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Tolerance, Rebound and Adaptation · Tutorial 189 of 598

Discrete Recurrence Models for Repeated-Dose Tachyphylaxis

Learn to track response loss across bolus doses, predict peak responses under different intervals, and interpret the recurrence’s long-run behavior.

Advanced 9 min read

What You'll Learn

  • Define a bounded tolerance state and distinguish its pre-dose and post-dose values.
  • Derive the repeated-dose recurrence from a bolus induction rule and exponential recovery.
  • Calculate peak responses over successive doses with explicit arithmetic.
  • Compare how dose intervals affect accumulated response loss.
  • Derive and interpret the periodic steady state under a constant schedule.
  • Recognize assumptions, limitations, and signs that the recurrence is inadequate.

From a Bolus Dose to a Dose-to-Dose Model

Tachyphylaxis is a rapid reduction in response during closely spaced drug administration. If the drug produces a similar exposure after each bolus but the peak response declines, a useful first model tracks a pharmacodynamic state that accumulates with dosing and recovers between doses. This gives a compact way to compare schedules without solving a continuous drug-concentration model anew at every dose.

The recurrence developed here represents one specific mechanism: each dose increases a bounded response-loss state, and that state then recovers exponentially until the next dose. The response at a dose is reduced in proportion to the state immediately before that dose. This is a modeling choice, not a universal definition of tolerance. The earlier material on quantifying tolerance as potency or efficacy change provides the broader distinction between those response changes; here, the focus is on tracking response loss dose by dose.

Definition: Let \(x_n\) be the fraction of response capacity lost immediately before dose \(n\), so \(0\leq x_n\leq 1\). Let \(x_n^+\) be the state immediately after that dose. Let \(q_n\) be the fraction of the remaining response capacity removed by dose \(n\), and let \(a_n\) be the fraction of the post-dose state that remains at the next dose. For a dose interval \(\tau_n\) and first-order recovery rate \(k_r\), \(a_n=\exp(-k_r\tau_n)\). Thus \(q_n\) and \(a_n\) are dimensionless, \(k_r\) has units of inverse time, and \(\tau_n\) has units of time.

The construction separates two events. The bolus produces an instantaneous increase in response loss, while recovery occurs during the interval that follows. The response loss is assumed to act as a multiplicative efficacy reduction: if \(E_0(D_n)\) is the peak response expected at dose \(D_n\) in the absence of this adaptation, the modeled peak response is \(R_n=E_0(D_n)(1-x_n)\). This expression assumes the same baseline exposure-response relation applies at each dose; it does not by itself describe changing concentrations or a shift in potency.

Deriving the Recurrence

At the bolus, the dose removes a fraction \(q_n\) of the response capacity that remains. Before the dose, the remaining capacity is \(1-x_n\). The dose-induced increase in loss is therefore \(q_n(1-x_n)\), giving the post-dose state

$$ x_n^+ = x_n+q_n(1-x_n)=q_n+(1-q_n)x_n. $$

Between doses, assume that response loss decays by first-order recovery, \(dx/dt=-k_r x\), with no further input. Over an interval \(\tau_n\), the solution is \(x(t+\tau_n)=x(t)\exp(-k_r\tau_n)\). Applying this to \(x_n^+\) gives the pre-dose state at the next bolus:

$$ x_{n+1}=a_n x_n^+ =a_n\left[q_n+(1-q_n)x_n\right], \qquad a_n=\exp(-k_r\tau_n). $$

This is the dose-to-dose recurrence. It requires \(0\leq q_n\leq 1\), \(k_r\geq 0\), and \(\tau_n\geq 0\), which ensure \(0\leq a_n\leq 1\). With an initial state in \([0,1]\), the post-dose and next pre-dose states also remain in \([0,1]\). In a simple application, the same bolus dose is repeated, so \(q_n=q\); the intervals may nevertheless vary.

Assumptions: The state is bounded between zero and one; each dose produces the specified instantaneous induction; recovery is first-order and acts only between doses; and there are no unmodeled inputs to the state. To interpret calculated peak responses, also assume the baseline peak \(E_0(D_n)\) is known or held constant and that the state scales it as \(1-x_n\). If exposure or the baseline response changes across doses, that change must not be mistaken for tachyphylaxis in this model.

Worked Example: Calculating Peak Responses Across Repeated Doses

Suppose a hypothetical bolus regimen has the same dose at each administration, with \(q=0.25\), and the fraction of post-dose response loss remaining at the next dose is \(a=0.80\). Assume the initial pre-dose state is \(x_1=0\), and the baseline peak response is \(E_0=80\) response units at every dose. Calculate the first four pre-dose states and peak responses.

For dose 1, \(x_1=0\), so \(R_1=80(1-0)=80\) response units. Immediately after the dose, \(x_1^+=0+0.25(1-0)=0.25\). Recovery to dose 2 gives \(x_2=0.80(0.25)=0.20\), so \(R_2=80(1-0.20)=64\) response units.

At dose 2, \(x_2^+=0.20+0.25(1-0.20)=0.20+0.20=0.40\). Thus \(x_3=0.80(0.40)=0.32\), and \(R_3=80(1-0.32)=54.4\) response units.

At dose 3, \(x_3^+=0.32+0.25(1-0.32)=0.32+0.17=0.49\). Therefore \(x_4=0.80(0.49)=0.392\), and \(R_4=80(1-0.392)=48.64\) response units. The calculations show a declining peak even though the dose and baseline response are held fixed. They are predictions under the stated model, not evidence that this mechanism is present in any particular clinical setting.

Constant Schedules and Their Long-Run Behavior

For a constant dose effect \(q\) and constant interval, \(a=\exp(-k_r\tau)\) is also constant. The recurrence can then be written as an affine map:

$$ x_{n+1}=aq+a(1-q)x_n. $$

The steady state \(x^*\) is the state that is unchanged from one pre-dose point to the next. Setting \(x_{n+1}=x_n=x^*\) gives \(x^*=aq+a(1-q)x^*\). Rearranging yields \(x^*[1-a(1-q)]=aq\). Since \(0\leq a\leq 1\) and \(0\leq q\leq 1\), the denominator \(1-a(1-q)\) is positive except when \(a=1\) and \(q=0\). For the nondegenerate case,

$$ x^*=\frac{aq}{1-a(1-q)}. $$

This fixed point is stable when \(a(1-q)<1\): the difference from the fixed point follows \(x_{n+1}-x^*=a(1-q)(x_n-x^*)\), so each iteration contracts that difference. If the baseline peak is constant, the predicted long-run peak response is \(E_0(1-x^*)\). The steady state is a periodic-dose prediction, not a claim that recovery has stopped; the state continues to rise after each dose and fall between doses.

Worked Example: Predicting the Long-Run Peak Under a Fixed Schedule

Use \(q=0.25\) and \(a=0.80\), with a baseline peak response of 80 response units. Find the steady-state pre-dose loss and the corresponding peak response.

Substitute into the fixed-point expression:

$$ x^*=\frac{(0.80)(0.25)}{1-(0.80)(1-0.25)} =\frac{0.20}{1-0.60} =\frac{0.20}{0.40} =0.50. $$

The predicted long-run peak is \(80(1-0.50)=40\) response units. The convergence can also be checked from the recurrence multiplier \(a(1-q)=(0.80)(0.75)=0.60\), which is less than one. Starting from \(x_1=0\), the pre-dose states approach 0.50: for example, \(x_2=0.20\), \(x_3=0.32\), and \(x_4=0.392\). Each difference from 0.50 is multiplied by 0.60 at the next step.

Comparing Dose Intervals

For a given recovery rate, a longer interval means a smaller \(a=\exp(-k_r\tau)\), so less response loss persists until the next dose. This generally reduces accumulation when the dose-induced fraction \(q\) is unchanged. The comparison below isolates the interval effect by holding the dose effect and baseline peak response fixed.

Worked Example: Comparing Two Intervals at the Third Dose

Consider two hypothetical schedules with \(q=0.25\), initial state \(x_1=0\), and baseline peak response \(E_0=80\) response units. Schedule A has \(a=0.80\) between doses; Schedule B has \(a=0.50\). Calculate each third-dose peak.

For Schedule A, the first post-dose state is \(x_1^+=0.25\), so \(x_2=(0.80)(0.25)=0.20\). The second post-dose state is \(x_2^+=0.20+0.25(0.80)=0.40\), giving \(x_3=(0.80)(0.40)=0.32\). The third-dose peak is \(80(1-0.32)=54.4\) response units.

For Schedule B, \(x_2=(0.50)(0.25)=0.125\). Then \(x_2^+=0.125+0.25(1-0.125)=0.125+0.21875=0.34375\), so \(x_3=(0.50)(0.34375)=0.171875\). The third-dose peak is \(80(1-0.171875)=66.25\) response units.

The smaller persistence factor in Schedule B produces less pre-dose loss and a larger predicted third-dose peak. This is a comparison at the same dose number, not at the same elapsed time: the schedules differ in their intervals. It does not establish that lengthening an interval will have this effect in a real treatment setting, where exposure, disease state, or other mechanisms may also change.

Using the Model Responsibly

For a variable schedule, calculate \(a_n=\exp(-k_r\tau_n)\) separately for each interval and apply \(x_{n+1}=a_n[q_n+(1-q_n)x_n]\) in order. The same logic handles different bolus effects \(q_n\), provided each \(q_n\) is specified on the model’s scale. A useful implementation check is to compute both the post-dose update and the subsequent recovery step explicitly for at least a few doses before automating the sequence.

The recurrence has several simplifying alternatives, each answering a somewhat different modeling need. A continuous-time state-variable model can represent induction throughout exposure rather than as an instantaneous jump, which may be more suitable when drug concentrations change gradually or the adaptation develops over the dosing interval. A richer PK/PD model can calculate concentration-dependent responses rather than treating \(E_0(D_n)\) as given. A potency-change formulation can represent a rightward shift in an exposure-response relation instead of the multiplicative efficacy loss used here. The discrete recurrence is attractive when doses are well separated as identifiable events and the main question is how the state carries from one dose to the next; it is not a substitute for those alternatives when their mechanisms or time courses matter.

Diagnostics should focus on whether the recurrence’s pattern is compatible with the observations and design. With comparable exposure at each dose, systematic peak-response residuals that change with dose number or interval may indicate that the induction or recovery rule is too simple. A state inferred outside \([0,1]\), or a predicted response below zero under this response scaling, signals an implementation error or incompatible model assumptions. If peaks differ alongside changes in exposure, the data do not by themselves distinguish pharmacokinetic changes from pharmacodynamic response loss. Sparse observations, uncertain baseline responses, and variation between individuals can also make the apparent dose-to-dose pattern misleading.

Common pitfall: It is tempting to apply the recovery factor \(a_n\) to the pre-dose state \(x_n\) and then add \(q_n\). That reverses the event order and gives \(x_{n+1}=a_nx_n+q_n\), which does not represent a dose that removes a fraction of the response capacity remaining at administration. The specified sequence is induction first, \(x_n^+=x_n+q_n(1-x_n)\), then recovery, \(x_{n+1}=a_nx_n^+\). Check the timing before implementing any recurrence.

The model’s scope is deliberately narrow. It assumes a single state, exponential recovery, a dose effect expressed as a fraction of remaining capacity, and a direct multiplicative link from pre-dose state to peak response. It does not describe post-dose response time courses, rebound after discontinuation, between-person variability, or the estimation of recovery parameters from washout observations. Its value is that the assumptions are explicit and the schedule predictions are transparent; when the observed behavior conflicts with them, the model should be revised rather than interpreted literally.

Check Your Understanding

Answer the questions using the definitions and assumptions in this tutorial. Brief solutions are provided after the questions.

  1. If \(q=0.20\), \(a=0.75\), and \(x_n=0.30\), calculate \(x_n^+\) and \(x_{n+1}\).
  2. With \(q=0.20\) and \(a=0.75\), calculate the constant-schedule steady-state pre-dose loss.
  3. Why does a longer interval reduce \(a\), when \(k_r>0\)?
  4. What is wrong with updating the state by \(x_{n+1}=a x_n+q\) under the induction rule used here?

Solutions. (1) \(x_n^+=0.30+0.20(1-0.30)=0.44\), and \(x_{n+1}=0.75(0.44)=0.33\). (2) \(x^*=(0.75)(0.20)/[1-(0.75)(0.80)]=0.15/0.40=0.375\). (3) Since \(a=\exp(-k_r\tau)\), increasing \(\tau\) decreases the exponential when \(k_r>0\), so a smaller fraction of the post-dose loss remains. (4) It adds \(q\) without scaling induction by the remaining capacity \(1-x_n\), and it places the dose update after recovery rather than applying the dose first and recovery second.

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