Why Interpolation Matters
Plasma concentrations are usually observed at discrete sampling times, but the effect compartment responds continuously. To calculate an effect-site trajectory between observations, one must specify what \(C_p(t)\) does between them. That choice is not merely a plotting convention: the effect-site concentration is a weighted history of plasma exposure, so changing the assumed curve can change both the calculated \(C_e(t)\) and the \(K_{e0}\) needed to match observed response information.
The effect-compartment equation is \(dC_e(t)/dt=K_{e0}[C_p(t)-C_e(t)]\), where \(K_{e0}>0\) has units of inverse time. Over an interval of length \(\Delta\), the solution from \(C_e(0)\) is
This expression separates the contribution of the starting effect-site concentration from the weighted plasma input during the interval. Plasma values closer to the end of the interval receive greater weight because they have had less time to equilibrate away. The formulas below assume \(K_{e0}\) is constant over the interval, plasma concentrations are nonnegative, and the chosen interpolation or PK model represents the unobserved concentration course adequately.
Updates Under Linear and Log-Linear Interpolation
Let the interval begin at time 0, with plasma concentrations \(C_{p,0}\) and \(C_{p,1}\) at times 0 and \(\Delta\). Under linear interpolation, \(C_p(s)=C_{p,0}+ms\), where \(m=(C_{p,1}-C_{p,0})/\Delta\). Substituting this expression into the effect-compartment solution gives
The last term follows by integrating the contribution of the slope: \(K_{e0}m\int_0^\Delta s e^{-K_{e0}(\Delta-s)}ds=m[\Delta-(1-e^{-K_{e0}\Delta})/K_{e0}]\). Thus the update is exact for the assumed straight-line plasma curve; it is not an approximation to the differential equation solution.
For log-linear interpolation, write \(C_p(s)=C_{p,0}e^{as}\), where \(a=\log(C_{p,1}/C_{p,0})/\Delta\), provided both endpoint concentrations are positive. Integrating the exponential input gives
If \(K_{e0}+a=0\), the fraction is evaluated by its limit, \( \Delta e^{-K_{e0}\Delta}\). The log-linear curve is exactly a single exponential over the interval. It is not generally the same curve as a straight line between the same observations.
Worked Example: Two Interpolations, Two Effect-Site Values
Suppose, hypothetically, that \(C_{p,0}=8.00\) mg/L and \(C_{p,1}=4.00\) mg/L two hours later. Let \(C_e(0)=3.00\) mg/L and \(K_{e0}=0.500\) h\(^{-1}\). Calculate the ending effect-site concentration under each endpoint interpolation.
Linear interpolation. Here \(\Delta=2\) h, \(m=(4-8)/2=-2.00\) mg/L/h, and \(e^{-K_{e0}\Delta}=e^{-1}=0.3679\). Substitution gives
Log-linear interpolation. The exponential rate is \(a=\log(4/8)/2=-0.3466\) h\(^{-1}\). Thus
The two answers differ by about 0.141 mg/L despite using the same endpoints and \(K_{e0}\). During a decline, the log-linear curve falls more quickly at first than the straight line in this example. Because effect-site concentration depends on the time course, not just the endpoints, the calculated values differ.
How Interpolation Changes an Estimated \(K_{e0}\)
When \(K_{e0}\) is estimated from concentration-effect information, the assumed plasma curve enters every interval update. An estimate obtained under one interpolation rule is conditional on that rule. The following calculation isolates this effect: imagine that effect-site concentrations at the interval endpoints are known for the calculation. In an actual analysis, they may instead be inferred through a response model, which adds further uncertainty and dependence on model assumptions.
Worked Example: Estimating \(K_{e0}\) Under Two Curves
Suppose, hypothetically, plasma concentration rises from 2.00 to 6.00 mg/L over one hour, starting \(C_e(0)=2.00\) mg/L. An independently specified ending effect-site value is \(C_e(1)=2.852\) mg/L. Find the \(K_{e0}\) that matches the endpoint under linear and log-linear interpolation.
For the linear curve, \(m=4.00\) mg/L/h. Substituting \(K_{e0}=0.500\) h\(^{-1}\) into the linear update gives
Therefore \(K_{e0}=0.500\) h\(^{-1}\) matches the stipulated endpoint under linear interpolation (to the displayed precision).
For the log-linear curve, \(a=\log(6/2)=1.0986\) h\(^{-1}\), and the endpoint update is
This equation is solved numerically because \(K_{e0}\) appears both in the exponential and elsewhere in the expression. At \(K_{e0}=0.620\) h\(^{-1}\), the right side is approximately 2.852 mg/L, so the log-linear estimate is about 0.620 h\(^{-1}\). The two estimates differ because the log-linear curve lies below the straight line during this rising interval. A larger equilibration rate is needed to match the same ending \(C_e\).
Model-Based Interpolation
If a suitable PK model is available, its predicted \(C_p(s)\) can be substituted into the effect-compartment integral. For a sum of exponential terms, \(C_p(s)=\sum_j A_j e^{-\lambda_j s}\), the update has a closed form:
When \(K_{e0}=\lambda_j\), the corresponding fraction is replaced by its limit, \(\Delta e^{-K_{e0}\Delta}\). More complex models can be evaluated by numerical integration. A model-based curve can incorporate the shape implied by absorption, distribution, and elimination, but it is only as appropriate as the PK model and its parameter estimates.
Worked Example: A Two-Exponential PK Curve
Consider a hypothetical plasma profile \(C_p(s)=6e^{-0.200s}+2e^{-1.00s}\) mg/L, with time in hours. Over two hours, it falls from 8.00 mg/L to \(6e^{-0.400}+2e^{-2.00}=4.293\) mg/L. Let \(C_e(0)=2.00\) mg/L and \(K_{e0}=0.500\) h\(^{-1}\).
The model-based update is
For comparison, linear interpolation between the same endpoints gives approximately 4.429 mg/L using the linear update. Log-linear interpolation gives approximately 4.311 mg/L, using \(a=\log(4.293/8)/2\approx-0.3113\) h\(^{-1}\). The model-based result differs because the two-exponential curve has a different shape from either endpoint-only curve. It is not inherently superior: its advantage depends on whether the PK model represents the plasma profile adequately.
Choosing, Checking, and Comparing the Approaches
The methods answer different modeling questions. Linear interpolation is transparent and handles zero endpoint concentrations, but a straight segment may poorly describe rapid changes. Log-linear interpolation is appropriate when concentration changes approximately exponentially within an interval, but it requires positive endpoints and can be misleading across absorption peaks or changing disposition phases. Model-based interpolation uses more PK structure and can link intervals coherently, but may impose structural assumptions and transmit PK parameter error into the effect-site calculation.
| Approach | Curve between observations | Useful when | Key caution |
|---|---|---|---|
| Linear | Straight line in concentration | Changes are gradual or only endpoint information is available | Can smooth over curvature or a peak |
| Log-linear | Exponential change in concentration | Exponential decline or rise is plausible in the interval | Undefined at a zero endpoint; may misrepresent mixed processes |
| Model-based | Predictions from a specified PK model | A supported PK model describes the observed profile | Model misspecification or uncertain parameters affect the input curve |
A useful check is to evaluate the interpolated plasma curve at intermediate times and confirm that it is plausible given the sampling schedule, dose timing, and observed data. Also verify that the calculated \(C_e\) remains nonnegative when inputs and initial conditions are nonnegative, and that \(C_e\) moves toward sustained \(C_p\) rather than away from it. A calculated \(C_e\) need not lie between the two plasma endpoints: it reflects prior exposure and can lag a rapidly changing \(C_p\).
Compare estimates under reasonable alternative input curves when observations are sparse or intervals span likely peaks, rapid absorption, or changes in elimination phase. A large shift in the \(K_{e0}\) estimate indicates that the data do not determine the result independently of the interpolation assumption. More frequent plasma sampling around rapid changes can reduce this dependence. Do not treat different interpolation-derived estimates as independent evidence for different physiology; they may simply reflect different assumed inputs.
Common Pitfall
Another practical error is to use the log-linear formula when an endpoint is zero or negative. The logarithm of the concentration ratio is then undefined. Use a defensible alternative representation, such as a suitable PK model or a justified linear segment, rather than silently adding an arbitrary constant to make the logarithm possible.
Check Your Understanding
Use the interval formulas and reasoning above. Answers with calculations are included below for self-checking.
- For positive plasma endpoints, what assumption makes log-linear interpolation exponential rather than straight in concentration?
- Why can two curves with identical endpoint concentrations produce different \(C_e(\Delta)\)?
- Under the assumptions of the worked estimation example, why was the log-linear \(K_{e0}\) estimate larger than the linear estimate?
- Can log-linear interpolation be used directly when one endpoint concentration is zero? Explain.
Solutions. (1) It assumes log concentration changes linearly with time, so \(C_p(s)=C_{p,0}e^{as}\). (2) The effect-site update integrates the entire intervening plasma curve, not only its endpoints; the exponential weighting gives different contributions to different times. (3) The rising log-linear curve in that example lies below the linear curve between the endpoints, so a larger \(K_{e0}\) was needed to reach the same specified ending effect-site concentration. (4) No. The logarithm of zero is undefined, so the log-linear rate cannot be calculated from those endpoints; another justified curve or a suitable PK model is required.