Worked Design: Meeting Peak and Trough Limits With an Infusion

Designing for Both Peak and Trough Limits

A dose that meets a daily exposure target need not satisfy separate peak and trough limits. For an intermittent infusion, dose amount sets the overall concentration scale, while infusion duration shapes the concentration fluctuation within each dosing interval. A useful design therefore asks two linked questions: for a proposed duration, is there any dose that meets both limits, and, if so, what doses are admissible?

The calculation below concerns a hypothetical drug under a deliberately simple pharmacokinetic model. The design target is a steady-state peak no greater than $C_{\max}$ and a steady-state trough no lower than $C_{\min}$. These are constraints on predicted concentrations, not evidence by themselves that the limits are clinically appropriate.

Model and assumptions: The drug follows a one-compartment model with linear elimination, constant clearance $CL$, and elimination rate constant $k$. Each dose $D$ is infused at a constant rate over duration $T$, repeated every $\tau$ hours, with $0<T\leq\tau$. The infusion rate is $D/T$. At steady state, the peak is the concentration at the end of an infusion, and the trough is the concentration immediately before the next infusion. The calculation assumes unchanged clearance and volume, correctly timed doses, and no additional concentration input or removal.

With constant infusion, concentration rises during the infusion toward the infusion-specific equilibrium; after infusion stops, it declines by first-order elimination. Thus, under these assumptions, the end-of-infusion concentration is the interval peak and the pre-infusion concentration is the interval trough. The model also presumes that a meaningful steady state has been reached. It does not describe between-person variability or uncertainty in the pharmacokinetic parameters.

Steady-State Peak and Trough

Let $C_{\text{peak}}$ denote the end-of-infusion concentration at steady state. The previous peak is one full dosing interval before the current peak, so its contribution remaining at the current peak is $C_{\text{peak}}e^{-k\tau}$. During the current infusion, the constant input adds $(D/T)/CL$ times the fraction $1-e^{-kT}$. Equating the current peak to those two contributions gives

C_{\text{peak}} = C_{\text{peak}}e^{-k\tau} + \frac{D/T}{CL}(1-e^{-kT}).

Solving this steady-state balance gives the peak per unit dose. The trough is the peak followed by elimination during the off-infusion time $\tau-T$.

C_{\text{peak}} = D\,P(T), \qquad P(T)=\frac{1-e^{-kT}}{CL\,T(1-e^{-k\tau})}, \qquad C_{\text{trough}} = D\,Q(T), \qquad Q(T)=P(T)e^{-k(\tau-T)}.

If $D$ is in mg, $CL$ in L/h, and $T$ in h, then both $P(T)$ and $Q(T)$ have units $(\text{mg/L})/\text{mg}$. Multiplication by dose in mg gives concentration in mg/L.

Design rule: For fixed $k$, $CL$, $\tau$, and $T$, the dose must satisfy both $D Q(T)\geq C_{\min}$ and $D P(T)\leq C_{\max}$. Equivalently, the admissible dose interval is $$ \frac{C_{\min}}{Q(T)}\leq D\leq\frac{C_{\max}}{P(T)}. $$ A dose exists only if the lower bound does not exceed the upper bound.

The lower dose bound protects the trough, while the upper bound protects the peak. Raising the dose increases both concentrations in direct proportion; it does not change their ratio. Consequently, changing dose alone cannot repair a peak-to-trough fluctuation that is too large for the specified limits.

Worked Example: Testing a Four-Hour Infusion

Suppose a hypothetical drug has $CL=3.00$ L/h and $k=0.200\ \text{h}^{-1}$, and is infused every $\tau=8.00$ h. The required steady-state trough is at least $4.00$ mg/L and the peak must not exceed $12.0$ mg/L. Evaluate a proposed infusion duration of $T=4.00$ h.

First calculate the peak coefficient:

P(4.00) = \frac{1-e^{-(0.200)(4.00)}}{(3.00)(4.00)\left(1-e^{-(0.200)(8.00)}\right)} = \frac{1-e^{-0.800}}{12.0(1-e^{-1.60})} \approx 0.0574979\ \frac{\text{mg/L}}{\text{mg}}.

The off-infusion time is $8.00-4.00=4.00$ h, so

Q(4.00) = P(4.00)e^{-(0.200)(4.00)} \approx (0.0574979)(0.449329) = 0.0258355\ \frac{\text{mg/L}}{\text{mg}}.

The trough constraint requires $D\geq 4.00/0.0258355\approx154.8$ mg, and the peak constraint requires $D\leq12.0/0.0574979\approx208.7$ mg. Thus the admissible interval is approximately 154.8 to 208.7 mg. A 250 mg dose is not admissible: its predicted peak is $250(0.0574979)\approx14.37$ mg/L, above the 12.0 mg/L limit. Its trough is $250(0.0258355)\approx6.459$ mg/L, which meets the trough limit but does not offset the peak violation. For comparison, a 180 mg dose predicts a peak of about 10.35 mg/L and a trough of about 4.650 mg/L, meeting both limits.

When Is Any Dose Feasible?

The dose interval is nonempty exactly when its lower endpoint is no greater than its upper endpoint. With positive $P(T)$ and $Q(T)$, this condition simplifies to a comparison between the steady-state peak-to-trough ratio and the allowed ratio of concentration limits.

\frac{C_{\min}}{Q(T)} \leq \frac{C_{\max}}{P(T)} \quad\Longleftrightarrow\quad \frac{P(T)}{Q(T)} \leq \frac{C_{\max}}{C_{\min}} \quad\Longleftrightarrow\quad e^{k(\tau-T)} \leq \frac{C_{\max}}{C_{\min}}.

This condition follows because $Q(T)=P(T)e^{-k(\tau-T)}$. It also shows why the duration matters: a longer off-infusion period allows more elimination between peak and trough, increasing their ratio. If $C_{\max}\geq C_{\min}>0$, the condition can be expressed as a minimum infusion duration:

T \geq \tau-\frac{\ln(C_{\max}/C_{\min})}{k}.

The result must still be interpreted within the permitted range $0<T\leq\tau$. If the calculated lower bound is below zero, this ratio condition places no additional positive-duration restriction; if it exceeds $\tau$, no permitted duration can satisfy it. Once a duration passes the ratio test, the dose interval gives the actual doses that meet both limits.

Worked Example: A Three-Hour Infusion

Use the same hypothetical drug and limits, but set $T=3.00$ h. The off-infusion time is $8.00-3.00=5.00$ h. The peak coefficient is

P(3.00) = \frac{1-e^{-0.600}}{(3.00)(3.00)(1-e^{-1.60})} = \frac{0.451188}{9.00(0.798103)} \approx0.0628140\ \frac{\text{mg/L}}{\text{mg}}.

Therefore,

Q(3.00) = (0.0628140)e^{-(0.200)(5.00)} = (0.0628140)e^{-1.00} \approx0.0231080\ \frac{\text{mg/L}}{\text{mg}}.

The trough requires $D\geq4.00/0.0231080\approx173.1$ mg, while the peak requires $D\leq12.0/0.0628140\approx191.0$ mg. The interval is nonempty, though narrower than in the four-hour design. A 180 mg dose gives a predicted peak of $180(0.0628140)\approx11.31$ mg/L and trough of $180(0.0231080)\approx4.159$ mg/L, satisfying both limits.

The ratio check agrees: $P(3.00)/Q(3.00)=e^{1.00}\approx2.718$, which is below $C_{\max}/C_{\min}=12.0/4.00=3.00$. Thus feasibility is possible, and the dose bounds confirm it.

Worked Example: Finding a Short Feasible Duration

For the same limits and pharmacokinetic parameters, find the minimum duration allowed by the peak-to-trough ratio condition, then check a practical duration of $2.60$ h.

The minimum duration from the ratio condition is

T_{\min} = 8.00-\frac{\ln(12.0/4.00)}{0.200} = 8.00-\frac{\ln(3.00)}{0.200} \approx2.507\ \text{h}.

At $T=2.60$ h, the off-infusion time is $5.40$ h. Substitution gives

P(2.60) = \frac{1-e^{-(0.200)(2.60)}}{(3.00)(2.60)(1-e^{-1.60})} \approx0.0651351\ \frac{\text{mg/L}}{\text{mg}}, \qquad Q(2.60) = (0.0651351)e^{-1.08} \approx0.0221196\ \frac{\text{mg/L}}{\text{mg}}.

The dose bounds are $D\geq4.00/0.0221196\approx180.8$ mg and $D\leq12.0/0.0651351\approx184.2$ mg. A 182 mg dose lies inside this interval: its peak is approximately $182(0.0651351)=11.85$ mg/L, and its trough is approximately $182(0.0221196)=4.026$ mg/L. The design is feasible but has little dose margin.

At $T=2.50$ h, the peak-to-trough ratio would be $e^{(0.200)(8.00-2.50)}=e^{1.10}\approx3.004$, slightly above the allowed ratio of 3.00. No dose could meet both limits at that duration. This comparison illustrates that rounding a duration near the boundary can change feasibility; calculations should use unrounded values before selecting practical administration times.

Checking and Interpreting a Design

A feasible mathematical interval is a starting point, not a complete regimen recommendation. Before interpreting one, verify that the units are coherent, that the proposed duration is no longer than the dosing interval, and that $k$, $CL$, and the limits refer to the same model and concentration units. Then calculate both concentrations using the same dose and compare each directly with its corresponding limit. Do not infer that meeting a trough automatically guarantees a safe peak, or vice versa.

A useful diagnostic is the ratio $C_{\text{peak}}/C_{\text{trough}}$. Under this model it equals $e^{k(\tau-T)}$, independent of dose. If a calculated ratio does not match this expression, an arithmetic, timing, or unit error is likely. Also check that the peak exceeds the trough when $T<\tau$, and that the difference disappears as the off-infusion time approaches zero. Such checks can catch errors before a dose is chosen.

The main alternative is to select the dose from one concentration constraint alone and then inspect the other concentration afterward. That approach is simpler, but it can fail: a dose chosen to meet the trough may exceed the peak ceiling, while a dose capped by the peak may leave the trough below its minimum. The interval method tests both constraints explicitly and reveals when no dose can work at a proposed duration.

A further design option is to change the dosing interval or infusion duration rather than dose alone. Here the interval is fixed, so duration is the available way to reduce peak-to-trough fluctuation. The calculation does not account for infusion practicality, dose rounding, parameter uncertainty, changes in clearance, or a clinical response that is not determined by these two concentration limits. In practice, a narrow admissible interval deserves particular caution: plausible changes in parameters or rounding could remove the margin. A population or individual model with appropriate uncertainty assessment may therefore be needed before implementation.

Common Mistakes

Using the wrong peak coefficient. For constant infusion, the denominator includes $CL\,T$ and the accumulation term $1-e^{-k\tau}$. Omitting either changes the predicted concentration per dose.
Using the infusion duration as the off-time. The trough is reached after $\tau-T$ hours of elimination, not after $T$ hours unless those times happen to be equal.
Adjusting dose to fix an impossible ratio. Dose scales peak and trough together. If $e^{k(\tau-T)}>C_{\max}/C_{\min}$, no dose can satisfy both limits; duration or interval must change.
Reporting a feasible dose without checking both concentrations. Substitute the selected dose into both $D P(T)$ and $D Q(T)$, with units, and verify each inequality.
Treating model predictions as clinical guarantees. The limits and model assumptions require clinical justification; a deterministic calculation does not represent individual variability or parameter uncertainty.

Key takeaway: For a proposed infusion duration, calculate the steady-state peak and trough coefficients, convert both concentration limits into dose bounds, and use their overlap to assess feasibility. If the bounds do not overlap, changing dose cannot solve the design; the concentration fluctuation itself must be reduced.

Check Your Understanding

Answer each question using the stated model. Full solutions follow the questions.

For fixed dose and duration, what determines the number of hours between the end-of-infusion peak and the next trough?
With $\tau=8.00$ h, $k=0.200\ \text{h}^{-1}$, $C_{\max}=12.0$ mg/L, and $C_{\min}=4.00$ mg/L, what is the minimum duration allowed by the ratio condition?
For $T=4.00$ h in the worked design, calculate the peak and trough for a 180 mg dose and state whether both limits are met.
Why can increasing dose not make a design feasible if its peak-to-trough ratio exceeds $C_{\max}/C_{\min}$?

Solutions. (1) The off-infusion time is $\tau-T$; the concentration declines for that duration before the next infusion begins. (2) $T_{\min}=8.00-\ln(12.0/4.00)/0.200=2.507$ h. (3) Using $P(4.00)=0.0574979$ and $Q(4.00)=0.0258355$, the peak is $180(0.0574979)=10.35$ mg/L and the trough is $180(0.0258355)=4.650$ mg/L. Both limits are met. (4) Dose multiplies both peak and trough by the same amount, so it does not change their ratio. If that ratio is too large, increasing dose enough to raise the trough necessarily raises the peak beyond its ceiling.