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Optimal Sampling for Individual PK Models · Tutorial 49 of 497

Deriving D-Optimal Schedules Under Time Constraints

Learn how earliest and latest sampling limits change a local D-optimal schedule, and how to derive and verify the resulting boundary choices.

Advanced 10 min read

What You'll Learn

  • State the assumptions behind a local D-optimal schedule with bounded sampling times.
  • Derive the Fisher information determinant for a one-compartment IV bolus model.
  • Show why the optimal times in that model lie at the earliest and latest allowable times.
  • Determine how many samples to allocate to each boundary in a fixed-size schedule.
  • Apply first-order boundary conditions and identify when a numerical result needs checking.
  • Compare analytic boundary derivations with numerical searches over candidate times.

Why time limits change an optimal schedule

A locally D-optimal schedule chooses sampling times to maximize the determinant of the Fisher information matrix at specified nominal parameter values. The determinant summarizes information about all parameters jointly: a larger value corresponds to a smaller local confidence-ellipsoid volume when the regularity assumptions for that approximation hold. In practice, a mathematically attractive time can be unavailable. A sample may not be collected before a minimum processing time, or the study visit may end before the latest informative time. The design problem is therefore an optimization over a feasible time interval, not over all possible times.

The key distinction is between an unconstrained optimum that happens to be feasible and a constrained optimum that is forced to a boundary. The latter is not automatically a failure of the design. It can be the correct consequence of the allowable window. To see exactly how this happens, consider a simple individual IV bolus model and then state the boundary conditions that apply more generally.

Definition: For a fixed number \(N\) of samples, a time-constrained local D-optimal schedule maximizes \(\det I(\mathbf{t};\boldsymbol{\theta}_0)\) over feasible sample times \(\mathbf{t}=(t_1,\ldots,t_N)\), at specified nominal parameter values \(\boldsymbol{\theta}_0\). With a common earliest time \(L\) and latest time \(U\), feasibility means \(L\leq t_i\leq U\) for every \(i\). A boundary is active when an optimal time equals its lower or upper allowable value.

Model, information, and assumptions

Use the one-compartment IV bolus model \(C(t)=D V^{-1}e^{-kt}\), where dose \(D\) is known, \(V\) is volume, and \(k\) is the elimination rate constant. For this derivation, assume the measured log concentration has independent, normally distributed errors with common known variance \(\sigma^2\). Thus the observation at time \(t_i\) is modeled as \(\log C(t_i)\) plus an error with mean zero and variance \(\sigma^2\). This constant-variance assumption on the log scale is part of this example; the resulting schedule need not remain optimal under a different error model.

Take the parameter vector to be \(\boldsymbol{\theta}=(\log V,\log k)\). The log-concentration mean is \(\log D-\log V-kt\). Its sensitivity row with respect to these parameters is \(\mathbf{s}(t)=(-1,-kt)\). Under the stated independent-error model, the Fisher information is the sum of sensitivity outer products divided by the error variance:

$$ I(\mathbf{t};\boldsymbol{\theta})= \frac{1}{\sigma^2}\sum_{i=1}^{N}\mathbf{s}(t_i)^{T}\mathbf{s}(t_i) = \frac{1}{\sigma^2} \begin{pmatrix} N & k\sum_i t_i\\ k\sum_i t_i & k^2\sum_i t_i^2 \end{pmatrix}. $$

The sensitivities depend on \(k\), so this is a local design: evaluate the information at a nominal value of \(k\). The formula also assumes the model is appropriate over the selected times, the observations are independent, and the stated variance is applicable to each sample. If the information matrix is singular, its determinant is zero and it cannot support local estimation of both parameters.

Why the optimum uses the allowable endpoints

For this model the determinant simplifies to a direct measure of the spread of the sampling times. Expanding the two-by-two determinant gives

$$ \det I = \frac{k^2}{\sigma^4} \left[ N\sum_i t_i^2-\left(\sum_i t_i\right)^2 \right] = \frac{k^2}{\sigma^4}\sum_{i<j}(t_i-t_j)^2. $$

The second equality follows by expanding the pairwise sum: each \(t_i^2\) appears \(N-1\) times, and each cross-product \(t_it_j\) appears with coefficient \(-2\). The expression is therefore \(N\sum_i t_i^2-(\sum_i t_i)^2\). With \(k\) and \(\sigma\) fixed, maximizing the determinant is equivalent to maximizing the sum of squared pairwise time differences.

This objective is convex as a function of any one time when the other times are held fixed: it is a sum of squared linear differences. A convex function on a closed interval attains a maximum at an endpoint (or is constant, in which case an endpoint is also a maximizer). Apply this fact to one time at a time: starting from any feasible schedule, replace each time by an endpoint that does not decrease the objective. Thus at least one maximizer has every time at \(L\) or \(U\). If \(m\) samples are at \(L\), the only nonzero pairwise differences are between the two groups, giving

$$ \det I= \frac{k^2}{\sigma^4}m(N-m)(U-L)^2. $$

For integer \(m\) between 0 and \(N\), \(m(N-m)\) is largest when the two group sizes are as equal as possible. This follows from \(m(N-m)=N^2/4-(m-N/2)^2\). Hence the optimal schedule places approximately half the samples at each allowable endpoint. If \(N\) is odd, either of the two nearest integer allocations is optimal. At least two distinct times are needed for a nonsingular information matrix for these two parameters.

Worked Example: Five Samples in a Common Time Window

Suppose five log-concentration samples are planned for a hypothetical individual IV bolus study. The earliest allowable time is \(L=0.5\) h and the latest is \(U=8.0\) h. At the nominal value \(k=0.20\ \text{h}^{-1}\), assume independent log-scale errors with \(\sigma=0.10\). Determine the D-optimal schedule under these assumptions.

The endpoint result applies because the model, parameterization, and common-variance assumptions above hold, and the determinant is positive for any schedule containing both endpoints. Since \(N=5\), the most balanced endpoint allocation is two samples at one endpoint and three at the other. The determinant is the same for either allocation:

$$ \det I = \frac{(0.20\ \text{h}^{-1})^2}{(0.10)^4} (2)(3)(8.0-0.5)^2 = \frac{0.0400\times 6\times 56.25}{0.0001} = 135{,}000. $$

Thus one optimal schedule is two samples at \(0.5\) h and three at \(8.0\) h; reversing the counts gives the same determinant. For comparison, an allocation of one sample at \(0.5\) h and four at \(8.0\) h has \(m(N-m)=4\), rather than 6, and determinant \(90{,}000\). Under this model and error assumption, the more balanced allocation provides the larger joint information determinant.

General boundary conditions for local designs

The endpoint result above is specific to the stated model and assumptions. For a more general PK model, information can vary nonlinearly with time, and the optimal schedule need not place every sample at an endpoint. A useful general check comes from the first-order conditions for constrained optimization. Write \(F(\mathbf{t})=\log\det I(\mathbf{t})\), provided \(I\) is positive definite. Maximizing \(F\) is equivalent to maximizing the determinant because the logarithm is strictly increasing.

For a small change \(dI\), the determinant identity \(\det(I+dI)=\det(I)\det(\mathrm{Id}+I^{-1}dI)\), where \(\mathrm{Id}\) is the identity matrix, implies \(d\log\det I=\operatorname{tr}(I^{-1}dI)\) to first order. Consequently,

$$ \frac{\partial F}{\partial t_i} = \operatorname{tr} \left( I^{-1}\frac{\partial I}{\partial t_i} \right). $$

For an interior time \(L<t_i<U\), a differentiable local optimum must have \(\partial F/\partial t_i=0\). At a lower boundary \(t_i=L\), only small increases are feasible, so a necessary first-order condition is \(\partial F/\partial t_i\leq 0\). At an upper boundary \(t_i=U\), only small decreases are feasible, so the necessary condition is \(\partial F/\partial t_i\geq 0\). These signs say that a feasible move into the interval cannot improve the objective to first order. They are necessary conditions, not proof of a global maximum for a general nonlinear design problem.

Worked Example: Checking Two Active Boundaries

Consider two samples in the same hypothetical IV bolus model, but suppose sample 1 is allowed only between 1 and 3 h and sample 2 only between 4 and 7 h. Let \(k=0.15\ \text{h}^{-1}\) and \(\sigma=0.12\) on the log scale. Since the first sample is always earlier than the second, the two-sample determinant is

$$ \det I=\frac{k^2}{\sigma^4}(t_2-t_1)^2. $$

The determinant increases with the squared time separation, so the greatest feasible separation is obtained at \(t_1=1\) h and \(t_2=7\) h. Its value is

$$ \det I= \frac{(0.15)^2(7-1)^2}{(0.12)^4} = \frac{0.0225\times36}{0.00020736} \approx 3{,}906. $$

The boundary signs confirm the result. Here \(F=\log\det I\), so \(\partial F/\partial t_1=-2/(t_2-t_1)=-1/3\ \text{h}^{-1}\), which is nonpositive at the lower bound for sample 1. Also, \(\partial F/\partial t_2=2/(t_2-t_1)=1/3\ \text{h}^{-1}\), which is nonnegative at the upper bound for sample 2. Moving either time into its allowed interval reduces the separation to first order and cannot improve the determinant.

Worked Example: Quantifying a Restricted Latest Time

Return to a hypothetical four-sample design with \(L=0.5\) h, \(k=0.20\ \text{h}^{-1}\), and \(\sigma=0.10\). Suppose an operational limit reduces the latest allowable time from 8.0 h to 5.5 h. The assumptions for the endpoint derivation are unchanged, so the balanced schedule has two samples at each boundary. Its determinant is

$$ \det I_{\text{restricted}} = \frac{(0.20)^2}{(0.10)^4}(2)(2)(5.5-0.5)^2 = \frac{0.0400\times4\times25}{0.0001} = 40{,}000. $$

With an 8.0 h latest time instead, the same four-sample endpoint allocation would give

$$ \det I_{\text{wider}} = \frac{(0.20)^2}{(0.10)^4}(2)(2)(8.0-0.5)^2 = 90{,}000. $$

The restricted design has \(40{,}000/90{,}000\approx0.444\), or about 44.4%, of the determinant under the wider window. This is a comparison of determinants, not a statement that every parameter variance increases by the reciprocal of 0.444. The determinant describes joint information; individual uncertainty depends on the full information matrix and its inverse.

How to diagnose a constrained solution

A boundary-valued schedule should be checked, not rejected merely because some times equal \(L\) or \(U\). For the illustrative one-compartment model, the derivation establishes that endpoint times are optimal under its assumptions. For a more general model, verify the time bounds, calculate the information matrix at the proposed schedule, and check whether it is positive definite. If a claimed D-optimal design has a singular information matrix, it cannot jointly identify all modeled parameters locally, regardless of how attractive its time placement appears.

Numerical optimization is useful when the objective does not have the simple endpoint form derived here. Numerical Search Over Candidate Sampling Times can compare feasible schedules on a grid, but a grid is an approximation to continuous-time optimization. A coarse grid can miss a narrow optimum or make a genuinely active boundary appear slightly interior. Conversely, an optimizer may report a boundary because it has reached a constraint, not because it has established global optimality. Check the analytic derivative signs at active bounds, refine the time grid if using one, and compare feasible alternatives and starting values when the objective may have multiple local maxima.

The analytic derivation and numerical search serve different roles. The derivation explains why a boundary must be active in the stated model and gives the exact allocation. A numerical search can handle models for which no such closed-form argument is available, but it should be checked against feasibility and local behavior. Neither approach removes the dependence of a local design on its nominal parameter values or validates a misspecified PK or observation model.

Common mistake and limitations

  • Mistake: “The unconstrained best time is outside the clinic window, so I will simply move that sample to the nearest boundary.” Correction: Sampling times contribute jointly to the information matrix. In a general model, moving one time can change the best placement of the others. Re-optimize the full feasible schedule or verify the relevant optimality conditions.
  • Mistake: “A time at a boundary means the numerical optimization failed.” Correction: A boundary can be the true constrained optimum. Confirm the feasible-direction derivative sign, and distinguish a justified active constraint from a numerical result that has not converged.

The closed-form endpoint and balanced-allocation result depends on a two-parameter one-compartment IV bolus model, a fixed sample count, a common allowable interval, and independent equal-variance errors on the log-concentration scale. It is not a general rule for oral absorption, multi-compartment models, different residual error assumptions, or additional operational constraints. Those changes alter the sensitivities, information matrix, or feasible set and can change the optimal schedule.

Check Your Understanding

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

  1. For the two-parameter IV bolus example, why does the determinant become zero if every sample is taken at the same time?
  2. With \(N=6\) and a common interval \([L,U]\), how many samples should be placed at each endpoint under the stated assumptions?
  3. At an upper time bound, what sign must the derivative of \(\log\det I\) have as a necessary first-order condition for a constrained maximum?
  4. For \(N=4\), \(k=0.10\ \text{h}^{-1}\), \(\sigma=0.10\), and a window of 1 to 6 h, calculate the determinant of the balanced endpoint schedule.
  5. Why is it not generally valid to optimize each sample time independently by clipping an unconstrained solution to the allowable interval?

Solutions. (1) Identical times give identical sensitivity rows, so the information matrix has rank one and cannot estimate both parameters jointly. (2) Use three samples at \(L\) and three at \(U\), maximizing \(m(6-m)\). (3) The derivative must be nonnegative: only a decrease in time is feasible, and it must not improve the objective to first order. (4) The balanced allocation gives \(\det I=(0.10)^2(2)(2)(6-1)^2/(0.10)^4=10{,}000\). (5) The determinant depends jointly on all sampling times; changing one time changes the information contributions and may alter the optimal placement of the others.

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