Deriving an Exact Lumped Compartment

Why an Exact Lump Can Be Useful

An organ-resolved model can track drug amounts in individual tissues, while a lumped model represents several tissues with one amount state. A reduction is exact only if the aggregate amount evolves according to a closed equation: its future change must be calculable from the aggregate state and the external input, without separately knowing how the amount is divided between tissues.

Here, consider two perfused tissues exposed to the same arterial drug concentration. The aim is to add their amount balances and determine when their combination can itself be represented as one perfusion-limited compartment. The key condition concerns the rate at which each tissue's amount is removed from its state.

Definition: An exact lumped compartment for two tissues is a single aggregate state whose equation reproduces the sum of the two tissue amounts for every time course of the shared arterial concentration and every allowed initial distribution of drug between the tissues. The aggregate equation must not require the separate tissue amounts.

Starting From Two Perfused-Tissue Balances

Assume tissue \(i\) has fixed volume \(V_i\), blood flow \(Q_i\), tissue amount \(A_i\), and tissue-to-blood partition coefficient \(K_i\), for \(i=1,2\). Its tissue concentration is \(A_i/V_i\). Under instantaneous partitioning at the venous end of the tissue, venous concentration is \(A_i/(K_iV_i)\). The arterial concentration \(C_a(t)\) is shared by both tissues.

The perfusion-limited balance says that the rate of change in tissue amount is blood flow multiplied by the arterial-to-venous concentration difference. This setup assumes well-mixed tissue compartments, constant flows and volumes, no inter-tissue exchange, and no separate metabolism or other local source or sink. These assumptions define the model being reduced; they are not consequences of the aggregation.

\frac{dA_i}{dt} =Q_i\left(C_a-\frac{A_i}{K_iV_i}\right) =Q_iC_a-\lambda_i A_i, \qquad \lambda_i=\frac{Q_i}{K_iV_i}.

The rate coefficient \(\lambda_i\) has units of inverse time. It governs the amount-dependent loss term for tissue \(i\); the input term \(Q_iC_a\) has units of amount per time. The equation can be used for a time-varying \(C_a(t)\), not only a constant arterial concentration.

Adding Amounts and Deriving the Exact Condition

Define the combined amount \(A=A_1+A_2\), total volume \(V=V_1+V_2\), and total flow \(Q=Q_1+Q_2\). Adding the two balance equations gives

\frac{dA}{dt} =QC_a-\lambda_1A_1-\lambda_2A_2.

The input term depends only on the common arterial concentration and the total flow. The loss term, however, depends on how the total amount is divided unless the two rate coefficients are equal. If \(\lambda_1=\lambda_2=\lambda\), the amount-dependent terms combine as \(\lambda A_1+\lambda A_2=\lambda A\), yielding a closed equation.

Theorem: Under the stated model assumptions, if \(\lambda_1=\lambda_2=\lambda\), then the aggregate amount obeys the exact equation \(\frac{dA}{dt}=QC_a-\lambda A\). For \(C=A/V\), the combined concentration equation is \(\frac{dC}{dt}=\frac{Q}{V}(C_a-C/K_{\mathrm{eff}})\), where \(K_{\mathrm{eff}}=\frac{Q}{\lambda V}\). If the tissues also share a common partition coefficient \(K\), equality of the rate coefficients is equivalent to \(Q_1/V_1=Q_2/V_2\), and \(K_{\mathrm{eff}}=K\).

The equivalent aggregate partition coefficient is a parameter that makes the summed equation take the familiar perfusion-limited form. It need not equal either tissue's partition coefficient when \(K_1\) and \(K_2\) differ. In the special case of a common \(K\), equal rate coefficients imply \(Q_i/(K V_i)=\lambda\), so the flows per unit volume are equal. Then \(Q/(\lambda V)=K\), and the aggregate retains that same partition coefficient.

A Short Justification of Exactness

The derivation establishes more than a convenient approximation. When \(\lambda_1=\lambda_2\), the summed equation contains only \(A\), \(C_a(t)\), and fixed aggregate parameters. Therefore, any two tissue-level states with the same total amount have the same instantaneous aggregate rate of change under the same arterial input. Applying the same equation forward in time gives the same aggregate trajectory, regardless of the initial split between tissues.

Conversely, if the rate coefficients differ, consider two possible allocations with the same total \(A\), such as \((A_1,A_2)\) and \((A_1+\delta,A_2-\delta)\). Their loss terms differ by

\lambda_1\delta-\lambda_2\delta =(\lambda_1-\lambda_2)\delta.

For unequal rate coefficients and a nonzero shift \(\delta\), the same aggregate amount can therefore have different rates of change. In general, no single equation based only on \(A\) and the shared input can reproduce both cases. This is the closure test: the aggregate state must determine its own derivative.

Worked Example: Two Tissues With a Common Partition Coefficient

Suppose two hypothetical tissues have \(V_1=2\) L, \(Q_1=0.4\) L/h, \(V_2=3\) L, and \(Q_2=0.6\) L/h. Both have \(K_1=K_2=2\). Check whether they form an exact lump, then calculate the aggregate concentration rate when \(C_a=1.5\) mg/L and the tissue amounts are \(A_1=4\) mg and \(A_2=6\) mg.

The rate coefficients are

\lambda_1=\frac{0.4\ \mathrm{L/h}}{(2)(2\ \mathrm{L})} =0.1000\ \mathrm{h}^{-1}, \qquad \lambda_2=\frac{0.6\ \mathrm{L/h}}{(2)(3\ \mathrm{L})} =0.1000\ \mathrm{h}^{-1}.

They are equal, so the reduction is exact. The combined amount, volume, and flow are \(A=4+6=10\) mg, \(V=2+3=5\) L, and \(Q=0.4+0.6=1.0\) L/h. The aggregate concentration is \(C=A/V=2.0\) mg/L, and \(K_{\mathrm{eff}}=2\). Substitution into the aggregate equation gives

\frac{dC}{dt} =\frac{1.0\ \mathrm{L/h}}{5\ \mathrm{L}} \left(1.5\ \mathrm{mg/L}-\frac{2.0\ \mathrm{mg/L}}{2}\right) =0.2000\ \mathrm{h}^{-1}(0.5\ \mathrm{mg/L}) =0.1000\ \mathrm{mg/(L\,h)}.

Thus, at this instant the combined concentration is increasing at \(0.1000\) mg/(L·h). Equivalently, the total amount is increasing at \(V(dC/dt)=5(0.1000)=0.5000\) mg/h.

Worked Example: Equal Rate Coefficients but Different Partitions

Consider hypothetical tissues with \(V_1=2\) L, \(Q_1=0.6\) L/h, \(K_1=1.5\), and \(V_2=4\) L, \(Q_2=1.6\) L/h, \(K_2=2.0\). Determine whether their amounts can be lumped exactly and find the equivalent aggregate parameters.

For the first tissue, \(\lambda_1=0.6/((1.5)(2))=0.2000\) h\(^{-1}\). For the second, \(\lambda_2=1.6/((2.0)(4))=0.2000\) h\(^{-1}\). The rate coefficients match, so the total amount satisfies \(dA/dt=2.2C_a-0.2A\), with \(Q=2.2\) L/h and \(V=6\) L.

The equivalent partition coefficient is

K_{\mathrm{eff}} =\frac{Q}{\lambda V} =\frac{2.2\ \mathrm{L/h}} {(0.2000\ \mathrm{h}^{-1})(6\ \mathrm{L})} =1.833.

For example, if \(C_a=1.0\) mg/L and \(A=8\) mg, then \(C=8/6=1.333\) mg/L and

\frac{dC}{dt} =\frac{2.2}{6} \left(1.0-\frac{1.333}{1.833}\right) =0.1000\ \mathrm{mg/(L\,h)}.

The lump is exact even though the tissue partition coefficients differ. The aggregate uses \(K_{\mathrm{eff}}=1.833\), rather than either tissue's coefficient, to represent the summed exchange equation.

When the Reduction Fails

A common mistake is to add the amounts and assume that the result is one compartment simply because both tissues receive blood from the same arterial concentration. A shared input is not enough: the loss term must also be determined by the total amount. Unequal \(\lambda_i\) values make the rate depend on the hidden allocation between tissues.

Worked Example: Unequal Rate Coefficients Prevent Closure

Take two hypothetical tissues with \(V_1=V_2=2\) L, \(K_1=K_2=2\), \(Q_1=0.4\) L/h, and \(Q_2=0.8\) L/h. Their rate coefficients are \(\lambda_1=0.1000\) h\(^{-1}\) and \(\lambda_2=0.2000\) h\(^{-1}\). Let \(C_a=1.0\) mg/L and compare two distributions with the same total amount \(A=10\) mg.

For \(A_1=4\) mg and \(A_2=6\) mg, the input is \((0.4+0.8)(1.0)=1.2\) mg/h and the loss is \((0.1)(4)+(0.2)(6)=1.6\) mg/h. Thus \(dA/dt=1.2-1.6=-0.4000\) mg/h.

For \(A_1=6\) mg and \(A_2=4\) mg, the input remains \(1.2\) mg/h, while the loss is \((0.1)(6)+(0.2)(4)=1.4\) mg/h. Now \(dA/dt=1.2-1.4=-0.2000\) mg/h. Both states have the same total amount but different total rates of change. Therefore, the total alone cannot define an exact one-compartment equation for this example.

Checks, Alternatives, and Limits

A practical exactness check is to calculate \(\lambda_i=Q_i/(K_iV_i)\) for each tissue, using consistent units. Equal values support the exact reduction under the model assumptions; approximate equality supports, at most, an approximation whose adequacy depends on the intended prediction. Also verify that both tissues receive the same \(C_a(t)\), and that unmodeled tissue-specific processes do not add different source or loss terms.

The main alternative is to retain two tissue states. That preserves the different rate coefficients and allows predictions to depend on the separate amounts. A one-compartment approximation may still be considered when rate coefficients are close, but closeness alone does not make the reduction exact. A more detailed connected organ network can represent distinct flows and tissue properties, at the cost of additional states and parameters.

The exact result is limited to the specified linear, perfusion-limited structure. If arterial inputs differ, flows or volumes vary with time, or local metabolism adds tissue-specific losses, the summed equation must be re-derived with those terms included. Exact lumping may still be possible under other special conditions, but equality of the two \(\lambda_i\) values by itself is not sufficient once the model structure changes.

Key takeaway: With a shared arterial input and the stated perfusion-limited balances, equal amount-loss rate coefficients make the sum of tissue amounts a closed, exact compartment. If those coefficients differ, the same total amount can have different derivatives depending on its distribution.

Check Your Understanding

Use the definitions and equations above to answer each question. Solutions are provided below the questions.

Two tissues have \(Q_1=0.3\) L/h, \(V_1=1.5\) L, \(K_1=2\), and \(Q_2=0.8\) L/h, \(V_2=4\) L, \(K_2=2\). Are their rate coefficients equal?
If two tissues have a common rate coefficient \(\lambda\), what equation does their total amount satisfy when they share arterial concentration \(C_a(t)\)?
For \(Q=1.2\) L/h, \(V=4\) L, and \(\lambda=0.15\) h\(^{-1}\), calculate \(K_{\mathrm{eff}}\).
Why can different tissue partition coefficients still permit an exact lump?
What observation demonstrates that an aggregate equation is not closed when the rate coefficients differ?

Solutions. (1) \(\lambda_1=0.3/((2)(1.5))=0.1000\) h\(^{-1}\), and \(\lambda_2=0.8/((2)(4))=0.1000\) h\(^{-1}\), so they are equal. (2) \(dA/dt=(Q_1+Q_2)C_a(t)-\lambda A\). (3) \(K_{\mathrm{eff}}=Q/(\lambda V)=1.2/((0.15)(4))=2.000\). (4) Exactness requires the loss terms to sum to \(\lambda A\); the individual \(K_i\) values can differ if the resulting rate coefficients are equal. The lump then uses \(K_{\mathrm{eff}}=Q/(\lambda V)\). (5) If two different tissue allocations have the same total amount and shared input but yield different derivatives of that total, the aggregate amount alone does not determine its rate of change.