Tutorials › Quantitative Systems Pharmacology (QSP) › Comparing DDE, Transit-Chain, and SSA Models of Cell Maturation

Delay and Stochastic Extensions · Tutorial 112 of 799

Comparing DDE, Transit-Chain, and SSA Models of Cell Maturation

Learn how fixed-delay, transit-chain, and SSA formulations represent cell maturation differently, and how those choices change predicted completion profiles and uncertainty.

Advanced 12 min read

What You'll Learn

  • Formulate a fixed maturation time as a delay equation and interpret its completion profile.
  • Derive the transit-chain completion-time distribution and its mean and variance.
  • Relate stochastic stage transitions to both the transit-chain equations and cell-to-cell variability.
  • Compare predicted completion counts for one hypothetical cell cohort.
  • Recognize when the models are alternatives and when their assumptions are not interchangeable.

One Maturation Process, Three Representations

A cell-maturation model can represent time in a precursor state in at least three ways: every cell spends exactly the same fixed time there; cells pass through a sequence of stages with random waiting times; or individual stage transitions are simulated as stochastic events. These choices may produce different completion profiles even when they share the same mean maturation time. Comparing the formulations is useful because a smooth average curve can conceal the distinction between a precise delay and variable single-cell durations.

We will use one simple process throughout: cells enter an immature compartment and eventually complete maturation. There is no cell death, proliferation, or return to an earlier stage. The input rate is \(u(t)\) cells per unit time, and \(M(t)\) denotes the cumulative number that have matured by time \(t\). For a cohort of \(N\) cells entering at time zero, the question is how many have matured by a given time. The assumptions are deliberately narrow so that differences in maturation-time assumptions are visible.

Definition: A fixed-delay model assigns every entering cell the same maturation duration \(\tau\). A transit-chain model represents maturation as passage through \(k\) sequential stages, each with an exponential waiting time of rate \(\lambda\). An SSA model simulates individual state transitions as random events using their specified propensities. If the SSA uses the same \(k\) stages and \(\lambda\) as the transit chain, it represents the stochastic version of that chain, not a separate fixed-duration assumption.

Fixed Delay: Completion Is a Shifted Input

With a fixed maturation duration \(\tau\), each cell entering at time \(s\) completes at time \(s+\tau\). Thus, after the initial delay, the completion rate is the input rate shifted by \(\tau\). If the input begins at time zero and there are no cells initially in the maturation process, the rate \(m(t)\) at which cells mature is

$$ m(t)= \begin{cases} 0, & 0\leq t<\tau,\\ u(t-\tau), & t\geq \tau. \end{cases} $$

This is a delay relation: the current output depends on an earlier input value. For continuous input, the number not yet matured at time \(t\) is the number that entered during the preceding \(\tau\) time units, with the interval restricted to times after input began:

$$ I(t)=\int_{\max(0,t-\tau)}^{t}u(s)\,ds. $$

For a cohort of \(N\) cells entering together at time zero, the completion profile is a jump: none have matured before \(\tau\), and all have matured at or after \(\tau\). A point cohort can be represented as a pulse input; its output is concentrated at the time \(\tau\). This idealized jump is not a numerical artifact. It follows from the assumption that all durations are identical.

Worked Example: A Cohort Under a Fixed Delay

Suppose 20 hypothetical cells enter maturation at time zero, and every cell has a fixed duration of 8 hours. Find the number matured by 7.9 hours, at 8 hours, and at 10 hours.

For this fixed-delay model, a cell is immature for \(t<8\) hours and mature for \(t\geq8\) hours. Therefore, before the delay expires,

$$ M(7.9)=0. $$

At 8 hours, all 20 cells reach completion together, so \(M(8)=20\). At 10 hours no additional cells have entered, so \(M(10)=20\). The completion rate for the cohort is concentrated at 8 hours rather than spread over a range of times. If a model uses a different convention for whether completion at the endpoint is counted, that convention must be stated; here, cells completing at exactly 8 hours are included in \(M(8)\).

Transit Chains: A Distribution of Maturation Times

In a transit chain, each cell passes through \(k\) stages in order. In every stage, it waits an exponentially distributed time with rate \(\lambda\) per hour, independently of its waiting times in other stages. If \(Z_i(t)\) is the number of cells in stage \(i\), the deterministic equations for the population amounts are

$$ \begin{aligned} \frac{dZ_1}{dt}&=u(t)-\lambda Z_1,\\ \frac{dZ_i}{dt}&=\lambda Z_{i-1}-\lambda Z_i,\qquad i=2,\ldots,k,\\ m(t)&=\lambda Z_k(t). \end{aligned} $$

These equations assume nonnegative input, no losses, and that the same transition rate applies to every stage. They describe a deterministic population trajectory. At the individual-cell level, the total maturation duration \(T\) is the sum of \(k\) independent exponential waiting times. This is an Erlang distribution, a special case of the gamma distribution. Its mean is \(k/\lambda\), and its variance is \(k/\lambda^2\). Choosing \(\lambda=k/\tau\) makes the mean duration equal to \(\tau\), but does not make each duration equal to \(\tau\).

For one cell entering at time zero, the probability of completion by time \(t\) is the Erlang cumulative distribution function:

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

For a cohort of \(N\) cells, the deterministic transit-chain model predicts a cumulative number \(N F_k(t)\) matured by time \(t\). That prediction is generally fractional because it is a population-level expected amount, not an assertion that a fraction of one cell has matured in a particular realization. The completion rate for a continuous input is the input convolved with the maturation-time density: each earlier input contributes according to the chance of completing at the current age.

Worked Example: A Four-Stage Chain with an Eight-Hour Mean

Consider 20 hypothetical cells entering at time zero. Use \(k=4\) stages and choose \(\lambda=0.5\) per hour, so the mean maturation time is \(k/\lambda=4/0.5=8\) hours. Calculate the deterministic chain prediction for the number matured by 8 hours.

Here \(\lambda t=0.5(8)=4\). Substituting into the Erlang cumulative distribution gives

$$ F_4(8) =1-e^{-4}\left(1+4+\frac{4^2}{2!}+\frac{4^3}{3!}\right) =1-e^{-4}(23.6667) \approx 0.5665. $$

The expected completion count is therefore

$$ 20F_4(8)=20(0.5665)=11.33\text{ cells}. $$

The transit-chain model predicts about 11.33 of the 20 cells matured by 8 hours. This differs from the fixed-delay prediction of 20 at that time, despite both formulations having an eight-hour maturation time in the relevant sense: the fixed-delay duration is exactly eight hours, whereas eight hours is the chain's mean duration.

SSA: Individual Random Events and Cohort Variability

For a stochastic simulation of the same \(k\)-stage chain, a cell in stage \(i\) transitions to stage \(i+1\) with propensity \(\lambda Z_i\) when there are \(Z_i\) cells in that stage. The final-stage transition is maturation, with propensity \(\lambda Z_k\). The SSA generates one event at a time according to the current propensities. For a single cell in a stage, this gives an exponential waiting time with rate \(\lambda\), as assumed by the transit-chain model.

The transit-chain equations and the SSA therefore have a precise relationship under these assumptions: the deterministic equations give the mean population trajectory of the linear stochastic stage process, while an individual SSA run gives one random realization. For a cohort of \(N\) independent cells, each with completion probability \(F_k(t)\), the number \(C(t)\) matured by time \(t\) has a binomial distribution:

$$ C(t)\sim\operatorname{Binomial}\!\left(N,F_k(t)\right),\qquad E[C(t)]=NF_k(t),\qquad \operatorname{Var}(C(t))=NF_k(t)\bigl(1-F_k(t)\bigr). $$

This binomial result requires independent cell trajectories and a fixed cohort size. Shared environmental fluctuations, cell interactions, or random input would change the dependence structure or cohort size, so this particular variance formula would not generally apply.

Worked Example: Completion Count Variability in SSA

Use the four-stage chain from the previous calculation, with 20 independently transitioning cells and \(F_4(8)\approx0.5665\). Find the mean and standard deviation of the number matured by 8 hours across repeated SSA realizations.

The mean is \(20(0.5665)=11.33\) cells. The variance is

$$ \operatorname{Var}(C) =20(0.5665)(1-0.5665) =20(0.5665)(0.4335) =4.911555 \approx4.912\text{ cells}^2. $$

Thus the standard deviation is \(\sqrt{4.911555}\approx2.216\) cells. An individual SSA run need not produce exactly 11.33 matured cells; it produces an integer count. The mean describes the average over repeated runs, and the standard deviation describes run-to-run variability under the stated independent-cell model. It is not a confidence interval for a parameter.

Why the Predictions Differ

The main distinction is the distribution of individual maturation times. In the fixed-delay model, that distribution has all its mass at \(\tau\). In a \(k\)-stage chain, it has mean \(k/\lambda\) and coefficient of variation \(1/\sqrt{k}\), where the coefficient of variation is the standard deviation divided by the mean. Thus, with mean set to \(\tau\), increasing \(k\) narrows the transit-time distribution, while any finite \(k\) retains nonzero duration variability. The fixed-delay profile is recovered as a limiting idealization as the number of stages grows with mean held fixed, not by choosing a small finite chain.

The SSA adds another distinction: a single run is noisy, while the deterministic chain gives its mean profile under the linear, independent-transition assumptions. With enough independent cells, relative variation in the cohort completion fraction typically becomes smaller. This does not mean a stochastic model and its deterministic mean are interchangeable for every question. If the outcome of interest depends on rare events, extinction, or cell-level variation, the mean trajectory alone can omit important features.

Comparison: A fixed-delay formulation is appropriate when the maturation duration is treated as exactly specified. A transit chain is useful when a smooth finite-dimensional model is needed and a distribution of durations is acceptable. An SSA is needed when individual event histories and run-to-run variability matter; when it uses the transit-chain transitions, its expected profile corresponds to that chain rather than to a fixed delay.

Assumptions, Diagnostics, and Common Pitfalls

Before comparing numerical profiles, align what the models mean by “maturation time.” Matching the mean is not enough to match the duration distribution. Also use the same initial cohort or input function, time units, and definition of completion. For a chain with \(k\) stages, specify the rate in every stage; setting \(\lambda=1/\tau\) would give a mean duration of \(k\tau\), not \(\tau\).

Several checks help reveal a model mismatch. A fixed-delay cohort should have no completions before \(\tau\), and all cells should have completed by \(\tau\) under the no-loss assumption. A finite-stage chain should have a spread of completion times, including some completions before and after its mean. In an SSA, completion counts should be integer-valued and individual runs should fluctuate around the deterministic mean. If the simulated average systematically differs from the chain's predicted mean, investigate whether the rates, initial conditions, and event definitions actually match; Monte Carlo variability alone can also make a finite collection of runs differ from the mean.

A common mistake is to set the chain's mean to eight hours and conclude that all cells mature at eight hours. The mean only summarizes the duration distribution: in the four-stage example, the probability of completion by eight hours is about 0.5665, not 1. Another mistake is to treat one SSA trajectory as the deterministic prediction. The correction is to distinguish the individual realization from its ensemble mean and to compare either matched mean profiles or full distributions, according to the scientific question.

These formulations also have limits. The simple comparison assumes no loss during maturation and constant stage-transition rates. Adding cell death, time-varying rates, or dependencies between cells changes the equations and may change the relationship between mean trajectories and individual variability. A fixed delay can be a poor representation when durations vary substantially; a short chain can be a poor approximation when the intended duration is nearly deterministic. The choice should reflect the biological assumption and the outcome of interest, not just computational convenience.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. A three-stage chain uses \(\lambda=0.75\) per hour. What are its mean maturation time and coefficient of variation?
  2. For a cohort of 10 cells under a fixed 5-hour delay, how many have matured by 4.9 hours and by 5 hours, using the endpoint convention in this tutorial?
  3. Why does matching a transit chain's mean duration to a fixed delay not make their completion profiles identical?
  4. Under what assumptions is the number matured by time \(t\) binomial in the SSA chain model?
  5. What does the deterministic transit-chain prediction represent relative to repeated SSA runs of the same linear stage process?

Solutions. (1) The mean is \(k/\lambda=3/0.75=4\) hours. The coefficient of variation is \(1/\sqrt{3}\approx0.5774\). (2) Zero have matured by 4.9 hours; all 10 have matured by 5 hours. (3) The fixed-delay duration is identical for all cells, whereas a finite transit chain sums exponential waiting times and therefore has a distribution with nonzero variance. (4) The cohort size is fixed, and cells transition independently with the same stage rates, giving each cell the same completion probability \(F_k(t)\). (5) It is the mean population trajectory under the stated assumptions; individual SSA runs fluctuate around it.

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