Schedule Design With Nonlinear Elimination

Why Saturable Elimination Changes Schedule Design

With linear elimination, the fraction of drug removed per unit time is proportional to concentration. Saturable elimination does not have this property: as concentration rises, the elimination rate approaches a maximum. Consequently, a dose increase can raise concentrations disproportionately, and repeated doses cannot be analyzed by adding independent single-dose profiles. Schedule design must account for how long each concentration trajectory takes to decline between administrations.

The derivation below concerns an idealized one-compartment model with repeated instantaneous bolus doses. It provides a direct way to examine how a dose and interval determine steady-state peak, trough, and AUC. The model is deliberately specific: schedules for more complex structural models generally require numerical integration and numerical solution of the corresponding periodic steady-state conditions.

Model and assumptions: Drug is administered as an instantaneous bolus of amount $D$ every $\tau$ time units into a well-mixed compartment of fixed volume $V$. Concentration is $C$, and elimination follows Michaelis–Menten kinetics with maximum elimination rate $V_{\max}$ (amount/time) and Michaelis constant $K_m$ (concentration). Between doses, $$ \frac{dC}{dt}=-\frac{V_{\max}}{V}\frac{C}{K_m+C}. $$ There is no input between boluses, no distribution delay, and no time variation in $V$, $V_{\max}$, or $K_m$. Each dose raises concentration by $D/V$.

Here $K_m$ is the concentration at which the elimination rate is half its maximum. At concentrations much smaller than $K_m$, elimination is approximately first order; at concentrations much larger than $K_m$, it is close to zero order. The latter means that an approximately fixed amount, rather than a fixed fraction, is eliminated per unit time.

Steady-State Peak, Trough, and Interval

Let $C_{\text{tr}}$ be the concentration immediately before a dose at steady state, and $C_{\text{pk}}$ the concentration immediately after that dose. The bolus produces the jump $C_{\text{pk}}=C_{\text{tr}}+D/V$. During the following interval, concentration declines continuously from $C_{\text{pk}}$ to $C_{\text{tr}}$. Separating variables in the model gives the elapsed time for this decline.

dt=-\frac{V}{V_{\max}}\frac{K_m+C}{C}\,dC

Integrating from the post-dose peak to the pre-dose trough yields the steady-state schedule equation:

\tau=\frac{V}{V_{\max}}\left[(C_{\text{pk}}-C_{\text{tr}}) +K_m\ln\left(\frac{C_{\text{pk}}}{C_{\text{tr}}}\right)\right], \qquad C_{\text{pk}}-C_{\text{tr}}=\frac{D}{V}.

The logarithmic term is the feature that distinguishes this relationship from the linear-elimination schedule equation. For fixed $D$, $V$, and model parameters, the equation can be solved for the trough at a specified interval. Conversely, if a target trough is specified, it gives the interval directly. The peak then follows from the bolus jump.

Finite steady-state condition: A finite periodic steady state with positive trough requires $$ \frac{D}{\tau}<V_{\max}. $$ The average dose rate must be strictly below the maximum elimination rate. If the dose rate equals or exceeds $V_{\max}$, the model cannot eliminate the administered amount over each interval while maintaining a finite periodic concentration profile.

The condition can also be seen from the schedule equation. Write $q=D/V$, so $C_{\text{pk}}=C_{\text{tr}}+q$. As $C_{\text{tr}}$ grows without bound, the logarithmic term tends to zero and the interval equation approaches $\tau=(V/V_{\max})q=D/V_{\max}$. A finite positive trough therefore requires $\tau>D/V_{\max}$, equivalent to the stated dose-rate condition.

Worked Schedule Calculations

Worked Example: Finding the Trough for a Fixed Interval

Hypothetical data. Suppose a one-compartment drug has $V=20$ L, $V_{\max}=20$ mg/h, and $K_m=2$ mg/L. A 100 mg bolus is administered every 12 h. Find the steady-state trough and peak, and check whether a finite steady state is possible.

The concentration increment per dose is $D/V=100/20=5$ mg/L. Substituting $C_{\text{pk}}=C_{\text{tr}}+5$ and the parameters into the schedule equation gives

12=1\left[5+2\ln\left(\frac{C_{\text{tr}}+5}{C_{\text{tr}}}\right)\right].

Thus $\ln((C_{\text{tr}}+5)/C_{\text{tr}})=3.5$, so the ratio is $e^{3.5}\approx33.115$. Solving for the trough,

C_{\text{tr}}=\frac{5}{e^{3.5}-1} =\frac{5}{32.115}\approx0.1557\text{ mg/L}.

The peak is $C_{\text{pk}}=0.1557+5=5.156$ mg/L. The dose rate is $100/12=8.333$ mg/h, below $V_{\max}=20$ mg/h, consistent with a finite steady state. The result is a low trough relative to the peak, but not the near-zero trough that a linear model matched to the low-concentration slope might predict. Saturation at higher concentrations slows the decline after the dose.

Worked Example: Choosing an Interval to Meet a Trough Constraint

Hypothetical design. Use the same $V=20$ L, $V_{\max}=20$ mg/h, and $K_m=2$ mg/L, but give 80 mg per bolus. Find the longest interval that keeps the steady-state trough at or above 1 mg/L.

The concentration increment is $80/20=4$ mg/L. The longest allowable interval occurs at the boundary $C_{\text{tr}}=1$ mg/L: a longer interval would allow further decline and violate the trough constraint. The peak at that boundary is $1+4=5$ mg/L. Therefore,

\tau_{\max}=\frac{20}{20}\left[(5-1)+2\ln\left(\frac{5}{1}\right)\right] =4+2\ln(5)\approx7.219\text{ h}.

At this schedule the dose rate is $80/7.219\approx11.08$ mg/h, below $V_{\max}=20$ mg/h. The finite-steady-state condition is satisfied. An interval shorter than 7.219 h gives a trough above 1 mg/L under this model; an interval longer than 7.219 h gives a trough below 1 mg/L.

The AUC over this steady-state interval can also be calculated by integrating concentration over time. From the differential equation, $C\,dt=-(V/V_{\max})(K_m+C)dC$, so

\mathrm{AUC}_{\tau} =\frac{V}{V_{\max}}\left[ K_m(C_{\text{pk}}-C_{\text{tr}}) +\frac{C_{\text{pk}}^2-C_{\text{tr}}^2}{2} \right].

For the boundary profile, this is $(20/20)[2(5-1)+(25-1)/2]=8+12=20.00$ mg·h/L. The mean concentration over the interval is $20.00/7.219=2.771$ mg/L.

Designing to an AUC Target

The AUC formula provides a second useful design relationship. For fixed $D$, the peak-trough difference is fixed, but both concentrations change when the interval changes. Thus AUC per interval is not generally a fixed multiple of dose independent of the concentration profile. Under nonlinear elimination, it is unsafe to assume that doubling dose doubles AUC at a fixed schedule.

Worked Example: Selecting an Interval for a Target AUC

Hypothetical design. Suppose $V=20$ L, $V_{\max}=20$ mg/h, $K_m=2$ mg/L, and the dose is fixed at 100 mg. Find the interval for which the steady-state AUC per interval is 25 mg·h/L.

The concentration increment is $D/V=5$ mg/L, so $C_{\text{pk}}=C_{\text{tr}}+5$. Since $V/V_{\max}=1$ L·h/mg, the AUC formula becomes

25=2(5)+\frac{(C_{\text{tr}}+5)^2-C_{\text{tr}}^2}{2} =22.5+5C_{\text{tr}}.

Solving gives $C_{\text{tr}}=(25-22.5)/5=0.5000$ mg/L, and hence $C_{\text{pk}}=5.500$ mg/L. Substituting these concentrations into the schedule equation,

\tau=\frac{20}{20}\left[5+2\ln\left(\frac{5.5}{0.5}\right)\right] =5+2\ln(11)\approx9.796\text{ h}.

The corresponding dose rate is $100/9.796\approx10.21$ mg/h, below $V_{\max}=20$ mg/h. The computed trough, peak, interval, and AUC therefore satisfy the model equations and the finite-steady-state condition. The mean concentration is $25/9.796\approx2.552$ mg/L. This is an interval-specific average; it does not imply that concentration is constant at that value.

Why the Schedule Equation Has One Feasible Trough

For a fixed dose increment $q=D/V>0$, define the interval implied by a candidate trough $x>0$:

\tau(x)=\frac{V}{V_{\max}}\left[q+K_m\ln\left(\frac{x+q}{x}\right)\right].

This expression decreases continuously with $x$, because

\frac{d\tau(x)}{dx} =\frac{V}{V_{\max}}K_m \left(\frac{1}{x+q}-\frac{1}{x}\right)<0.

As $x$ approaches zero, $\tau(x)$ tends to infinity. As $x$ tends to infinity, $\tau(x)$ approaches $Vq/V_{\max}=D/V_{\max}$. Therefore, for any specified interval satisfying $\tau>D/V_{\max}$, there is exactly one positive trough solving the equation. This gives both a feasibility check and a useful diagnostic for numerical root-finding: a valid solution must be positive, and increasing the interval must lower the trough.

Check average input.
For bolus amount $D$ and interval $\tau$, verify $D/\tau<V_{\max}$. If not, this model has no finite periodic steady state.

Set the concentration jump.
Use $C_{\text{pk}}=C_{\text{tr}}+D/V$. Do not treat the peak and trough as independent design variables.

Apply the relevant constraint.
Use the interval equation for a specified peak or trough, and the AUC equation when the target is exposure per interval.

Check the resulting schedule.
Verify units, the bolus jump, the target, and the finite-steady-state condition. For numerical solutions, confirm the result by substituting it back into the equations.

Diagnostics, Alternatives, and Limitations

A linear one-compartment model is the main analytic alternative. It assumes elimination rate is proportional to concentration and produces exponential decline and linear superposition. It is simpler, but it cannot represent saturation across the concentration range. When concentrations remain well below $K_m$, the Michaelis–Menten model approximately reduces to first-order elimination with rate constant $V_{\max}/(VK_m)$. At concentrations comparable to or above $K_m$, that approximation can misstate accumulation and the required interval.

For this model, common diagnostic checks include confirming that calculated concentration declines between doses, that the peak exceeds the trough by exactly $D/V$, and that the dose rate is below $V_{\max}$. A fitted or simulated profile that appears to reach a finite periodic state despite $D/\tau\geq V_{\max}$ signals either a model mismatch or a calculation error. Likewise, a negative calculated trough is not a meaningful schedule: check the target, units, equations, and numerical root.

When a model has multiple compartments, time-varying parameters, non-instantaneous input, or additional elimination pathways, the closed-form interval and AUC expressions here may not apply. The usual alternative is to integrate the model through repeated dosing until a periodic profile is reached, or solve directly for the periodic state, then evaluate the schedule constraints numerically. Such calculations still require checking feasibility, units, and agreement between the simulated profile and the intended assumptions.

Common mistake: treating $V_{\max}$ as a first-order rate constant and assuming that every dose adds the same independent concentration profile. $V_{\max}$ has units of amount per time, not inverse time, and saturable elimination makes the decline depend on the current concentration. The correction is to use the nonlinear differential equation (or its integrated schedule relationship) and to check $D/\tau<V_{\max}$ before interpreting a steady-state result.

Key takeaway: Under repeated bolus dosing with Michaelis–Menten elimination, the interval depends logarithmically on the peak-to-trough ratio, and a finite periodic profile requires average dose rate below $V_{\max}$. Use the nonlinear schedule and AUC equations rather than linear accumulation rules when saturation is material.

Check Your Understanding

Answer each question using the one-compartment bolus model described above. Solutions follow the questions.

For $D=60$ mg every 6 h with $V_{\max}=10$ mg/h, is a finite periodic steady state possible?
If a bolus raises concentration by 3 mg/L, and the trough is 1 mg/L, what is the peak?
For $V=10$ L, $V_{\max}=10$ mg/h, $K_m=1$ mg/L, and a trough of 1 mg/L with a 20 mg dose, calculate the interval.
Why does a longer interval lower the steady-state trough when dose and model parameters remain fixed?

Solutions. (1) The dose rate is $60/6=10$ mg/h, equal to $V_{\max}$, not strictly below it; a finite periodic steady state is not possible in this model. (2) The peak is $1+3=4$ mg/L. (3) The dose concentration increment is $20/10=2$ mg/L, so the peak is 3 mg/L. The interval is $(10/10)[(3-1)+1\ln(3/1)]=2+\ln(3)\approx3.099$ h. (4) The integrated schedule relationship is strictly decreasing in trough for fixed dose increment; equivalently, the profile must decline for a longer time before the next dose, so its pre-dose concentration is lower.