative to the error model; treat a single profile as a diagnostic, not a complete design evaluation.

Check Your Understanding

Try each question, then compare your reasoning with the brief solutions.

For the hypothetical individual with \(CL=4\) L/h and \(V=20\) L, what is \(x\) at 10 h, and what is the sign of the log-volume sensitivity?
At what time does the absolute log-clearance sensitivity peak for this individual? State the reasoning.
Why is the log-volume sensitivity zero at 5 h in this model?
With proportional residual error of 10%, which of the 2, 5, and 10 h samples has the largest standardized log-clearance sensitivity? Give its magnitude.
Why can a sample at the peak of one parameter’s sensitivity still be insufficient for estimating that parameter precisely?

Solutions. (1) \(x=(4/20)(10)=2\), so \((x-1)C\) is positive. (2) The peak occurs at \(x=1\), hence \(t=1/k=5\) h. (3) At \(x=1\), the log-volume sensitivity is \((x-1)C=0\). (4) At 10 h, \(x=2\); the standardized magnitude is \(|-x|/0.1=20\). (5) The peak describes a local response to that parameter alone. Noise and overlap with the effects of other parameters can still limit precision.