Lag Times Under Repeated and Staggered Dosing

Why Dose Timing Needs Care

With repeated dosing, a dose is not necessarily absorbed as soon as it is administered. A lag time delays the start of its absorption, while the dose’s first-order absorption continues after that start. If another dose is administered before the earlier dose has finished absorbing, both doses can contribute to the concentration at the same time. The calculation therefore has two parts: determine when each dose begins to enter the system, then add the concentration contributions that are active at the sample time.

This tutorial uses a one-compartment disposition model with first-order absorption and elimination. The focus is on how lag times and dose schedules determine the input times and how to calculate the resulting overlapping concentrations. It is not necessary for the doses to have the same administration time pattern or the same lag.

Input Time and Concentration Contribution

For dose \(j\), let \(a_j\) be its administration time and \(L_j\) its lag time. Its input begins at time \(\tau_j=a_j+L_j\). Before that time, the dose contributes no drug to the central compartment. Afterward, its contribution follows the first-order absorption profile.

Definition: The input time of dose \(j\) is \(\tau_j=a_j+L_j\), the administration time plus the lag time for that dose. A common lag uses the same \(L_j=L\) for every dose; dose-specific lags allow \(L_j\) to differ across doses. The elapsed time since input began is \(s_j=t-\tau_j\) at observation time \(t\).

Let \(D_j\) be the dose, \(F\) the bioavailability fraction, \(V\) the apparent volume of distribution, \(k_a\) the first-order absorption rate constant, and \(k_e\) the first-order elimination rate constant. For \(s_j>0\), dose \(j\)'s concentration contribution is

C_j(t)=\frac{F D_j k_a}{V(k_a-k_e)} \left(e^{-k_e s_j}-e^{-k_a s_j}\right).

For \(t\leq \tau_j\), define \(C_j(t)=0\). At \(t=\tau_j\), the post-input formula also equals zero, so the two parts join continuously. The dose’s concentration contribution has concentration units, such as mg/L, when \(D_j\) and \(V\) use compatible mass and volume units. The rate constants have units of inverse time, and the lag and elapsed times use the same time unit.

This formula assumes linear one-compartment disposition; first-order absorption beginning at \(\tau_j\); constant \(F,V,k_a,\) and \(k_e\) across the doses being summed; and no saturation or dose-dependent change in those quantities. In particular, it assumes \(k_a\ne k_e\). Under these assumptions, dose contributions add by linear superposition:

C(t)=\sum_j C_j(t).

A common lag shifts every dose’s input time by the same amount but does not change the interval between input times when administration is regularly spaced. A dose-specific lag can change the spacing and order of the input times. For either schedule, calculate the input time for each dose rather than assuming that all doses begin absorption together or at fixed intervals.

Why Contributions Add

The sum follows from the linear disposition model. Once dose \(j\)'s input starts, its amount in the absorption depot is \(D_j e^{-k_a s}\), so its input rate is \(F k_a D_j e^{-k_a s}\). For the central amount \(A_j\), elimination is proportional to amount, giving

\frac{dA_j}{ds}=F k_aD_j e^{-k_a s}-k_e A_j,\qquad A_j(0)=0.

Multiplying by the integrating factor \(e^{k_e s}\) and integrating from 0 to \(s\) gives

\begin{aligned} \frac{d}{ds}\left(A_j e^{k_e s}\right) &=F k_aD_j e^{(k_e-k_a)s},\\ A_j(s)e^{k_e s} &=F k_aD_j\int_0^s e^{(k_e-k_a)u}\,du\\ &=\frac{F k_aD_j}{k_a-k_e} \left(1-e^{(k_e-k_a)s}\right). \end{aligned}

Multiplying by \(e^{-k_e s}\) and dividing by \(V\) yields the concentration formula above. The differential equation is linear in \(A_j\) and in the input rate. Therefore, adding two dose-specific solutions gives the solution for the combined input: the total concentration at a time is the sum of all dose contributions at that time. This remains true when absorption overlaps; overlap does not require a special correction factor.

Worked Calculations

In the examples, the hypothetical parameter values are \(F=0.80\), \(D_j=100\) mg for each dose, \(V=25\) L, \(k_a=1.2\ \mathrm{h}^{-1}\), and \(k_e=0.20\ \mathrm{h}^{-1}\). Thus the coefficient in the exponential-difference formula is

\frac{F D_j k_a}{V(k_a-k_e)} =\frac{(0.80)(100\ \mathrm{mg})(1.2\ \mathrm{h}^{-1})} {(25\ \mathrm{L})(1.2-0.20)\ \mathrm{h}^{-1}} =3.84\ \mathrm{mg/L}.

The coefficient is already the multiplier of the exponential difference. It is not multiplied by \(k_a\) again.

Worked Example: Repeated Doses With a Common Lag

Three hypothetical 100 mg doses are administered at 0, 12, and 24 h. Every dose has a common lag of 2 h. Find the total concentration at 26.5 h using the parameter values above.

Find the input times. Adding 2 h to each administration time gives input times \(\tau_1=2\), \(\tau_2=14\), and \(\tau_3=26\) h. At 26.5 h, the elapsed times since input are 24.5, 12.5, and 0.5 h. All three doses have begun absorption.

Calculate and sum the contributions. Using the coefficient 3.84 mg/L and retaining the exponentials until the arithmetic is complete:

\begin{aligned} C_1&=3.84\left(e^{-0.20(24.5)}-e^{-1.2(24.5)}\right)\\ &\approx3.84(0.00744658-0.00000000000017) =0.02859\ \mathrm{mg/L},\\ C_2&=3.84\left(e^{-0.20(12.5)}-e^{-1.2(12.5)}\right)\\ &\approx3.84(0.0820850-0.000000306) =0.3152\ \mathrm{mg/L},\\ C_3&=3.84\left(e^{-0.20(0.5)}-e^{-1.2(0.5)}\right)\\ &\approx3.84(0.9048374-0.5488116) =1.367\ \mathrm{mg/L}. \end{aligned}

Therefore \(C(26.5)=0.02859+0.3152+1.367\approx1.711\ \mathrm{mg/L}\). The recent dose contributes the most at this time, while the earlier doses still contribute residual concentrations.

Worked Example: Staggered Doses With Different Lags

Three hypothetical 100 mg doses are administered at 0, 4, and 8 h, with dose-specific lags of 2, 1, and 0 h, respectively. Find the total concentration at 10 h.

Determine which doses have begun input. The input times are \(0+2=2\), \(4+1=5\), and \(8+0=8\) h. At 10 h, the elapsed times are 8, 5, and 2 h. All three doses are contributing.

Calculate each contribution. With the same parameter values, the calculations are

\begin{aligned} C_1&=3.84(e^{-0.20(8)}-e^{-1.2(8)})\\ &\approx3.84(0.2018965-0.00006773) =0.7750\ \mathrm{mg/L},\\ C_2&=3.84(e^{-0.20(5)}-e^{-1.2(5)})\\ &\approx3.84(0.3678794-0.00247875) =1.403\ \mathrm{mg/L},\\ C_3&=3.84(e^{-0.20(2)}-e^{-1.2(2)})\\ &\approx3.84(0.6703200-0.09071795) =2.226\ \mathrm{mg/L}. \end{aligned}

Adding the contributions gives \(C(10)\approx0.7750+1.403+2.226=4.404\ \mathrm{mg/L}\). The doses were administered 4 h apart, but their input times are 3 h and then 3 h apart because their lags differ. The calculation uses the actual input times, not an assumed regular input interval.

Worked Example: Overlap With a Dose Not Yet Absorbing

Three hypothetical 100 mg doses are administered at 0, 2, and 4 h, all with a common 3 h lag. Find the total concentration at 6 h.

The input times are 3, 5, and 7 h. At 6 h, the first two doses have elapsed input times of 3 and 1 h; the third has not begun input, so its contribution is zero. The active contributions are

\begin{aligned} C_1&=3.84(e^{-0.20(3)}-e^{-1.2(3)})\\ &\approx3.84(0.5488116-0.02732372) =2.003\ \mathrm{mg/L},\\ C_2&=3.84(e^{-0.20(1)}-e^{-1.2(1)})\\ &\approx3.84(0.8187308-0.3011942) =1.987\ \mathrm{mg/L},\\ C_3&=0. \end{aligned}

Thus \(C(6)\approx2.003+1.987+0=3.990\ \mathrm{mg/L}\). The first dose continues to contribute while absorption from the second dose is also underway. The third dose’s administration alone does not create a concentration contribution before its lag has elapsed.

Assumptions, Checks, and Common Pitfalls

Before applying these calculations, confirm that the time scale and model match the data: the recorded times must be measured from a consistent origin; the lag must be defined relative to administration; and the specified dose, lag, and rate constants must be appropriate for each dose. The formula also presumes the same linear disposition applies throughout. If dose size changes bioavailability or clearance, for example, simple addition of contributions with one shared parameter set may not be justified.

Check every input time. Write \(a_j\), \(L_j\), and \(\tau_j=a_j+L_j\) in a table before calculating. This makes a time-origin error or an unintended change in dose order easier to catch.
Use only active contributions. If \(t\leq\tau_j\), set \(C_j(t)=0\); do not evaluate the post-input formula with a negative elapsed time.
Check units and coefficient. The exponent arguments must be dimensionless, so a time in hours must pair with rate constants in \(\mathrm{h}^{-1}\). The multiplier in this formulation already includes \(k_a\); multiplying it by \(k_a\) again gives an incorrect coefficient and units.
Check the schedule, not just dose count. A common lag preserves the intervals between regularly spaced doses. Different lags need not. Use each actual input time when determining which doses overlap.

A lag shifts absorption onset; it does not make absorption slower. As emphasized in the earlier material on separating lag time from slow absorption, these features have different roles in the model: \(L_j\) changes the time at which a dose starts, whereas \(k_a\) determines the absorption rate after that start. If a fitted lag is being used to compensate for a mismatched absorption rate, the model may reproduce some observations while misrepresenting the timing mechanism.

The principal alternative within this calculation is to represent every dose with an explicit input time rather than impose one common lag. A common lag is parsimonious when a shared delay is defensible; dose-specific lags or direct input times allow more flexible schedules but require more information to estimate reliably. Neither choice removes the need to sum overlapping contributions under the linear model. These calculations also have limits: they do not account for nonlinear disposition, changing bioavailability, or a distribution of delays within a dose.

Key takeaway: For each dose, calculate its input time, determine whether input has begun by the observation time, evaluate that dose’s concentration contribution if active, and sum the contributions. A common lag shifts a schedule uniformly; dose-specific lags can alter the spacing and overlap.

Check Your Understanding

Use the first-order absorption model and the parameter values \(F=0.80\), \(D=100\) mg, \(V=25\) L, \(k_a=1.2\ \mathrm{h}^{-1}\), and \(k_e=0.20\ \mathrm{h}^{-1}\). The concentration coefficient is 3.84 mg/L.

A dose is given at 3 h with a 2 h lag. What is its input time, and does it contribute at 4 h?
Two doses are given at 0 and 6 h with a common 1 h lag. What are their input times and their separation?
A dose has been absorbing for 1 h. Calculate its concentration contribution using the stated parameters.
Two doses have input times of 2 and 5 h. At 4 h, which dose contributions should be included?
Why is a common lag not interchangeable with a lower \(k_a\)?

Solutions. (1) The input time is \(3+2=5\) h, so there is no contribution at 4 h. (2) The input times are \(0+1=1\) h and \(6+1=7\) h; their separation remains \(7-1=6\) h. (3) With \(s=1\) h, \(C=3.84(e^{-0.20}-e^{-1.2})=3.84(0.8187308-0.3011942)\approx1.987\ \mathrm{mg/L}\). (4) At 4 h, the first dose has been absorbing for 2 h and contributes; the second has not begun input and contributes zero. (5) A lag changes the input start time, while \(k_a\) controls the absorption rate after input begins, so they describe different model features.