Representing Delayed and Prolonged Input
A concentration profile that rises late or remains elevated for several hours can reflect different absorption processes. A hard delay before absorption begins, input spread uniformly over a known duration, and input distributed across transit stages are not interchangeable assumptions. Each can produce a delayed or prolonged profile, but each implies a different input rate over time.
To compare these models, begin with a shared disposition assumption: a single well-mixed compartment with volume \(V\), linear elimination rate constant \(k\), and clearance \(CL=kV\). Let \(D\) be the administered dose and \(F\) the fraction reaching the systemic circulation, so the systemic amount is \(FD\). Concentration is \(C(t)\), in mg/L when dose is in mg and volume is in L. The equations below describe one dose, no baseline concentration, and no input from other doses. The parameter values are treated as fixed for the individual; population and random-effects models are not needed to define these absorption functions.
The common disposition equation is \(dC/dt=R_{\mathrm{in}}(t)/V-kC(t)\), where \(R_{\mathrm{in}}(t)\) is systemic input in amount per time. Thus, given a specified input rate and \(C(0)=0\), the concentration is \(C(t)=V^{-1}\int_0^t R_{\mathrm{in}}(s)e^{-k(t-s)}\,ds\). This convolution accounts for elimination of each amount absorbed at time \(s\) before the observation at time \(t\).
Lagged First-Order Input
A lag-time model commonly shifts a first-order absorption process by a fixed lag \(t_{\mathrm{lag}}\). Its absorption rate is zero before the lag and then declines exponentially, with rate constant \(k_a\). This assumes no systemic absorption before the lag and first-order absorption afterward. It does not assume that absorption occurs uniformly over a duration.
For \(t>t_{\mathrm{lag}}\), write \(\tau=t-t_{\mathrm{lag}}\). When \(k_a\ne k\), integrating the input against the disposition function gives the concentration below. Before or at the lag, the concentration is zero under the stated assumptions.
If \(k_a=k\), the expression with \(k_a-k\) in its denominator is not used; its limit is \(C(t)=(FD/V)k\tau e^{-k\tau}\). The lag specifies a sharp onset, while \(k_a\) controls the subsequent rate of absorption.
Worked Example: Concentration After a Fixed Lag
Suppose \(D=100\) mg, \(F=0.80\), \(V=20\) L, \(k=0.50\ \mathrm{h}^{-1}\), \(k_a=1.00\ \mathrm{h}^{-1}\), and \(t_{\mathrm{lag}}=1.00\) h. Calculate the predicted concentration at \(t=3.00\) h. Assume the one-compartment linear disposition and lagged first-order input described above.
The time since absorption began is \(\tau=3.00-1.00=2.00\) h, and \(FD/V=(0.80)(100\ \mathrm{mg})/(20\ \mathrm{L})=4.00\ \mathrm{mg/L}\). Substitution gives \(C(3)=4.00[1.00/(1.00-0.50)](e^{-0.50(2.00)}-e^{-1.00(2.00)})\ \mathrm{mg/L}\). Using \(e^{-1}=0.367879\) and \(e^{-2}=0.135335\), this is \(8.00(0.232544)=1.86035\ \mathrm{mg/L}\), or **1.860 mg/L** to four significant figures.
Delayed Zero-Order Input
In a delayed zero-order model, systemic input begins at \(t_{\mathrm{lag}}\) and continues at a constant rate for duration \(T\). The rate must be \(FD/T\) for the total absorbed amount to equal \(FD\). This represents prolonged input spread uniformly across the specified interval; it does not represent an instantaneous dose at the interval's start.
Let \(u=t-t_{\mathrm{lag}}\). During input, the constant input rate and elimination yield \(C(t)=(FD/(TCL))(1-e^{-ku})\). After input ends, the concentration at the end of the interval declines by elimination alone. These expressions follow by solving the disposition equation over each interval, using the concentration at the interval boundary as the next initial value.
Worked Example: Constant Input Over a Defined Duration
Use \(D=100\) mg, \(F=0.80\), \(CL=10\) L/h, and \(V=20\) L, so \(k=CL/V=0.50\ \mathrm{h}^{-1}\). Suppose input starts after a \(1.00\)-h lag and is zero-order for \(T=4.00\) h. Calculate the concentration at \(t=3.00\) h, while input is ongoing, and at \(t=7.00\) h, after input has ended.
The systemic amount is \(FD=80.0\) mg and the input rate is \(FD/T=80.0/4.00=20.0\) mg/h. At \(t=3.00\) h, \(u=2.00\) h, which lies within the input interval. Therefore \(C(3)= [80.0\ \mathrm{mg}/(4.00\ \mathrm{h}\times10.0\ \mathrm{L/h})](1-e^{-0.50(2.00)})\). The prefactor is \(2.00\ \mathrm{mg/L}\), so \(C(3)=2.00(1-0.367879)=1.26424\ \mathrm{mg/L}\), or **1.264 mg/L**.
At \(t=7.00\) h, \(u=6.00\) h, two hours after the end of input. The concentration at the end of input is \(2.00(1-e^{-0.50(4.00)})=1.72933\ \mathrm{mg/L}\). Applying the subsequent elimination factor gives \(C(7)=1.72933e^{-0.50(2.00)}=0.6362\ \mathrm{mg/L}\).
Transit Input and Its Mean Transit Time
A transit model represents input distributed over time rather than delayed by a sharp cutoff. Use the convention that \(N\) is the number of transit compartments before absorption, and let \(m=N+1\). Under the standard equal-rate transit construction, the input rate has a gamma-shaped time profile. The rate constant \(k_{\mathrm{tr}}\) is related to the mean transit time \(MTT\) by \(k_{\mathrm{tr}}=m/MTT\). Here \(MTT\) is the mean time from dose entry into the transit process to absorption.
This input rate integrates to \(FD\): the gamma integral \(\int_0^\infty t^{m-1}e^{-k_{\mathrm{tr}}t}dt=(m-1)!/k_{\mathrm{tr}}^m\) cancels the rate's constant factor. For \(N\geq1\), input starts at zero and increases before declining; unlike a hard lag, it is positive for every \(t>0\). A transit model can be shifted by an additional explicit lag if the scientific model requires a period with exactly no input.
The concentration is obtained by applying the same one-compartment disposition convolution. For the \(N=1\) case, \(m=2\), and \(k_{\mathrm{tr}}\ne k\), integration gives the following useful closed form. The general case can be calculated from the convolution or the corresponding transit-compartment differential equations.
Worked Example: One Transit Compartment Before Absorption
Let \(D=100\) mg, \(F=0.80\), \(V=20\) L, \(k=0.50\ \mathrm{h}^{-1}\), \(N=1\), and \(MTT=2.00\) h. Calculate \(C(3.00\ \mathrm{h})\). Assume the equal-rate transit model and one-compartment linear disposition.
Here \(m=N+1=2\), so \(k_{\mathrm{tr}}=m/MTT=2/2.00=1.00\ \mathrm{h}^{-1}\), and \(k_{\mathrm{tr}}-k=0.50\ \mathrm{h}^{-1}\). Substituting \(FD/V=4.00\ \mathrm{mg/L}\) and \(t=3.00\) h gives \(C(3)=4.00(1.00^2/0.50^2)e^{-1.50}[1-(1+1.50)e^{-1.50}]\ \mathrm{mg/L}\). Retaining precision, \(e^{-1.50}=0.22313016\), so the bracket is \(1-2.50(0.22313016)=0.44217460\). The prefactor is \(4.00(4.00)(0.22313016)=3.57008\ \mathrm{mg/L}\). Thus \(C(3)=3.57008(0.44217460)=1.57860\ \mathrm{mg/L}\), or **1.579 mg/L**.
Comparing the Models and Checking Their Assumptions
The three examples used the same dose, bioavailability, volume, and elimination rate, but different absorption assumptions. At \(t=3.00\) h, the lagged first-order model predicted \(1.860\ \mathrm{mg/L}\), the delayed zero-order model predicted \(1.264\ \mathrm{mg/L}\), and the transit model predicted \(1.579\ \mathrm{mg/L}\). These are model-based predictions, not evidence that any one model is correct. The models also differ in what their parameters mean.
| Model | Input assumption | Distinctive timing parameter(s) | Profile feature |
|---|---|---|---|
| Lagged first-order | No input before the lag; exponential input afterward | \(t_{\mathrm{lag}}, k_a\) | Sharp onset followed by declining absorption rate |
| Delayed zero-order | Constant input over a finite interval | \(t_{\mathrm{lag}}, T\) | Sharp onset and end of input; concentration can rise during input |
| Transit | Input distributed across transit stages | \(MTT, N\) | Gradual rise and fall in input rate; no hard lag unless added |
Use parameter meanings, not just visual resemblance, to choose among them. A known administration process that releases drug at an approximately constant rate for a measured duration supports a zero-order assumption. A clearly defined onset followed by a decaying release process may support a lagged first-order assumption. A gradual, distributed delay without a sharp onset can motivate a transit model. Sparse early samples may not distinguish these shapes: several models can predict similar concentrations at a few observation times.
A particularly common mistake is to call \(MTT\) a hard lag. In a transit model, \(MTT\) is the mean time through the distributed transit process, not the time at which input first becomes possible. For \(N=1\), for example, the rate is proportional to \(t e^{-k_{\mathrm{tr}}t}\): it begins at zero, becomes positive immediately after time zero, and peaks at \(t=1/k_{\mathrm{tr}}\). The correction is to describe a hard no-input period with an explicit \(t_{\mathrm{lag}}\), and to use \(MTT\) and \(N\) to describe the transit distribution.
These models also have limitations. The one-compartment equations assume instantaneous mixing and linear elimination; the input profiles do not establish the biological mechanism by themselves. The zero-order duration, lag, transit count, and transit time may be difficult to distinguish from concentration data if sampling is sparse or misses the early rise and later decline. If model diagnostics show systematic residual patterns around onset or input termination, reconsider the assumed input shape and the sampling information available. Do not add parameters solely because a more flexible curve can follow a small set of observations.
State \(V\), \(CL\), and \(k=CL/V\), along with the dose, bioavailability, and units.
State whether there is a hard lag, a constant-rate interval, or a transit distribution; define all associated parameters.
Use the appropriate concentration equation, check the observation's timing relative to input, and confirm that total input is \(FD\).
Check Your Understanding
Answer each question, then compare with the solutions below.
- A lagged first-order model has \(FD/V=3.00\ \mathrm{mg/L}\), \(k=0.40\ \mathrm{h}^{-1}\), \(k_a=1.20\ \mathrm{h}^{-1}\), and \(t_{\mathrm{lag}}=0.50\) h. Calculate concentration at \(t=2.00\) h.
- A zero-order input has \(FD=60.0\) mg, \(CL=12.0\) L/h, \(k=0.30\ \mathrm{h}^{-1}\), and duration \(T=2.00\) h. What is the concentration one hour after input begins?
- For a transit model with \(N=2\) and \(MTT=6.00\) h, calculate the fraction of \(FD\) absorbed by \(t=4.00\) h.
- In one sentence, explain why \(MTT=2\) h does not mean that a transit model has no absorption before two hours.
Solutions. (1) Here \(\tau=2.00-0.50=1.50\) h. The formula gives \(3.00[1.20/(1.20-0.40)](e^{-0.40(1.50)}-e^{-1.20(1.50)})=4.50(0.548812-0.165299)=1.72581\ \mathrm{mg/L}\), or **1.726 mg/L**. (2) The ongoing-input formula gives \(FD/(TCL)=60.0/(2.00\times12.0)=2.50\ \mathrm{mg/L}\), so \(C=2.50(1-e^{-0.30(1.00)})=0.64796\ \mathrm{mg/L}\), or **0.6480 mg/L**. (3) Here \(m=N+1=3\) and \(k_{\mathrm{tr}}=3/6.00=0.500\ \mathrm{h}^{-1}\). The transit input fraction accumulated by time \(t\) is \(1-e^{-k_{\mathrm{tr}}t}[1+k_{\mathrm{tr}}t+(k_{\mathrm{tr}}t)^2/2]\). At four hours this is \(1-e^{-2}(1+2+2)=1-5(0.135335)=0.323324\), or **32.33% of \(FD\)**. (4) \(MTT\) is the mean time through a distributed transit process, whose input can begin before its mean time; it is not a hard lag.