Why Recursive Updates Are Useful
In an indirect response model, a drug changes the rate at which a response is produced, lost, or both. The response itself usually does not jump merely because concentration or a drug-effect multiplier changes. Instead, its rate of change changes, and the response then moves toward a new level over time. When exposure is treated as constant between successive times, the response can be updated exactly from one interval to the next.
This approach is useful when concentrations are available at discrete times, when a dosing change creates distinct exposure intervals, or when calculating a predicted response at observation times. It avoids treating each observation as a new initial condition: the endpoint of one interval is the starting response for the next.
Model and Assumptions
Let the interval boundaries be \(t_0,t_1,\ldots,t_J\), with interval duration \(\Delta_j=t_{j+1}-t_j>0\). On interval \(j\), suppose the production multiplier \(m_{\mathrm{in},j}\) and loss multiplier \(m_{\mathrm{loss},j}\) are constant. A general turnover model for response \(R(t)\) is
Here \(k_{\mathrm{in}}\) is the baseline production rate, in response units per time, and \(k_{\mathrm{out}}\) is the baseline fractional loss rate, in inverse time. The multipliers are dimensionless. A multiplier of 1 means that the corresponding rate is at baseline; a multiplier below 1 reduces that rate, while a multiplier above 1 increases it. For example, inhibition of production can be represented by \(m_{\mathrm{in},j}<1\).
The exact update below assumes \(k_{\mathrm{in}}\) and \(k_{\mathrm{out}}\) are constant, the multipliers do not vary inside an interval, and the response follows the stated turnover equation throughout that interval. The multipliers may be computed from a concentration using an effect function, such as a Hill function, but their values must be fixed over the interval for this update to apply exactly. The model also assumes no separate instantaneous jump in response at a dosing or concentration change.
Deriving the Interval Update
On interval \(j\), define the effective loss rate \(\lambda_j=k_{\mathrm{out}}m_{\mathrm{loss},j}\) and, when \(\lambda_j>0\), the interval equilibrium \(R_{\mathrm{eq},j}=k_{\mathrm{in}}m_{\mathrm{in},j}/\lambda_j\). These definitions rewrite the differential equation as
The equilibrium is the response at which production and loss balance for the fixed interval multipliers. To solve the equation, subtract that constant equilibrium from the response. Since \(R_{\mathrm{eq},j}\) is constant on this interval, the difference \(D(t)=R(t)-R_{\mathrm{eq},j}\) obeys \(dD/dt=-\lambda_jD\). Its solution is \(D(t)=D(t_j)e^{-\lambda_j(t-t_j)}\), as can be checked by differentiation and by evaluating at \(t=t_j\). Therefore, if \(R_j=R(t_j)\), then
This is an exact update under the interval assumptions, not an approximation from a small time step. If \(R_j\) is above the interval equilibrium, the exponential term shrinks the difference and the response moves downward; if it is below, the response moves upward. The amount of movement depends on both the interval duration and the effective loss rate.
If \(m_{\mathrm{loss},j}=0\), then \(\lambda_j=0\), so the equilibrium formula involving division by \(\lambda_j\) cannot be used. The differential equation instead has constant slope \(k_{\mathrm{in}}m_{\mathrm{in},j}\), giving \(R_{j+1}=R_j+k_{\mathrm{in}}m_{\mathrm{in},j}\Delta_j\). This is also the limit of the general update as \(\lambda_j\) approaches zero.
Worked Example: Updating After a Production-Effect Change
Suppose \(k_{\mathrm{in}}=3.00\) response units per hour and \(k_{\mathrm{out}}=0.400\) per hour. The initial response is \(R_0=26.66\) response units. Production is at baseline for the first 1.50 hours and is then reduced to a multiplier of 0.800 for the next 1.00 hour. Loss remains at baseline. Calculate the response at both boundaries after the initial time.
For the first interval, \(m_{\mathrm{in},0}=1\), \(m_{\mathrm{loss},0}=1\), so \(\lambda_0=(0.400)(1)=0.400\) per hour and \(R_{\mathrm{eq},0}=(3.00)(1)/0.400=7.50\) response units. The exact update is
For the second interval, the production multiplier is 0.800, so \(R_{\mathrm{eq},1}=(3.00)(0.800)/0.400=6.00\) response units. Using the unrounded first endpoint, \(R_1\approx18.01523\), the next update is
Thus the predicted response is about 18.02 response units at 1.50 hours and 14.05 response units at 2.50 hours. The production change alters the equilibrium and subsequent slope, but the response at the switching time is still the endpoint of the first interval. It is not reset to the new equilibrium of 6.00.
Changes in Loss and in Exposure-Driven Effects
The same update handles a change in loss. As long as the loss multiplier remains positive, reducing it while holding positive production fixed raises the interval equilibrium; at zero loss, there is no finite interval equilibrium. Increasing the loss multiplier makes the response approach a lower equilibrium more quickly. These are distinct effects: the multiplier affects both the equilibrium through the production-to-loss ratio and the rate of approach through \(\lambda_j\).
Worked Example: A Change in the Loss Multiplier
Let \(k_{\mathrm{in}}=8.00\) response units per hour and \(k_{\mathrm{out}}=0.500\) per hour. Start at \(R_0=16.00\) response units. Production stays at baseline. The loss multiplier is 0.600 for 1.00 hour, then 1.50 for 1.00 hour. Find the response at the end of each interval.
In the first interval, \(\lambda_0=(0.500)(0.600)=0.300\) per hour and \(R_{\mathrm{eq},0}=8.00/0.300=26.6667\) response units. Therefore,
The reduced loss rate moves the response upward toward 26.67 response units; it does not make the response jump to that value. In the second interval, \(\lambda_1=(0.500)(1.50)=0.750\) per hour and \(R_{\mathrm{eq},1}=8.00/0.750=10.6667\) response units. Carrying forward the unrounded first endpoint gives
The second interval has greater loss and a lower equilibrium, so the response falls toward that level. At the boundary, \(R(t)\) remains about 18.76 response units; only its derivative changes.
If concentration is constant within each interval, an effect function can provide the multiplier used in the update. For instance, for inhibition of production with maximum fractional inhibition \(I_{\max}\), half-maximal concentration \(IC_{50}\), and Hill coefficient \(\gamma\), one possible multiplier is \(m_{\mathrm{in}}(C)=1-I_{\max}C^\gamma/(IC_{50}^\gamma+C^\gamma)\). Once a concentration \(C_j\) is assigned to interval \(j\), evaluate the effect function to obtain that interval's multiplier, then apply the recurrence. This procedure does not require concentration itself to be a response state in the turnover equation; it only requires the assumed interval multiplier to be known.
Worked Example: Converting Interval Concentrations to Recursive Updates
Consider production inhibition with \(k_{\mathrm{in}}=10.0\) response units per hour, \(k_{\mathrm{out}}=0.500\) per hour, \(R_0=20.0\) response units, \(I_{\max}=0.800\), \(IC_{50}=2.00\) mg/L, and Hill coefficient \(\gamma=1\). Treat concentration as 2.00 mg/L over the first 1.00 hour and 6.00 mg/L over the following 2.00 hours. These are hypothetical interval concentrations. Calculate the final response.
For the first interval, the inhibition fraction is \(0.800(2.00)/(2.00+2.00)=0.400\), so the production multiplier is \(m_{\mathrm{in},0}=1-0.400=0.600\). Loss remains at baseline, giving \(\lambda_0=0.500\) per hour and \(R_{\mathrm{eq},0}=10.0(0.600)/0.500=12.0\) response units. Thus
For the second interval, the inhibition fraction is \(0.800(6.00)/(2.00+6.00)=0.600\), so \(m_{\mathrm{in},1}=0.400\), \(R_{\mathrm{eq},1}=10.0(0.400)/0.500=8.00\) response units, and the duration is 2.00 hours. Carrying the unrounded first endpoint forward,
The concentration changes at 1.00 hour, but the response there is still about 16.85 response units. The response then moves toward the lower equilibrium associated with the stronger production inhibition.
Using the Recurrence and Checking Its Behavior
A practical calculation proceeds in order: specify the interval boundaries, assign the constant multipliers for each interval, calculate that interval's effective rate and equilibrium, update the response, and use the unrounded endpoint as the next starting value. This is a recursive calculation because each update depends on the preceding one.
Use the modeled response at \(t_0\), based on the stated initial condition or preceding history.
For each interval, calculate \(\lambda_j=k_{\mathrm{out}}m_{\mathrm{loss},j}\); if \(\lambda_j>0\), calculate \(R_{\mathrm{eq},j}=k_{\mathrm{in}}m_{\mathrm{in},j}/\lambda_j\).
Apply the exact exponential update, retain adequate numerical precision, and set the endpoint equal to the next interval's starting response.
Several checks help detect errors. With nonnegative production and loss multipliers, nonnegative initial response, and positive effective loss, the update is a weighted average of \(R_j\) and \(R_{\mathrm{eq},j}\), because \(0<e^{-\lambda_j\Delta_j}<1\). It must therefore lie between those two values. If it does not, check the exponent sign, units, and interval duration. If all multipliers equal 1 and the response begins at the baseline equilibrium \(k_{\mathrm{in}}/k_{\mathrm{out}}\), the response should remain there. At a switching boundary, a computed response jump usually indicates that the new equilibrium was mistakenly substituted for the carried response.
The piecewise-constant assumption is a modeling choice, not a property guaranteed by having observations at discrete times. If concentration changes materially within an interval, holding its effect multiplier fixed may give a poor representation. One diagnostic is to compare predictions after subdividing intervals or using a more appropriate concentration profile; large changes in predicted responses indicate sensitivity to the interval approximation. The exact update is exact for the assumed constant coefficients, but cannot correct misspecified within-interval exposure.
Alternatives, Limitations, and a Common Pitfall
| Approach | What it does | Main limitation or use |
|---|---|---|
| Piecewise-constant exact update | Uses the exponential recurrence on each interval | Exact when the interval coefficients are constant; requires interval multipliers |
| Numerical integration | Approximates the response trajectory with time steps | Can handle varying inputs, but accuracy depends on step size and solver settings |
| Instantaneous equilibrium approximation | Sets the response equal to the current interval equilibrium | Neglects turnover delay and is generally poor when intervals are not long relative to the response time scale |
The recursive method is especially convenient for predictions at interval boundaries and for verifying calculations from a numerical solver under constant coefficients. It provides the response at those boundaries directly, but does not by itself describe a concentration that varies continuously within an interval. It also relies on the chosen turnover model and known multipliers; exact arithmetic under a model does not establish that the model is biologically adequate.
Common mistake: A learner may calculate a new interval equilibrium at a dosing change and set the response immediately to that value. This confuses the balance point under the new rates with the response already present at the switching time. The correction is to keep the preceding interval's endpoint as the next starting value, then use the new equilibrium only in the next exponential update. An instantaneous response jump would require a separate, explicitly modeled mechanism.
Check Your Understanding
Use the interval update and show the key calculation for each problem.
- For \(k_{\mathrm{in}}=4.00\) response units per hour, \(k_{\mathrm{out}}=0.500\) per hour, \(m_{\mathrm{in}}=0.500\), and \(m_{\mathrm{loss}}=1\), find the interval equilibrium.
- If \(R_j=10.0\) response units, \(R_{\mathrm{eq},j}=20.0\) response units, and \(\lambda_j\Delta_j=0.500\), calculate \(R_{j+1}\).
- At a boundary, the response is 12.0 response units and the new interval equilibrium is 8.00 response units. Does the response instantly become 8.00 under this model? Explain.
- What update applies if the effective loss rate is zero but production is constant at 3.00 response units per hour for 2.00 hours, starting from 5.00 response units?
- State one reason the exact recurrence may not accurately represent a long interval with changing concentration.
Solutions. (1) The effective loss rate is \(0.500(1)=0.500\) per hour, so the equilibrium is \(4.00(0.500)/0.500=4.00\) response units. (2) \(R_{j+1}=20.0+(10.0-20.0)e^{-0.500}=13.93\) response units. (3) No. The response remains 12.0 response units at the boundary; 8.00 is the new equilibrium toward which it subsequently moves. (4) With zero loss, \(R_{j+1}=R_j+k_{\mathrm{in}}m_{\mathrm{in}}\Delta=5.00+(3.00)(2.00)=11.00\) response units. (5) A single fixed multiplier may poorly represent the effect of concentration changes inside the interval, so the exact update for that fixed multiplier may not match the trajectory under the actual varying exposure.