Representing Schedule-Dependent Effects in Dynamic Combination Models

Why Treatment Schedule Belongs in a Dynamic Model

Two regimens can use the same drugs and total treatment durations but produce different responses if their doses occur at different times. A drug can alter the amount of available target, trigger compensatory pathway adaptation, or leave behind an effect that persists after dosing ends. A second drug given while those states are changing may therefore act in a different biological context than the same drug given later.

A schedule-dependent model makes this timing explicit. The examples below use normalized, hypothetical quantities and simplified drug inputs. They illustrate how to formulate and calculate state trajectories; they are not parameter estimates or predictions for a particular therapy.

A Dynamic Model for Target Turnover, Adaptation, and Persistence

Definition: A schedule-dependent combination model represents dosing schedules as time-varying inputs and describes how those inputs change biological states and response over time. In the illustrative model here, $u_A(t)$ and $u_B(t)$ indicate whether drugs A and B are exerting their modeled effects at time $t$. The states are $T(t)$, target availability relative to baseline; $H(t)$, a compensatory adaptation, where $H=0$ is baseline; and $P(t)$, a persistent inhibitory effect, where $P=0$ means no persistent inhibition. The normalized pathway output is $$ Y(t)=T(t)\,[1+H(t)]\,[1-P(t)]\,[1-e_Bu_B(t)]. $$ Here $e_B$ is the fractional inhibition produced by B while its modeled input is on. Lower $Y$ means lower pathway output relative to the untreated baseline $Y=1$.

One possible state model is

\begin{aligned} \frac{dT}{dt} &= k_T(1-T)-q_Au_A(t)T,\\ \frac{dH}{dt} &= r_Hu_A(t)-k_HH,\\ \frac{dP}{dt} &= s_Pu_A(t)-k_PP. \end{aligned}

The target turns over toward its baseline value of 1 at rate $k_T$, while A reduces its availability at rate $q_A$. A induces adaptation at rate $r_H$, which relaxes toward baseline at rate $k_H$. It also induces the persistent inhibitory state $P$ at rate $s_P$, which decays at rate $k_P$. The model treats B as a direct, schedule-controlled inhibition of pathway output. All rate constants have units of inverse time; $r_H$ and $s_P$ have units of state per time. The inputs $u_A$ and $u_B$ are dimensionless.

For the calculations here, the inputs are idealized as either 0 or 1, parameters are constant across the modeled period, and the initial states are $T(0)=1$, $H(0)=0$, and $P(0)=0$. Parameters are assumed to keep the states in biologically interpretable ranges, in particular $T\geq 0$ and $0\leq P\leq 1$. Adaptation can increase pathway output: if $Y>1$, the model predicts output above the untreated baseline. These assumptions make a transparent teaching model, not a universal description of combination therapy.

Solving the State Equations Across a Dosing Interval

When A is on constantly over an interval, each equation has a closed-form solution. For target availability, let $u_A=1$ and suppose $T(0)=T_0$. Rearranging gives $dT/dt=k_T-(k_T+q_A)T$. The equilibrium is $k_T/(k_T+q_A)$; subtracting this equilibrium from $T$ gives a quantity that decays exponentially at rate $k_T+q_A$. Thus,

T(t)=\frac{k_T}{k_T+q_A} +\left(T_0-\frac{k_T}{k_T+q_A}\right)e^{-(k_T+q_A)t}.

For constant A input, the other two solutions follow by integrating their linear equations:

\begin{aligned} H(t)&=H_0e^{-k_Ht}+\frac{r_H}{k_H}(1-e^{-k_Ht}),\\ P(t)&=P_0e^{-k_Pt}+\frac{s_P}{k_P}(1-e^{-k_Pt}). \end{aligned}

When A turns off, the forcing terms become zero. Then $H$ and $P$ decay exponentially, while $T$ recovers toward 1:

H(t+\tau)=H(t)e^{-k_H\tau},\qquad P(t+\tau)=P(t)e^{-k_P\tau},\qquad T(t+\tau)=1-[1-T(t)]e^{-k_T\tau}.

These expressions also show why a dosing schedule must be implemented as a sequence of intervals: the state at the end of one interval is the initial condition for the next. It is not reset to baseline when a dose ends.

Worked Example: State Changes During a One-Day A Course

Suppose A is given from day 0 to day 1, with $k_T=1$ per day, $q_A=1$ per day, $r_H=0.4$ per day, $k_H=1$ per day, $s_P=0.3$ per day, and $k_P=0.5$ per day. Start at $T=1$, $H=P=0$. Calculate the states and pathway output at day 1, before any B effect.

For target availability, the on-treatment equilibrium is $1/(1+1)=0.5$. Therefore,

T(1)=0.5+(1-0.5)e^{-2} =0.5+0.5(0.1353) =0.5677.

For adaptation and the persistent state,

\begin{aligned} H(1)&=\frac{0.4}{1}(1-e^{-1}) =0.4(0.6321)=0.2528,\\ P(1)&=\frac{0.3}{0.5}(1-e^{-0.5}) =0.6(0.3935)=0.2361. \end{aligned}

With $u_B=0$, substitute these states into the output equation:

Y(1)=0.5677(1+0.2528)(1-0.2361) \approx 0.5433.

The model predicts pathway output about 0.5433 times baseline, or about 45.67% below baseline, at day 1. This is the combined consequence of reduced target availability, compensatory adaptation, and persistent inhibition in this particular model. It is not the effect of any one state in isolation.

Comparing Concurrent and Sequential Treatment

The same equations can represent different schedules simply by changing the input functions. Consider A on from day 0 to day 1. In a concurrent regimen, B is also on from day 0 to day 1. In a sequential regimen, B is on from day 1 to day 2 instead. Let $e_B=0.5$, so B halves output while its input is on. To compare response during B treatment, evaluate each regimen at the midpoint of its B interval: day 0.5 for concurrent treatment and day 1.5 for sequential treatment.

Worked Example: Response Midway Through Each Schedule

Use the parameters and initial states from the preceding example. At day 0.5, A has been on for half a day. At day 1.5, A has been off for half a day. Calculate the pathway output at each B-treatment midpoint.

At day 0.5, the on-treatment solutions give

\begin{aligned} T(0.5)&=0.5+0.5e^{-1}=0.6839,\\ H(0.5)&=0.4(1-e^{-0.5})=0.1574,\\ P(0.5)&=0.6(1-e^{-0.25})=0.1327. \end{aligned}

B is on in the concurrent regimen, so

Y_{\mathrm{concurrent}}(0.5) =0.6839(1.1574)(0.8673)(1-0.5) \approx 0.3433.

For the sequential regimen, the day-1 states are $T=0.5677$, $H=0.2528$, and $P=0.2361$. After half a day of recovery without A,

\begin{aligned} T(1.5)&=1-(1-0.5677)e^{-0.5}\approx 0.7378,\\ H(1.5)&=0.2528e^{-0.5}=0.1534,\\ P(1.5)&=0.2361e^{-0.25}\approx 0.1839. \end{aligned}

B is on at day 1.5, giving

Y_{\mathrm{sequential}}(1.5) =0.7378(1.1534)(0.8161)(1-0.5) \approx 0.3472.

At these specified midpoints, the model predicts output of about 0.3433 of baseline during concurrent treatment and 0.3472 during sequential treatment. The corresponding reductions from baseline are about 65.67% and 65.28%. The small difference arises because the states evolve between the two B-treatment windows. This comparison is a midpoint snapshot, not a claim that the integrated response over each full regimen has the same ranking; that requires evaluating output across time and summarizing it with a specified endpoint.

Opposing State Deviations and a Timing Check

When two state changes act in opposite directions, their combined contribution should be calculated from the output equation, not inferred by inspecting either state alone. The following reduced illustration isolates that point. Suppose a pulse of A establishes deviations $x(0)=0.4$ and $h(0)=-0.4$ in two multiplicative factors, $1+x$ and $1+h$. Suppose both deviations decay at 1 per day after the pulse, including during a one-day period in which the drug is present. This is an impulse-triggered approximation, distinct from the continuously forced states in the preceding model.

Worked Example: Carrying Decay Through the Exposure Day

Let $k$ be the number of drug-free recovery days after the one-day A exposure. At the end of that exposure, one day has already elapsed since the deviations were established. Therefore, at the end of exposure and after $k$ recovery days,

x=h_{\mathrm{magnitude}}=0.4e^{-(1+k)}.

The two factors are $1+x$ and $1-x$. Their product is

(1+x)(1-x)=1-x^2 =1-[0.4e^{-(1+k)}]^2 =1-0.16e^{-2(1+k)}.

Immediately after the one-day exposure, $k=0$, so

1-0.16e^{-2} =1-0.16(0.1353) \approx 0.9783.

After two additional recovery days, $k=2$, and the product is

1-0.16e^{-6} =1-0.16(0.002479) \approx 0.9996.

The deviations are already reduced during the one-day exposure, so setting $k=0$ must not be treated as the time before that exposure. Omitting the exposure day would incorrectly give $1-0.4^2=0.84$ at $k=0$, rather than approximately 0.9783.

Choosing a Representation and Checking Its Behavior

A static combination summary evaluates effects at a fixed condition or endpoint. Such a summary can be useful for comparing combinations at a specified dose, but it does not by itself encode the order of doses, state recovery, or a persistent effect after dosing stops. A dynamic model instead carries states forward across treatment intervals. A pulse-response or exponential-decay representation can be a simpler alternative when data support a compact description of lingering effects; the forced differential-equation model is more explicit about how states accumulate during exposure and recover afterward.

The schedule input in this tutorial is an idealized effect input, not a model of drug concentration. It assumes that the modeled input turns on and off as specified. Real schedules may require measured or modeled exposure profiles, and an input that remains pharmacologically active after the last dose should not be switched off merely because dosing ended. Choosing that input representation is separate from deciding whether the biological state model is adequate.

Several checks help detect a model or implementation that is going wrong:

Check limits and units. With $u_A=0$, target availability should approach 1, while $H$ and $P$ should approach 0. The exponents must be dimensionless, so each rate constant is multiplied by time in matching units.
Check continuity at schedule changes. States should carry their values from one interval into the next. A sudden jump in a state at a dose boundary is inappropriate unless the model explicitly includes an instantaneous change.
Inspect trajectories, not just endpoint output. A negative target amount or $P>1$ may make the model biologically implausible or produce a negative factor $1-P$. Strong adaptation can instead raise output above baseline, which may be plausible but should be checked against the intended biology.
Vary time resolution. Numerical solutions should give stable outputs when the integration step is reduced, especially near dose changes and when turnover is fast.
Check the observation time and endpoint. A comparison at the end of dosing, at a fixed time after dosing, and over the full course can give different answers. State which is being compared.

A common mistake is to assume that identical drugs and total durations imply identical responses, or to apply a decay factor only to the drug-free interval. In a state model, the response depends on the entire input history, and each state continues evolving during every interval specified by its equation. Carrying the states through exposure, washout, and recovery prevents both errors.

Key takeaway: Represent schedule as time-varying input, propagate target and adaptation states across every interval, and calculate response at explicitly defined times. A schedule comparison is meaningful only when its model assumptions and response summary are stated.

Check Your Understanding

Answer each question, then check the brief solutions below.

If A is off for an interval of length $\tau$, how does a persistent state $P$ change?
With the parameters in the first worked example, what is $H$ after day 1 followed by one drug-free day?
Why can a static comparison at one dose combination fail to distinguish concurrent from sequential schedules?
In the paired-deviation example, what is the state-factor product immediately after one exposure day if the deviations start at magnitude 0.4 and decay at 1 per day?

Solutions. (1) It decays as $P(t+\tau)=P(t)e^{-k_P\tau}$, because its induction input is zero during the interval. (2) $H(2)=0.2528e^{-1}\approx0.0930$. (3) A static comparison does not carry the changing biological states through time or encode dose order; the dynamic model does. (4) The exposure day contributes one day of decay, so the product is $1-0.16e^{-2}\approx0.9783$, not 0.84.