What Washout Changes
During treatment, an indirect-response model describes a response altered through production, loss, or both. After the drug effect has been removed, the response does not generally jump back to baseline. It continues from its value at washout and returns toward baseline according to the untreated production and loss rates. The key distinction is between the response level at washout, which depends on the preceding treatment history, and the recovery rate after washout, which has a simple form under complete removal of the effect.
Let \(R(t)\) denote the response, \(R_0\) its untreated baseline, and \(k_{\mathrm{out}}\) the untreated response-loss rate constant. At baseline equilibrium, \(k_{\mathrm{in}}=k_{\mathrm{out}}R_0\), where \(k_{\mathrm{in}}\) is the untreated production rate. Let \(t_w\) be the time of complete washout and \(R_w=R(t_w)\) the response at that time. Complete washout here means that the drug effect on production or loss is zero for all \(t\geq t_w\). This is a model assumption; it need not mean that the drug concentration is literally zero.
The Same Recovery Equation for Four Mechanisms
The four canonical models differ in how a drug changes the response during exposure. For illustration, let \(I\) denote a constant fractional inhibition and \(S\) a constant fractional stimulation, with \(0\leq I<1\) and \(S\geq0\). Under these effects, their drug-on equations are as follows. The equations also show why the response at washout can differ substantially by mechanism and treatment duration.
| Drug effect during treatment | Drug-on equation |
|---|---|
| Inhibition of production | \(dR/dt=k_{\mathrm{in}}(1-I)-k_{\mathrm{out}}R\) |
| Inhibition of loss | \(dR/dt=k_{\mathrm{in}}-k_{\mathrm{out}}(1-I)R\) |
| Stimulation of production | \(dR/dt=k_{\mathrm{in}}(1+S)-k_{\mathrm{out}}R\) |
| Stimulation of loss | \(dR/dt=k_{\mathrm{in}}-k_{\mathrm{out}}(1+S)R\) |
At complete washout, \(I=0\) and \(S=0\). In each case, the equation therefore reduces to \(dR/dt=k_{\mathrm{in}}-k_{\mathrm{out}}R\). Subtract the baseline equilibrium equation \(0=k_{\mathrm{in}}-k_{\mathrm{out}}R_0\), and define the deviation \(D(t)=R(t)-R_0\). Then \(dD/dt=-k_{\mathrm{out}}D\), with \(D(t_w)=R_w-R_0\). The resulting recovery equation is common to all four mechanisms. The mechanism matters for determining \(R_w\); after full washout, it does not alter the recovery rate in this model.
Derivation and Conditions
The deviation equation can be solved with an integrating factor. This method also covers the special case \(R_w=R_0\), when the initial deviation is zero.
Since \(k_{\mathrm{in}}=k_{\mathrm{out}}R_0\), subtracting the baseline equation from the untreated response equation gives \(dD/dt=-k_{\mathrm{out}}D\).
Multiply by \(e^{k_{\mathrm{out}}t}\). The product rule gives \(d(e^{k_{\mathrm{out}}t}D)/dt=e^{k_{\mathrm{out}}t}(dD/dt+k_{\mathrm{out}}D)=0\).
The product \(e^{k_{\mathrm{out}}t}D(t)\) is constant, so \(e^{k_{\mathrm{out}}t}D(t)=e^{k_{\mathrm{out}}t_w}D(t_w)\). Rearrangement gives the recovery trajectory below.
This derivation requires constant untreated \(k_{\mathrm{in}}\) and \(k_{\mathrm{out}}\) after washout, baseline equilibrium under those rates, and complete removal of the drug effect at \(t_w\). It also assumes no additional input, disease progression, or other process changes the response dynamics after washout. If \(R_w=R_0\), the equation gives \(R(t)=R_0\) at all subsequent times; no division by the initial deviation is needed.
Worked Recovery Calculations
Worked Example: Stimulation of Production Followed by Washout
Suppose the untreated baseline is \(R_0=50.0\) response units and \(k_{\mathrm{out}}=0.200\ \mathrm{h}^{-1}\). A constant stimulation of production raises the drug-on equilibrium to \(125\) response units. The response starts at baseline, and the drug effect is completely removed after \(1.20\) h. Calculate the response at washout and 5.00 h later.
During stimulation of production, \(dR/dt=k_{\mathrm{in}}(1+S)-k_{\mathrm{out}}R\), so the drug-on equilibrium is \(125\) response units and the response approaches it at rate \(k_{\mathrm{out}}\). Starting from 50.0, the washout value is
Thus \(R_w=66.00\) response units to two decimal places. After washout, the production stimulation is zero, so the untreated recovery equation applies. At 5.00 h after washout,
The response is therefore \(55.89\) response units 5.00 h after washout. It has moved toward baseline but has not reached it. The calculation uses the unrounded washout value until the final result.
Worked Example: Inhibition of Loss Before Washout
Suppose \(R_0=60.0\) response units and \(k_{\mathrm{out}}=0.150\ \mathrm{h}^{-1}\). A constant inhibition of loss has \(I=0.600\) for 4.00 h, starting from baseline. At washout, the inhibition disappears completely. Find \(R_w\) and the response 2.00 h later.
During treatment, the loss rate is \(k_{\mathrm{out}}(1-I)=0.150(0.400)=0.0600\ \mathrm{h}^{-1}\). The drug-on equilibrium is \(k_{\mathrm{in}}/[k_{\mathrm{out}}(1-I)]=R_0/(1-I)=150\) response units. Starting from \(R_0\), the response at washout is
After washout, use the untreated \(k_{\mathrm{out}}\), not the smaller drug-on loss rate. Two hours later,
The response is \(74.23\) response units. The drug-on inhibited-loss rate determines the preceding response and hence \(R_w\); following complete washout, the deviation decays at \(0.150\ \mathrm{h}^{-1}\).
Worked Example: Comparing the Four Mechanisms After Washout
Consider four hypothetical treatment scenarios with the same untreated baseline, \(R_0=100\) response units, and \(k_{\mathrm{out}}=0.100\ \mathrm{h}^{-1}\). Suppose the measured responses at complete washout are 80, 140, 130, and 70 response units for inhibition of production, inhibition of loss, stimulation of production, and stimulation of loss, respectively. Calculate each response 10.0 h later.
For any of the four scenarios, \(e^{-(0.100)(10.0)}=e^{-1}=0.3678794\). Applying the same recovery equation separately:
- Inhibition of production: \(100+(80-100)(0.3678794)=92.6424\), or \(92.64\) response units.
- Inhibition of loss: \(100+(140-100)(0.3678794)=114.7152\), or \(114.7\) response units.
- Stimulation of production: \(100+(130-100)(0.3678794)=111.0364\), or \(111.0\) response units.
- Stimulation of loss: \(100+(70-100)(0.3678794)=88.9636\), or \(88.96\) response units.
The absolute responses differ because their washout values differ. Each scenario retains the same proportion, \(0.3679\), of its initial deviation from baseline after 10.0 h. These washout values are hypothetical inputs to the recovery calculation; no claim about a particular drug or study is implied.
When the Untreated Rate Does and Does Not Apply
Under the stated assumptions, recovery follows the untreated \(k_{\mathrm{out}}\) rate constant for all four canonical mechanisms. This is true whether \(R_w\) is above or below baseline, and also when it equals baseline. For \(R_w>R_0\), the response declines toward baseline; for \(R_w<R_0\), it rises toward baseline. In either case, the signed difference \(R(t)-R_0\) is multiplied by \(e^{-k_{\mathrm{out}}(t-t_w)}\).
The result changes if the drug effect has not actually disappeared. For example, a persistent inhibition of production leaves the production rate below baseline; a persistent inhibition or stimulation of loss leaves the loss-rate coefficient different from \(k_{\mathrm{out}}\). The resulting equation is not the untreated recovery equation. If drug concentration declines gradually after dosing stops, the effect may also change over time rather than vanish at one instant. In those cases, treating \(t_w\) as complete washout and fitting a single untreated exponential can misstate the recovery path.
The post-washout formula also depends on the response state being described by the model. A delayed change in another process, a changing baseline, or an added input can make the response depart from the predicted trajectory even if dosing has stopped. The introductory material on indirect-response state equations establishes the production-and-loss structure; for this recovery result, the relevant check is whether that structure has returned to its untreated form.
Diagnostics and Common Pitfalls
A useful diagnostic is to plot the signed deviation from baseline, \(R(t)-R_0\), against time since washout. Under the model, the deviation retains its sign and its magnitude shrinks exponentially. If the deviation crosses baseline, grows in magnitude, or has a clear changing decay pattern, at least one assumption may be inadequate. Possible explanations include residual drug effect, an incorrect baseline, measurement noise, or a response process not represented by the model. Such patterns do not by themselves identify which explanation is responsible.
A common mistake is to use the drug-on rate constant for post-washout recovery because that rate governed the response just before dosing stopped. That is incorrect when washout restores the untreated model: inhibition or stimulation of loss may have altered the drug-on rate, but after complete removal of that effect the coefficient is again \(k_{\mathrm{out}}\). Another mistake is to say that the response reaches baseline at a finite time. For a nonzero initial deviation, the exponential approaches zero as time increases; it does not equal zero at a finite time.
| Situation after dosing stops | Recovery implication |
|---|---|
| Drug effect is zero; untreated rates and baseline remain constant | The deviation follows \(e^{-k_{\mathrm{out}}(t-t_w)}\). |
| Drug effect persists at a constant level | The response generally follows a drug-modified equation, not untreated recovery. |
| Drug effect changes gradually as concentration declines | A single constant-rate recovery equation may not describe the trajectory. |
| Baseline or other production/loss processes change | The assumed untreated equilibrium and recovery equation may no longer apply. |
The principal alternative to a single untreated exponential is to represent the post-dose drug effect explicitly, using the relevant concentration-dependent production or loss term. That approach can be more appropriate when exposure or effect persists, but it requires specifying how the effect evolves after dosing. The simple washout equation is valuable when complete effect removal is defensible; it should not be used merely because administration has ended.
Check Your Understanding
Answer each question, then compare with the solutions below.
- After complete washout, \(R_0=40\) response units, \(R_w=70\) response units, and \(k_{\mathrm{out}}=0.100\ \mathrm{h}^{-1}\). What is the predicted response 10.0 h later?
- What does the model predict if the response is exactly at baseline at washout?
- Why does inhibition of loss during treatment not imply that the inhibited loss rate governs recovery after complete washout?
- A response crosses from below baseline to above baseline after washout. Is that trajectory consistent with the stated recovery equation? Give one possible reason it might be observed.
Solutions. 1. \(R=40+(70-40)e^{-0.100(10.0)}=40+30(0.3678794)=51.0364\), or \(51.04\) response units. 2. If \(R_w=R_0\), the initial deviation is zero and the equation predicts \(R(t)=R_0\) for all \(t\geq t_w\). 3. Complete washout removes the inhibition, restoring the untreated loss coefficient \(k_{\mathrm{out}}\); the drug-on coefficient applies only while inhibition remains. 4. No. With a fixed baseline, constant untreated rates, and complete washout, the signed deviation is multiplied by a positive exponential and cannot change sign. A changing baseline or a residual or additional effect could contribute to such a pattern.