Transit Compartments for Distributed Absorption Delays

Why Use a Chain of Compartments?

An oral dose may not begin entering the systemic circulation immediately, yet a single fixed lag time can be too rigid a description of the delay. A chain of transit compartments represents a sequence of stages between dose administration and uptake. Drug moves from one stage to the next by first-order transfer, and the final transit stage feeds an absorption compartment that releases drug by first-order uptake. The chain can spread input over time without requiring a single instant at which all absorption begins.

This tutorial focuses on the structure and amount accounting of that chain. The compartments are model states: their amounts are not necessarily distinct anatomical locations. Their purpose is to represent delayed progression toward uptake. The equations can be combined with a disposition model, but here the emphasis is on how much drug reaches the uptake step and when.

Definition: A transit-compartment chain is a sequence of \(N\) compartments through which drug moves in order, with each transit compartment transferring drug to the next at a first-order rate. In the model used here, the last transit compartment feeds a separate absorption compartment, from which drug enters the systemic circulation by first-order uptake.

Model Structure and Assumptions

Let \(T_i(t)\) be the amount of drug, in mg, in transit compartment \(i\), where \(i=1,\ldots,N\). Let \(G(t)\) be the amount in the final absorption compartment, also in mg. The transit rate constant is \(k_{\mathrm{tr}}\), and the first-order uptake rate constant is \(k_a\); both have units of inverse time, such as \(\mathrm{h}^{-1}\). Let \(D_{\mathrm{av}}\) be the dose amount available to this chain. If an oral dose is \(D\) mg and its bioavailability is \(F\), one may set \(D_{\mathrm{av}}=FD\) mg, provided \(F\) is treated as the fraction available for systemic uptake.

For a single dose at time zero, the dose is placed in the first transit compartment: \(T_1(0)=D_{\mathrm{av}}\). All remaining transit and absorption amounts are initially zero. Drug transfers from each transit compartment to the next at the same rate constant \(k_{\mathrm{tr}}\); the last transit compartment feeds \(G\). Uptake from \(G\) occurs at rate \(k_aG(t)\).

\begin{aligned} \frac{dT_1}{dt} &= -k_{\mathrm{tr}}T_1,\\ \frac{dT_i}{dt} &= k_{\mathrm{tr}}T_{i-1}-k_{\mathrm{tr}}T_i, \qquad i=2,\ldots,N,\\ \frac{dG}{dt} &= k_{\mathrm{tr}}T_N-k_aG,\\ R_{\mathrm{in}}(t) &= k_aG(t). \end{aligned}

Here \(R_{\mathrm{in}}(t)\) is the systemic input rate, in mg per unit time. A disposition model can receive this as an input; for example, systemic drug may then be distributed and eliminated. Elimination is not part of the transit-chain equations above. Keeping the input model distinct from disposition makes the amount entering the systemic circulation explicit.

The basic model assumes a single dose, constant rate parameters, complete transfer between consecutive stages, no loss from the transit chain, and first-order uptake from \(G\). Equal transit rate constants are a deliberate structural choice. The model is linear in amounts: doubling \(D_{\mathrm{av}}\) doubles every amount and the input rate when the rate constants remain unchanged. In a population model, parameter values may differ among individuals, but each individual’s chain still follows the stated amount equations.

How the Amount Balances Work

Each equation has an inflow-minus-outflow form. For example, \(T_2\) gains drug from \(T_1\) at rate \(k_{\mathrm{tr}}T_1\) and loses drug to \(T_3\) at rate \(k_{\mathrm{tr}}T_2\). The first compartment has no upstream inflow after the dose is placed there, so it only loses drug. The absorption compartment gains the outflow from \(T_N\) and loses drug through uptake.

A short mass-balance derivation checks that transfers neither create nor destroy drug. Define \(B(t)\) as the cumulative amount that has entered the systemic circulation, with \(B(0)=0\) and \(dB/dt=k_aG\). Add the derivatives of all transit amounts, \(G\), and \(B\). Every internal transfer appears once as an outflow and once as an inflow, so it cancels:

\frac{d}{dt}\left(\sum_{i=1}^{N}T_i+G+B\right) =-k_{\mathrm{tr}}T_1 +\sum_{i=2}^{N}(k_{\mathrm{tr}}T_{i-1}-k_{\mathrm{tr}}T_i) +(k_{\mathrm{tr}}T_N-k_aG)+k_aG=0.

With the stated initial conditions, \(\sum_{i=1}^{N}T_i(t)+G(t)+B(t)=D_{\mathrm{av}}\) throughout the dose’s passage through the chain. This identity is a useful implementation check. It does not say that all drug has entered the systemic circulation at any particular finite time; some may remain in the transit or absorption compartments.

Worked Amount Tracing

Worked Example: Following a Dose Through Three Transit Compartments

Suppose a hypothetical available dose of 100 mg enters a chain with \(N=3\), \(k_{\mathrm{tr}}=0.5\ \mathrm{h}^{-1}\), and \(k_a=1.0\ \mathrm{h}^{-1}\). Initially, \(T_1=100\) mg and \(T_2=T_3=G=B=0\) mg. To make the arithmetic transparent, use an explicit Euler step of \(\Delta t=1\) h. For each interval, calculate all rates from the amounts at the start of that interval, then add rate times \(\Delta t\) to each amount. This is a coarse numerical approximation, not an exact solution.

From 0 to 1 h, the outflow from \(T_1\) is \(0.5(100)=50\) mg/h. Thus \(T_1\) falls by \(50\) mg and \(T_2\) gains \(50\) mg. All other amounts remain zero because their starting amounts are zero. At 1 h, \(T_1=50\), \(T_2=50\), \(T_3=0\), \(G=0\), and \(B=0\) mg.

From 1 to 2 h, \(T_1\) transfers \(0.5(50)=25\) mg, while \(T_2\) transfers \(0.5(50)=25\) mg. The latter outflow goes to \(T_3\). Therefore, at 2 h, \(T_1=25\), \(T_2=50+25-25=50\), \(T_3=25\), \(G=0\), and \(B=0\) mg.

From 2 to 3 h, \(T_1\) transfers \(12.5\) mg and \(T_2\) transfers \(25\) mg. The outflow from \(T_3\) is \(0.5(25)=12.5\) mg, which enters \(G\). Since \(G\) was zero at the start of the step, its uptake over this Euler step is zero. At 3 h, the amounts are \(T_1=12.5\), \(T_2=37.5\), \(T_3=37.5\), \(G=12.5\), and \(B=0\) mg. They sum to 100 mg, as required by the mass balance.

From 3 to 4 h, the starting amounts imply outflows of \(6.25\) mg from \(T_1\), \(18.75\) mg from \(T_2\), \(18.75\) mg from \(T_3\), and \(12.5\) mg from \(G\), all over the one-hour step. The final 12.5 mg is systemic input, so \(B\) becomes 12.5 mg. The updated amounts are \(T_1=6.25\), \(T_2=25.0\), \(T_3=37.5\), \(G=18.75\), and \(B=12.5\) mg. Their sum is again 100 mg.

The example shows the sequence of transfers, not a precise prediction of a concentration profile. A smaller integration step or an accurate ODE solver would be needed for numerical predictions; the chosen one-hour step is intentionally coarse.

Initial Input and the Shape of the Delay

At the instant the dose is placed in \(T_1\), no drug is in \(G\). Consequently the systemic input rate is initially zero, even though the dose is present in the chain. For \(N\geq 1\), \(R_{\mathrm{in}}(0)=k_aG(0)=0\). The initial movement is through the transit stages, and uptake begins only after drug has reached \(G\).

The order of the chain helps explain the early-time behavior. Starting with \(T_1(0)=D_{\mathrm{av}}\), the first transit outflow is initially \(k_{\mathrm{tr}}D_{\mathrm{av}}\). For a chain of \(N\) transit compartments, the first nonzero contribution to uptake requires passage through \(N\) transit transfers and then first-order uptake. Repeated integration of the initial rates gives the local leading term for cumulative systemic input:

B(t)=\frac{D_{\mathrm{av}}k_{\mathrm{tr}}^{N}k_a}{(N+1)!}\,t^{N+1} +\text{terms of higher order in }t \qquad\text{as }t\to 0.

This is a local approximation near time zero, not a complete input profile. It shows why adding transit stages can make early input rise more gradually. It does not by itself identify the best number of stages or establish how a population parameter should be estimated.

Worked Example: Checking the Initial Rates and Early Input

Consider a hypothetical 80 mg available dose, \(N=2\), \(k_{\mathrm{tr}}=0.4\ \mathrm{h}^{-1}\), and \(k_a=1.2\ \mathrm{h}^{-1}\). At time zero, \(T_1=80\) mg, \(T_2=G=0\) mg. Substitution into the equations gives:

\(\frac{dT_1}{dt}=-0.4(80)=-32\) mg/h, and \(\frac{dT_2}{dt}=0.4(80)-0.4(0)=32\) mg/h. Also, \(\frac{dG}{dt}=0.4(0)-1.2(0)=0\) mg/h and \(R_{\mathrm{in}}(0)=1.2(0)=0\) mg/h. The initial loss from \(T_1\) exactly matches the initial gain in \(T_2\); neither the absorption compartment nor systemic input changes at that instant.

For \(N=2\), the leading early-input term is \(B(t)\approx \frac{80(0.4)^2(1.2)}{3!}t^3=2.56t^3\) mg when \(t\) is in hours and the time is sufficiently close to zero. For example, at \(t=0.1\) h the approximation is \(2.56(0.1)^3=0.00256\) mg. This tiny local value is consistent with the two transit transfers and uptake still needing time to occur; it is not intended as an accurate approximation at later times.

Worked Example: Interpreting a Change in Transit Rate

Worked Example: Comparing Early Stage Progression

Suppose a hypothetical available dose of 120 mg is placed in \(T_1\), with \(N=3\), \(k_a=1.0\ \mathrm{h}^{-1}\), and all other amounts initially zero. Compare \(k_{\mathrm{tr}}=0.25\ \mathrm{h}^{-1}\) with \(k_{\mathrm{tr}}=0.50\ \mathrm{h}^{-1}\). At time zero, the transit outflow from \(T_1\) is \(k_{\mathrm{tr}}(120)\): it is 30 mg/h in the first case and 60 mg/h in the second. In both cases, the systemic input rate is zero because \(G(0)=0\).

For a concrete approximate comparison, use one explicit Euler step of \(\Delta t=0.5\) h. At \(k_{\mathrm{tr}}=0.25\ \mathrm{h}^{-1}\), 15 mg leaves \(T_1\), so after the step \(T_1=105\) mg and \(T_2=15\) mg; \(T_3\) and \(G\) remain zero. At \(k_{\mathrm{tr}}=0.50\ \mathrm{h}^{-1}\), 30 mg leaves \(T_1\), giving \(T_1=90\) mg and \(T_2=30\) mg. Neither calculation predicts systemic uptake in that first step under this starting state.

The larger rate moves more amount out of the first stage over this interval. It does not mean that the dose is absorbed twice as fast at all times: later uptake depends on amounts throughout the chain and on \(k_a\) as well as \(k_{\mathrm{tr}}\). These Euler values illustrate the first transfer only and should not be treated as precise model predictions.

Implementation Checks, Alternatives, and Limitations

When implementing the model, verify that each transfer is used with the same magnitude and opposite signs in its source and destination equations. Check units: \(k_{\mathrm{tr}}T_i\) and \(k_aG\) must both be amount per time. For a single dose with no loss inside the chain, the sum of transit amounts, \(G\), and cumulative input should remain equal to \(D_{\mathrm{av}}\), apart from numerical integration error. If calculated amounts become negative, or the total changes substantially, inspect the equations and solver settings; a step size that is too large can also make a numerical approximation unreliable.

The main alternatives describe different assumptions about the delay. A lag-time model shifts the start of input by a specified time, whereas a chain allows amounts to progress through stages before uptake. A single first-order absorption compartment has no preceding transit chain and therefore represents a simpler absorption process. A constant-rate input represents a different input shape. These are not interchangeable descriptions: the model should reflect the intended mechanism and be assessed against the observed data.

A transit chain also has limitations. It assumes a sequence of first-order transfers and, in this formulation, the same transit rate for every stage. It does not independently represent losses, variable transfer rates across stages, or multiple absorption routes. A chain with more stages adds structure and parameters that may be difficult to distinguish from each other using limited observations, particularly when early samples are sparse. A good numerical fit alone does not prove that the stages correspond to literal anatomical compartments.

Common pitfall: “There is drug in the first transit compartment at time zero, so systemic input must also be positive at time zero.” This is incorrect for the stated model. Input is \(R_{\mathrm{in}}=k_aG\), and \(G(0)=0\), so \(R_{\mathrm{in}}(0)=0\). Transit-compartment amounts are not themselves systemic input; only uptake from \(G\) contributes to it.

Check Your Understanding

Try each question, then check the solutions below.

For \(N=2\), identify the amount that determines systemic input and give the input rate.
A chain has \(T_1=40\) mg, \(T_2=10\) mg, \(G=5\) mg, \(k_{\mathrm{tr}}=0.2\ \mathrm{h}^{-1}\), and \(k_a=0.8\ \mathrm{h}^{-1}\). What are the instantaneous rates of change of \(T_1\), \(T_2\), and \(G\)?
Why should cumulative input be included when checking total amount conservation?
If the amount in \(T_1\) is positive but \(G=0\), is systemic input positive? Explain.

Solutions. 1. The systemic input rate is determined by \(G\): \(R_{\mathrm{in}}=k_aG\). 2. \(\frac{dT_1}{dt}=-0.2(40)=-8\) mg/h; \(\frac{dT_2}{dt}=0.2(40)-0.2(10)=6\) mg/h; and \(\frac{dG}{dt}=0.2(10)-0.8(5)= -2\) mg/h. The negative rate for \(G\) means it is currently losing amount overall. 3. Drug that has entered the systemic circulation has left the absorption compartments, so without cumulative input the sum of the remaining compartment amounts would decrease even though no drug was lost from the modelled chain. 4. No. Since \(R_{\mathrm{in}}=k_aG\) and \(G=0\), the instantaneous systemic input rate is zero.