Repeated IV Bolus Dosing in One Compartment

Why Repeated Doses Accumulate

After a single intravenous bolus in a one-compartment model, concentration declines exponentially. With repeated boluses, each new dose adds to the drug that remains from earlier doses. The concentration is therefore not just the response to the most recent dose: it is the sum of the contributions from all doses already administered. This sum, called superposition, gives a direct way to calculate concentrations between doses and to understand accumulation over time.

The formulas here describe a hypothetical individual with fixed model parameters. They assume the one-compartment IV bolus model and its exponential elimination, as introduced in the earlier material on the One-Compartment IV Bolus Model. They do not assume that concentration is at steady state unless that condition is stated explicitly.

Definition: Let \(D\) be the dose (mg), \(V\) the apparent volume of distribution (L), \(k\) the first-order elimination rate constant (\(\text{h}^{-1}\)), and \(\tau\) the time between doses (h). For identical boluses at times \(0,\tau,2\tau,\ldots\), define \(C_n(t)\) as the concentration at time \(t\) after the \(n\)th dose, where \(0\leq t<\tau\). The dose at time zero of this local time scale is the most recent dose. The concentration immediately after each bolus includes that bolus; immediately before a bolus, it does not.

The model requires \(D>0\), \(V>0\), \(k>0\), and \(\tau>0\). It also assumes instantaneous delivery into the compartment, the same dose and interval on each occasion, constant \(V\) and \(k\), and linear pharmacokinetics over the concentration range considered. In particular, each dose contributes independently to the concentration according to the same exponential disposition function. Changes in dose, interval, elimination, or volume require a corresponding change in the calculation.

Concentration After a Finite Number of Doses

One dose contributes \(D/V\) immediately after administration and \( (D/V)e^{-kt}\) after \(t\) hours. At time \(t\) after the most recent dose, the dose given one interval earlier has been declining for \(t+\tau\) hours; the dose before that has been declining for \(t+2\tau\) hours. Adding the contributions from \(n\) doses gives

C_n(t)=\frac{D}{V}\sum_{j=0}^{n-1}e^{-k(t+j\tau)} =\frac{D}{V}e^{-kt}\sum_{j=0}^{n-1}q^j,\qquad q=e^{-k\tau}.

Here \(q\) is the fraction remaining after one dosing interval: because \(k>0\) and \(\tau>0\), \(0<q<1\). The finite geometric-series identity \(\sum_{j=0}^{n-1}q^j=(1-q^n)/(1-q)\) gives an equivalent form:

C_n(t)=\frac{D}{V}e^{-kt}\frac{1-q^n}{1-q}.

This expression describes the concentration between the \(n\)th and \((n+1)\)th doses. At the next dose, concentration jumps upward by \(D/V\), and the dose count in the expression increases by one. Just before that next dose, the concentration is the preceding dose's profile evaluated at \(t=\tau\).

Worked Example: Concentration After Three Doses

Suppose hypothetical IV boluses of \(100\) mg are given every \(8\) h to a one-compartment system with \(V=20\) L and \(k=0.150\ \text{h}^{-1}\). Calculate concentration \(4\) h after the third dose, given at time \(16\) h.

The assumptions for superposition are that the three doses are equal, the interval is \(8\) h, and \(V\) and \(k\) remain constant. The dose concentration increment is \(D/V=100/20=5.00\) mg/L. At time \(20\) h, the three doses have ages \(20\), \(12\), and \(4\) h, respectively. Thus

C_3(4)=5.00\left(e^{-0.150(20)}+e^{-0.150(12)}+e^{-0.150(4)}\right)

Substituting the exponential values, \(e^{-3}=0.0497871\), \(e^{-1.8}=0.165299\), and \(e^{-0.6}=0.548812\), gives

C_3(4)=5.00(0.0497871+0.165299+0.548812) =5.00(0.763898)=3.81949\ \text{mg/L}.

The concentration is therefore \(3.819\) mg/L, to four significant figures. It is below the post-dose concentration because four hours of elimination have occurred since the third bolus.

Steady State and the Accumulation Factor

Repeated dosing approaches a steady periodic profile: at the same time after a dose, concentrations converge as more doses are given. This is a limiting profile, not a claim that concentration is constant throughout an interval. In the finite-dose expression, \(q^n\) tends to zero as \(n\) increases because \(0<q<1\). Therefore, at time \(t\) after a dose, the steady-state concentration is

C_{\mathrm{ss}}(t)=\lim_{n\to\infty}C_n(t) =\frac{D}{V}\frac{e^{-kt}}{1-q} =\frac{D}{V}\frac{e^{-kt}}{1-e^{-k\tau}},\qquad 0\leq t<\tau.

At the same time \(t\) after a dose, a single isolated dose would produce \((D/V)e^{-kt}\). Thus the steady-state accumulation factor relative to that single-dose concentration is

R_{\mathrm{acc}}=\frac{C_{\mathrm{ss}}(t)}{(D/V)e^{-kt}} =\frac{1}{1-e^{-k\tau}}=\frac{1}{1-q}.

This factor applies at a matched time after dosing; it is not a universal multiplier for concentrations compared at different points in the dosing interval. It also does not imply that the body accumulates drug without limit. For \(k>0\), the residual fraction from each successively earlier dose shrinks geometrically, and the total converges.

Steady-State Peaks, Troughs, and Fluctuation

For \(k>0\), concentration decreases continuously between boluses. Consequently, the maximum in a steady-state interval occurs immediately after a dose, and the minimum occurs immediately before the next dose. These are often called \(C_{\max,\mathrm{ss}}\) and \(C_{\min,\mathrm{ss}}\), respectively. Evaluating the steady-state profile at the interval endpoints gives

C_{\max,\mathrm{ss}}=\frac{D/V}{1-e^{-k\tau}}, \qquad C_{\min,\mathrm{ss}}=C_{\max,\mathrm{ss}}e^{-k\tau} =\frac{D/V}{e^{k\tau}-1}.

Their ratio follows directly from exponential decline across one interval:

\frac{C_{\max,\mathrm{ss}}}{C_{\min,\mathrm{ss}}} =\frac{1}{e^{-k\tau}}=e^{k\tau}.

The ratio depends on \(k\tau\), not on the dose or volume. Increasing the dose scales both peak and trough by the same factor under the linear model, while changing the interval or elimination rate changes the degree of fluctuation. A larger \(k\tau\) means a larger peak-to-trough ratio.

Worked Example: Steady-State Peak, Trough, and Ratio

Suppose hypothetical \(200\)-mg boluses are given every \(12\) h, with \(V=40\) L and \(k=0.100\ \text{h}^{-1}\). The model assumptions are constant parameters, instantaneous equal boluses, and linear elimination. The dose increment is \(D/V=200/40=5.00\) mg/L, and \(q=e^{-0.100(12)}=e^{-1.2}=0.3011942\).

The steady-state peak is

C_{\max,\mathrm{ss}}=\frac{5.00}{1-0.3011942} =\frac{5.00}{0.6988058}=7.15506\ \text{mg/L}.

The trough is the peak multiplied by the fraction remaining over \(12\) h:

C_{\min,\mathrm{ss}}=7.15506(0.3011942) =2.15506\ \text{mg/L}.

The peak-to-trough ratio is \(7.15506/2.15506=3.32012\). Independently, \(e^{k\tau}=e^{1.2}=3.32012\), confirming the ratio. To three significant figures, the peak is \(7.16\) mg/L and the trough is \(2.16\) mg/L; using unrounded values, their ratio is \(3.320\).

The peak exceeds the trough by \(5.00\) mg/L, equal to the dose increment \(D/V\). This is expected: the concentration falls during the interval and then increases instantaneously by \(D/V\) at the next bolus.

How Quickly Does the Profile Approach Steady State?

The finite-dose and steady-state expressions also quantify how far the concentration has progressed toward its limiting value. At a fixed time after a dose,

\frac{C_n(t)}{C_{\mathrm{ss}}(t)}=1-q^n.

The fraction of the steady-state concentration attained after \(n\) doses is therefore \(1-q^n\), and the fraction still missing is \(q^n\). This follows by dividing the finite-dose expression by the steady-state expression; the common factors \((D/V)e^{-kt}\) cancel. Because \(q\) is the fraction remaining per interval, the missing fraction is the residual contribution that would come from doses earlier than those already given.

Worked Example: Four Doses Compared With Steady State

Consider hypothetical \(80\)-mg boluses every \(12\) h, with \(V=20\) L and \(k=0.0800\ \text{h}^{-1}\). Calculate the concentration immediately after the fourth dose and compare it with the steady-state peak. The constant-dose, constant-parameter, linear one-compartment assumptions make the finite geometric sum applicable.

Here \(D/V=4.00\) mg/L and \(q=e^{-0.0800(12)}=e^{-0.96}=0.3828929\). Immediately after the fourth dose, \(t=0\), so

$$ C_4(0)=4.00(1+q+q^2+q^3) $$

Using \(q^2=e^{-1.92}=0.146607\) and \(q^3=e^{-2.88}=0.0561348\),

C_4(0)=4.00(1+0.3828929+0.146607+0.0561348) =4.00(1.585635)=6.34254\ \text{mg/L}.

The steady-state peak is \(C_{\max,\mathrm{ss}}=4.00/(1-0.3828929)=6.48185\) mg/L. The fraction attained after four doses is \(1-q^4=1-e^{-3.84}=1-0.0214936=0.978506\), or \(97.85\%\). As a check, \(6.34254/6.48185=0.978506\). The fourth-dose peak is close to, but still below, the steady-state peak.

Calculation Checks, Alternatives, and Limitations

A useful check is to calculate a concentration in two ways: sum the individual dose contributions and use the geometric-series form. They should agree before rounding. Other checks include confirming that \(q\) lies between zero and one, that a concentration declines between boluses, and that each bolus increases concentration by exactly \(D/V\) under this model. At steady state, the peak-minus-trough difference should equal \(D/V\), and the peak-to-trough ratio should equal \(e^{k\tau}\).

A recursive calculation is a practical alternative to writing the entire sum. If \(C_{\mathrm{pre},n}\) is the concentration just before dose \(n\), then the post-dose concentration is \(C_{\mathrm{pre},n}+D/V\); after one interval, the next pre-dose concentration is that post-dose value multiplied by \(e^{-k\tau}\). This recursion and the superposition formula are mathematically equivalent. A numerical simulation can also propagate concentrations over time, but the closed-form sum is simpler when the assumptions hold and provides an independent check on simulation output.

These expressions are not appropriate without modification if doses differ, intervals vary, parameters change over time, or concentration-dependent processes violate linearity. A measured profile that does not show the expected exponential decline between boluses, or that cannot be reconciled with the dose-related jump and the assumed parameters, is a reason to review dose times, concentrations, units, and model adequacy. A discrepancy is not by itself proof of a particular alternative model; data and design must be examined.

Using the wrong time since dose. Each prior dose has a different age. The \(j\)th earlier contribution is multiplied by \(e^{-k(t+j\tau)}\), not by the same exponential as the most recent dose.
Mixing pre-dose and post-dose concentrations. The current bolus contributes \(D/V\) immediately after administration, but is not included in the concentration immediately before it.
Rounding too early. Calculate with unrounded \(q\) and intermediate concentrations, then round the reported result. Rounded intermediate values can produce an incorrect final digit or an apparent mismatch in the peak-to-trough ratio.
Treating steady state as a flat concentration. Steady state means the profile repeats from interval to interval; it still declines between doses and rises at each bolus.

Key takeaway: For repeated equal IV boluses in a linear one-compartment model, add the exponentially declining contribution of every administered dose. The resulting geometric series gives finite-dose concentrations and, in the limit, the steady-state profile, accumulation factor, peak, and trough.

Check Your Understanding

Answer each question, then compare with the solutions below.

For doses given every \(\tau\) hours, what does \(q=e^{-k\tau}\) represent?
A dose has \(D/V=3.00\) mg/L and \(q=0.25\). What is its steady-state peak concentration?
For \(k=0.0800\ \text{h}^{-1}\) and \(\tau=10\) h, write the steady-state peak-to-trough ratio.
After three doses, what fraction of the steady-state concentration is attained at a fixed time after dosing, in terms of \(q\)?
Why must the current dose be included for a post-dose concentration but excluded for the immediately pre-dose concentration?

Solutions. (1) \(q\) is the fraction of concentration remaining after one interval under first-order elimination. (2) \(C_{\max,\mathrm{ss}}=3.00/(1-0.25)=4.00\) mg/L. (3) The ratio is \(e^{k\tau}=e^{0.800}=2.226\). (4) The fraction is \(1-q^3\), from \(C_n(t)/C_{\mathrm{ss}}(t)=1-q^n\). (5) An IV bolus produces an instantaneous increase of \(D/V\); immediately before administration that increase has not occurred, whereas immediately afterward it has.