D-Optimal Sampling for One-Compartment IV Bolus PK

From a Concentration Model to a Sampling Design

For an individual receiving a known IV bolus dose, sampling times determine how much the observed concentration data can distinguish clearance from volume. The design question is not simply where concentrations are largest: it is which schedule provides information about both parameters together. D-optimality answers this question by maximizing the determinant of the information matrix at specified nominal parameter values.

The design developed here is local: its information is calculated at a chosen pair of clearance and volume values. We first specify an idealized concentration and observation model, derive the sensitivity of concentration to each parameter, and then obtain a closed-form result for two samples. The result depends on the error model and on the allowed sampling window.

Model and assumptions: A known dose \(D\) is administered as an ideal IV bolus at time zero. The one-compartment concentration prediction is \(C(t)=\frac{D}{V}\exp(-CLt/V)\), where \(CL>0\) is clearance, \(V>0\) is volume, and \(t\geq 0\) is time after the bolus. At each scheduled time \(t_i\), the observed concentration is \(Y_i=C(t_i;CL,V)+\varepsilon_i\), with independent errors \(\varepsilon_i\sim N(0,\sigma^2)\). The error variance \(\sigma^2\) is known and constant across times. Dose, sampling times, and the nominal values of \(CL\) and \(V\) are treated as fixed.

This idealized model excludes distribution compartments, absorption, infusion duration, and changes in clearance or volume during observation. The independent, constant-variance error assumption matters: it gives each concentration the same residual variance and determines the weights in the information matrix. A different residual error model would generally give a different design criterion as a function of time.

Sensitivities and the Individual Information Matrix

Write \(k=CL/V\), the elimination rate constant, and let \(\boldsymbol{\theta}=(CL,V)^\mathsf{T}\). The derivative of the predicted concentration with respect to each component of \(\boldsymbol{\theta}\) measures how much a small parameter change alters the prediction at a particular time. These derivatives form the sensitivity vector.

\frac{\partial C(t)}{\partial CL} =-\frac{t}{V}C(t), \qquad \frac{\partial C(t)}{\partial V} =\frac{C(t)}{V}(-1+kt), \qquad \mathbf{g}(t)= \begin{pmatrix} -\frac{t}{V}C(t)\\[2pt] \frac{-1+kt}{V}C(t) \end{pmatrix}.

The first derivative follows by differentiating the exponent \(-CLt/V\) with respect to \(CL\). For the second, both the prefactor \(D/V\) and the exponent depend on \(V\): their contributions are \(-C(t)/V\) and \(C(t)CLt/V^2\), respectively. Omitting either contribution gives an incorrect sensitivity.

Under the stated Gaussian model, the log likelihood is a sum of squared residual terms with variance \(\sigma^2\). Its expected information for \(\boldsymbol{\theta}\) is the sum of the sensitivity outer products, each scaled by \(1/\sigma^2\). For \(n\) observations, the individual Fisher information matrix is therefore:

\mathcal{I}(\boldsymbol{\theta}) =\frac{1}{\sigma^2}\sum_{i=1}^{n} \mathbf{g}(t_i)\mathbf{g}(t_i)^\mathsf{T}.

The matrix is positive semidefinite. It has full rank for two parameters only when the scheduled sensitivities span two distinct directions. A large sensitivity at one time is not sufficient if the other measurements provide essentially the same direction of information.

Worked Example: Calculating a Concentration Sensitivity

Suppose the hypothetical dose is \(D=100\) mg, with nominal \(CL=5\) L/h and \(V=20\) L. Consider a sample at \(t=2\) h. The nominal rate constant is \(k=5/20=0.25\) h\(^{-1}\), and the predicted concentration is:

C(2)=\frac{100\ \mathrm{mg}}{20\ \mathrm{L}}\exp[-(0.25\ \mathrm{h}^{-1})(2\ \mathrm{h})] =5\exp(-0.5)=3.033\ \mathrm{mg/L}.

The clearance sensitivity is \(-tC/V=-(2/20)(3.033)=-0.3033\), in concentration units per unit of clearance. The volume sensitivity is \(C(-1+kt)/V=3.033(-1+0.5)/20=-0.07582\), in concentration units per unit of volume. The negative volume derivative here reflects the net effect of increasing \(V\) at this time: it lowers the initial concentration and changes the elimination rate.

The Determinant for Two Sampling Times

For exactly two samples, write their times as \(t_1<t_2\) and form a \(2\times2\) matrix \(G\) whose rows are \(\mathbf{g}(t_1)^\mathsf{T}\) and \(\mathbf{g}(t_2)^\mathsf{T}\). Then \(\mathcal{I}=G^\mathsf{T}G/\sigma^2\). Since \(\det(G^\mathsf{T}G)=\det(G)^2\), the information determinant is the squared area spanned by the two sensitivity vectors, scaled by the observation variance.

The determinant of \(G\) simplifies because the terms involving \(kt_1t_2\) cancel:

\det(G) =\frac{C(t_1)C(t_2)}{V^2} \left[t_1-t_2\right], \qquad \det\{\mathcal{I}(\boldsymbol{\theta})\} =\frac{C(t_1)^2C(t_2)^2}{\sigma^4V^4}(t_2-t_1)^2.

Substituting the concentration model gives a form that makes the time trade-off explicit:

\det\{\mathcal{I}(\boldsymbol{\theta})\} = \frac{D^4}{\sigma^4V^8} \exp[-2k(t_1+t_2)](t_2-t_1)^2.

For fixed \(D,\sigma,V\), and nominal \(k\), maximizing this determinant is equivalent to maximizing \(\exp[-2k(t_1+t_2)](t_2-t_1)^2\). The exponential factor penalizes later total sampling time because predicted concentrations decline. The squared time separation rewards distinct sampling times because they produce less redundant sensitivity directions. D-optimality balances both effects by maximizing the determinant, or equivalently its logarithm.

Rank condition: With two parameters and two observations, the information determinant is positive exactly when the two sensitivity vectors are linearly independent. In this model, for finite positive concentrations and \(\sigma^2>0\), that requires \(t_1\ne t_2\). Repeating a time gives zero determinant and cannot identify both parameters from these two observations under the local linear information calculation.

Worked Example: Comparing Two Two-Sample Schedules

Use the hypothetical values \(D=100\) mg, \(CL=5\) L/h, \(V=20\) L, and \(\sigma=0.2\) mg/L, so \(k=0.25\) h\(^{-1}\). Compare schedules \((0,4)\) h and \((1,3)\) h. The common factor in the determinant expression is \(D^4/(\sigma^4V^8)=2.441\) when consistent units are used. For \((0,4)\), the time-dependent factor is \(\exp[-2(0.25)(0+4)](4-0)^2=16\exp(-2)=2.165\). Thus the determinant is approximately \(2.441(2.165)=5.286\) in the corresponding information units.

For \((1,3)\), the factor is \(\exp[-2(0.25)(1+3)](3-1)^2=4\exp(-2)=0.5413\), giving a determinant of approximately \(2.441(0.5413)=1.322\). The first schedule has four times the determinant of the second. Both schedules have the same sum of times, and hence the same exponential factor, but the first has twice the time separation, which is squared in the determinant. This comparison is local to the stated nominal parameters and error model.

Optimizing Times Within a Sampling Window

For two samples constrained to \(L\leq t_1<t_2\leq U\), where \(0\leq L<U\), take the logarithm of the determinant and discard terms constant in the times. The objective is:

\ell(t_1,t_2)=-2k(t_1+t_2)+2\log(t_2-t_1).

For fixed \(t_2\), its derivative with respect to \(t_1\) is \(-2k-2/(t_2-t_1)\), which is strictly negative. Thus the determinant increases as \(t_1\) moves earlier, and the optimum has \(t_1=L\). With \(t_1=L\), the derivative with respect to \(t_2\) is \(-2k+2/(t_2-L)\). It is positive when \(t_2-L<1/k\), zero at \(t_2-L=1/k\), and negative when \(t_2-L>1/k\). Consequently, the maximizing schedule is:

Two-sample local D-optimal schedule: Under the stated model and fixed nominal \(k=CL/V\), with \(0\leq L<U\), choose \(t_1=L\) and \(t_2=\min(L+1/k,\ U)\). The result is conditional on the assumed sampling window; it does not assert that either boundary time is operationally feasible.

Worked Example: Applying a Finite Sampling Window

Suppose the same hypothetical values \(CL=5\) L/h and \(V=20\) L give \(k=0.25\) h\(^{-1}\), and the permitted window is \(0.5\) to \(6\) h. The unconstrained separation indicated by the derivative is \(1/k=4\) h. The first sample is placed at the earliest allowed time, \(t_1=0.5\) h, and \(t_2=\min(0.5+4,6)=4.5\) h.

The predicted concentrations are \(C(0.5)=5\exp(-0.125)=4.412\) mg/L and \(C(4.5)=5\exp(-1.125)=1.623\) mg/L. With \(\sigma=0.2\) mg/L, the determinant is:

\det(\mathcal{I}) =\frac{(4.412)^2(1.623)^2}{(0.2)^4(20)^4}(4.5-0.5)^2 \approx 3.21.

The optimum follows from the derivative and the specified time bounds, not from comparing these rounded concentrations. If the upper bound had been \(3.5\) h instead, the optimum would be \((0.5,3.5)\) h because the preferred four-hour separation would not fit within the window.

More Observations, Diagnostics, and Limitations

For more than two observations, the information remains the sum of sensitivity outer products. For a two-parameter model, its determinant can also be written as a sum over pairs:

\det(\mathcal{I}) =\frac{1}{\sigma^4} \sum_{i<j} \left[ \frac{C(t_i)C(t_j)}{V^2}(t_j-t_i) \right]^2.

Each pair contributes its squared sensitivity area. This identity follows by applying the determinant formula for a product of a two-column sensitivity matrix and its transpose, or by expanding the two-by-two determinant. It shows why duplicated times add information but do not create a new direction on their own. The two-sample closed-form optimum does not, by itself, solve the schedule-selection problem for an arbitrary number of samples.

A useful diagnostic is to inspect the sensitivity vectors or the eigenvalues of the information matrix at the proposed nominal values. A zero determinant signals rank deficiency; a very small determinant or a very small eigenvalue indicates weak information in at least one parameter direction. Also verify that the calculated concentrations and sensitivities use consistent units, that times satisfy the actual sampling constraints, and that the assumed residual variance is reasonable across the relevant concentration range.

The determinant criterion is not the only possible design objective. D-optimality targets the joint volume of the local parameter uncertainty region; other alphabetic criteria target different summaries of uncertainty or prediction. A schedule preferred by one criterion need not be preferred by another. In addition, this calculation is local: if the true \(CL/V\) differs substantially from the nominal value, the time separation \(1/k\) suggested by the derivation can change.

The result also idealizes bolus administration and permits an earliest sample at the lower window boundary. In practice, a zero-time sample may be impossible, and collection, assay, or model limitations may exclude some times. The calculation does not represent interindividual variation, correlated errors, parameter uncertainty in the nominal values, or model misspecification. Those features require corresponding changes to the information calculation and can change the preferred design.

Common pitfall: It is tempting to choose two widely separated times solely because the determinant contains \((t_2-t_1)^2\). That ignores the declining concentrations in the exponential factor. Conversely, choosing only the earliest possible times ignores the loss of sensitivity-direction separation. The determinant balances both terms; under the stated two-sample assumptions, the preferred separation is \(1/k\), subject to the available window.

Check Your Understanding

Answer each question using the model and assumptions stated above. Concise solutions are included after the questions.

For \(D=80\) mg, \(CL=4\) L/h, and \(V=16\) L, calculate \(k\) and the nominal concentration at \(t=2\) h.
Why does taking both samples at the same time give zero determinant for two parameters?
If \(k=0.5\) h\(^{-1}\) and the sampling window is 1 to 5 h, what two-sample schedule maximizes the determinant?
With the same \(k=0.5\) h\(^{-1}\) but a window of 1 to 2.5 h, what is the maximizing schedule?
Name two assumptions whose violation could make the derived design inappropriate.

Solutions. (1) \(k=CL/V=4/16=0.25\) h\(^{-1}\). The concentration is \((80/16)\exp[-(0.25)(2)]=5\exp(-0.5)=3.033\) mg/L. (2) The two sensitivity rows are identical, so their matrix has rank at most one; its determinant and the information determinant are zero. (3) The preferred separation is \(1/k=2\) h, so \(t_1=1\) h and \(t_2=\min(3,5)=3\) h. (4) The same preferred separation cannot fit after \(t_1=1\) h; therefore the upper boundary is used, giving \((1,2.5)\) h. (5) Examples include independent errors, constant known additive variance, an ideal bolus, and a correctly specified one-compartment model. Violating any of these can change the sensitivity weighting or the concentration predictions and hence the information determinant.