Why Parameterization Matters
A transit absorption model represents drug input as passage through a sequence of first-order steps. Its parameters can be expressed as a transit rate constant, a mean transit time, and a number of transit compartments. These quantities are linked: choosing two determines the third. However, the number of compartments does more than set a bookkeeping detail. At the same mean transit time, it changes how concentrated or dispersed the absorption times are.
This tutorial makes that relationship explicit. The key is to distinguish the average time to systemic input from a fixed delay applied to every molecule. A transit model spreads input over time; its mean does not say that input begins at that time.
The \(N+1\) steps comprise transfer through the \(N\) transit compartments and the final first-order transfer into systemic circulation. When \(N=0\), there are no transit compartments and the model reduces to direct first-order input. In a transit-chain model, \(N\) is an integer; an implementation that uses a continuous gamma-shape parameter is a different parameterization and should not be interpreted as a literal fractional number of compartments.
Deriving the Relationship and the Input-Time Distribution
Assume that each step has the same rate constant \(k_{tr}\), that the dose enters the first step at time zero, and that each molecule proceeds through every step in sequence. For a first-order step, the waiting time is exponentially distributed with rate \(k_{tr}\), so its mean is \(1/k_{tr}\) and its variance is \(1/k_{tr}^{2}\). The total time \(T\) to systemic input is the sum of \(N+1\) such waiting times. Linearity of expectation therefore gives \(E(T)=(N+1)/k_{tr}\); independence of the waiting times gives \(\operatorname{Var}(T)=(N+1)/k_{tr}^{2}\).
The full distribution follows from convolution. Write \(m=N+1\). For one step, the waiting-time density is \(f_1(t)=k_{tr}e^{-k_{tr}t}\) for \(t\geq 0\). If the sum of \(m\) waiting times has density \(f_m(t)=k_{tr}^{m}t^{m-1}e^{-k_{tr}t}/(m-1)!\), convolution with one more exponential waiting time gives
Starting from \(m=1\), this establishes the density for every positive integer \(m\). With \(m=N+1\), the time to input has an Erlang distribution (a gamma distribution with integer shape):
The cumulative fraction of the dose that has reached systemic circulation by time \(t\), when there is no loss in the transit chain, is
The formula follows by integrating the density; repeated integration by parts gives the complementary probability \(P(T>t)=e^{-k_{tr}t}\sum_{j=0}^{N}(k_{tr}t)^j/j!\). The mode, or time at which the density is highest, is \(N/k_{tr}\) when \(N\geq1\). Differentiating the log density gives \(N/t-k_{tr}=0\), hence \(t=N/k_{tr}\). For \(N=0\), the exponential density is highest at time zero.
Worked Example: Convert a Target MTT to a Transit Rate
Suppose a hypothetical model uses \(N=3\) transit compartments and a target \(MTT\) of 3.6 hours. Calculate \(k_{tr}\), the standard deviation of input time, and the mode.
There are \(N+1=4\) sequential steps. Rearranging the mean relationship gives
The variance and standard deviation are
For \(N\geq1\), the mode is \(N/k_{tr}=3/(1.111\ \mathrm{h}^{-1})=2.700\) hours. Thus this parameterization has a mean input time of 3.6 hours and a most frequent input time of 2.7 hours. Neither quantity is a fixed onset delay.
What Changing N Does at a Fixed MTT
For a fixed \(MTT\), increasing \(N\) increases the number of steps and therefore requires a larger \(k_{tr}\) at each step. The coefficient of variation of input time is
Consequently, more steps make the distribution relatively narrower around its mean. This does not mean that all input occurs at \(MTT\), or that the model has a hard delay before absorption starts. For any finite \(N\), the Erlang density has support from time zero onward. A larger \(N\) changes the probability of early input and the distribution's shape; it does not create a fixed lag.
Worked Example: Calculate MTT From the Rate and Compartment Number
In a hypothetical model, let \(N=5\) and \(k_{tr}=0.75\ \mathrm{h}^{-1}\). Find \(MTT\), the standard deviation, and the mode.
There are six sequential steps, so
The input-time variance and standard deviation are
Since \(N=5\), the mode is \(5/(0.75\ \mathrm{h}^{-1})=6.667\) hours. The rate constant has units of inverse hours, and dividing a dimensionless step count by that rate yields hours, as required for \(MTT\).
Worked Example: Compare Input by the Mean Time Under Three Parameterizations
Consider three hypothetical transit models, each with \(MTT=6.0\) hours, but with \(N=0\), \(N=2\), or \(N=5\). Calculate each rate constant and the fraction of dose reaching systemic circulation by 6 hours. Assume all dose passes through the chain and use the Erlang cumulative distribution above.
For \(N=0\), \(k_{tr}=1/6=0.1667\ \mathrm{h}^{-1}\). Here \(k_{tr}t=1\), so
For \(N=2\), \(k_{tr}=3/6=0.5000\ \mathrm{h}^{-1}\), and \(k_{tr}t=3\). Therefore
For \(N=5\), \(k_{tr}=6/6=1.000\ \mathrm{h}^{-1}\), and \(k_{tr}t=6\). Thus
All three models have the same mean input time, but their fractions of dose input by that time differ: 63.21%, 57.68%, and 55.43%, respectively. These are model-implied fractions by 6 hours, not probabilities that the model parameters are correct. The calculation illustrates why \(MTT\) alone does not determine the shape of the absorption input.
Using the Parameterization in an NLME Model
An NLME model may describe between-subject variation on \(MTT\) while holding \(N\) fixed. For example, if an individual's \(MTT_i\) is represented as a typical value multiplied by \(\exp(\eta_i)\), then the corresponding individual transit rate must be calculated using the same \(N\):
If \(N\) is fixed, this can also be written \(k_{tr,i}=k_{tr,\mathrm{typ}}\exp(-\eta_i)\), where \(k_{tr,\mathrm{typ}}=(N+1)/MTT_{\mathrm{typ}}\). Thus a positive individual deviation on log \(MTT\) corresponds to a lower individual transit rate. This inverse relationship matters when interpreting parameter estimates and when coding the model: computing the rate from the individual's \(MTT\) preserves the intended relationship.
A useful parameterization choice is often to estimate \(MTT\) and fix \(N\), or to estimate \(MTT\) while comparing a small set of plausible integer values of \(N\). Estimating both \(k_{tr}\) and \(MTT\) as independent parameters while also fixing \(N\) is redundant, because one is determined exactly by the other. If both are entered freely, the model has a structural inconsistency rather than two independent pieces of information.
| Quantity or approach | What it determines | Interpretive caution |
|---|---|---|
| \(MTT\) and \(N\) | Determine \(k_{tr}=(N+1)/MTT\) | The same MTT with different N gives different input-time spread. |
| \(k_{tr}\) and \(N\) | Determine \(MTT=(N+1)/k_{tr}\) | A rate constant alone does not specify the mean without the number of steps. |
| A fixed lag time | Shifts input by a specified fixed duration | It is not interchangeable with a distributed transit time. |
| Direct first-order input | Is the \(N=0\) boundary case of this chain | It has an exponential input-time distribution, not a fixed delay. |
Assumptions, Diagnostics, and Common Errors
The conversions and distributional results above assume identical first-order rates across steps, a fixed integer \(N\), and sequential passage without competing routes or loss before systemic input. They also describe time to input, not the subsequent concentration profile. A disposition model and any bioavailability or dose-scaling terms determine how systemic input produces measured concentrations.
When checking a model implementation or interpreting estimates, verify the following:
- Check units. If \(MTT\) is in hours, \(k_{tr}\) must be in \(\mathrm{h}^{-1}\). A value in days combined with a rate in \(\mathrm{h}^{-1}\) produces a factor-of-24 error.
- Check the step count. Under this convention, there are \(N+1\) transitions. Using \(N/k_{tr}\) as the mean omits the final step; \(N/k_{tr}\) is the mode when \(N\geq1\), not the mean.
- Check the implied distribution. A predicted input profile inconsistent with the intended early-versus-late input may indicate that \(N\) or \(k_{tr}\) was interpreted incorrectly. Compare model-implied input fractions at relevant times, not just the mean.
- Check parameter redundancy. With \(N\) fixed, \(k_{tr}\) and \(MTT\) cannot vary independently. A model that treats both as unconstrained parameters can be poorly specified.
- Check practical information in the data. Concentration observations may not distinguish changes in \(MTT\) from changes in \(N\), especially if early input is sparsely observed. A stable fit does not by itself prove the chosen number of compartments is uniquely supported.
A common mistake is to say, “The MTT is 6 hours, so absorption starts at 6 hours.” That reasoning treats an average as a hard threshold. The correction is that the transit-time distribution begins at time zero and has mean 6 hours; the fraction of dose entering by any chosen time must be obtained from the distribution, using both \(MTT\) and \(N\). A fixed lag time, by contrast, imposes a threshold by model construction.
The transit parameterization is useful because it gives a clear mean-time scale and a separate shape control through \(N\). Its limitation is that this shape is constrained by a chain of identical first-order steps. Real input processes may not follow that structure, and the observed concentration data can provide limited information about the detailed input-time distribution. Parameter values should therefore be interpreted as features of the specified model, not as direct observations of molecular transit.
Check Your Understanding
Answer each question, then compare with the solutions below.
- A hypothetical model has \(N=2\) and \(k_{tr}=0.40\ \mathrm{h}^{-1}\). What is its MTT?
- A model has \(MTT=5.0\) hours and \(N=4\). Calculate \(k_{tr}\). What is the mode of its input-time distribution?
- Two models have the same MTT, but one has a larger \(N\). Which has the smaller coefficient of variation in input time, and why?
- Why is \(MTT\) not a fixed lag time?
Solutions. (1) There are \(N+1=3\) steps, so \(MTT=3/(0.40\ \mathrm{h}^{-1})=7.50\) hours. (2) \(k_{tr}=(4+1)/(5.0\ \mathrm{h})=1.00\ \mathrm{h}^{-1}\); the mode is \(N/k_{tr}=4/1.00=4.00\) hours. (3) The model with larger \(N\) has the smaller coefficient of variation because \(CV(T)=1/\sqrt{N+1}\). (4) MTT is the expected time to systemic input, whereas the transit-time distribution has support from time zero; it does not prevent input before MTT.