Why the Dosing Schedule Is Part of the Simulation
A concentration-time profile is conditional on both the model and the dose history supplied to it. A fitted structural model describes how drug enters and leaves the body; the dosing schedule specifies when and how much drug enters. In this tutorial, the schedule is fixed: every simulated profile uses the specified dose amounts, routes, and administration times. The simulation calculates the consequences of that regimen; it does not select or adjust a dose.
This distinction matters in pharmacometrics because a concentration prediction can be wrong even when the structural model is implemented correctly. A dose entered at the wrong time, an omitted prior dose, or an infusion treated as a bolus changes the input to the model. Simulated profiles then answer a different question from the one intended.
A dose event and a sampling time are different things. Dose events define drug input. Sampling times specify when the model output is recorded. A simulation may use the same dose history but evaluate concentration at a denser grid, at different observation times, or at both. Changing the output grid does not change the regimen.
Representing Dose Events and Model Input
For a schedule with \(J\) events, let \(t_j\) be the time of event \(j\), and let \(D_j\) be its dose amount. For an infusion, also specify the infusion duration \(T_j\), so the infusion rate is \(R_j=D_j/T_j\) when the rate is constant. Route and units must be explicit: for example, a dose in milligrams, time in hours, and concentration in milligrams per liter.
The model's input is a function of time determined by those events. For a one-compartment model with intravenous input, volume \(V\), clearance \(CL\), and concentration \(C(t)\), define \(k=CL/V\). Between dose events and infusion boundaries, concentration obeys the elimination equation. During an infusion, input adds a rate term:
Here \(R(t)\) is zero when no infusion is running and equals the prescribed rate while an infusion is running. For a bolus, concentration increases instantaneously by \(D_j/V\) at \(t_j\); it does not increase gradually over an infusion duration. Repeated events contribute to the concentration that remains from earlier events. The state is carried forward through the schedule rather than reset to zero at each new dose.
Why Infusion Timing Requires Care
For a one-compartment model initially at zero concentration, a constant-rate infusion beginning at time zero produces an increasing concentration while drug is entering. Solving the differential equation over the infusion duration gives the concentration at the end of the infusion. Once input stops, concentration declines exponentially. These two pieces are then combined with any contributions remaining from earlier doses.
For a dose of \(D\) infused over \(T\), \(R=D/T\). The first expression applies only while the infusion is running; the second applies after it ends. To justify the first expression, multiply \(dC/dt+kC=R/V\) by the integrating factor \(e^{kt}\). Integrating from zero to \(t\), using \(C(0)=0\), gives \(e^{kt}C(t)=(R/V)(e^{kt}-1)/k\). Since \(Vk=CL\), this is \(C(t)=(R/CL)(1-e^{-kt})\). After the infusion, the input is zero, so \(dC/dt=-kC\), whose solution starts from the end-of-infusion concentration and decays exponentially.
Worked Example: Two Infusions and the Correct Event Boundary
Hypothetical schedule and model. A one-compartment model has \(V=20\) L and \(CL=4\) L/h. There are two 100 mg intravenous infusions: one begins at hour 0 and one at hour 12. Each lasts 2 hours at a constant rate. Calculate the model concentration at hour 13, ignoring residual error.
First calculate \(k=CL/V=4/20=0.2\ \text{h}^{-1}\), and \(R=100/2=50\) mg/h for each infusion. The infusion that began at hour 12 is still running at hour 13: its elapsed time is \(13-12=1\) hour. Its scheduled end is hour \(12+2=14\), so \(13-14=-1\) hour. The negative result means that the infusion ends one hour in the future, not that it ended 11 hours ago.
The concentration contribution from the infusion that began at hour 12 is therefore \( (R/CL)(1-e^{-k(1)})= (50/4)(1-e^{-0.2})=12.5(0.18127)=2.266\) mg/L.
The infusion that began at hour 0 ended at hour 2. Its concentration at the end of that infusion was \(12.5(1-e^{-0.4})=4.121\) mg/L. At hour 13, it has been declining for \(13-2=11\) hours, so its remaining contribution is \(4.121e^{-0.2(11)}=4.121e^{-2.2}=0.4565\) mg/L.
By superposition, the contributions add in this linear model: \(C(13)=2.266+0.4565=2.722\) mg/L. The prior infusion ended 11 hours earlier; the infusion that began at hour 12 has not ended yet. This calculation depends on respecting each event's own start and end times.
Repeated Doses and the Carryover of Concentration
For an intravenous bolus in a one-compartment model, each dose adds an immediate concentration increment \(D_j/V\), followed by exponential elimination. For times after the bolus, the contribution of event \(j\) is \(D_j/V\) multiplied by the elimination factor for the elapsed time. With several doses, sum the contributions from all doses already administered. This is a direct way to check whether a simulation has retained the appropriate dose history.
Worked Example: Repeated Boluses on an Unchanged Schedule
Hypothetical schedule and model. A subject receives 100 mg intravenous boluses at hours 0, 8, and 16. Let \(V=30\) L and \(CL=3\) L/h. Find the concentration at hour 24, with no residual error.
The elimination rate constant is \(k=CL/V=3/30=0.1\ \text{h}^{-1}\). The elapsed times since the three doses at hour 24 are 24, 16, and 8 hours. Thus,
Using \(e^{-2.4}=0.09072\), \(e^{-1.6}=0.2019\), and \(e^{-0.8}=0.4493\), the result is \(3.333(0.09072+0.2019+0.4493)=3.333(0.7419)=2.473\) mg/L.
All three doses enter the calculation because none occurs after hour 24. A simulation that used only the most recent bolus would return \(3.333e^{-0.8}=1.498\) mg/L and would omit carryover from the two earlier doses. The dose schedule was not changed; the difference is whether its full history was represented.
Same Schedule, Different Parameter Values
A fixed regimen does not imply that every simulated subject has the same profile. The model parameters determine the response to the common input. For example, a subject with faster elimination may have lower concentrations at a given time after the same sequence of boluses. This is consistent with simulations using subject-specific parameter values: the dose-event record remains fixed while the parameter values used to calculate the response differ.
Worked Example: Comparing Profiles Without Changing Doses
Hypothetical comparison. Use the 100 mg intravenous boluses at hours 0, 8, and 16 and \(V=30\) L from the preceding example. Compare predicted concentration at hour 24 for two parameter sets: \(CL=3\) L/h and \(CL=6\) L/h. Both use the same fixed schedule, and residual error is omitted.
For \(CL=3\) L/h, \(k=3/30=0.1\ \text{h}^{-1}\), giving \(C(24)=2.473\) mg/L as calculated above.
For \(CL=6\) L/h, \(k=6/30=0.2\ \text{h}^{-1}\). The dose times and elapsed times are unchanged, so \(C(24)=(100/30)(e^{-4.8}+e^{-3.2}+e^{-1.6})\). The exponential values are approximately \(0.008230\), \(0.04076\), and \(0.2019\), respectively. Therefore \(C(24)=3.333(0.2509)=0.8363\) mg/L.
The higher clearance produces the lower concentration at hour 24 in this example. The calculation changes the model parameter, not the prescribed doses or their timing. These are conditional model predictions; they are not evidence that changing clearance would cause the same result in every clinical setting.
Oral Doses and Multiple Input Events
For first-order absorption after an oral dose, each dose generates an input that is spread over time rather than delivered instantaneously. In a one-compartment model with bioavailability \(F\), absorption rate constant \(k_a\), elimination rate constant \(k\), and volume \(V\), the contribution of a dose \(D_j\), for \(t\geq t_j\) and \(k_a\ne k\), is \(FD_jk_a/[V(k_a-k)]\) multiplied by the difference between the elimination and absorption exponentials. The repeated-dose profile is the sum of contributions from dose events already given.
Worked Example: Two Oral Doses With First-Order Absorption
Hypothetical schedule and model. Oral doses of 100 mg are given at hours 0 and 8. Set \(F=1\), \(V=20\) L, \(k_a=1\ \text{h}^{-1}\), and \(k=0.2\ \text{h}^{-1}\). Calculate the concentration at hour 12, with residual error omitted.
The elapsed times since the doses are 12 and 4 hours. For each dose, the coefficient is \(FDk_a/[V(k_a-k)]=(1)(100)(1)/[20(1-0.2)]=6.25\) mg/L.
The first dose contributes \(6.25(0.09072-0.000006)=0.5669\) mg/L. The second contributes \(6.25(0.4493-0.01832)=2.694\) mg/L. Thus \(C(12)=0.5669+2.694=3.261\) mg/L. The second oral dose does not instantaneously add \(100/V\) to concentration; its contribution reflects absorption over time. The calculation retains both scheduled doses.
Implementation Checks, Alternatives, and Limitations
Before interpreting a simulated profile, verify the inputs and the resulting trajectory. In particular, check that dose amounts, time units, route, infusion duration, and time origin match the intended regimen. For each infusion, verify the interval from start to end; for every predicted time, determine whether each event has not yet begun, is ongoing, or has ended. Confirm that earlier doses contribute when the model implies carryover, and that concentrations remain nonnegative under the model and inputs used.
A useful diagnostic is to inspect the concentration immediately before and after a dose event. A bolus can produce an instantaneous change in concentration; a finite-rate infusion should not be represented by that same jump. For numerical differential-equation simulations, evaluate the solver around infusion starts and stops and check that reducing the time step or tightening tolerances does not materially change the predicted profile. Implausible jumps, negative concentrations, or unexpected discontinuities can indicate an event-coding, unit, or numerical problem.
The main alternative is to simulate a different dose history, such as a regimen with different amounts or administration times. That is a different scenario, not a harmless change to the evaluation grid. A further distinction is whether predictions use one parameter set, subject-specific parameter values, or parameter uncertainty draws. Those choices affect the source of variation in profiles; none should silently alter the fixed schedule when the target question is conditional on that regimen.
A fixed-schedule simulation is only as relevant as its assumptions. It treats the specified dose history as the input, so it cannot represent omitted, delayed, or partially taken doses unless those events are explicitly encoded. It also inherits the structural model's assumptions about disposition and input, and its predictions may be unreliable outside the data or regimen conditions that support that model. A numerically smooth profile is not proof that either the model or dose history is clinically appropriate.
- Wrong reasoning: “At hour 13, the infusion that started at hour 12 and lasts 2 hours ended 11 hours ago.” This confuses elapsed time since the start with time relative to the end.
- Correction: The scheduled end is hour \(12+2=14\), and \(13-14=-1\) hour. The infusion is still running and ends one hour later. An infusion that began at hour 0 and ended at hour 2, by contrast, ended \(13-2=11\) hours before hour 13.
- Another common error: Treating an observation time as an additional dosing time. Sampling records the model output; it does not add drug to the system.
Check Your Understanding
Use the stated model assumptions and show the key arithmetic for each calculation.
- A 60 mg intravenous bolus is given at hour 0. If \(V=20\) L and \(k=0.1\ \text{h}^{-1}\), what is the concentration at hour 5, ignoring residual error?
- An infusion begins at hour 6 and lasts 3 hours. At hour 8, is it running, and how much time remains?
- For 100 mg boluses at hours 0 and 10, let \(V=25\) L and \(k=0.2\ \text{h}^{-1}\). Find the concentration at hour 12.
- In one or two sentences, explain how two simulations can use the same fixed dosing schedule but produce different concentration profiles.
Solutions. (1) The bolus contribution is \(D/V=60/20=3\) mg/L at time zero. At hour 5, \(C(5)=3e^{-0.1(5)}=3e^{-0.5}=1.820\) mg/L. (2) The infusion ends at hour \(6+3=9\). Since hour 8 is before hour 9, it is running, with \(9-8=1\) hour remaining. (3) At hour 12, the elapsed times are 12 and 2 hours. Thus \(C(12)=(100/25)(e^{-0.2(12)}+e^{-0.2(2)})=4(0.09072+0.6703)=3.044\) mg/L. (4) The dose amounts and event times can be identical while parameter values differ, for example, producing different elimination rates. The structural model then maps the same input schedule to different predicted profiles.