Repeated Oral Dosing With First-Order Absorption

Why Repeated Oral Dosing Changes the Concentration Profile

After an oral dose, drug enters the body over time rather than appearing instantaneously in the central compartment. If another dose is given before the first dose has been fully absorbed and eliminated, each dose contributes to the concentration observed after later doses. The resulting profile is not obtained by multiplying the entire single-dose curve by one universal accumulation factor: absorption and elimination contribute separate terms, each accumulating at its own rate.

The objective here is to calculate the concentration within a dosing interval after the profile has reached steady state, including its peak time, peak concentration, and trough concentration. These calculations describe a one-compartment structural model with first-order absorption and linear elimination; they do not establish that this model is suitable for every drug or data set.

Definition: Let $D$ be the oral dose, $F$ the fraction of the dose reaching the systemic circulation, $V$ the apparent volume of distribution, $k_a$ the first-order absorption rate constant, $k_e$ the first-order elimination rate constant, and $\tau$ the dosing interval. Assume $k_a>k_e>0$, constant $F$, and $k_a\ne k_e$. Time $t$ is measured from a dose and satisfies $0\le t<\tau$ within a dosing interval. The steady-state concentration is the concentration profile that repeats from one dosing interval to the next after repeated dosing.

Building the Steady-State Equation

For one oral dose, the one-compartment concentration at time $t\ge0$ is

C_1(t)=\frac{FDk_a}{V(k_a-k_e)} \left(e^{-k_e t}-e^{-k_a t}\right).

This expression has an elimination term and an absorption term. At steady state, previous doses contribute additional terms to both. A dose given $j\tau$ earlier, where $j=0,1,2,\ldots$, contributes the single-dose profile evaluated at $t+j\tau$. Summing these contributions gives

C_{\mathrm{ss}}(t)= \frac{FDk_a}{V(k_a-k_e)} \left[ \frac{e^{-k_e t}}{1-e^{-k_e\tau}} - \frac{e^{-k_a t}}{1-e^{-k_a\tau}} \right].

The denominators follow from geometric series. For example, the elimination contributions are proportional to $e^{-k_e t}\{1+e^{-k_e\tau}+e^{-2k_e\tau}+\cdots\}$, whose sum is $e^{-k_e t}/(1-e^{-k_e\tau})$. The absorption contributions sum in the same way, with $k_a$ replacing $k_e$. Since both rate constants and $\tau$ are positive, each series converges.

The factors $1/(1-e^{-k_e\tau})$ and $1/(1-e^{-k_a\tau})$ describe accumulation in the elimination and absorption terms, respectively. They generally differ. Therefore, a single accumulation multiplier applied to the entire single-dose curve will not in general give the steady-state curve.

Result: At steady state, the trough occurs at the interval boundary, $t=0$ (equivalently, just before the next dose at $t=\tau$). Its concentration is $$ C_{\mathrm{trough,ss}}= \frac{FDk_a}{V(k_a-k_e)} \left[ \frac{1}{1-e^{-k_e\tau}} - \frac{1}{1-e^{-k_a\tau}} \right]. $$

Finding the Steady-State Peak

The peak occurs where the concentration’s time derivative is zero. Differentiating the steady-state equation gives

\frac{dC_{\mathrm{ss}}(t)}{dt} = \frac{FDk_a}{V(k_a-k_e)} \left[ -\frac{k_e e^{-k_e t}}{1-e^{-k_e\tau}} + \frac{k_a e^{-k_a t}}{1-e^{-k_a\tau}} \right].

Set this derivative to zero and rearrange:

e^{(k_a-k_e)t} = \frac{k_a(1-e^{-k_e\tau})} {k_e(1-e^{-k_a\tau})}.

Taking the natural logarithm and dividing by the rate difference yields the peak time:

t_{\max,\mathrm{ss}} = \frac{ \ln\!\left[ \dfrac{k_a(1-e^{-k_e\tau})} {k_e(1-e^{-k_a\tau})} \right] }{k_a-k_e}.

The rate difference is outside the logarithm. The logarithm’s argument is dimensionless, as required. Substituting this time into $C_{\mathrm{ss}}(t)$ gives $C_{\max,\mathrm{ss}}$. Under the stated assumptions, the derivative is positive at $t=0$ and negative at $t=\tau$: $k/(1-e^{-k\tau})$ increases with $k>0$, whereas $k/(e^{k\tau}-1)$ decreases with $k>0$. Continuity therefore gives a zero in $(0,\tau)$; after rearrangement, the derivative equation’s left side is a strictly increasing exponential in $t$, so that zero is unique. The profile rises from its interval-boundary trough, reaches this interior maximum, and returns to the trough at the next interval boundary.

Worked Example: Peak and Trough With an Eight-Hour Interval

Consider a hypothetical regimen with $D=100$ mg, $F=1$, $V=40$ L, $k_a=1.0$ h$^{-1}$, $k_e=0.20$ h$^{-1}$, and $\tau=8$ h. These values satisfy $k_a>k_e>0$, and the model assumes the same dose is given every eight hours.

The concentration multiplier is $$ \frac{FDk_a}{V(k_a-k_e)} = \frac{(1)(100\ \mathrm{mg})(1.0\ \mathrm{h}^{-1})} {(40\ \mathrm{L})(0.80\ \mathrm{h}^{-1})} =3.125\ \mathrm{mg/L}. $$ The peak time is $$ t_{\max,\mathrm{ss}} = \frac{\ln\!\left[ \dfrac{1.0(1-e^{-0.20(8)})} {0.20(1-e^{-1.0(8)})} \right]}{1.0-0.20} = \frac{\ln(3.9919)}{0.80\ \mathrm{h}^{-1}} \approx1.730\ \mathrm{h}. $$ The value $3.9919$ is the dimensionless logarithm argument; it is not divided by $0.80$ inside the logarithm.

At this time, $e^{-0.20t}\approx0.7075$ and $e^{-1.0t}\approx0.1772$. Therefore, $$ C_{\max,\mathrm{ss}} = 3.125\left[ \frac{0.7075}{1-e^{-1.6}} - \frac{0.1772}{1-e^{-8}} \right] \approx2.216\ \mathrm{mg/L}. $$ For the trough, use the interval-boundary expression: $$ C_{\mathrm{trough,ss}} = 3.125\left[ \frac{1}{1-e^{-1.6}}- \frac{1}{1-e^{-8}} \right] \approx0.7895\ \mathrm{mg/L}. $$ Thus the hypothetical steady-state profile peaks about 1.730 h after a dose and falls to about 0.7895 mg/L at the interval boundary.

How the Dosing Interval Affects the Profile

A shorter interval leaves less time for concentrations to decline between doses. The next example uses the same hypothetical drug parameters and dose as above but shortens the interval. The point is to recompute the accumulation terms, not to assume that a peak or trough scales by a fixed multiplier when $\tau$ changes.

Worked Example: Recomputing for a Four-Hour Interval

Use $D=100$ mg, $F=1$, $V=40$ L, $k_a=1.0$ h$^{-1}$, and $k_e=0.20$ h$^{-1}$, but now let $\tau=4$ h. The concentration multiplier remains $3.125$ mg/L. The peak-time calculation is $$ t_{\max,\mathrm{ss}} = \frac{\ln\!\left[ \dfrac{1.0(1-e^{-0.20(4)})} {0.20(1-e^{-1.0(4)})} \right]}{0.80\ \mathrm{h}^{-1}} = \frac{\ln(2.8047)}{0.80\ \mathrm{h}^{-1}} \approx1.289\ \mathrm{h}. $$

At $t=1.289$ h, $e^{-0.20t}\approx0.7727$ and $e^{-1.0t}\approx0.2755$. Hence $$ C_{\max,\mathrm{ss}} = 3.125\left[ \frac{0.7727}{1-e^{-0.8}} - \frac{0.2755}{1-e^{-4}} \right] \approx3.509\ \mathrm{mg/L}. $$ The trough is $$ C_{\mathrm{trough,ss}} = 3.125\left[ \frac{1}{1-e^{-0.8}}- \frac{1}{1-e^{-4}} \right] \approx2.492\ \mathrm{mg/L}. $$ Compared with the eight-hour regimen, this hypothetical four-hour regimen has a higher trough and a higher peak, while the peak occurs earlier within the interval. The reported values are concentrations, not dose-normalized quantities.

For a final calculation, the peak concentration can be found for a different set of parameters by applying the same expressions with their own units. This also illustrates why a stated bioavailability below one scales the concentration multiplier.

Worked Example: A Regimen With Incomplete Bioavailability

Suppose a hypothetical regimen has $D=150$ mg, $F=0.80$, $V=30$ L, $k_a=1.2$ h$^{-1}$, $k_e=0.30$ h$^{-1}$, and $\tau=6$ h. The rate ordering and positive interval satisfy the assumptions. The multiplier is $$ \frac{FDk_a}{V(k_a-k_e)} = \frac{(0.80)(150\ \mathrm{mg})(1.2\ \mathrm{h}^{-1})} {(30\ \mathrm{L})(0.90\ \mathrm{h}^{-1})} =5.333\ \mathrm{mg/L}. $$ The peak time is $$ t_{\max,\mathrm{ss}} = \frac{\ln\!\left[ \dfrac{1.2(1-e^{-0.30(6)})} {0.30(1-e^{-1.2(6)})} \right]}{0.90\ \mathrm{h}^{-1}} \approx1.340\ \mathrm{h}. $$ At that time, $e^{-0.30t}\approx0.6689$ and $e^{-1.2t}\approx0.2003$. Substitution gives $$ C_{\max,\mathrm{ss}} = 5.333\left[ \frac{0.6689}{1-e^{-1.8}} - \frac{0.2003}{1-e^{-7.2}} \right] \approx3.21\ \mathrm{mg/L}. $$ The trough is $$ C_{\mathrm{trough,ss}} = 5.333\left[ \frac{1}{1-e^{-1.8}}- \frac{1}{1-e^{-7.2}} \right] \approx1.052\ \mathrm{mg/L}. $$

Assumptions, Diagnostics, and Alternatives

These equations are appropriate only when their structural assumptions are defensible. They assume one well-mixed compartment; oral input governed by a single first-order absorption process; linear elimination; fixed doses and intervals; constant bioavailability; and parameters that do not change over time. The formula also assumes $k_a\ne k_e$. If the two rates are equal or nearly equal, the displayed expressions can be numerically unstable or require a limiting form; do not treat direct substitution into a near-zero denominator as reliable.

Diagnostics should compare the model’s predicted profile with the timing and shape of observed concentrations, not merely check whether calculated concentrations are positive. Potential warning signs include a systematic mismatch in the observed time of peak, residual patterns that change across the dosing interval, or consistent underprediction of late concentrations. Such patterns can indicate that the assumed absorption or disposition structure is inadequate, although no single diagnostic pattern identifies a unique alternative model. Confirm dose times, interval, units, and whether samples were collected at steady state before attributing a mismatch to model structure.

The repeated oral formula is a superposition result: it adds contributions from doses under linear kinetics. A numerical implementation can instead add a finite number of dose contributions, which is useful for irregular dosing histories or checking the geometric-series calculation. Repeated IV bolus and constant-rate infusion models are other structural alternatives, but they represent different input patterns and should not be substituted for oral absorption without a scientific rationale. The present model does not cover absorption lag times, zero-order or transit-compartment input, multiple absorption pathways, or nonlinear elimination.

Common pitfall: A frequent calculation error is to divide by $k_a-k_e$ inside the logarithm when computing $t_{\max,\mathrm{ss}}$. The correct expression is $\ln(R)/(k_a-k_e)$, where $R$ is the dimensionless ratio in the peak-time formula. Dividing $R$ by a rate difference before taking its logarithm changes the result and makes the logarithm’s argument carry units. Another error is to use one accumulation multiplier for the whole oral profile; the absorption and elimination terms have distinct denominators.

Check the regimen and units.
Confirm $k_a>k_e>0$, a positive fixed dosing interval, and consistent dose, volume, and time units.

Compute the multiplier.
Evaluate $FDk_a/[V(k_a-k_e)]$, retaining concentration units.

Find the peak time and concentrations.
Use the dimensionless ratio in the logarithm, then substitute $t_{\max,\mathrm{ss}}$ into the steady-state equation and calculate the boundary trough separately.

Check Your Understanding

Answer each question, then compare with the brief solutions below.

Which two terms accumulate separately in the steady-state oral concentration equation?
For the eight-hour example, is $t_{\max,\mathrm{ss}}$ the peak time measured from the dose or from the preceding trough?
Why is it incorrect to put $k_a-k_e$ inside the logarithm in the peak-time formula?
What model assumption is violated if elimination is nonlinear over the concentration range?

Solutions. (1) The elimination exponential and absorption exponential accumulate separately, with denominators $1-e^{-k_e\tau}$ and $1-e^{-k_a\tau}$, respectively. (2) It is measured from the dose that begins the interval; in the example it is about 1.730 h. (3) The logarithm argument must be dimensionless, and the rate difference divides the logarithm after it is evaluated. (4) The linear-elimination assumption is violated, so the stated superposition formula does not generally apply.