Dimensionless Sensitivity Coefficients

Why Normalize a Local Sensitivity?

A local sensitivity derivative describes how an output changes when a parameter changes near a specified parameter set. Its numerical value depends on the units used. For example, a derivative with respect to clearance measured in liters per hour will have a different numerical value from the derivative with respect to the same clearance measured in milliliters per minute. Comparing such raw derivatives across parameters can therefore be misleading: their magnitudes reflect both model response and the chosen measurement scales.

A dimensionless sensitivity coefficient addresses this scale issue by comparing relative changes rather than absolute changes. It is a local quantity: it describes the model response near one parameter value and one output value, not a universal effect across all doses, times, or parameter ranges.

Definition: Let $y$ be a differentiable model output depending on a parameter $\theta$. At a point where $\theta \ne 0$ and $y \ne 0$, the dimensionless local sensitivity coefficient of $y$ with respect to $\theta$ is $$ S_{\theta}^{y} = \frac{\theta}{y}\frac{\partial y}{\partial \theta}. $$ It is dimensionless because the units of $\theta/y$ cancel the units of $\partial y/\partial\theta$. It expresses the local relative change in $y$ per relative change in $\theta$.

The derivative $\partial y/\partial\theta$ is evaluated at the parameter set and conditions of interest, holding the other model inputs fixed. In PBPK work, $y$ might be a concentration at a specified time, an area under a concentration–time curve, or another defined model output. The coefficient is meaningful only when those choices, as well as the evaluated parameter values, are stated.

Deriving the Relative-Change Interpretation

Differentiability gives a first-order approximation for a small change $\Delta\theta$:

\Delta y \approx \frac{\partial y}{\partial\theta}\Delta\theta.

When both $\theta$ and $y$ are nonzero, divide by $y$ and express the parameter change relative to $\theta$. Rearranging yields the central interpretation:

\frac{\Delta y}{y} \approx \frac{1}{y}\frac{\partial y}{\partial\theta}\Delta\theta = \left(\frac{\theta}{y}\frac{\partial y}{\partial\theta}\right) \frac{\Delta\theta}{\theta} = S_{\theta}^{y}\frac{\Delta\theta}{\theta}.

Thus, for a sufficiently small 1% relative increase in $\theta$, the output's relative change is approximately $S_{\theta}^{y}\%$. This is a first-order approximation, not an exact prediction for a finite change. Its accuracy depends on how well the local linear approximation represents the model over that change.

If both parameter and output are positive, the coefficient can also be written as a logarithmic derivative:

S_{\theta}^{y} = \frac{\partial \log y}{\partial \log\theta}.

This form follows from the chain rule: $\partial \log y/\partial\log\theta=(\partial\log y/\partial y)(\partial y/\partial\theta)(\partial\theta/\partial\log\theta)=(1/y)(\partial y/\partial\theta)\theta$. The ratio-of-relative-changes definition remains useful even when logarithms are not defined, provided $\theta$ and $y$ are nonzero.

Worked Example: Normalizing a Clearance Derivative

Suppose a hypothetical PBPK model predicts a concentration $C_{\max}=2.00$ mg/L at a clearance $CL=5.00$ L/h. At this point, the model derivative is $\partial C_{\max}/\partial CL=-0.300$ (mg/L)/(L/h). Calculate the dimensionless coefficient and interpret a small 1% increase in clearance.

The definition gives

S_{CL}^{C_{\max}} = \frac{CL}{C_{\max}} \frac{\partial C_{\max}}{\partial CL} = \frac{5.00\ \mathrm{L/h}}{2.00\ \mathrm{mg/L}} \left[-0.300\ \frac{\mathrm{mg/L}}{\mathrm{L/h}}\right] = -0.750.

The units cancel, and the coefficient is dimensionless. For a 1% relative increase in clearance, $\Delta CL/CL=0.01$, so

\frac{\Delta C_{\max}}{C_{\max}} \approx (-0.750)(0.01)=-0.00750.

The predicted first-order relative change is therefore about $-0.750\%$, corresponding to an approximate change of $(-0.00750)(2.00\ \mathrm{mg/L})=-0.0150$ mg/L. The resulting approximate concentration is $1.985$ mg/L. Because both clearance and concentration are positive here, the negative coefficient agrees with the negative derivative: locally, increasing clearance decreases the concentration.

Comparing Parameters on Different Scales

The main use of normalization is to make local relative responses interpretable even when parameters have different units or typical magnitudes. A coefficient of $0.50$ means approximately a 0.50% relative output increase for a 1% relative parameter increase, locally. It does not mean that the output changes by 0.50 units, and it does not by itself establish that the parameter is precisely known or clinically important.

Worked Example: Comparing Intrinsic Clearance and Bioavailability

Consider a hypothetical model with AUC $=8.00$ mg·h/L at intrinsic clearance $CL_{\mathrm{int}}=12.0$ L/h and bioavailability $F=0.600$. At this same parameter set, suppose the local derivatives are $\partial AUC/\partial CL_{\mathrm{int}}=-0.200$ (mg·h/L)/(L/h) and $\partial AUC/\partial F=10.0$ mg·h/L per unit $F$. Calculate both normalized coefficients.

For intrinsic clearance,

S_{CL_{\mathrm{int}}}^{AUC} = \frac{12.0\ \mathrm{L/h}}{8.00\ \mathrm{mg\cdot h/L}} \left[-0.200\ \frac{\mathrm{mg\cdot h/L}}{\mathrm{L/h}}\right] = -0.300.

For bioavailability,

S_F^{AUC} = \frac{0.600}{8.00\ \mathrm{mg\cdot h/L}} \left[10.0\ \frac{\mathrm{mg\cdot h/L}}{\text{unit }F}\right] = 0.750.

The coefficients are dimensionless. Near this parameter set, a 1% relative increase in intrinsic clearance corresponds to an approximate 0.300% relative decrease in AUC. A 1% relative increase in bioavailability corresponds to an approximate 0.750% relative increase in AUC. These comparisons use relative changes in each parameter; comparing the two raw derivatives alone would mix their distinct units and scales.

The result is local and conditional on the same output definition and model conditions. It does not say what happens for a large parameter change, nor does it account for uncertainty or dependence among parameters. Those are different questions from the normalized local response calculated here.

Sign, Units, and Interpretation

The coefficient's sign needs care. Its defining relation can be rearranged as

\frac{\partial y}{\partial\theta} = S_{\theta}^{y}\frac{y}{\theta}.

Therefore, the coefficient's sign indicates the direction of output change as the parameter increases when $\theta/y>0$, as in the common case of positive parameter and output values. If $\theta/y<0$, the coefficient's sign is opposite the derivative's sign. When signs are unusual, use the derivative or the rearranged relation rather than interpreting the coefficient's sign alone.

Worked Example: A Negative Output and a Positive Derivative

For a hypothetical model output $y=-1.00$ output units at $\theta=2.00$ (dimensionless), suppose $\partial y/\partial\theta=0.400$ output units per unit $\theta$. Find the normalized coefficient and check its direction of change.

The coefficient is

S_{\theta}^{y} = \frac{2.00}{-1.00}(0.400) = -0.800.

A negative coefficient does not mean the output decreases numerically as $\theta$ increases in this example. Indeed, the derivative is positive, and the identity confirms this:

\frac{\partial y}{\partial\theta} = (-0.800)\frac{-1.00}{2.00} = 0.400.

For a 1% increase in $\theta$, the first-order relative output change is $\Delta y/y\approx(-0.800)(0.01)=-0.00800$. Since $y=-1.00$, this implies $\Delta y\approx(-0.00800)(-1.00)=0.00800$ output units: the output becomes less negative, approximately $-0.992$. This illustrates why the coefficient's sign must not be equated with the direction of numerical change unless $\theta/y>0$.

The coefficient is invariant to ordinary changes of units. To see this, let the same quantities be represented as $\theta^*=a\theta$ and $y^*=b y$, where $a$ and $b$ are positive unit-conversion constants. The chain rule gives $\partial y^*/\partial\theta^*=(b/a)(\partial y/\partial\theta)$. Consequently,

\frac{\theta^*}{y^*}\frac{\partial y^*}{\partial\theta^*} = \frac{a\theta}{b y}\frac{b}{a}\frac{\partial y}{\partial\theta} = \frac{\theta}{y}\frac{\partial y}{\partial\theta}.

The numerical coefficient is unchanged, although the raw derivative changes with the units. This invariance assumes a linear unit conversion; it does not make the coefficient invariant to changing the model output or evaluation point.

Assumptions, Diagnostics, and Limitations

The definition and first-order interpretation require a finite derivative at the point of interest, as well as nonzero $\theta$ and $y$. The approximation also requires a sufficiently small perturbation that local linear behavior is informative. In practice, report the parameter set, output, time or summary endpoint, and relevant model conditions alongside the coefficient. A coefficient without that context can be ambiguous.

Normalization becomes unstable when $y$ is very close to zero: division by $y$ can produce a very large coefficient even when the absolute output change is small. At $y=0$, this coefficient is undefined. Likewise, it is undefined at $\theta=0$, even if the raw derivative exists. In either situation, the raw derivative may still be informative, or the output and parameter may need a scientifically justified alternative reference scale. An arbitrary scale should not be presented as though it were a relative sensitivity.

A very large coefficient is not automatically evidence of a model error. It may reflect a genuinely strong local relative response, a small output denominator, or behavior near a threshold or other region of rapid change. Check whether the coefficient is finite and whether its magnitude is plausible for the specified point; check the underlying output and derivative; and verify the signs and units in the calculation. A result that changes substantially across nearby operating conditions should be described as point-specific, not as a single stable property of the model.

The raw derivative and the normalized coefficient answer complementary questions. The raw derivative gives an absolute output change per unit parameter change and can be preferable when the output or parameter is zero or near zero. The dimensionless coefficient gives a relative output response per relative parameter change and permits comparisons across measurement scales when both quantities are nonzero. Neither replaces the other, and neither alone summarizes response over a broad parameter range.

Common Mistake

A common error is to say, “The coefficient is negative, so increasing the parameter decreases the output.” That conclusion is justified when $\theta/y>0$, especially for positive parameters and outputs, but not in every case. From $\partial y/\partial\theta=S_{\theta}^{y}y/\theta$, the derivative's sign depends on the signs of the coefficient and $y/\theta$. Always check that relation before stating the direction of change.

Key takeaway: The dimensionless local sensitivity coefficient is the parameter-to-output scale ratio multiplied by the output derivative. It describes a local relative response, is unchanged by linear unit conversions, and requires nonzero parameter and output values. Interpret its sign with the signs of both quantities in view.

Check Your Understanding

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

A positive parameter and positive output have $S_{\theta}^{y}=0.40$. What is the approximate relative output change for a 2% relative parameter increase?
If $y=5.0$ output units, $\theta=10$ parameter units, and $\partial y/\partial\theta=-0.20$ output units per parameter unit, calculate $S_{\theta}^{y}$.
Can the normalized coefficient be calculated when $y=0$? What related quantity might still be defined?
A coefficient is negative, but $\theta/y<0$. What can you conclude about the derivative's sign?
Why can a raw derivative be unsuitable for comparing two parameters measured in different units?

Solutions. (1) The first-order relative change is approximately $0.40(2\%)=0.80\%$, an increase because both values are positive and the coefficient is positive. (2) $S_{\theta}^{y}=(10/5.0)(-0.20)=-0.400$, a dimensionless coefficient. (3) No: the definition divides by $y$, so it is undefined at zero output. The raw derivative may still exist and describe absolute local change. (4) The derivative has the opposite sign from the coefficient because $y/\theta<0$; thus a negative coefficient implies a positive derivative. (5) Raw derivatives carry units of output per parameter, so differing parameter units and scales affect their numerical values. Normalized coefficients compare relative changes when the parameters and outputs are nonzero.