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Optimal Sampling for Individual PK Models · Tutorial 63 of 497

Worked Design of an Oral Sparse Sampling Schedule

Build and compare candidate oral sparse-sampling schedules by calculating parameter sensitivities and weighted information under a one-compartment, first-order absorption model.

Advanced 11 min read

What You'll Learn

  • Specify the oral concentration model and assumptions used for a local design.
  • Calculate absorption and elimination sensitivities on the log-concentration scale.
  • Compare feasible three-time schedules using sensitivity and information determinants.
  • Incorporate unequal assay variances into the information calculation.
  • Recognize when a nominal D-optimal schedule is unreliable or infeasible.

From an Oral PK Model to a Sampling Schedule

An oral sparse-sampling design must place observations where they collectively distinguish the parameters of interest. Early concentrations can be informative about absorption, later concentrations about elimination, and all observations contribute to the concentration scale. A schedule concentrated in one part of the profile may measure concentrations precisely yet provide little separation between parameter effects.

This worked design uses a one-compartment model with first-order absorption and elimination. The purpose is not to claim that one schedule is universally optimal; it is to show how a candidate time grid, model sensitivities, assay precision, and feasibility rules combine in a local design comparison.

Model and assumptions: After a single oral dose, assume the concentration at time \(t>0\) is $$ C(t)=S\frac{k_a}{k_a-k_e}\left(e^{-k_e t}-e^{-k_a t}\right), $$ where \(k_a>k_e>0\) are the first-order absorption and elimination rate constants and \(S=DF/V\) is the dose-bioavailability-volume scale. The dose \(D\) is known; the data identify \(S\), rather than separating \(F\) and \(V\). The model assumes linear, time-invariant disposition; first-order absorption without a lag time; a single compartment; and no dose-dependent or time-varying parameters. For design calculations, use the stated nominal parameter values and assume independent log-concentration errors with known variances. All scheduled concentrations must be measurable, and the candidate schedules compared below have the same number of samples.

For each observation, use the log-concentration sensitivity row for the parameter vector \((\log S,\log k_a,\log k_e)\). The sensitivity is the derivative of the model-predicted log concentration with respect to each log parameter. It is dimensionless and describes the local proportional change in concentration associated with a proportional parameter change.

$$ g(t)=\left(1,\ a(t),\ b(t)\right), \qquad a(t)=-\frac{k_e}{k_a-k_e} +\frac{k_a t}{e^{(k_a-k_e)t}-1}, \qquad b(t)=\frac{k_e}{k_a-k_e} -\frac{k_e t}{1-e^{-(k_a-k_e)t}}. $$

The first component is 1 because \(C(t)\) is proportional to \(S\). The other components describe how the profile changes with absorption and elimination, respectively. Their values depend on the nominal rates and the sample time, so this is a local design calculation.

Why the Sensitivity Formula Has This Form

Write \(\Delta=k_a-k_e\). Taking the logarithm of the concentration model gives $$ \log C(t)=\log S+\log k_a-\log\Delta-k_e t+ \log\left(1-e^{-\Delta t}\right). $$

Differentiate with respect to \(\log k_a\), which means multiplying the derivative with respect to \(k_a\) by \(k_a\). The derivative of \(-\log\Delta\) contributes \(-k_a/\Delta\); the final logarithm contributes \(k_a t/(e^{\Delta t}-1)\). Together with the derivative of \(\log k_a\), this gives \(a(t)=-k_e/\Delta+k_a t/(e^{\Delta t}-1)\). Differentiating with respect to \(\log k_e\) means multiplying the derivative with respect to \(k_e\) by \(k_e\). The \(-\log\Delta\) term contributes \(k_e/\Delta\), and the final two time-dependent terms contribute \(-k_e t/(1-e^{-\Delta t})\), giving \(b(t)\) above. These derivatives are finite for \(t>0\) and distinct positive rates; the equal-rate case is not the model considered here.

Worked Example: Calculating Sensitivity at Two Hours

Suppose a hypothetical individual has nominal \(k_a=1.20\ \mathrm{h}^{-1}\) and \(k_e=0.120\ \mathrm{h}^{-1}\). At \(t=2\ \mathrm{h}\), \(\Delta=1.08\ \mathrm{h}^{-1}\). Calculate the two rate sensitivities.

For absorption, $$ a(2)=-\frac{0.120}{1.08} +\frac{(1.20)(2)}{e^{(1.08)(2)}-1} =-0.1111+\frac{2.40}{e^{2.16}-1} \approx -0.1111+0.3129=0.2018. $$

For elimination, $$ b(2)=\frac{0.120}{1.08} -\frac{(0.120)(2)}{1-e^{-(1.08)(2)}} =0.1111-\frac{0.240}{1-e^{-2.16}} \approx 0.1111-0.2713=-0.1602. $$

Thus the row is approximately \((1,0.2018,-0.1602)\). The positive absorption sensitivity and negative elimination sensitivity indicate the directions of local change at this time; neither value is itself a parameter estimate or a measure of uncertainty.

Constructing and Comparing Feasible Schedules

Consider a hypothetical candidate grid of \(0.5,\ 2,\ 4,\ 8,\) and \(12\ \mathrm{h}\), with three samples per individual. Use the nominal rates above. Suppose a clinic can collect only three samples and requires at least \(1.5\ \mathrm{h}\) between samples. The four illustrative schedules below satisfy that spacing rule and have the same sample count. The table rounds sensitivities for display; determinant calculations use the unrounded values from the formulas.

Time (h)Scale sensitivityAbsorption sensitivity \(a(t)\)Elimination sensitivity \(b(t)\)
0.510.7269-0.03269
210.2018-0.1602
41-0.04642-0.3754
81-0.1094-0.8491
121-0.1111-1.329

Stack the three sensitivity rows for a schedule into a \(3\times3\) matrix \(G\). If the log-concentration errors are independent normal with a common known variance \(\sigma^2\), the expected Fisher information is \(I=G^\mathsf{T}G/\sigma^2\). A full-rank \(G\) is required to distinguish all three parameters locally. The D-optimal comparison chooses the feasible schedule with the largest \(\det(I)\); this corresponds to the smallest local joint uncertainty volume under the information approximation.

For three rows, \(\det(I)=(\det G)^2/(\sigma^2)^3\). Since all candidate schedules here have three observations and the same variance, ranking them by \(\det(I)\) is equivalent to ranking by \(|\det G|\). Geometrically, \(|\det G|\) is the volume spanned by the sensitivity rows: rows that are nearly redundant produce a small determinant.

Worked Example: Comparing Four Candidate Schedules

Suppose the hypothetical log-concentration error variance is \(\sigma^2=0.04\) for every observation. Let A use \((0.5,2,8)\ \mathrm{h}\), B use \((0.5,2,12)\ \mathrm{h}\), C use \((0.5,8,12)\ \mathrm{h}\), and D use \((0.5,4,12)\ \mathrm{h}\). For three rows whose first component is 1, subtracting the first row from each of the other two leaves the determinant unchanged. For D, this gives, using displayed rounded differences, $$ |\det(G_D)|\approx (-0.7733)(-1.2962)-(-0.8380)(-0.3427) =0.71516886\approx0.7152. $$

Applying the same determinant calculation to the unrounded sensitivity rows gives approximately \(0.3221\) for A, \(0.5738\) for B, \(0.3999\) for C, and \(0.7152\) for D. With \((\sigma^2)^3=(0.04)^3=0.000064\), the information determinants are: $$ \det(I_A)\approx\frac{0.3221^2}{0.000064}=1{,}621,\quad \det(I_B)\approx\frac{0.5738^2}{0.000064}=5{,}144, $$ $$ \det(I_C)\approx\frac{0.3999^2}{0.000064}=2{,}499,\quad \det(I_D)\approx\frac{0.7152^2}{0.000064}=7{,}992. $$

Among these four feasible candidates, D has the largest information determinant and is therefore the D-optimal choice for the stated model, nominal rates, parameterization, and error assumption. This conclusion applies only to the candidates compared; it does not establish that D is the best schedule among all possible times or under different constraints.

Weighting Unequal Assay Precision

If observation \(i\) has log-concentration error variance \(\sigma_i^2\), define \(W\) as the diagonal matrix with entries \(1/\sigma_i^2\). The information becomes \(I=G^\mathsf{T}WG\). For a square three-row matrix \(G\), $$ \det(I)=\det(G)^2\det(W) =\frac{\det(G)^2}{\sigma_1^2\sigma_2^2\sigma_3^2}. $$

Thus an observation with greater variance contributes less information, but it is not generally valid to decide whether to include or remove a sample by considering its variance alone: the sensitivity direction matters too. The determinant formula makes that joint role explicit for this three-observation, three-parameter case.

Worked Example: Reweighting Schedule D for Assay Precision

Keep schedule D at \((0.5,4,12)\ \mathrm{h}\), but suppose the hypothetical log-error variances at those respective times are \(0.04,\ 0.01,\) and \(0.04\). The middle observation is more precise. The variance product is $$ (0.04)(0.01)(0.04)=0.000016. $$

Using \(|\det G_D|\approx0.7152\), so that \(\det(G_D)^2\approx0.5115\), gives $$ \det(I_D)\approx\frac{0.5115}{0.000016}=31{,}970. $$

This is four times the equal-variance information determinant for D, approximately \(7{,}992\), because one variance was reduced by a factor of four while the other two were unchanged. The result assumes the variances are known and apply to the stated time points; it is not a general guarantee that improving one assay will make D optimal among all schedules.

Diagnostics, Alternatives, and Limitations

Before accepting a result, check that the design calculation represents a viable experiment. Confirm that the nominal rates satisfy \(k_a>k_e>0\), that each time is permitted by clinic windows and spacing rules, and that predicted concentrations are measurable at every selected time. Confirm the assumed residual variance scale as well: a variance for log concentrations is not interchangeable with a variance for untransformed concentrations. If some observations are expected to be below quantification, treating them as ordinary precise log-concentration measurements is inappropriate.

A very small \(|\det G|\), or a zero determinant, signals near-redundancy or rank deficiency in the sensitivity rows. That means some parameter combinations are weakly distinguished, even if the observed concentrations may be spread over a wide numerical range. Inspecting parameter correlations or the eigenstructure of the information matrix can show which combinations are poorly identified. Also compare the local result across plausible nominal values: because the sensitivities change with \(k_a\) and \(k_e\), a schedule that performs well at one nominal individual may not perform well at another.

D-optimality is not the only defensible criterion. A-optimality targets an overall summary of parameter variances, while c-optimality targets uncertainty in a specified parameter combination. These can rank schedules differently from D-optimality, especially when a particular quantity such as half-life matters more than joint precision in all model parameters. A practical design may also use a compromise criterion or require minimum precision for a specific parameter. Such choices should be stated before ranking schedules.

The calculation is a local, model-based approximation. It assumes the structural model is adequate and uses fixed nominal values and specified error variances. It does not account automatically for model misspecification, uncertain nominal values, dose deviations, informative missingness, or censoring. The single-compartment oral model also cannot represent every absorption pattern or distribution phase. A high determinant under this model cannot compensate for a poor model or an infeasible sampling plan.

Common pitfall: Choosing the schedule with the latest sample because it appears to capture elimination best is not a valid D-optimal argument. A late time may add a useful elimination direction, but it can also have low measurable concentration or contribute a sensitivity row nearly redundant with another. Calculate the weighted information for the feasible schedules and check quantifiability and parameter identifiability; do not infer optimality from one time point or one sensitivity alone.
1
Specify the model and target parameters.
State the concentration model, nominal parameter values, residual-error scale, and parameters to be estimated.
2
Define the feasible candidate grid.
Apply sample-count, spacing, clinic-window, and quantifiability constraints before ranking schedules.
3
Calculate and compare information.
Evaluate the sensitivity rows and weighted information for every candidate, then compare the selected criterion.
4
Stress-test the design.
Check rank, weak parameter directions, plausible nominal values, and whether the model and error assumptions are credible.

Check Your Understanding

Try these questions, then check the solutions below.

  1. For a three-sample design with a common log-error variance, why does maximizing \(|\det G|\) rank schedules the same way as maximizing \(\det(I)\)?
  2. Using the equal-variance comparison, which of schedules A–D is D-optimal, and what is its approximate information determinant?
  3. If one of three independent log-error variances is halved while the other two are unchanged, by what factor does \(\det(I)\) change for the same square sensitivity matrix?
  4. What does a determinant of zero indicate about a three-parameter design with three observations?

Solutions. (1) Since \(I=G^\mathsf{T}G/\sigma^2\), \(\det(I)=\det(G)^2/(\sigma^2)^3\); the common positive denominator and square preserve the ranking by \(|\det G|\). (2) D has the largest value, approximately \(7{,}992\). (3) The variance product is halved, so its reciprocal doubles and \(\det(I)\) doubles. (4) The sensitivity rows are linearly dependent, so the information matrix is singular and the three parameters cannot all be distinguished locally from those observations.

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