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Dose Regimens and Titration Algorithms · Tutorial 251 of 497

Schedule Optimization With Peak and Trough Constraints

Learn to convert steady-state peak and trough limits into dose and interval constraints, identify infeasible windows, and select the least-frequent feasible regimen under a linear pharmacokinetic model.

Advanced 10 min read

What You'll Learn

  • Define a frequency-minimizing regimen optimization problem using dose and dosing interval as decision variables.
  • Derive the steady-state peak and trough for repeated bolus dosing with first-order elimination.
  • Calculate the feasible dose range at a fixed interval and verify candidate regimens against both concentration limits.
  • Determine the largest feasible interval and its corresponding dose.
  • Diagnose infeasible limits and recognize when model assumptions or numerical rounding undermine a proposed regimen.

From Concentration Limits to a Schedule

A regimen can satisfy an average-exposure target and still produce an excessive peak or an inadequate trough. When both extremes matter, choosing a dose without considering the interval can be misleading: increasing the interval generally makes the peak-to-trough swing larger, while changing the dose shifts both concentrations. Schedule optimization makes these trade-offs explicit.

Here the objective is to reduce administration frequency while keeping the steady-state peak below an upper limit and the steady-state trough above a lower limit. We use a repeated intravenous bolus regimen with a constant dose and interval. This is a deliberately specified optimization problem: if “burden” instead means dose amount, pill count, or a weighted combination of dose and frequency, the objective must be changed and may lead to a different solution.

Definition: Let \(D\) be the dose administered at each occasion, \(\tau\) the time between doses, \(k\) the first-order elimination rate constant, \(V\) the apparent volume of distribution, and \(F\) the fraction of the dose reaching the systemic circulation. Let \(L\) and \(U\) be the required lower and upper concentration limits, with \(0<L<U\). The frequency-minimizing problem is to minimize \(1/\tau\), the number of administrations per unit time, subject to \(C_{\min,ss}\geq L\) and \(C_{\max,ss}\leq U\).

The model assumes linear, time-invariant pharmacokinetics; instantaneous bolus input; first-order elimination; identical doses at a fixed interval; and evaluation at steady state. The parameters \(F\), \(V\), and \(k\) are treated as known and fixed. Concentrations are assumed to be measured in the same units as \(L\) and \(U\), and time units must be consistent with \(k\). The derivation does not account for between-person variability, parameter uncertainty, dose constraints, or administration-time restrictions.

Steady-State Constraints and the Feasible Dose Range

For a one-compartment model, one dose contributes an initial concentration \(FD/V\), followed by exponential decline. At steady state, the residual concentrations from earlier doses accumulate. Summing those contributions gives the concentration immediately after a dose (the peak) and immediately before the next dose (the trough):

$$ C_{\max,ss}=\frac{FD/V}{1-e^{-k\tau}}, \qquad C_{\min,ss}=C_{\max,ss}e^{-k\tau} =\frac{FD/V}{e^{k\tau}-1}. $$

For a fixed interval, the trough requirement places a lower bound on dose, while the peak requirement places an upper bound. Rearranging each inequality gives:

$$ D\geq \frac{V}{F}L(e^{k\tau}-1), \qquad D\leq \frac{V}{F}U(1-e^{-k\tau}). $$

Thus a feasible dose exists only if the lower dose bound does not exceed the upper dose bound. Dividing the steady-state peak by the trough gives a particularly useful test: the dose and the scale factor \(FD/V\) cancel.

$$ \frac{C_{\max,ss}}{C_{\min,ss}}=e^{k\tau}. $$

If the trough is at least \(L\) and the peak is at most \(U\), their ratio cannot exceed \(U/L\). Therefore the interval must satisfy \(e^{k\tau}\leq U/L\). This condition is also sufficient under the stated model: when it holds, the lower dose bound is no greater than the upper bound, so at least one dose satisfies both concentration constraints.

Theorem: Under the repeated-bolus model and assumptions above, a feasible dose exists at interval \(\tau\) if and only if \(e^{k\tau}\leq U/L\). Equivalently, \(0<\tau\leq \ln(U/L)/k\). At any feasible interval, the allowable dose range is \(\frac{V}{F}L(e^{k\tau}-1)\leq D\leq\frac{V}{F}U(1-e^{-k\tau})\).

Why the Longest Feasible Interval Is Optimal

The objective \(1/\tau\) decreases as \(\tau\) increases, so minimizing administration frequency is equivalent to maximizing the interval. The theorem gives a finite upper bound on that interval. At its largest feasible value, \(e^{k\tau}=U/L\), so the trough and peak constraints meet exactly when the dose is chosen at either end of the feasible dose range.

$$ \tau^*=\frac{\ln(U/L)}{k}, \qquad D^*=\frac{V}{F}L\left(e^{k\tau^*}-1\right) =\frac{V}{F}(U-L). $$

At this solution, \(C_{\min,ss}=L\) and \(C_{\max,ss}=U\). The equality is a consequence of the selected objective and the model: any longer interval violates the ratio condition, and at the longest feasible interval there is only one dose that meets both limits. In practice, an exactly boundary-touching solution may be fragile to parameter uncertainty or rounding; that limitation should be considered before treating the mathematical optimum as a clinical recommendation.

Worked Example: Finding a Feasible Dose at a Fixed Interval

Suppose, hypothetically, that \(k=0.12\ \mathrm{h}^{-1}\), \(V/F=20\ \mathrm{L}\), and the required steady-state window is \(2.00\) to \(6.00\ \mathrm{mg/L}\). Assess an interval of \(8.00\ \mathrm{h}\), then check a dose of \(70.0\ \mathrm{mg}\). Assume the repeated-bolus model applies and the values of \(k\) and \(V/F\) are fixed.

First calculate \(e^{k\tau}=e^{0.12(8.00)}=e^{0.96}=2.612\). Since \(U/L=6.00/2.00=3.00\) and \(2.612<3.00\), a feasible dose range exists. Its lower bound is:

$$ D_{\mathrm{low}}=20(2.00)(2.612-1)=64.47\ \mathrm{mg}. $$

The upper bound is:

$$ D_{\mathrm{high}}=20(6.00)(1-e^{-0.96}) =120(1-0.3829)=74.05\ \mathrm{mg}. $$

The proposed \(70.0\ \mathrm{mg}\) dose lies within \(64.47\) to \(74.05\ \mathrm{mg}\). For a direct concentration check, \(FD/V=D/(V/F)=70.0/20=3.500\ \mathrm{mg/L}\). Therefore:

$$ C_{\min,ss}=\frac{3.500}{e^{0.96}-1} =\frac{3.500}{1.612}=2.172\ \mathrm{mg/L}, \qquad C_{\max,ss}=\frac{3.500}{1-e^{-0.96}} =\frac{3.500}{0.6171}=5.672\ \mathrm{mg/L}. $$

Both values meet the specified limits: the trough exceeds \(2.00\ \mathrm{mg/L}\), and the peak is below \(6.00\ \mathrm{mg/L}\). This conclusion is conditional on the stated model and parameter values; it is not a claim about an actual drug or patient.

Worked Example: Optimizing Frequency Under a Concentration Window

Consider a hypothetical repeated-bolus regimen with \(k=0.100\ \mathrm{h}^{-1}\), \(V/F=20.0\ \mathrm{L}\), and limits \(L=2.00\ \mathrm{mg/L}\) and \(U=6.00\ \mathrm{mg/L}\). Assume all the model conditions stated above. Find the interval that minimizes administration frequency and the corresponding dose.

The largest feasible interval is the boundary value:

$$ \tau^*=\frac{\ln(6.00/2.00)}{0.100} =\frac{\ln(3.00)}{0.100} =10.99\ \mathrm{h}. $$

At this interval, \(e^{k\tau^*}=3.00\). The dose that sets the trough to the lower limit is:

$$ D^*=\frac{V}{F}L(e^{k\tau^*}-1) =20.0(2.00)(3.00-1) =80.0\ \mathrm{mg}. $$

The trough is \(C_{\min,ss}=(80.0/20.0)/(3.00-1)=2.00\ \mathrm{mg/L}\). The peak is \(C_{\max,ss}=(80.0/20.0)/(1-1/3.00)=6.00\ \mathrm{mg/L}\). The resulting frequency is \(0.100/\ln(3.00)=0.09102\) administrations per hour. Any interval longer than \(10.99\ \mathrm{h}\) would require a peak-to-trough ratio greater than \(U/L=3.00\), so no dose could satisfy both limits.

Worked Example: Detecting an Infeasible Interval

Suppose a hypothetical regimen has \(k=0.100\ \mathrm{h}^{-1}\), \(L=2.00\ \mathrm{mg/L}\), and \(U=5.00\ \mathrm{mg/L}\). A proposed interval is \(10.99\ \mathrm{h}\). Assume the same linear repeated-bolus model. Determine whether adjusting the dose can make this interval feasible.

The required ratio at this interval is \(e^{k\tau}=e^{0.100(10.99)}\), approximately \(3.00\). But the allowed peak-to-trough ratio is only \(U/L=5.00/2.00=2.50\). Since \(3.00>2.50\), the interval is infeasible for every dose: increasing or decreasing \(D\) scales peak and trough together and cannot change their ratio.

The longest feasible interval is \(\ln(2.50)/0.100=9.163\ \mathrm{h}\). The proposed \(10.99\ \mathrm{h}\) interval exceeds that limit. The correction is to shorten the interval or reconsider the specified concentration window or model; selecting a different dose alone cannot resolve the conflict.

Implementation, Diagnostics, and Alternatives

For a practical calculation, first confirm the units and model inputs, then test interval feasibility before searching over doses. This order avoids wasting effort on dose adjustment when the peak-to-trough ratio already makes the requested schedule impossible.

1
Specify the objective and constraints.
State that burden means administration frequency, identify \(L\) and \(U\), and define the allowed decision variables \(D\) and \(\tau\).
2
Check interval feasibility.
Calculate \(e^{k\tau}\) and compare it with \(U/L\). If \(e^{k\tau}>U/L\), no dose can satisfy both limits.
3
Calculate the dose range.
For a feasible interval, calculate the trough-based lower dose bound and the peak-based upper dose bound. Confirm that the lower bound is no greater than the upper bound.
4
Optimize and verify.
For frequency minimization, set \(\tau=\ln(U/L)/k\), calculate the boundary dose, and recompute both steady-state concentrations using the original formulas.

A useful diagnostic is to compare the dose bounds as well as the ratio condition. If the bounds appear reversed despite \(e^{k\tau}\leq U/L\), check for a sign error in \(1-e^{-k\tau}\), inconsistent time units, or a peak/trough formula used in the wrong order. An interval in hours requires \(k\) in inverse hours. Rounding should be done after feasibility is checked: rounding an optimal interval upward can make it infeasible, and rounding a boundary dose can place a concentration just outside its limit.

The analytic solution is preferable to a numerical search when the stated model applies: it shows directly why the optimum occurs at the interval boundary and provides a dose verification. A numerical grid or general optimizer can be useful when the concentration equations are more complicated, but it should recover this result in the simple model and should still report constraint residuals. A different formulation, such as fixing the interval and choosing any feasible dose, answers a dose-selection question rather than minimizing frequency. A dose-minimization objective is also different: without additional restrictions or a clearly specified burden function, it is not interchangeable with minimizing \(1/\tau\).

Common Mistakes and Limits of the Result

  • Trying to fix an infeasible interval by changing dose. Under linear repeated bolus dosing, dose cancels from \(C_{\max,ss}/C_{\min,ss}\). If \(e^{k\tau}>U/L\), no dose can satisfy both limits.
  • Using the peak bound as if it were the trough bound. The trough constraint gives the lower dose bound; the peak constraint gives the upper dose bound. Reversing them can falsely suggest feasibility.
  • Calling a boundary solution robust. At the frequency optimum, both limits are met exactly. Model error or uncertainty in \(k\), \(V\), or \(F\) can move actual concentrations beyond the window.
  • Applying the result outside its assumptions. Nonlinear elimination, delayed or prolonged input, time-varying clearance, or non-steady-state conditions change the concentration relationships and require the relevant model to be optimized instead.
Key takeaway: With equal repeated bolus doses and first-order elimination, the steady-state peak-to-trough ratio depends on the interval, not the dose. The longest feasible interval is \(\ln(U/L)/k\); at that boundary, the dose places the trough at \(L\) and the peak at \(U\).

Check Your Understanding

Answer each question using the repeated-bolus assumptions in this tutorial. Full solutions follow.

  1. For \(k=0.100\ \mathrm{h}^{-1}\), \(L=1.50\ \mathrm{mg/L}\), \(U=5.00\ \mathrm{mg/L}\), and \(\tau=6.00\ \mathrm{h}\), is a feasible dose possible? If so, calculate the allowable dose range when \(V/F=15.0\ \mathrm{L}\).
  2. For \(k=0.0800\ \mathrm{h}^{-1}\), \(L=2.00\ \mathrm{mg/L}\), and \(U=8.00\ \mathrm{mg/L}\), what is the largest feasible interval?
  3. At \(k=0.120\ \mathrm{h}^{-1}\) and \(\tau=8.00\ \mathrm{h}\), can any dose at that interval satisfy limits \(L=3.00\ \mathrm{mg/L}\) and \(U=7.00\ \mathrm{mg/L}\)? Explain.
  4. Why does the frequency-minimizing solution generally place both peak and trough at their limits?

Solutions. (1) \(e^{k\tau}=e^{0.600}=1.822\), while \(U/L=3.333\), so the interval is feasible. The dose range is \(15.0(1.50)(1.822-1)=18.50\ \mathrm{mg}\) to \(15.0(5.00)(1-e^{-0.600})=33.84\ \mathrm{mg}\). (2) \(\tau_{\max}=\ln(8.00/2.00)/0.0800=\ln(4.00)/0.0800=17.33\ \mathrm{h}\). (3) No. The maximum allowed peak-to-trough ratio is \(U/L=7.00/3.00=2.333\), but the interval requires \(e^{0.120(8.00)}=e^{0.96}=2.612\). Since \(2.612>2.333\), changing dose cannot make the interval feasible. (4) Frequency minimization selects the largest feasible interval. At that boundary \(e^{k\tau}=U/L\), so satisfying the trough lower bound forces the peak to reach \(U\); equivalently, satisfying the peak upper bound forces the trough to equal \(L\).

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