Deriving the Transit-Time Input Distribution

From Sequential Delays to an Input Distribution

A single first-order absorption step gives an input profile with a single exponential waiting-time distribution. A sequence of steps gives a different pattern: material must complete each step before it can enter the systemic circulation. Even if every step has the same rate constant, the total delay is distributed across the whole chain. Deriving that distribution makes clear how the chain length shapes the absorption flux.

The derivation depends on a specific counting convention. Here, \(N\) is the total number of sequential first-order steps from the initial dose to systemic entry. Thus, \(N=1\) is one first-order step, while \(N=3\) means that material passes through three steps before entering the systemic circulation. Some models count only the compartments before a separate absorption step; that convention shifts the shape parameter by one. State the convention before interpreting \(N\).

Definition: Suppose a dose \(D\) enters the first step at time zero. Each of the \(N\) sequential steps has the same first-order rate constant \(k_{\mathrm{tr}}>0\), and there are no losses between steps. Let \(T\) denote the time from the initial dose to systemic entry. Then \(T\) has a gamma distribution with integer shape \(N\) and rate \(k_{\mathrm{tr}}\). Its density is \(k_{\mathrm{tr}}^N t^{N-1}e^{-k_{\mathrm{tr}}t}/(N-1)!\) for \(t\geq 0\). The systemic input rate is this density multiplied by the systemically available dose \(FD\), where \(F\) is the fraction of the dose that eventually enters systemically.

The rate parameter \(k_{\mathrm{tr}}\) has units of inverse time, and the transit-time density has units of inverse time. The density describes the distribution of entry times for dose, not an amount entering per unit time by itself. Multiplying by \(FD\), an amount, gives the systemic input rate in units such as mg/h.

Deriving the Distribution From the Chain

Let \(A_j(t)\) be the amount in step \(j\) at time \(t\), for \(j=1,\ldots,N\). The initial conditions are \(A_1(0)=D\) and \(A_j(0)=0\) for \(j>1\). Amount leaves each step at rate \(k_{\mathrm{tr}}\); all of that amount enters the next step, except for the final step, whose outflow enters the systemic circulation.

\frac{dA_1}{dt}=-k_{\mathrm{tr}}A_1,\qquad \frac{dA_j}{dt}=k_{\mathrm{tr}}A_{j-1}-k_{\mathrm{tr}}A_j \quad (j=2,\ldots,N).

The first equation gives \(A_1(t)=D e^{-k_{\mathrm{tr}}t}\). In general, the solution for step \(j\) is \(A_j(t)=D e^{-k_{\mathrm{tr}}t}(k_{\mathrm{tr}}t)^{j-1}/(j-1)!\). This can be checked by substitution. Differentiating this expression gives an outflow term \(-k_{\mathrm{tr}}A_j\) and, for \(j>1\), an inflow term equal to \(k_{\mathrm{tr}}A_{j-1}\); at time zero it also satisfies the stated initial condition.

The final step empties at rate \(k_{\mathrm{tr}}A_N(t)\). Multiplying that outflow by \(F\) gives the systemic input rate:

R_{\mathrm{abs}}(t) =FD\,k_{\mathrm{tr}}A_N(t) =FD\,\frac{k_{\mathrm{tr}}^N t^{N-1}e^{-k_{\mathrm{tr}}t}}{(N-1)!}, \qquad t\geq 0.

This formula has a direct probability interpretation. Each step has an independent exponential waiting time with rate \(k_{\mathrm{tr}}\). The total transit time \(T\) is the sum of \(N\) such waiting times. The density of the sum is the convolution of the individual exponential densities, yielding the same gamma density as the compartment derivation. This interpretation relies on the model's first-order, time-invariant rates and on the absence of losses between steps.

Justification: The density integrates to one because \(\int_0^\infty t^{N-1}e^{-k_{\mathrm{tr}}t}\,dt=(N-1)!/k_{\mathrm{tr}}^N\). Therefore, integrating the input rate over all time gives \(FD\): the model eventually delivers the systemically available fraction of the dose, not more or less.

For integer \(N\), the fraction of the dose that has completed the chain by time \(t\) is the gamma cumulative distribution:

P(T\leq t) =1-e^{-x}\sum_{j=0}^{N-1}\frac{x^j}{j!}, \qquad x=k_{\mathrm{tr}}t.

One way to obtain this expression is to integrate the density from zero to \(t\), substitute \(u=k_{\mathrm{tr}}s\), and integrate \(u^{N-1}e^{-u}\) by parts repeatedly. Each integration reduces the power of \(u\), leaving the finite sum shown above. The cumulative systemic amount is \(FD\,P(T\leq t)\).

Worked Calculations

Worked Example: Three Steps and the Input at Two Hours

Consider a hypothetical dose \(D=100\) mg, with \(F=0.80\), passing through \(N=3\) steps at \(k_{\mathrm{tr}}=0.60\ \mathrm{h}^{-1}\). Calculate the systemic input rate at \(t=2\) h and the cumulative systemic amount by that time.

Here, \(x=k_{\mathrm{tr}}t=(0.60\ \mathrm{h}^{-1})(2\ \mathrm{h})=1.20\). The available dose is \(FD=(0.80)(100\ \mathrm{mg})=80\ \mathrm{mg}\). For \(N=3\), the density is \(k_{\mathrm{tr}}^3t^2e^{-k_{\mathrm{tr}}t}/2!\), so:

\begin{aligned} R_{\mathrm{abs}}(2) &=(80\ \mathrm{mg})\, \frac{(0.60\ \mathrm{h}^{-1})^3(2\ \mathrm{h})^2e^{-1.20}}{2!}\\ &=80\ \mathrm{mg}\times \frac{(0.216\ \mathrm{h}^{-3})(4\ \mathrm{h}^2)(0.3012)}{2}\\ &\approx 10.41\ \mathrm{mg\,h}^{-1}. \end{aligned}

For the cumulative fraction, the sum stops at \(N-1=2\):

\begin{aligned} P(T\leq 2) &=1-e^{-1.20}\left(1+1.20+\frac{1.20^2}{2}\right)\\ &=1-e^{-1.20}(2.92)\\ &\approx 0.1205. \end{aligned}

Thus, the cumulative systemic amount is \(80\ \mathrm{mg}\times0.1205\approx9.64\) mg. The flux is an instantaneous rate at two hours; the cumulative amount is the integrated input up to two hours. They are different quantities with different units.

Worked Example: How Four Steps Shape the Peak

For a hypothetical \(D=120\) mg dose with \(F=0.75\), suppose \(N=4\) and \(k_{\mathrm{tr}}=0.50\ \mathrm{h}^{-1}\). Find the time of maximum input rate and calculate the rate at \(t=4\) h.

For \(N>1\), take the logarithm of the density, ignoring constants that do not depend on time: \((N-1)\log(t)-k_{\mathrm{tr}}t\). Its derivative is \((N-1)/t-k_{\mathrm{tr}}\). Setting this to zero gives the mode \(t_{\mathrm{mode}}=(N-1)/k_{\mathrm{tr}}\); the second derivative, \(-(N-1)/t^2\), is negative, confirming a maximum. Here \(t_{\mathrm{mode}}=3/(0.50\ \mathrm{h}^{-1})=6\) h.

At four hours, \(x=(0.50)(4)=2\), and \(FD=(0.75)(120\ \mathrm{mg})=90\) mg. Substitution gives:

\begin{aligned} R_{\mathrm{abs}}(4) &=90\ \mathrm{mg}\times \frac{(0.50\ \mathrm{h}^{-1})^4(4\ \mathrm{h})^3e^{-2}}{3!}\\ &=90\ \mathrm{mg}\times \left(\frac{(0.50\ \mathrm{h}^{-1})(2)^3}{6}e^{-2}\right)\\ &\approx 8.120\ \mathrm{mg\,h}^{-1}. \end{aligned}

The rate at four hours is below its maximum because four hours precedes the calculated six-hour mode. This calculation illustrates how increasing the number of sequential steps changes the profile's shape; the total eventual systemic input remains \(FD=90\) mg.

Worked Example: Cumulative Input for a Two-Step Chain

Suppose a hypothetical \(150\) mg dose has \(F=0.80\), and passes through \(N=2\) steps at \(k_{\mathrm{tr}}=0.40\ \mathrm{h}^{-1}\). Calculate the cumulative systemic amount and the input rate at \(t=5\) h.

The available dose is \(FD=120\) mg and \(x=(0.40\ \mathrm{h}^{-1})(5\ \mathrm{h})=2\). For two steps, the cumulative fraction is \(1-e^{-x}(1+x)\), so:

\begin{aligned} P(T\leq5) &=1-e^{-2}(1+2)\\ &=1-(0.1353353)(3)\\ &\approx0.5940,\\ \text{cumulative systemic amount} &=(120\ \mathrm{mg})(0.5940) \approx71.28\ \mathrm{mg}. \end{aligned}

The density for \(N=2\) is \(k_{\mathrm{tr}}^2t e^{-k_{\mathrm{tr}}t}\). Therefore:

\begin{aligned} R_{\mathrm{abs}}(5) &=(120\ \mathrm{mg})(0.40\ \mathrm{h}^{-1})^2(5\ \mathrm{h})e^{-2}\\ &=(120)(0.16)(5)(0.1353353)\ \mathrm{mg\,h}^{-1}\\ &\approx12.99\ \mathrm{mg\,h}^{-1}. \end{aligned}

The cumulative amount is about \(71.28\) mg by five hours, while the instantaneous input rate at five hours is about \(12.99\) mg/h.

What Chain Length Changes—and What It Does Not

With \(k_{\mathrm{tr}}\) fixed, the chain length controls the distribution's shape. For \(N=1\), the density is exponential and has its maximum at time zero. For \(N>1\), the density starts at zero, rises to a peak at \((N-1)/k_{\mathrm{tr}}\), and then declines. Increasing \(N\) moves that peak later under this fixed-rate comparison. These changes redistribute input over time; they do not change the total input, which remains \(FD\) under the model assumptions.

This result is useful because it gives a mechanistic way to represent distributed delays without assigning every dose unit one fixed lag time. A fixed lag shifts entry by the same duration, whereas the chain produces a range of transit times. A single first-order step is the \(N=1\) special case. More elaborate input assumptions, such as a constant input over a specified duration, have different rate shapes and should not be treated as interchangeable with this chain.

The distribution describes the input to the systemic circulation. It does not by itself specify the resulting concentration profile: that also depends on disposition after systemic entry. If systemic availability varies with time or material is lost between steps, multiplying the normalized density by a single \(F\) may no longer describe the input correctly.

Assumptions, Diagnostics, and a Common Pitfall

The derivation assumes an initially empty chain except for the dose in the first step, identical constant first-order rates, forward movement only, and no loss between steps. The factor \(F\) is a constant scale for eventual systemic availability. The model also treats the dose as entering the chain at one time; a dose delivered over an extended period would require accounting for that input history.

When fitting or checking a model, inspect whether the implied input has the shape the data can support. For example, systematic early residuals may indicate that the model's predicted onset is too delayed, while systematic residuals around the peak may indicate that the predicted peak timing or breadth is inappropriate. Such patterns do not uniquely identify the cause: disposition misspecification, sparse sampling, or other input mechanisms can produce similar discrepancies. Check the assumed chain convention and units as well as the fit.

Off-by-one shape: With the convention here, \(N\) steps give shape \(N\) and the denominator is \((N-1)!\). If \(N\) counts only compartments before a separate final absorption step, the total number of steps is different.
Density mistaken for a fraction: The density has units of inverse time and is not the fraction absorbed by a given time. Integrate it, or use the cumulative distribution, to obtain a fraction.
Forgetting the dose scale: The density integrates to one. The systemic input rate is \(FD\) times that density, not the density alone.
Over-interpreting the chain: A good fit does not prove that the physical absorption process consists of literal, identical anatomical compartments. The chain is a model assumption that generates a useful input-time distribution.

A practical distinction from alternative input assumptions is that this model has a specific gamma-shaped input profile generated by sequential first-order steps. A fixed lag-time model shifts an input without spreading the delay, and a constant-rate input over a duration has a different shape. Choosing among them requires considering the model assumptions and whether the observed concentration-time data can distinguish the resulting profiles; none should be selected solely because it is mathematically convenient.

Key takeaway: Under the stated convention, the total time through \(N\) identical first-order steps has a gamma density with shape \(N\) and rate \(k_{\mathrm{tr}}\). Multiplying that density by \(FD\) gives the systemic input rate, while its cumulative distribution gives the fraction of available dose entered by a specified time.

Check Your Understanding

Answer each question, then compare with the worked solutions below.

With the convention in this tutorial, what are the gamma shape and factorial denominator for a chain of five steps?
For \(N=3\) and \(k_{\mathrm{tr}}=0.75\ \mathrm{h}^{-1}\), find the time of maximum input rate.
Why is the density at a particular time not itself the fraction of dose that has entered systemically by that time?
For \(N=1\), \(k_{\mathrm{tr}}=0.50\ \mathrm{h}^{-1}\), and \(t=2\) h, calculate the fraction of the available dose that has completed the chain.

Solutions. (1) The shape is \(N=5\), and the density denominator is \((N-1)!=4!=24\). (2) Since \(N>1\), \(t_{\mathrm{mode}}=(N-1)/k_{\mathrm{tr}}=2/(0.75\ \mathrm{h}^{-1})=2.667\) h. (3) The density has units of inverse time and describes transit time per unit time; a fraction is cumulative and requires integrating the density up to the specified time. (4) For \(N=1\), \(P(T\leq t)=1-e^{-k_{\mathrm{tr}}t}\), so \(P(T\leq2)=1-e^{-1}\approx0.6321\). Thus about 63.21% of the available dose has completed the chain by two hours; the systemic amount is that fraction multiplied by \(FD\).