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Absorption Models · Tutorial 60 of 797

Lag-Time Models for Delayed Input

Learn how a fixed absorption delay shifts first-order input and how to calculate concentrations at specified times after a delayed dose.

Intermediate 9 min read

What You'll Learn

  • Define lag time relative to the nominal dose time.
  • Derive the one-compartment concentration profile after a fixed delay.
  • Calculate concentrations before and after the lag using consistent units.
  • Find how lag time shifts the time of peak concentration.
  • Check the assumptions behind a fixed-lag model and recognize common pitfalls.

Why Model an Absorption Lag?

An oral dose need not begin entering the systemic circulation at the recorded dosing time. A delay can arise before absorption starts, so a model that begins first-order input immediately may predict concentrations too early. A fixed absorption lag represents this delay by shifting the start of input while leaving the subsequent absorption process first order.

The central modeling idea is simple: before the lag has elapsed, the dose contributes no systemic input; after it has elapsed, the usual first-order absorption process begins. This distinction matters because a lag shifts the profile in time, whereas the absorption and disposition rate constants determine its shape once input begins.

Definition: A fixed absorption lag time, denoted \(t_{\mathrm{lag}}\), is a nonnegative time interval between the nominal dose time and the start of absorption. For a single dose given at time zero, first-order absorption starts at \(t=t_{\mathrm{lag}}\). “Fixed” means that the delay is a specified duration for the profile being modeled; it does not mean that the lag is necessarily known without uncertainty.

Model and Assumptions

Consider one oral dose \(D\) given at time zero. Let \(F\) be the fraction of the dose reaching the systemic circulation, \(V\) the apparent volume of distribution, \(k_a\) the first-order absorption rate constant, and \(k_e\) the first-order elimination rate constant. Both rate constants have units of inverse time; \(D\) is an amount and \(V\) is a volume. Assume a one-compartment disposition model with concentration \(C(t)\) equal to the amount in the central compartment divided by \(V\).

The profile below assumes one dose, no drug in the body at time zero, no input before the lag ends, and first-order absorption after that point. It also assumes linear disposition and constant \(F\), \(V\), \(k_a\), and \(k_e\). The model is not intended to represent a distributed delay or input at a constant rate.

Model: Let \(\tau=t-t_{\mathrm{lag}}\) be elapsed time since absorption began. Before or at the end of the lag, \(C(t)=0\). For \(t>t_{\mathrm{lag}}\) and \(k_a\ne k_e\), $$ C(t)=\frac{F D k_a}{V(k_a-k_e)} \left(e^{-k_e\tau}-e^{-k_a\tau}\right), \qquad \tau=t-t_{\mathrm{lag}}. $$ Here \(t\), \(t_{\mathrm{lag}}\), and \(\tau\) use the same time units.

The depot amount remaining after absorption begins is \(D e^{-k_a\tau}\), so the systemic input rate is \(F k_a D e^{-k_a\tau}\). If \(A(t)\) denotes the central-compartment amount, its rate of change after the lag is the systemic input rate minus elimination:

$$ \frac{dA}{d\tau}+k_e A=F k_a D e^{-k_a\tau}, \qquad A(0)=0. $$

This initial condition is important: at the moment absorption begins, the model has not yet delivered any of the dose to the central compartment. To solve the equation, multiply both sides by the integrating factor \(e^{k_e\tau}\), then integrate from zero to \(\tau\):

$$ \begin{aligned} \frac{d}{d\tau}\left(Ae^{k_e\tau}\right) &=F k_a D e^{-(k_a-k_e)\tau},\\ Ae^{k_e\tau} &=F k_a D\int_0^\tau e^{-(k_a-k_e)s}\,ds\\ &=\frac{F k_a D}{k_a-k_e} \left(1-e^{-(k_a-k_e)\tau}\right). \end{aligned} $$

Multiplying by \(e^{-k_e\tau}\) gives \(A=F D k_a(e^{-k_e\tau}-e^{-k_a\tau})/(k_a-k_e)\). Dividing by \(V\) gives the concentration formula. If \(k_a=k_e=k\), the integral instead becomes \(\int_0^\tau 1\,ds=\tau\), so \(C(t)=F D k\tau e^{-k\tau}/V\) after the lag.

Calculating a Shifted Profile

Worked Example: Translate the Dose Time Into Absorption Time

Suppose a hypothetical single dose is given at time zero with \(t_{\mathrm{lag}}=1.0\) h. The absorption rate constant is \(k_a=0.60\ \mathrm{h}^{-1}\), and the elimination rate constant is \(k_e=0.10\ \mathrm{h}^{-1}\). Identify the elapsed absorption time at a sample collected at \(t=0.5\) h and at a sample collected at \(t=3.0\) h.

At \(t=0.5\) h, the nominal time is still within the lag: \(0.5<1.0\). The model therefore predicts \(C(0.5\ \mathrm{h})=0\) from this dose.

At \(t=3.0\) h, absorption has been under way for \(\tau=3.0-1.0=2.0\) h. The concentration formula must use \(\tau=2.0\) h in its exponentials, not the nominal time \(t=3.0\) h. The lag changes the time argument; it does not change either rate constant.

For a single dose with the same parameters and disposition, the lagged profile is the zero-lag profile evaluated at \(t-t_{\mathrm{lag}}\), once \(t>t_{\mathrm{lag}}\). Before then, the lagged profile is zero. Consequently, changing only \(t_{\mathrm{lag}}\) translates the post-lag curve horizontally; it does not alter its peak height or its shape as a function of elapsed absorption time.

Worked Example: Calculate Concentrations at Two Sampling Times

For a hypothetical oral dose, let \(D=500\) mg, \(F=0.80\), \(V=100\) L, \(k_a=0.60\ \mathrm{h}^{-1}\), \(k_e=0.10\ \mathrm{h}^{-1}\), and \(t_{\mathrm{lag}}=1.0\) h. Calculate concentrations at \(t=3.0\) h and \(t=4.0\) h. The lag condition is satisfied at both times, and \(k_a\ne k_e\), so the stated formula applies.

First calculate the coefficient: $$ \frac{F D k_a}{V(k_a-k_e)} =\frac{(0.80)(500\ \mathrm{mg})(0.60\ \mathrm{h}^{-1})} {(100\ \mathrm{L})(0.50\ \mathrm{h}^{-1})} =4.80\ \mathrm{mg/L}. $$

At \(t=3.0\) h, \(\tau=3.0-1.0=2.0\) h. Thus $$ \begin{aligned} C(3.0\ \mathrm{h}) &=4.80\left(e^{-(0.10)(2.0)}-e^{-(0.60)(2.0)}\right)\ \mathrm{mg/L}\\ &=4.80\left(e^{-0.2}-e^{-1.2}\right)\ \mathrm{mg/L}\\ &=4.80(0.8187-0.3012)\ \mathrm{mg/L}\\ &=2.484\ \mathrm{mg/L}. \end{aligned} $$

At \(t=4.0\) h, \(\tau=4.0-1.0=3.0\) h. Therefore $$ \begin{aligned} C(4.0\ \mathrm{h}) &=4.80\left(e^{-(0.10)(3.0)}-e^{-(0.60)(3.0)}\right)\ \mathrm{mg/L}\\ &=4.80\left(e^{-0.3}-e^{-1.8}\right)\ \mathrm{mg/L}\\ &=4.80(0.7408-0.1653)\ \mathrm{mg/L}\\ &=2.762\ \mathrm{mg/L}. \end{aligned} $$

These are model-predicted concentrations for the hypothetical parameter values, not measurements from a study. In each calculation the time since absorption began, rather than time since dosing, enters the exponentials.

How the Lag Changes the Peak Time

For \(k_a\ne k_e\), differentiating the post-lag concentration with respect to \(\tau\) gives a factor proportional to \(-k_e e^{-k_e\tau}+k_a e^{-k_a\tau}\). Setting this derivative to zero yields \(k_a e^{-k_a\tau}=k_e e^{-k_e\tau}\). Taking logarithms and solving gives the time from absorption onset to the peak:

$$ \tau_{\max}=\frac{\ln(k_a/k_e)}{k_a-k_e}, \qquad t_{\max}=t_{\mathrm{lag}}+\tau_{\max}. $$

For positive, unequal rates, \(\tau_{\max}\) is positive. The lag adds to the time of the peak but does not otherwise enter the expression for \(\tau_{\max}\). This result is a useful check on the model's time shift.

Worked Example: Find the Peak Time and a Concentration

Consider a separate hypothetical dose with \(D=200\) mg, \(F=0.75\), \(V=40\) L, \(k_a=0.80\ \mathrm{h}^{-1}\), \(k_e=0.20\ \mathrm{h}^{-1}\), and \(t_{\mathrm{lag}}=2.0\) h. Calculate the peak time and the concentration at \(t=3.5\) h.

The time from absorption onset to the peak is $$ \tau_{\max} =\frac{\ln(0.80/0.20)}{0.80-0.20} =\frac{\ln 4}{0.60} =2.310\ \mathrm{h}. $$ Therefore the peak occurs at \(t_{\max}=2.0+2.310=4.310\) h after the dose.

At \(t=3.5\) h, \(\tau=3.5-2.0=1.5\) h. The coefficient is $$ \frac{F D k_a}{V(k_a-k_e)} =\frac{(0.75)(200\ \mathrm{mg})(0.80\ \mathrm{h}^{-1})} {(40\ \mathrm{L})(0.60\ \mathrm{h}^{-1})} =5.00\ \mathrm{mg/L}. $$ Hence $$ \begin{aligned} C(3.5\ \mathrm{h}) &=5.00\left(e^{-(0.20)(1.5)}-e^{-(0.80)(1.5)}\right)\ \mathrm{mg/L}\\ &=5.00(e^{-0.3}-e^{-1.2})\ \mathrm{mg/L}\\ &=5.00(0.7408-0.3012)\ \mathrm{mg/L}\\ &=2.198\ \mathrm{mg/L}. \end{aligned} $$ The sampling time is before the predicted peak, since \(3.5<4.310\) h.

Checks, Alternatives, and Limitations

A useful first diagnostic is whether the model's implied timing is compatible with the observations and the dosing record. If a substantial measured concentration is present before the proposed lag ends, a single-dose model with no baseline concentration and no other input cannot explain it. If early samples are sparse, however, a lag may be weakly informed by the data; absence of an observed early concentration is not by itself proof that a true delay occurred.

The main simple alternative is the same first-order absorption model with \(t_{\mathrm{lag}}=0\). It predicts input immediately after dosing. A fixed lag instead imposes no input during a specified initial interval, then uses the same first-order input. When comparing these formulations, inspect early-time predictions and residual patterns as well as the overall profile. Systematic early prediction errors can indicate that neither assumed onset time is adequate, or that some other model assumption fails.

A fixed lag is a simplified representation. It assumes a sharply defined onset, rather than a gradual or distributed delay, and it does not by itself explain the biological mechanism of delayed input. Further, with few observations near the onset, different combinations of timing and absorption behavior may produce similar observed profiles. The concentration equation describes the consequences of a specified lag; it does not guarantee that the data identify that lag precisely.

Common pitfall: Substituting the nominal time \(t\) into the exponential terms after adding a lag gives the wrong profile. The absorption clock starts at \(t_{\mathrm{lag}}\), so use \(\tau=t-t_{\mathrm{lag}}\) for \(t>t_{\mathrm{lag}}\), and predict zero contribution from that dose at earlier times under the stated assumptions.

Before applying the formula, check that the dose is a single dose at time zero, that the sample is after the lag, that time and rate units agree, and that the unequal-rate formula is being used only when \(k_a\ne k_e\). The special equal-rate limit avoids division by zero when the two rates coincide. Finally, confirm that the calculated coefficient has concentration units: \(F\) is dimensionless, and the dose divided by volume is an amount per volume.

Check Your Understanding

Use the model assumptions above. The solutions are included so you can check both the time shift and the arithmetic.

  1. A single dose has \(t_{\mathrm{lag}}=1.5\) h. What is the model-predicted contribution from this dose at \(t=1.0\) h?
  2. If \(t=5.0\) h and \(t_{\mathrm{lag}}=1.5\) h, what elapsed absorption time belongs in the exponentials?
  3. For \(k_a=0.50\ \mathrm{h}^{-1}\) and \(k_e=0.10\ \mathrm{h}^{-1}\), calculate the time from absorption onset to the peak.
  4. For \(F=1\), \(D=100\) mg, \(V=20\) L, \(k_a=0.50\ \mathrm{h}^{-1}\), \(k_e=0.10\ \mathrm{h}^{-1}\), and \(t_{\mathrm{lag}}=1.0\) h, calculate \(C(3.0\ \mathrm{h})\).
  5. A sample has a positive concentration before the proposed lag ends. Is that compatible with the stated single-dose model if there is no baseline concentration or other input?

Solutions. (1) Since \(1.0<1.5\) h, the predicted contribution is zero. (2) \(\tau=5.0-1.5=3.5\) h. (3) \(\tau_{\max}=\ln(0.50/0.10)/(0.50-0.10)=\ln(5)/0.40=4.024\) h after absorption onset. (4) Here \(\tau=3.0-1.0=2.0\) h and the coefficient is \((1)(100)(0.50)/[(20)(0.40)]=6.25\ \mathrm{mg/L}\). Thus \(C=6.25(e^{-0.2}-e^{-1.0})=6.25(0.8187-0.3679)=2.818\ \mathrm{mg/L}\). (5) No. Under the stated assumptions the model predicts no contribution from the dose before its lag ends, so that observation signals a mismatch with at least one assumption or input specification.