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Repeated Intravenous Dosing and Infusion Schedules

Learn to represent repeated intravenous schedules as time-dependent inputs and calculate concentrations under a clearly stated linear disposition model.

Intermediate 10 min read

What You'll Learn

  • Represent repeated boluses and finite-duration infusions as separate components of an input function.
  • Calculate concentration at a specified time by summing contributions from doses already given.
  • Distinguish dose amount, infusion rate, and total administered amount using consistent units.
  • Combine a loading dose with delayed or immediate maintenance infusion.
  • Check schedule boundaries, units, and model assumptions when implementing repeated inputs.
  • Recognize when a one-compartment calculation cannot substitute for a full PBPK model.

Introduction

A repeated intravenous regimen is a sequence of inputs into the body, not a single dose with a single rate. A model must represent when each dose enters, whether it enters instantaneously or over an interval, and how those inputs combine with drug already in the system. This matters in PBPK models because the input schedule determines the amount delivered to the venous blood or another specified model compartment at each time.

The earlier material on intravenous bolus and infusion inputs establishes the basic distinction: a bolus is an instantaneous amount, while an infusion delivers drug at a rate over time. Here, those inputs are assembled into repeated schedules and used to calculate concentrations at specified times. The worked concentration calculations use a one-compartment linear model to make the schedule arithmetic transparent; a PBPK model generally uses the same input schedule in its compartmental differential equations, but its concentration response depends on the full model.

Definition: An intravenous input schedule specifies the time and amount of every bolus and the start time, stop time, and rate of every infusion. A bolus dose has units of amount, such as mg. An infusion rate has units of amount per time, such as mg/h. A finite infusion's administered amount is its rate multiplied by its duration.

Representing a Repeated Schedule

Let \(u(t)\) denote the input rate in amount per unit time. A bolus at time \(t_i\) with amount \(D_i\) can be represented mathematically by \(D_i\delta(t-t_i)\), where \(\delta\) is the Dirac delta: its integral over an interval containing \(t_i\) is one. An infusion beginning at \(a_j\), ending at \(b_j\), and delivered at constant rate \(R_j\) contributes \(R_j\) between those times and zero outside them. Thus a schedule containing both types of input can be written as

$$ u(t)=\sum_i D_i\delta(t-t_i)+ \sum_j R_j\,\mathbf{1}_{[a_j,b_j)}(t). $$

Here, \(\mathbf{1}_{[a_j,b_j)}(t)\) equals one when \(a_j\leq t<b_j\), and zero otherwise. The half-open interval is a convenient convention for bookkeeping: an infusion is active at its start and inactive at its stop. The amount delivered is unaffected by the value assigned at a single endpoint. If two infusions overlap, their rates add during the overlap.

For repeated equal boluses every \(\tau\) time units, starting at time zero, the bolus part is \(\sum_{n=0}^{N-1}D\delta(t-n\tau)\) when \(N\) doses are scheduled. For repeated infusions, each infusion must have its own start and stop times. Do not describe a two-hour infusion every eight hours as a continuous infusion lasting eight hours: the drug input is zero during the six-hour gap.

Concentration Calculations Under Linear One-Compartment Disposition

For the calculations below, assume instantaneous mixing in a single apparent distribution volume \(V\), constant first-order elimination with rate constant \(k\), and linear disposition. Let \(A(t)\) be the amount in the compartment and \(C(t)=A(t)/V\) its concentration. Between instantaneous boluses, the model is

$$ \frac{dA(t)}{dt}=u(t)-kA(t), \qquad C(t)=\frac{A(t)}{V}. $$

The assumptions mean that the elimination rate is proportional to the amount, with no saturation or time-varying clearance, and that each input is mixed into the compartment immediately. The elimination half-life is \(\ln(2)/k\). For this one-compartment model, clearance is \(CL=kV\). These assumptions are a transparent calculation model, not a claim that all PBPK concentration-time profiles are one-compartment profiles.

Result: Under linear one-compartment disposition, a bolus \(D_i\) at time \(t_i\) contributes \(D_i e^{-k(t-t_i)}/V\) to concentration at \(t\geq t_i\). A constant-rate infusion \(R\) from \(a\) to \(b\) contributes, for \(t\geq a\), $$ C_{\mathrm{inf}}(t)=\frac{R}{CL} \left[e^{-k(t-\min(t,b))}-e^{-k(t-a)}\right], $$ where \(CL=kV\). This expression applies both during infusion (\(a\leq t<b\)) and after it ends (\(t\geq b\)).

The infusion expression follows by integrating the input delivered at each time \(s\) and allowing each part to decay until \(t\). More explicitly, define \(q=\min(t,b)\). The amount present at \(t\) from the infusion is

$$ A_{\mathrm{inf}}(t)=\int_a^q R e^{-k(t-s)}\,ds =\frac{R}{k}\left[e^{-k(t-q)}-e^{-k(t-a)}\right]. $$

Dividing by \(V\), and using \(CL=kV\), gives the stated concentration expression. If \(t\) is during the infusion, \(q=t\) and the expression reduces to \(R/CL[1-e^{-k(t-a)}]\). If \(t\) is after the infusion, \(q=b\), so the accumulated amount declines after the input stops. Because the model is linear, contributions from all boluses and infusions can be added to obtain the total concentration.

Worked Example: Repeated Intravenous Boluses

Hypothetical data. A drug is given as three 100 mg boluses at times 0, 8, and 16 h. Assume \(V=20\) L and \(k=0.200\ \mathrm{h}^{-1}\). Calculate the concentration at 18 h. The model assumes instantaneous mixing, constant \(V\), and first-order elimination; no dose is given at 18 h.

Each dose contributes its remaining amount at 18 h. The elapsed times since the three doses are 18, 10, and 2 h, so

$$ C(18)=\frac{100}{20} \left[e^{-0.200(18)}+e^{-0.200(10)}+e^{-0.200(2)}\right]\ \mathrm{mg/L}. $$

Substituting the exponential values gives

$$ C(18)=5\left[0.02732+0.1353+0.6703\right] =5(0.8330)=4.165\ \mathrm{mg/L}. $$

The concentration is therefore approximately 4.165 mg/L at 18 h under the stated model. The most recent dose contributes the most, but both earlier doses still contribute; omitting them would underpredict the concentration.

Intermittent Infusions and Loading-Plus-Maintenance Schedules

An intermittent infusion has a positive rate only during each infusion interval. At a time between infusions, its contribution continues to decline by elimination, even though the input rate is zero. To calculate concentration at any selected time, identify each infusion that has started, use the appropriate during-infusion or post-infusion expression, and sum the contributions. This accounts for accumulation without assuming that the system is already at steady state.

Worked Example: Two Intermittent Infusions

Hypothetical data. A patient receives 120 mg over 2 h every 8 h, with infusions starting at 0 and 8 h. Assume \(V=30\) L and \(k=0.100\ \mathrm{h}^{-1}\), with linear first-order elimination and instantaneous mixing. Find the concentration at 9 h, one hour into the second infusion.

Each infusion rate is \(R=120/2=60\) mg/h. The first infusion ended at 2 h, so at 9 h its concentration contribution is

$$ C_1(9)=\frac{60}{0.100(30)} \left[e^{-0.100(9-2)}-e^{-0.100(9-0)}\right] =20(0.4966-0.4066)=1.800\ \mathrm{mg/L}. $$

The second infusion is still running at 9 h, so its contribution is

$$ C_2(9)=\frac{60}{0.100(30)} \left[1-e^{-0.100(9-8)}\right] =20(1-0.9048)=1.903\ \mathrm{mg/L}. $$

Adding the two contributions gives \(C(9)=1.800+1.903=3.703\ \mathrm{mg/L}\) (approximately 3.704 mg/L before rounding intermediate values). The first infusion must be included even though it has stopped; the second must be treated as still in progress.

A loading dose is an initial amount intended to raise concentration promptly. A maintenance input then supplies drug to offset elimination over time. In the linear one-compartment model, a loading dose that would produce target concentration \(C_{\mathrm{target}}\) immediately is \(D_L=VC_{\mathrm{target}}\). A continuous maintenance infusion that would maintain that concentration at steady state has rate \(R=CL\,C_{\mathrm{target}}\), because its input rate must equal the elimination rate \(CL\,C_{\mathrm{target}}\). These relationships depend on the stated one-compartment, linear assumptions.

Worked Example: Loading Dose With Delayed Maintenance Infusion

Hypothetical data. Suppose \(V=25\) L and \(k=0.120\ \mathrm{h}^{-1}\), so \(CL=kV=3.00\) L/h. A 200 mg loading bolus is given at time zero. A maintenance infusion begins at 2 h at 24 mg/h. Calculate concentration at 6 h. Assume the same linear, one-compartment model and constant parameters throughout.

The loading dose contributes

$$ C_L(6)=\frac{200}{25}e^{-0.120(6)} =8e^{-0.72}=3.894\ \mathrm{mg/L}. $$

The maintenance infusion has been running for 4 h at the calculation time. Its contribution is

$$ C_M(6)=\frac{24}{3.00}\left[1-e^{-0.120(6-2)}\right] =8(1-e^{-0.48}) =8(0.3812)=3.050\ \mathrm{mg/L}. $$

Thus \(C(6)=3.894+3.050=6.944\ \mathrm{mg/L}\). The maintenance rate equals \(CL\) times 8 mg/L, but the concentration is below 8 mg/L at 6 h because the infusion began after the loading dose and has not yet replaced all drug eliminated during the delay and subsequent interval.

Assumptions, Checks, and Alternative Representations

The one-compartment calculations require a specified \(V\), a specified elimination rate \(k\), linear superposition, and a clearly defined input schedule. A full PBPK model need not have a single concentration response of this form. Its input may be assigned to a particular blood or tissue compartment, and the resulting concentrations are obtained by solving the PBPK system of differential equations. The schedule construction remains useful, but the simple exponential formulas should not be substituted for the PBPK model unless the one-compartment assumptions are appropriate.

Before interpreting a calculated concentration, check the input and the time point. For a finite infusion, verify that \(R(b-a)\) equals the intended dose. Confirm that \(R\) is in amount/time, \(V\) is in volume, \(k\) is in inverse time, and the resulting concentration is amount/volume. At a requested time, include all earlier inputs that still contribute, use the partial-infusion formula if the time falls within an infusion, and do not include future doses.

A bolus is most naturally represented as a discrete jump in amount, while a finite infusion is represented as a rate over an interval. Some numerical implementations instead approximate a bolus by a very short infusion. That can be adequate only if the approximation preserves the intended dose and its duration is negligible for the time scale being modeled. For a PBPK ordinary differential equation solver, explicitly adding the bolus to the state at its administration time often represents the instantaneous event more directly.

Useful diagnostic checks include plotting the input rate alongside the predicted concentration, confirming that the input rate returns to zero during gaps, and checking that bolus events produce state jumps rather than prolonged input. A concentration curve that rises during a scheduled gap may still be correct if input from earlier doses is accumulating, but it should not be caused by an accidentally continuous rate. A sudden concentration discontinuity at an infusion stop is usually inconsistent with the one-compartment model: when the infusion ends, the input rate changes, but the amount and concentration remain continuous.

1
List every event.
Record each bolus time and amount and each infusion's start, stop, and rate. Check that each rate multiplied by duration gives the intended infusion amount.
2
Locate the calculation time.
For each infusion, determine whether it has not started, is active, or has ended. Exclude future doses and retain contributions from past doses.
3
Calculate and combine contributions.
Use the bolus or infusion expression that matches each event and add the results only when the disposition model is linear.
4
Check units and behavior.
Confirm concentration units, verify that infusion input stops at the correct time, and check that concentration changes continuously at infusion boundaries under the stated model.

A frequent mistake is to treat a repeated dose as though only the most recent dose affects concentration. That reasoning ignores residual drug from earlier doses. Under linear disposition, the correction is to sum the surviving contributions from every dose already administered, as in the bolus example. Another mistake is to multiply an infusion rate by the dosing interval rather than by the infusion duration; that would incorrectly count the gap as drug delivery. In a nonlinear model, even summing separately calculated dose contributions may fail because superposition no longer holds.

Key takeaway: A repeated intravenous regimen is a timed collection of bolus amounts and infusion rates. Build the input schedule first; then calculate or simulate the concentration using a disposition model whose assumptions match the intended application.

Check Your Understanding

Try each problem, then compare your reasoning with the solutions below. Unless stated otherwise, assume a one-compartment model with linear first-order elimination and instantaneous mixing.

  1. A 50 mg bolus is given at 0 h and another at 6 h. With \(V=10\) L and \(k=0.100\ \mathrm{h}^{-1}\), what is the concentration at 8 h?
  2. An infusion delivers 90 mg over 3 h. What is its rate, and how much is delivered if it is stopped after 2 h?
  3. An infusion runs from 0 to 2 h at 30 mg/h. Describe the input rate from 0 to 2 h and from 2 to 5 h, and state what happens to concentration immediately at 2 h in the stated model.
  4. With \(V=20\) L and \(k=0.150\ \mathrm{h}^{-1}\), what loading bolus corresponds to an immediate target concentration of 5 mg/L? What continuous infusion rate corresponds to maintaining that concentration at steady state?
  5. Why is it not generally valid to add separate dose contributions when disposition is nonlinear?

Solutions. (1) The dose ages at 8 h are 8 and 2 h. Thus \(C(8)=(50/10)[e^{-0.100(8)}+e^{-0.100(2)}]=5(0.4493+0.8187)=6.340\ \mathrm{mg/L}\). (2) The rate is \(90/3=30\) mg/h. If stopped after 2 h, the delivered amount is \(30(2)=60\) mg. (3) The input rate is 30 mg/h for \(0\leq t<2\) h and zero for \(2\leq t<5\) h. Concentration remains continuous at 2 h; its rate of change changes when the infusion input stops. (4) The loading amount is \(VC_{\mathrm{target}}=20(5)=100\) mg. Since \(CL=kV=0.150(20)=3.00\) L/h, the maintenance infusion rate is \(CL C_{\mathrm{target}}=3.00(5)=15.0\) mg/h. (5) Nonlinear disposition means the response to combined inputs is not generally the sum of the responses to each input considered separately. Dose contributions must then be simulated together under the nonlinear model.