Selecting a Dose Interval for a Concentration Window

Choosing a Regimen for a Concentration Window

A dose that produces a suitable average concentration may still produce an unacceptably high peak or a low trough. For repeated bolus dosing, the dose amount and the interval between doses jointly determine these extremes. Selecting them together makes the design question precise: can the steady-state peak remain below an upper bound while the trough stays above a lower bound?

This tutorial considers the idealized one-compartment bolus model. It uses the steady-state peak and trough expressions from the earlier material on calculating accumulation and peak–trough fluctuation, then derives the dose and interval conditions needed to keep both concentrations within a specified window. These are deterministic model-based conditions: they apply to the parameter values supplied, not automatically to every patient.

Model and assumptions: A dose $D$ is administered instantaneously into a one-compartment system every $\tau$ time units. The dose enters the systemic circulation completely, the volume of distribution $V$ is constant, and elimination is linear and first order with rate constant $k=CL/V>0$. Doses and elimination continue on the same schedule long enough for steady state to be reached. There is no absorption phase, time-varying clearance, or dose-dependent pharmacokinetics.

Let $C_{\max,ss}$ denote the concentration immediately after a dose at steady state, and $C_{\min,ss}$ the concentration just before the next dose. For a bolus regimen, these are the peak and trough over each interval. Suppose the acceptable window is from $L$ to $U$ concentration units, where $0<L<U$. The design requirement is $C_{\min,ss}\geq L$ and $C_{\max,ss}\leq U$.

Steady-state expressions: Under the stated model, $$ C_{\max,ss}=\frac{D/V}{1-e^{-k\tau}}, \qquad C_{\min,ss}=\frac{D/V}{e^{k\tau}-1} =C_{\max,ss}e^{-k\tau}. $$

The peak expression includes accumulation from the current dose and all previous doses; the trough is the peak decayed over one interval. These formulas require $k>0$ and $\tau>0$. The peak exceeds the trough for any positive interval. Therefore, requiring the trough to be at least $L$ and the peak no greater than $U$ ensures that the entire concentration profile between them lies within the window.

Dose Range at a Chosen Interval

For a specified interval, the peak constraint places an upper limit on the dose, while the trough constraint places a lower limit on it. Rearranging the trough condition gives

\frac{D/V}{e^{k\tau}-1}\geq L \quad\Longleftrightarrow\quad D\geq LV(e^{k\tau}-1).

Similarly, rearranging the peak condition gives

\frac{D/V}{1-e^{-k\tau}}\leq U \quad\Longleftrightarrow\quad D\leq UV(1-e^{-k\tau}).

Thus, for a chosen $\tau$, the dose must lie between these limits. A feasible dose exists only when the lower limit does not exceed the upper limit.

Feasibility result: For the specified concentration window and a fixed positive interval, an admissible dose exists exactly when $$ LV(e^{k\tau}-1)\leq UV(1-e^{-k\tau}), $$ which is equivalent to $$ \tau\leq \frac{\ln(U/L)}{k}. $$ At the maximum feasible interval, both concentration constraints are met exactly; there is only one admissible dose.

To see the equivalence, write $x=e^{k\tau}>1$. Since $1-e^{-k\tau}=(x-1)/x$, the feasibility inequality becomes $L(x-1)\leq U(x-1)/x$. Dividing by the positive quantity $x-1$ gives $L\leq U/x$, or $x\leq U/L$. Taking logarithms yields $k\tau\leq\ln(U/L)$, as stated. Notice that $V$ cancels: the longest feasible interval depends on the elimination rate and the ratio of the concentration bounds, not on volume. Volume still affects the dose required.

Worked Example: Finding a Dose Range at an Eight-Hour Interval

Consider a hypothetical intravenous bolus regimen with $CL=4.0$ L/h and $V=40$ L. The desired steady-state window is 2.0 to 8.0 mg/L. Find the allowable dose range at $\tau=8.0$ h, then check a dose of 140 mg.

First, $k=CL/V=4.0/40=0.100$ h$^{-1}$. At this interval, $e^{k\tau}=e^{0.8}=2.2255$, and $1-e^{-k\tau}=1-e^{-0.8}=0.55067$.

The trough constraint gives the minimum dose: $D_{\min}=LV(e^{k\tau}-1)=(2.0\text{ mg/L})(40\text{ L})(2.2255-1)=98.04$ mg. The peak constraint gives the maximum dose: $D_{\max}=UV(1-e^{-k\tau})=(8.0\text{ mg/L})(40\text{ L})(0.55067)=176.2$ mg. Therefore, any dose from about 98.04 to 176.2 mg satisfies both nominal constraints at this interval.

For $D=140$ mg, $D/V=3.50$ mg/L. The predicted peak is $3.50/0.55067=6.356$ mg/L. The predicted trough is $3.50/(2.2255-1)=2.856$ mg/L. Both are within the 2.0–8.0 mg/L window, so this dose and interval meet the stated model-based design criterion.

Finding Feasible Intervals for a Fixed Dose

Sometimes the dose is fixed by a tablet strength, protocol, or other practical constraint, and the interval is the quantity to select. The same two concentration requirements can be rearranged to give an interval range. From the peak constraint,

\tau\geq -\frac{1}{k}\ln\left(1-\frac{D}{UV}\right),

provided $0<D/(UV)<1$. If $D/(UV)\geq1$, even the limiting peak as $\tau$ becomes very large, $D/V$, is not below $U$; no positive interval satisfies the peak limit. From the trough constraint,

\tau\leq \frac{1}{k}\ln\left(1+\frac{D}{LV}\right).

The peak therefore supplies a minimum interval: dosing too frequently causes accumulation that pushes the peak above $U$. The trough supplies a maximum interval: dosing too infrequently lets the concentration fall below $L$. A fixed-dose solution exists only if the resulting lower interval bound is no greater than the upper bound.

Worked Example: Selecting an Interval for a Fixed Dose

Use the same hypothetical $CL=4.0$ L/h, $V=40$ L, and concentration window of 2.0–8.0 mg/L, but now fix the dose at 120 mg. Calculate the feasible interval range and check a six-hour interval.

Here $k=0.100$ h$^{-1}$, and $D/(UV)=120/(8.0\times40)=0.375$. The peak constraint requires $\tau\geq-\ln(1-0.375)/0.100=-\ln(0.625)/0.100=4.700$ h.

The trough constraint requires $\tau\leq\ln(1+120/(2.0\times40))/0.100=\ln(2.5)/0.100=9.163$ h. Thus, intervals from about 4.700 to 9.163 h meet both nominal constraints for this dose.

At $\tau=6.0$ h, $e^{-k\tau}=e^{-0.6}=0.54881$ and $e^{k\tau}-1=e^{0.6}-1=0.82212$. The peak is $(120/40)/(1-0.54881)=6.649$ mg/L, and the trough is $(120/40)/0.82212=3.649$ mg/L. The peak is below 8.0 mg/L and the trough is above 2.0 mg/L, so the six-hour schedule is feasible under the model.

When No Dose Can Work

The maximum feasible interval is a useful first check before calculating a dose. For the example window and elimination rate above, it is $\ln(8.0/2.0)/0.100=13.86$ h. An interval longer than this cannot meet both bounds, regardless of the dose amount. Increasing the dose enough to restore the trough would force the peak above the upper limit.

Worked Example: Diagnosing an Infeasible Interval

Keep $CL=4.0$ L/h, $V=40$ L and the 2.0–8.0 mg/L window, but consider $\tau=16.0$ h. Determine whether any dose can meet both requirements.

Here $k\tau=1.6$, $e^{1.6}=4.9530$, and $1-e^{-1.6}=0.79810$. The trough requires $D\geq(2.0)(40)(4.9530-1)=316.2$ mg. The peak requires $D\leq(8.0)(40)(0.79810)=255.4$ mg.

The lower required dose exceeds the upper allowable dose, so no dose satisfies both constraints at 16.0 h. This is not fixed by choosing a compromise dose: every dose large enough to meet the trough limit violates the peak limit, and every dose low enough to meet the peak limit violates the trough limit. The interval must be shortened, or the specified window or model assumptions must be reconsidered.

A Practical Selection and Diagnostic Sequence

Specify the window and model inputs.
Record $L$, $U$, $CL$, and $V$ in compatible units; calculate $k=CL/V$. Confirm that the intended regimen is reasonably represented by instantaneous boluses and linear one-compartment elimination.

Check the interval before choosing a dose.
Calculate $\tau_{\max}=\ln(U/L)/k$. If the proposed interval exceeds this value, no dose can keep both the trough and peak within the window.

Calculate the admissible dose or interval range.
For a selected interval, calculate the dose lower and upper bounds. For a fixed dose, calculate the minimum interval from the peak constraint and the maximum interval from the trough constraint.

Verify the selected regimen using both concentrations.
Substitute the selected dose and interval into $C_{\max,ss}$ and $C_{\min,ss}$. Check units and confirm $C_{\min,ss}\geq L$ and $C_{\max,ss}\leq U$.

If a calculated regimen appears inconsistent, check whether $CL/V$ was used for $k$, whether the interval and clearance use the same time unit, and whether the trough expression was evaluated as $D/V$ divided by $e^{k\tau}-1$. A unit mismatch can change $k\tau$ substantially; reversing the peak and trough formulas can lead to a plausible-looking but incorrect result. A very narrow dose range or interval range also signals little design margin, even if the nominal inequalities hold.

The two common design choices are complementary. If the interval is fixed, the allowable dose range directly shows whether a practical dose is possible. If the dose is fixed, the interval range shows how often it must be given to balance peak and trough constraints. Neither calculation alone determines which feasible regimen is preferable in practice; the selected schedule must also be suitable for the intended setting and administration constraints.

Assumptions, Limitations, and a Common Pitfall

These calculations describe a steady-state profile for specified parameter values. They presume that clearance and volume are known well enough for the design calculation, and that the model captures the concentration-time behavior relevant to the target window. If absorption is appreciable, elimination is not linear, or distribution is not adequately represented by one compartment, the bolus formulas may not describe the observed peak and trough. The calculations also do not establish that a patient has reached steady state or that the bounds are clinically appropriate.

A frequent error is to calculate the peak constraint and conclude that the regimen is feasible because the peak is below $U$. That conclusion ignores the trough: a lower dose or longer interval may satisfy the peak limit while allowing the concentration to fall below $L$. The correction is to check both inequalities, either by verifying both predicted concentrations or by confirming that the allowable dose or interval range is nonempty. Likewise, “within the window” is not a guarantee for every individual when pharmacokinetic parameters vary; it is a conditional prediction under the specified model and parameter values.

Key takeaway: At a chosen interval, the dose must be high enough for the trough but low enough for the peak. A dose exists only when $\tau\leq\ln(U/L)/k$; for a fixed dose, the peak and trough constraints define a minimum and maximum interval that must overlap.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

For $k=0.20$ h$^{-1}$ and a window of 1.0–5.0 mg/L, what is the longest interval for which some dose can satisfy both bounds?
At a fixed interval, which concentration constraint supplies the lower dose limit, and which supplies the upper dose limit? Explain briefly.
For $D=80$ mg, $V=20$ L, $k=0.10$ h$^{-1}$, and $\tau=5.0$ h, calculate the steady-state peak and trough.
For $D=80$ mg, $V=20$ L, $k=0.10$ h$^{-1}$, and a window of 1.0–4.0 mg/L, find the feasible interval range.
A dose meets the peak limit but produces a trough below the lower bound. Is the regimen feasible? What should be checked or changed?

Solutions. (1) $\tau_{\max}=\ln(U/L)/k=\ln(5.0/1.0)/0.20=8.047$ h. At this boundary, one dose produces peak 5.0 mg/L and trough 1.0 mg/L. (2) The trough condition $C_{\min,ss}\geq L$ gives the lower dose limit $D\geq LV(e^{k\tau}-1)$; the peak condition $C_{\max,ss}\leq U$ gives the upper limit $D\leq UV(1-e^{-k\tau})$. (3) $D/V=4.0$ mg/L, $e^{-0.5}=0.60653$, and $e^{0.5}-1=0.64872$. Thus $C_{\max,ss}=4.0/(1-0.60653)=10.17$ mg/L and $C_{\min,ss}=4.0/0.64872=6.166$ mg/L. (4) The peak requires $\tau\geq-\ln(1-80/(4.0\times20))/0.10=-\ln(0)/0.10$, which is impossible: $D/V=U$, and for every finite positive interval the peak exceeds $D/V=4.0$ mg/L. There is no feasible interval. (5) No. Both bounds must hold. Check the trough calculation, units, and model inputs; if they are correct, consider an interval or dose change and re-evaluate both concentrations.