Why Titration Step Size Depends on Sensitivity
A dose increase of 10 mg/day does not imply a particular response increase on its own. The expected response change depends on how dose changes exposure and how sensitive response is to exposure at the current operating point. A modest dose step may produce a large response change where the exposure–response curve is steep, but little change where it is nearly flat.
This tutorial develops a local calculation for translating a specified response increment into a dose step. It complements the earlier material on writing titration rules as state-update equations: here the focus is how to choose the size of the dose change from a model’s local sensitivity. A calculated step is a model-based expectation, not a guarantee that an individual’s response will change by exactly the requested amount.
Units are part of the calculation. If response is measured in response-score points and exposure in mg·h/L, then \(S_E\) has units of response-score points per mg·h/L. If dose is measured in mg/day, \(S_D\) has units of response-score points per mg/day. Consequently, dividing a response increment by \(S_D\) gives a dose increment in mg/day.
The exposure metric must be stated and used consistently. It might be an average concentration, an AUC over a defined interval, or another model-based exposure summary. The dose-to-exposure mapping must correspond to the same regimen and exposure definition. The calculation does not mean that a dose increment and a response increment have matching numerical values: they are different quantities with different units.
Deriving the Local Dose-Step Formula
Suppose the current dose is \(D_0\), with predicted response \(R(D_0)\). For a small proposed step \(\Delta D\), a Taylor expansion around \(D_0\) gives
for some \(\xi\) between \(D_0\) and \(D_0+\Delta D\), when the second derivative exists on that interval. Subtracting the current prediction shows that the response change is approximately \(S_D\Delta D\) when the second-order term is small. Setting this approximate change equal to the desired increment \(\Delta R\) and solving yields
This derivation also identifies the approximation’s main condition: curvature must be small enough over the proposed step that the second-order contribution does not materially alter the predicted response change. A derivative at one point describes local sensitivity, not necessarily sensitivity across a large dose range.
Signs matter. If response increases with dose, then \(S_D>0\); a desired response increase has \(\Delta R>0\) and gives a positive dose step. If response decreases with dose, a desired response decrease has \(\Delta R<0\), and the quotient again gives a positive step only when the sign of the desired change and slope are treated consistently. The sign of \(\Delta D\) says whether dose should increase or decrease.
Worked Example: A Linear Exposure–Response Region
Consider a hypothetical regimen with dose \(D\) in mg/day and exposure \(E(D)=0.08D\) mg·h/L. In the local region of interest, suppose the modeled response is \(R(E)=20+0.50E\) response-score points. At a current dose of 250 mg/day, calculate the approximate dose step for a desired increase of 4 response-score points.
First calculate the current exposure and response:
The exposure–response slope is \(dR/dE=0.50\) points per mg·h/L, and the dose–exposure slope is \(dE/dD=0.08\) (mg·h/L)/(mg/day). Therefore,
The proposed dose is 350 mg/day. Under the stated model, \(E(350)=28\) mg·h/L and \(R(28)=20+0.50(28)=34\) points, an increase of 4 points. Here the model is linear over the step, so the calculation is exact under its assumptions. The 4-point response increment is not a 4 mg/day dose change; the corresponding dose change is 100 mg/day.
Nonlinear Curves and the Limits of a Local Step
In many models, sensitivity changes with exposure. For example, an Emax response curve becomes progressively flatter as exposure approaches saturation. A derivative-based step can still be useful, but a proposed step should be checked against the model’s predicted response at the new dose. If the local calculation is intended to be approximate, this check reveals whether curvature makes the approximation poor.
Worked Example: Checking a Step on an Emax Curve
Suppose a hypothetical response model is \(R(E)=100E/(4+E)\), where exposure \(E\) is in mg/L and response is a score in points. Let \(E(D)=0.05D\) mg/L, with dose \(D\) in mg/day. At 80 mg/day, calculate a local step for a 5-point response increase, then compare it with the dose change that gives exactly 5 points under the model.
At 80 mg/day, exposure is \(E=0.05(80)=4\) mg/L and response is \(R(4)=100(4)/(4+4)=50\) points. Differentiating the exposure–response function gives
At \(E=4\), this slope is \(400/64=6.25\) points per mg/L. Since \(dE/dD=0.05\) (mg/L)/(mg/day), the response–dose slope is \(6.25(0.05)=0.3125\) points per mg/day. Thus the local step is
The local calculation proposes 96 mg/day. At that dose, exposure is 4.8 mg/L, and the model predicts \(R(4.8)=100(4.8)/(8.8)=54.55\) points. The actual model-predicted increase is therefore about 4.55 points, not 5.
For an exact 5-point increase, the target response is 55 points. Solve \(100E/(4+E)=55\): \(100E=220+55E\), so \(E=220/45=4.889\) mg/L. The dose producing this exposure is \(D=4.889/0.05=97.78\) mg/day, an exact model-based increase of 17.78 mg/day. The local step was smaller because the curve’s slope decreases as exposure rises; the tangent at the current exposure overestimates how much response each additional unit of exposure will produce along the full step.
Estimating Sensitivity From Model Predictions
A fitted model may be available as a prediction function or numerical program rather than a convenient symbolic derivative. Sensitivity can then be estimated with a finite difference: evaluate predicted response at nearby doses and divide the response difference by the dose difference. A centered difference uses predictions on both sides of the current dose and can reduce one-sided approximation error, provided both doses are in a scientifically meaningful region.
Worked Example: A Centered Difference From Predictions
A hypothetical model predicts responses of 42 points at 40 mg/day, 47 points at 50 mg/day, and 51.5 points at 60 mg/day. Estimate the local slope at 50 mg/day and calculate a local dose step for a desired response increase of 2 points.
The centered-difference estimate uses the predictions 10 mg/day below and above the current dose:
The approximate step is \(2/0.475=4.211\) mg/day. This predicts a new dose near 54.21 mg/day. As a basic check on local behavior, the one-sided slopes are \((47-42)/10=0.500\) and \((51.5-47)/10=0.450\) points per mg/day. Their similarity supports—but does not prove—that the centered slope is a reasonable local summary. The model prediction at 54.21 mg/day should be evaluated directly when available; the centered-difference calculation alone does not establish that the response increase will be exactly 2 points.
Assumptions, Diagnostics, and Alternatives
The calculation is meaningful only when its inputs represent the same modeling context. In particular, the exposure definition, regimen, response scale, and current operating point must be clear. The dose–exposure derivative should reflect the relevant dose-to-exposure relationship; assuming exposure is proportional to dose without justification can produce a misleading step. The response model and exposure mapping should also be differentiable locally, with a slope that is sufficiently far from zero for division to be stable.
Useful diagnostics include comparing slopes just below and above the current dose, evaluating the full model prediction at the proposed new dose, and checking the size of the second-order change when curvature is available. A very large proposed step relative to the dose range used to estimate the local slope is a warning that the calculation is extrapolating a local quantity. A slope near zero is also a warning: dividing by a small number can produce an impractically large step, while small errors in the estimated slope can cause large changes in the calculated step.
Sensitivity itself may be uncertain because model parameters, individual exposure, and response measurements are uncertain. If the estimated slope varies substantially across plausible parameter values or individuals, a single step computed from a point estimate should not be described as reliably achieving a fixed response increment. The calculated change is an expected model-based change under the stated inputs, not an individual-level promise.
| Approach | What it uses | Strength and limitation |
|---|---|---|
| Local derivative step | Current local slope and desired response increment | Quick and interpretable for small steps; can be inaccurate on a curved response surface. |
| Exact model inversion | The response target and the full response–dose function | Uses the modeled curve across the whole step; requires a solvable, appropriate model and may have no feasible solution. |
| Fixed dose increment | A prespecified dose change | Simple to state, but does not correspond to a constant expected response increment when sensitivity varies. |
A practical choice is to use the local derivative to identify a candidate step, then use the full model to check the predicted response at the candidate dose. If a specific response target must be met under the model and inversion is feasible, solving \(R(E(D))=R(D_0)+\Delta R\) is more direct than relying on a tangent approximation. Neither method removes uncertainty in the model or guarantees the outcome for a particular patient.
Check Your Understanding
Work each problem, then compare with the solutions below.
- A model has \(dR/dE=2\) points per mg/L and \(dE/dD=0.1\) (mg/L)/(mg/day). What local dose step corresponds to a 3-point response increase?
- If \(S_D=-0.25\) points per mg/day, what dose change locally corresponds to a desired response change of \(-2\) points? State whether dose increases or decreases.
- Why can a local derivative step be inaccurate for a large step on a saturating response curve?
- A local slope is estimated as 0.02 points per mg/day. What warning arises when calculating the dose step for a 4-point increase?
Solutions. (1) \(S_D=(2)(0.1)=0.2\) points per mg/day, so \(\Delta D\approx3/0.2=15\) mg/day. (2) \(\Delta D\approx(-2)/(-0.25)=8\) mg/day, so dose increases; both dose and response decrease under the negative local slope. (3) The slope changes across a saturating curve, so the tangent at the starting point may not represent the average response change across the full interval; evaluate the model at the proposed dose or solve for the target directly. (4) The proposed step is \(4/0.02=200\) mg/day, potentially large. A small slope makes the calculation sensitive to slope error, and the local approximation may not hold over 200 mg/day.