Why Identifiability Depends on the Observations
A one-compartment IV bolus model has only two disposition parameters, but the number of parameters alone does not tell us whether the data determine them. The observation times matter, as does whether the dose or one of the parameters is already known. The key question is whether distinct parameter values can produce exactly the same ideal concentration observations.
The introductory material on structural and practical identifiability distinguishes uniqueness under an ideal model from the precision achievable with finite, noisy data. Here we apply that distinction to the one-compartment IV bolus model, assuming a known dose and linear elimination.
Let \(A(t)\) be the amount of drug in the compartment at time \(t\), and let \(C(t)=A(t)/V\). The amount-balance equation and initial condition are
Solving this first-order equation gives the concentration-time profile. Define the elimination rate constant \(k=CL/V\), with units of inverse time.
Thus the profile contains an amplitude \(D/V\) and a decay rate \(k=CL/V\). Known dose allows the amplitude to determine volume; once volume and the rate are known, clearance follows. But a single concentration at a positive time generally does not separately reveal both amplitude and rate.
What the Model Identifies
With known dose, the initial concentration is \(C(0)=D/V\). Therefore, an actual observation at time zero identifies \(V\), and a later positive concentration then identifies \(k\), hence \(CL\). If there is no time-zero observation, two concentrations at distinct times can identify both parameters: their ratio determines \(k\), and either concentration then determines \(V\).
For positive concentrations under this model, the ratio formula gives a positive elimination rate when \(C(t_2)<C(t_1)\). The recovered amplitude and clearance are then
These equations establish uniqueness for two distinct-time, noiseless observations when \(D\) is known and the observations follow the assumed model. They do not guarantee that estimates from noisy observations will be precise.
Worked Identifiability Cases
Worked Example: An Initial and a Later Concentration
Suppose a hypothetical known dose of \(400\) mg produces an ideal initial concentration \(C(0)=8.00\) mg/L and, at \(4.00\) h, \(C(4)=8e^{-0.2}=6.54985\) mg/L. Find \(V\), \(k\), and \(CL\).
The initial concentration gives the volume directly:
The concentration ratio determines the elimination rate:
Therefore, \(CL=kV=(0.05000\ \text{h}^{-1})(50.0\ \text{L})=2.500\) L/h. The dose, initial concentration, and later concentration are mutually consistent with the stated profile.
Worked Example: Two Positive-Time Concentrations
A hypothetical known dose is \(500\) mg. Exact model concentrations are \(C(2)=12.5e^{-0.2}=10.2312\) mg/L and \(C(7)=12.5e^{-0.7}=6.20731\) mg/L, with times in hours. Neither observation is at time zero. Determine the parameters.
The observations are at distinct times, and the later concentration is smaller, as required by this positive-elimination model. Their ratio gives
Using the exact profile values, the amplitude is \(C(2)e^{k(2)}=12.5\) mg/L. Thus
Both parameters are uniquely determined in this idealized example. With rounded concentrations treated as exact rather than as rounded displays of the stated model values, the computed estimates would differ slightly.
A different result holds if only one positive-time concentration is available. With both \(V\) and \(CL\) unknown, one observation supplies one equation for two unknowns. For any admissible \(V\) that makes \(D/V>C(t)\), the corresponding rate and clearance can be chosen as
Many such parameter pairs can reproduce the same observation. In contrast, knowing \(V\) makes a single positive-time concentration sufficient to determine \(k\) and \(CL\), provided the concentration is compatible with the model.
Worked Example: Known Volume and One Later Concentration
Suppose a hypothetical \(450\) mg dose is given, \(V=30.0\) L is known, and the concentration at \(3.00\) h is \(15e^{-0.3}=11.1123\) mg/L. The predicted initial concentration is \(D/V=15.0\) mg/L, so the observation is positive and below the initial concentration, consistent with positive elimination.
The rate and clearance are
The units of clearance are L/h because the rate has units h\(^{-1}\) and volume has units L.
Why Known Clearance Does Not Always Identify Volume
It may seem that if clearance is known, one positive-time concentration should determine volume. However, at a fixed \(t>0\), the concentration as a function of volume is
As \(V\) increases from values near zero, this concentration first increases and then decreases. To locate its maximum, differentiate its logarithm:
The derivative is positive for \(V<CLt\) and negative for \(V>CLt\), so the maximum occurs at \(V=CLt\). The concentration tends to zero at both extremes: as \(V\to0^+\), the exponential tends to zero faster than \(1/V\) grows; as \(V\to\infty\), the factor \(D/V\) tends to zero. Consequently, a positive observation below the maximum corresponds to two positive volumes, the maximum corresponds to one, and a concentration above it is incompatible with the assumed model and known clearance.
Worked Example: One Concentration, Two Possible Volumes
Consider a hypothetical known dose \(D=600\) mg, known clearance \(CL=3.00\) L/h, and one concentration \(C(4)=15.0\) mg/L at \(t=4.00\) h. The maximum possible concentration at that time occurs at \(V=CLt=(3.00)(4.00)=12.0\) L and equals
Since \(15.0\) mg/L is positive and below this maximum, there are two volume solutions. Set \(x=CLt/V=12.0\ \text{L}/V\). The concentration equation becomes
This equation has one solution on each side of \(x=1\), approximately \(x=0.4894\) and \(x=1.781\). The corresponding volumes are \(V=12.0/x\), approximately \(24.52\) L and \(6.737\) L. Substitution into \(C(4)=(600/V)e^{-12/V}\) gives approximately \(15.0\) mg/L for either volume. The single observation therefore does not uniquely identify volume, despite known dose and clearance.
Diagnostics, Alternatives, and Limitations
Before interpreting an estimate, check whether the observations could have arisen under the model. With positive \(CL\), concentrations must decline over time; an exact increase between two times contradicts the assumed profile. If \(V\) is known, a positive-time concentration must be less than \(D/V\). If \(CL\) is known, the concentration cannot exceed \(D/(e\,CLt)\) at a positive time. These are model compatibility checks, not proof that the model is correct.
The algebraic two-observation calculation is a transparent identifiability argument, not usually a preferred way to estimate parameters from noisy data. One alternative is to fit the concentration equation directly to all observations, using an error model appropriate to the measurements. Another is to work with log concentrations, for which the ideal relationship is linear in time:
A straight-line representation can clarify how slope and intercept relate to \(k\) and \(V\), but a log transformation changes the error structure. It should not be treated as interchangeable with fitting concentrations on their original scale unless its assumptions are appropriate. Neither fitting method can repair structural non-identifiability: data cannot uniquely estimate two parameters if different parameter pairs imply the same observations.
Practical identifiability can also be weak even when the ideal model is structurally identifiable. For example, when concentration measurements are noisy or the informative part of a profile is poorly represented, small data changes may produce large changes in estimates. Model misspecification, assay limitations, uncertain dose, and an unobserved early profile can also undermine inference. If dose is not known, the amplitude \(D/V\) alone cannot separate dose from volume; external information or additional assumptions are needed.
The central distinction is between identifying the profile’s amplitude and rate and identifying the underlying parameters. With known dose, amplitude determines \(V\), while \(k\) and \(V\) together determine \(CL\). A time-zero concentration or two distinct-time concentrations can supply those pieces under the stated assumptions; one later concentration generally cannot when both disposition parameters are unknown.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- With known dose \(300\) mg and an ideal concentration \(C(0)=10.0\) mg/L, what is \(V\)?
- Two ideal observations are \(C(1)=9.00\) mg/L and \(C(4)=6.66\) mg/L. What is the implied \(k\), approximately?
- If \(V\) is known and a positive-time concentration is less than \(D/V\), which parameters can one such observation identify?
- For known \(D\) and \(CL\) at \(t>0\), how many positive volumes can produce a concentration strictly between zero and the maximum?
Solutions. (1) \(V=D/C(0)=300/10.0=30.0\) L. (2) \(k=\ln(9.00/6.66)/(4-1)\approx0.100\) h\(^{-1}\). (3) It identifies \(k=\ln[(D/V)/C(t)]/t\), and then \(CL=kV\). (4) Two: the concentration-volume curve increases to its maximum and then decreases, taking each strictly intermediate positive value once on either side of \(V=CLt\).