Transit-Time Distributions for Gastric Emptying

Why Represent Gastric Emptying as a Distribution?

A single gastric emptying rate can be useful when the aim is to describe a simple decline in gastric content. But it implies a particular residence-time pattern: some material exits very early, while a long tail remains. A meal or drug dose may instead be represented as passing through several sequential transit stages before it leaves the stomach. This creates a distribution of residence times whose spread can be controlled separately from its mean.

The model developed here is a mathematical representation of gastric residence, not a claim that the stomach contains a known number of discrete anatomical compartments. It is useful when the timing of material leaving the stomach matters to subsequent gastrointestinal input. The introductory material on first-order and lagged gastric emptying establishes those simpler descriptions; the focus here is how a sequential chain generates a richer time course.

Definition: In an \(N\)-stage sequential transit model, material begins in stage 1 and moves successively through \(N\) compartments. Transfer from each stage occurs by a first-order process with the same rate constant \(k_{\mathrm{tr}}\) (units of inverse time). Material leaves the gastric transit system from stage \(N\). The gastric residence time is the elapsed time between entering stage 1 and exiting stage \(N\).

Model Equations and Assumptions

Let \(A_i(t)\) be the amount of material in stage \(i\) at time \(t\), for \(i=1,\ldots,N\), and let \(A_0\) be the amount initially placed in stage 1. The basic model assumes that all material starts in that first stage, that every stage has the same constant transfer rate, and that there is no branching, return flow, or loss within the chain. Thus, for \(t\geq 0\),

\frac{dA_1}{dt}=-k_{\mathrm{tr}}A_1,\qquad \frac{dA_i}{dt}=k_{\mathrm{tr}}A_{i-1}-k_{\mathrm{tr}}A_i \quad (i=2,\ldots,N), \qquad A_1(0)=A_0,\quad A_i(0)=0\ (i>1).

These equations are linear and time-invariant: the rate is proportional to the amount currently in a stage, and the proportionality constant does not change with time or amount. Interpreting the resulting distribution as the gastric residence-time distribution requires an additional modeling assumption: the tracked material follows these same transfers and is not absorbed, degraded, or otherwise removed while in the chain. If gastric loss is being represented separately, the output from the chain is only the amount that reaches its exit, not necessarily the amount ultimately absorbed.

Each stage has an exponential waiting time with rate \(k_{\mathrm{tr}}\). Under the model, those successive waiting times are independent, so total residence time is their sum. This already suggests that the number of stages controls the distribution: adding stages adds waiting periods rather than simply shifting every exit time by the same fixed delay.

Deriving the Residence-Time Distribution

Solving the compartment equations gives the amount in each stage:

A_i(t)=A_0e^{-k_{\mathrm{tr}}t} \frac{(k_{\mathrm{tr}}t)^{i-1}}{(i-1)!}, \qquad i=1,\ldots,N.

The expression can be verified directly for stage 1. For each later stage, the solution follows by integrating its inflow from the preceding stage against the exponential loss from the current stage. In particular, if the expression holds for stage \(i-1\), then

A_i(t)=\int_0^t k_{\mathrm{tr}}A_{i-1}(u)e^{-k_{\mathrm{tr}}(t-u)}\,du =A_0e^{-k_{\mathrm{tr}}t}\frac{(k_{\mathrm{tr}}t)^{i-1}}{(i-1)!}.

The exit flux is the rate of transfer out of the final stage. Dividing that flux by \(A_0\) gives the probability density of residence time, denoted \(f_T(t)\):

J_{\mathrm{out}}(t)=k_{\mathrm{tr}}A_N(t), \qquad f_T(t)=\frac{J_{\mathrm{out}}(t)}{A_0} =\frac{k_{\mathrm{tr}}^N t^{N-1}e^{-k_{\mathrm{tr}}t}}{(N-1)!}, \quad t\geq 0.

This is a gamma distribution with shape \(N\) and rate \(k_{\mathrm{tr}}\). The corresponding cumulative fraction exited by time \(t\) is the integral of the density, while the fraction still in the transit system is the sum of the amounts in its stages divided by \(A_0\):

F_T(t)=1-e^{-k_{\mathrm{tr}}t} \sum_{j=0}^{N-1}\frac{(k_{\mathrm{tr}}t)^j}{j!}, \qquad S_T(t)=\frac{\sum_{i=1}^{N}A_i(t)}{A_0} =e^{-k_{\mathrm{tr}}t}\sum_{j=0}^{N-1}\frac{(k_{\mathrm{tr}}t)^j}{j!}.

The two expressions satisfy \(F_T(t)+S_T(t)=1\). This is also the model’s mass-balance statement: the fraction that has exited plus the fraction still in the chain equals the initial fraction, provided no other loss is included.

Result: For \(N\) identical sequential stages with rate \(k_{\mathrm{tr}}\), mean residence time is \(N/k_{\mathrm{tr}}\), variance is \(N/k_{\mathrm{tr}}^2\), and standard deviation is \(\sqrt{N}/k_{\mathrm{tr}}\). The coefficient of variation is \(1/\sqrt{N}\). The mean and variance have units of time and time squared, respectively.

These moments also follow from the sum-of-waiting-times interpretation. Each stage contributes an exponential waiting time with mean \(1/k_{\mathrm{tr}}\) and variance \(1/k_{\mathrm{tr}}^2\). Means and variances of independent waiting times add, giving \(N/k_{\mathrm{tr}}\) and \(N/k_{\mathrm{tr}}^2\). This derivation makes the role of the assumptions visible: if stage rates differ or waiting times are dependent, these particular formulas need not apply.

Worked Example: Calculating a Three-Stage Residence Distribution

Suppose a hypothetical gastric transit model has \(N=3\) stages and a mean residence time of \(2.40\) h. Find the rate constant, standard deviation, and fraction exited by \(2.00\) h. Assume the three stages have identical constant rates and no material is lost in the chain.

From \(\tau=N/k_{\mathrm{tr}}\), where \(\tau\) denotes the mean residence time, \(k_{\mathrm{tr}}=N/\tau=3/2.40=1.25\ \mathrm{h}^{-1}\). The standard deviation is \(\sqrt{N}/k_{\mathrm{tr}}=\sqrt{3}/1.25=1.386\ \mathrm{h}\).

At \(t=2.00\) h, \(k_{\mathrm{tr}}t=1.25(2.00)=2.50\). Therefore,

F_T(2.00)=1-e^{-2.50} \left(1+2.50+\frac{2.50^2}{2}\right) =1-(0.082085)(6.625)=0.4562.

Under this hypothetical model, approximately \(45.6\%\) of the material has exited the chain by 2.00 h. This is a model-based cumulative fraction, not a claim that every portion of the material follows an identical fixed residence time.

How the Number of Stages Changes the Distribution

The effect of \(N\) depends on what is held fixed. To compare distribution shapes at the same mean \(\tau\), set \(k_{\mathrm{tr}}=N/\tau\). Then increasing \(N\) decreases the coefficient of variation to \(1/\sqrt{N}\): residence times become less dispersed relative to their mean. By contrast, if \(k_{\mathrm{tr}}\) is held fixed, increasing \(N\) increases the mean as well as the spread. Those are different comparisons and should not be conflated.

Worked Example: Comparing Stage Numbers at a Fixed Mean

Consider hypothetical transit models with the same mean residence time, \(\tau=4.00\) h, but with \(N=1\), \(N=3\), or \(N=8\). For each model, calculate \(k_{\mathrm{tr}}\), the standard deviation, and the fraction exited by 2.00 h. The assumptions are identical stage rates within each model, constant rates, and no loss before exit.

The rates are \(k_{\mathrm{tr}}=N/\tau\): \(0.250\ \mathrm{h}^{-1}\), \(0.750\ \mathrm{h}^{-1}\), and \(2.00\ \mathrm{h}^{-1}\), respectively. The standard deviations are \(\tau/\sqrt{N}\): \(4.00\) h, \(2.309\) h, and \(1.414\) h.

For \(N=1\), the cumulative fraction at 2.00 h is \(1-e^{-0.250(2.00)}=1-e^{-0.5}=0.3935\). For \(N=3\), \(k_{\mathrm{tr}}t=1.50\), so \(F_T(2.00)=1-e^{-1.50}(1+1.50+1.50^2/2)=0.1912\). For \(N=8\), \(k_{\mathrm{tr}}t=4.00\), giving \(F_T(2.00)=1-e^{-4.00}\sum_{j=0}^{7}4.00^j/j!=0.0511\).

All three distributions have a 4.00 h mean, yet their early exit fractions differ substantially. In this comparison, more stages concentrate residence times more closely around the common mean. The \(N=1\) model has an exponential distribution with its largest density at time zero; the larger-\(N\) distributions place progressively less probability on very early exits.

A useful limiting interpretation follows from the variance. If the mean \(\tau\) is fixed, then the variance is \(\tau^2/N\), which approaches zero as \(N\) grows. Thus the distribution approaches a fixed residence time at \(\tau\). A finite chain remains distributed: it does not impose a perfectly deterministic delay.

Exit Flux and Mass-Balance Checks

The residence-time density describes the timing of exit; multiplying it by the initial amount gives the amount exiting per unit time. The compartments also show how much material remains within the chain at any time. These quantities offer a direct check that the model is implemented consistently.

Worked Example: Checking Exit and Remaining Amounts

A hypothetical model starts with \(A_0=100\) mg in stage 1, has \(N=2\), and uses \(k_{\mathrm{tr}}=0.500\ \mathrm{h}^{-1}\). Calculate the amounts in both stages and the exit flux at \(t=1.00\) h, then check the cumulative amount exited.

Using the stage solutions, \(A_1(1)=100e^{-0.5}=60.65\) mg and \(A_2(1)=100e^{-0.5}(0.5)(1.00)=30.33\) mg. The exit flux is \(J_{\mathrm{out}}(1)=k_{\mathrm{tr}}A_2(1)=0.500(30.33)=15.16\ \mathrm{mg\,h}^{-1}\).

For two stages, the fraction exited by 1.00 h is \(1-e^{-0.5}(1+0.5)=0.09020\), so the cumulative amount exited is \(100(0.09020)=9.020\) mg. The amount remaining in the stages is \(60.65+30.33=90.98\) mg. The sum, \(9.020+90.98=100.0\) mg, agrees with the initial amount up to rounding.

Choosing and Checking a Representation

A single first-order emptying model is the \(N=1\) case: its residence-time distribution is exponential, with mean \(1/k_{\mathrm{tr}}\) and coefficient of variation 1. A lagged first-order model adds a fixed period before first-order emptying begins. That can represent a prescribed onset delay, but it does not create the same progressively narrower distribution around a fixed mean as adding transit stages. A sequential chain is appropriate when a distributed delay is useful and its shape should be adjustable through \(N\).

The chain has limitations. Identical constant rates impose a specific gamma-shaped distribution, and the integer stage count is a structural choice rather than a direct measurement of anatomical regions. A poor fit to an observed time course may reflect the wrong distributional shape or violated assumptions, not simply a need to increase \(N\). In particular, a rate that changes over time, different material fractions with different emptying behavior, or loss within the stomach is not represented by this basic model.

Several diagnostics help identify implementation or interpretation problems:

Check units. \(k_{\mathrm{tr}}\) must have units of inverse time; \(k_{\mathrm{tr}}t\) must be dimensionless; the exit flux must have units of amount per time.
Check mass balance. With no internal loss, the cumulative exit amount plus the total amount in all stages should equal \(A_0\) at every time, apart from numerical tolerance.
Check the time origin and initial conditions. If material is initially spread across stages or begins after a delay, the residence distribution differs from the one derived here.
Check what is held fixed when changing \(N\). Holding \(k_{\mathrm{tr}}\) fixed changes the mean; holding \(\tau\) fixed changes the rate to \(N/\tau\).
Check the model against the intended process. A systematic mismatch in early or late emptying can indicate that the gamma-shaped distribution or the constant-rate chain is inadequate.

Common pitfall: “Adding transit compartments makes gastric emptying slower.” This is not necessarily true. At fixed \(k_{\mathrm{tr}}\), more stages increase the mean residence time; at fixed mean \(\tau\), the mean stays unchanged and the distribution becomes less dispersed. State the comparison condition before interpreting a change in \(N\).

Check Your Understanding

Use the stated chain-model assumptions unless a question asks you to identify an assumption that may fail.

For \(N=4\) stages and \(k_{\mathrm{tr}}=0.80\ \mathrm{h}^{-1}\), find the mean and standard deviation of residence time.
At fixed mean residence time, what happens to the coefficient of variation as \(N\) increases?
For \(N=2\) and \(k_{\mathrm{tr}}=1.00\ \mathrm{h}^{-1}\), write the fraction exited by time \(t\).
Why must a comparison of stage numbers state whether \(k_{\mathrm{tr}}\) or the mean residence time is held fixed?

Solutions. (1) The mean is \(N/k_{\mathrm{tr}}=4/0.80=5.00\) h. The standard deviation is \(\sqrt{N}/k_{\mathrm{tr}}=2/0.80=2.50\) h. (2) It decreases as \(1/\sqrt{N}\), because the coefficient of variation is \(1/\sqrt{N}\). (3) \(F_T(t)=1-e^{-t}(1+t)\), for \(t\geq0\), since \(k_{\mathrm{tr}}t=t\) and \(N=2\). (4) Holding \(k_{\mathrm{tr}}\) fixed makes the mean \(N/k_{\mathrm{tr}}\) increase with \(N\); holding the mean fixed instead requires \(k_{\mathrm{tr}}=N/\tau\). The resulting comparisons answer different questions.