When Transit Chains and Fixed Delays Give Different Predictions

Why the Same Mean Delay Can Give Different Outputs

A maturation process may have a characteristic average duration without every cell taking that exact amount of time. A fixed-delay model assigns each entrant the same completion time; a transit chain represents completion times that vary across entrants. Matching the mean duration therefore does not generally make the models predict the same onset or output profile. The difference matters when interpreting early responses, timing thresholds, or the rate at which mature cells appear.

The key comparison is between the completion-time distributions, not merely between their means. A fixed delay has zero variation in completion time and no completions before the delay expires. A finite transit chain has a spread of completion times and, mathematically, a positive probability of completion at every positive time after entry. Its output usually starts gradually rather than switching on at one exact time.

Model Definitions and Assumptions

Definition: In a fixed-delay maturation model with delay \(\tau\), an entrant at time \(s\) completes at time \(s+\tau\). If entrants arrive at rate \(u(t)\), and there is no loss during maturation, the completion rate is \(y_{\mathrm{fixed}}(t)=u(t-\tau)\). In a \(k\)-stage transit chain, an entrant passes through \(k\) sequential stages, each with transition rate \(\lambda\). Its completion time \(T\) is the sum of \(k\) independent exponential waiting times with rate \(\lambda\), so \(T\) has an Erlang distribution with mean \(k/\lambda\) and variance \(k/\lambda^2\).

The comparison below assumes entrants are not lost, duplicated, or diverted while maturing; all chain stages have the same rate; and the delay or stage-rate parameters do not vary across entrants. For a continuous input, initial conditions or input history must also be specified. Unless stated otherwise, the examples start with no cells already partway through maturation and no input before time zero. These assumptions isolate the effect of completion-time variability.

To match a chain to a fixed delay \(\tau\) by mean duration, choose \(\lambda=k/\tau\). Its variance is then \(\tau^2/k\), whereas the fixed delay has variance zero. Increasing \(k\) concentrates the chain's completion times more tightly around \(\tau\); for any finite \(k\), however, the chain remains variable and has no hard minimum completion time.

From Stage Transitions to Completion Profiles

Let \(f_k(a)\) be the density of the time \(a\) from entry to completion, and let \(F_k(a)=P(T\leq a)\) be its cumulative distribution function. The chain completion rate is \(\lambda\) times the number in its final stage. Equivalently, each cohort contributes according to its completion-time density, so for an input rate \(u(t)\) that is zero before time zero,

y_{\mathrm{chain}}(t)=\int_0^t u(s)f_k(t-s)\,ds =\int_0^t u(t-a)f_k(a)\,da .

Here \(s\) is entry time and \(a=t-s\) is the age of that cohort. The density has units of inverse time, so integrating the product of the entry rate and density over time gives an output rate in the same units as \(u(t)\). For a finite cohort of \(N\) cells entering at time zero, the cumulative number completed by age \(t\) is \(NF_k(t)\), and the instantaneous completion rate is \(Nf_k(t)\).

The Erlang density follows directly from adding the stage waiting times. For one stage, \(f_1(t)=\lambda e^{-\lambda t}\) for \(t\geq0\). If the sum of \(j\) stage times has density \(\lambda^j t^{j-1}e^{-\lambda t}/(j-1)!\), adding one more independent exponential waiting time gives

f_{j+1}(t)=\int_0^t \frac{\lambda^j s^{j-1}e^{-\lambda s}}{(j-1)!} \lambda e^{-\lambda(t-s)}\,ds =\frac{\lambda^{j+1}e^{-\lambda t}}{(j-1)!} \int_0^t s^{j-1}\,ds =\frac{\lambda^{j+1}t^je^{-\lambda t}}{j!}.

Thus \(f_k(t)=\lambda^k t^{k-1}e^{-\lambda t}/(k-1)!\) for \(t\geq0\). Integrating this density, or differentiating the expression below and checking its value at zero, gives

F_k(t)=1-e^{-\lambda t}\sum_{j=0}^{k-1}\frac{(\lambda t)^j}{j!}, \qquad t\geq0.

This formula makes the onset difference explicit. For finite \(k\), \(F_k(t)>0\) for every \(t>0\), even though the probability may be very small at early times. Near zero, \(F_k(t)\) is approximately \((\lambda t)^k/k!\), so the chain's output for a step input ramps up from zero. In contrast, a fixed delay produces exactly zero output before \(\tau\).

Worked Example: A Step Input and the Onset of Output

Suppose maturation begins with no cells in transit, and a constant input of 100 cells per day starts at time zero. Compare a fixed delay of 2 days with a four-stage chain having the same mean. For the chain, \(\lambda=4/2=2\) per day.

At day 1, the fixed-delay model predicts zero output because the delay has not expired. The chain's completion fraction is

F_4(1)=1-e^{-2}\left(1+2+\frac{2^2}{2!}+\frac{2^3}{3!}\right) =1-(0.1353353)(6.3333333) =0.1428765\approx0.1429.

For a step input, the chain output at time \(t\) is \(100F_4(t)\) cells per day: it is the sum of contributions from all cohorts that have entered since time zero. Thus at day 1 it predicts \(100(0.1428765)=14.28765\), or about 14.29 cells per day. This is positive, unlike the fixed-delay prediction.

At day 2, use \(\lambda t=4\):

F_4(2)=1-e^{-4}\left(1+4+\frac{4^2}{2!}+\frac{4^3}{3!}\right) =1-(0.01831564)(23.6666667) \approx0.5665.

The chain therefore predicts about \(100(0.5665)=56.65\) cells per day at day 2. With the convention that the fixed-delay output switches on at \(t=\tau\), it predicts 100 cells per day from day 2 onward. Both models eventually produce the 100 cells per day entering the system, but their transient outputs differ substantially.

Finite Cohorts and the Shape of the Output

For a brief input, it is especially important to distinguish the cumulative number completed from the instantaneous completion rate. A fixed delay shifts the input profile without changing its shape. A transit chain spreads that profile over time by averaging contributions from entrants of different ages. The next example uses a finite-duration input to compare those output rates directly.

Worked Example: A Two-Hour Input Pulse

Suppose 50 cells per hour enter maturation from hour 0 through hour 2, for 100 cells in total. Compare a fixed delay of 8 hours with a four-stage chain of mean 8 hours. The chain rate is \(\lambda=4/8=0.5\) per hour. At hour 9, an entrant from time \(s\) has age \(9-s\), so the chain output rate is

y_{\mathrm{chain}}(9) =50\int_0^2 f_4(9-s)\,ds =50\{F_4(9)-F_4(7)\}.

For this rate, \(F_4(a)=1-e^{-0.5a}\sum_{j=0}^3(0.5a)^j/j!\). At ages 9 and 7 hours, respectively, the arguments \(0.5a\) are 4.5 and 3.5:

F_4(9)=1-e^{-4.5}(30.8125)\approx0.6577040, \qquad F_4(7)=1-e^{-3.5}(17.7708333)\approx0.4633673.

Using these unrounded values, \(y_{\mathrm{chain}}(9)=50(0.6577040-0.4633673)\approx9.717\) cells per hour. The fixed-delay output is the original pulse shifted by 8 hours: it is 50 cells per hour from hour 8 through hour 10, so at hour 9 it predicts 50 cells per hour. The chain spreads the same total 100 cells across a wider interval. Its lower output at hour 9 does not indicate cell loss; it reflects the distribution of completion times.

Matching Means Does Not Match Early Completion

The chain order \(k\) changes the spread of completion times while the mean can remain fixed. A one-stage chain has an exponential completion-time distribution and a long right tail. Higher-order chains concentrate more completion times around their common mean. A fixed delay is the limiting idealization with no spread. The following cohort calculation shows why the choice matters when a question concerns completion before the mean time.

Worked Example: Same Mean, Different Early Completion

Consider 100 cells entering at time zero. Compare a fixed delay of 8 days with transit chains of mean 8 days and orders \(k=1\) and \(k=4\). At day 4, the fixed-delay model predicts no completed cells, since \(4<8\).

For \(k=1\), \(\lambda=1/8=0.125\) per day, so the fraction completed by day 4 is

F_1(4)=1-e^{-0.5}=0.3934693, \qquad 100F_1(4)\approx39.35\ \text{cells}.

For \(k=4\), \(\lambda=4/8=0.5\) per day and \(\lambda t=2\). Therefore

F_4(4)=1-e^{-2}\left(1+2+\frac{2^2}{2!}+\frac{2^3}{3!}\right) \approx0.1428765, \qquad 100F_4(4)\approx14.29\ \text{cells}.

Both chains have mean completion time 8 days, but the more dispersed one-stage chain predicts more early completions. The fixed-delay model predicts none before day 8. The instantaneous rates also differ: for \(k=1\), \(f_1(4)=0.125e^{-0.5}\), giving about 7.582 cells per day; for \(k=4\), \(f_4(4)=0.5^4(4^3)e^{-2}/3!\), giving about 9.022 cells per day. A cumulative count and an instantaneous rate answer different questions, so one should not infer one from the other.

Checks, Alternatives, and Limitations

A useful comparison begins by stating what is being matched. Matching the mean delay alone leaves the variance, early completion probability, and output shape different. A fixed delay is appropriate when maturation is genuinely close to deterministic at the scale of interest. A transit chain is useful when a distributed delay is plausible and a finite set of ordinary differential equations is desired. A general distributed-delay model can represent other completion-time distributions, but it requires specifying their kernel rather than assuming an Erlang form.

Several checks help detect a mistaken or poorly resolved implementation:

Check onset. A fixed delay with zero prehistory must have zero output before \(\tau\). A finite transit chain may have output at every \(t>0\); a numerical trace that appears exactly zero early may simply be below plotting or solver tolerance.
Check mass balance. With no loss and no initial cells in transit, cumulative completions plus cells still in transit must equal cumulative entrants. A mismatch suggests a coding, initialization, or numerical-integration problem.
Check units and timing. The chain rate \(\lambda\) has units of inverse time, and its mean is \(k/\lambda\). A mistaken time unit changes both onset and output. The density \(f_k\) has units of inverse time; \(F_k\) is dimensionless.
Resolve rapid changes. A fixed-delay step input creates a discontinuity at the shifted input time. A solver or output grid that steps over the discontinuity can obscure its shape. Transit-chain outputs can also require fine resolution when the order is high and the profile is sharply concentrated.

Neither representation is universally superior. The fixed-delay model makes an exact lag transparent, but its deterministic completion time may be unrealistic and its shifted discontinuities can complicate numerical integration. A finite chain gives smooth, distributed timing and can be integrated as a system of ordinary differential equations, but its Erlang shape is a restriction and it predicts some early completion at every positive time. For fixed mean \(\tau\), its variance \(\tau^2/k\) decreases as the order rises; only in the limit of increasing order does the completion-time distribution concentrate at \(\tau\). If early output is central to the scientific question, selecting a chain order from the mean alone is not enough: the implied dispersion must also be plausible.

Common Mistake

Common pitfall: “The fixed-delay model and the chain both have an 8-day mean, so they should predict the same number of cells by day 4.” This reasoning confuses equality of means with equality of distributions. In the cohort example, the fixed delay predicts 0 completions by day 4, while the four-stage chain predicts about 14.29 and the one-stage chain about 39.35. Compare the full completion-time distribution whenever timing before or around the mean matters.

Key takeaway: A fixed delay shifts an input profile by one exact duration. A transit chain convolves the input with a completion-time distribution. Even when their means match, the chain's dispersion changes the onset, early completion, and shape of output.

Check Your Understanding

Answer each question, then compare with the solutions below.

A three-stage chain has mean duration 6 hours. What is its stage rate, and what is its completion-time variance?
For a four-stage chain with mean 8 days, what fraction of a cohort is predicted to have completed by day 4? How many of a 200-cell cohort is that?
A constant input begins at time zero. Explain why a finite transit chain can have positive output before its mean delay, while a matched fixed delay cannot.
A pulse of 30 cells per hour runs from hour 0 to hour 1. Write an expression for the chain output rate at hour 5 in terms of the completion CDF \(F\).

Solutions. (1) The stage rate is \(\lambda=k/\tau=3/6=0.5\) per hour. The variance is \(k/\lambda^2=3/(0.5)^2=12\) hours squared. (2) Here \(\lambda=4/8=0.5\) per day, so \(\lambda t=2\). Thus \(F_4(4)=1-e^{-2}(1+2+2^2/2!+2^3/3!)\approx0.1428765\). The expected completed count is \(200(0.1428765)\approx28.58\) cells. (3) A finite chain sums positive exponential waiting times, whose sum has positive probability of being less than any positive time; a fixed delay assigns completion time exactly equal to its delay, so it has no completions earlier. (4) Entrants at times 0 through 1 have ages 5 through 4 at hour 5, so \(y_{\mathrm{chain}}(5)=30\int_0^1 f(5-s)\,ds=30\{F(5)-F(4)\}\) cells per hour.