Tutorials › Quantitative Systems Pharmacology (QSP) › Solving a Tumor Growth and Kill Model by Hand

Oncology: Tumor Growth and Therapy Response · Tutorial 269 of 799

Solving a Tumor Growth and Kill Model by Hand

Learn to solve piecewise tumor-growth and kill equations by hand, carry values accurately across treatment changes, and interpret the resulting trajectories.

Advanced 10 min read

What You'll Learn

  • Derive the exponential solution for a constant net tumor growth rate.
  • Combine interval-specific rates while carrying tumor burden continuously through switches.
  • Check rate units, signs, and continuity at treatment changes.
  • Calculate tumor burden and time-to-target values from hypothetical parameters.
  • Interpret zero, positive, and negative net rates and identify model limitations.

Why Solve the Model by Hand?

A hand solution makes the assumptions and the effect of each treatment interval visible. It provides a direct check on whether a predicted tumor burden is consistent with the growth and killing rates, and it can expose errors in units, signs, or treatment-switch calculations before those errors are carried into a more elaborate analysis. The calculations here use an exponential model with rates that are constant within each interval.

Model and Assumptions

Let \(V(t)\) denote tumor volume at time \(t\), measured, for example, in \(\mathrm{cm}^3\). During an untreated interval, assume tumor volume grows exponentially at a constant per-capita rate \(r\). During a treated interval, assume treatment adds a constant per-capita killing rate \(k\). Both rates have units of inverse time, such as \(\mathrm{day}^{-1}\). The model treats growth and killing as proportional to the current volume, so their net effect is the difference between the two rates.

Definition: On an interval with constant untreated growth rate \(r\) and constant treatment-associated killing rate \(k\), the tumor-growth and kill model is \(dV/dt=(r-k)V\). During an untreated interval, \(k=0\). The net rate is \(g=r-k\), in units of inverse time. A treatment switch changes the rate used after the switch, but, unless an instantaneous volume change is explicitly specified, it does not itself change \(V\).

The hand solution assumes that volume is positive at the start of each interval, the stated rates remain constant over that interval, and no separate process causes an instantaneous change in volume. In a multi-interval example, the untreated growth rate may be specified separately for different intervals; this represents a model assumption, not an effect generated automatically by the treatment-kill term. The simple model also assumes that the same net rate applies at every tumor volume within a given interval.

Deriving the Solution on One Interval

For a constant net rate \(g\), separate the variables and integrate from the start of an interval, \(t_a\), to any time \(t\) within it. The starting volume is \(V(t_a)\), and the solution follows by exponentiating both sides:

$$ \frac{dV}{V}=g\,dt \qquad\Longrightarrow\qquad \ln\!\left(\frac{V(t)}{V(t_a)}\right)=g(t-t_a) \qquad\Longrightarrow\qquad V(t)=V(t_a)e^{g(t-t_a)}. $$

The exponent must be dimensionless: \(g\) has units of inverse time and \(t-t_a\) has units of time. If \(g>0\), the volume increases; if \(g<0\), it decreases; and if \(g=0\), it remains constant. These statements concern the model's trajectory, not a guarantee that real tumors follow exponential behavior indefinitely.

For several consecutive intervals, use the volume at the end of one interval as the starting volume for the next. If the rate is \(g_i\) on the interval from \(t_{i-1}\) to \(t_i\), then after \(m\) intervals the solution is

$$ V(t_m)=V(t_0)\exp\!\left(\sum_{i=1}^{m}g_i(t_i-t_{i-1})\right). $$

This expression is obtained by multiplying the exponential factors from the individual intervals. It is often more accurate to add the exponents first and round only the final volume.

Switch Times, Continuity, and Interpretation

Suppose the rate changes at time \(t_s\), from \(g_1\) before the switch to \(g_2\) after it. With no instantaneous volume change, the value at the switch is the end of the first interval and the start of the second:

$$ V(t_s^-)=V(t_s)=V(t_s^+). $$

The derivative can change at \(t_s\), because the rate changes, even though the volume remains continuous. This distinction is important: changing the slope of the trajectory is not the same as imposing a jump in tumor volume. If a model includes an instantaneous procedure or other discrete volume change, that jump must be stated and calculated separately.

1
List intervals and units.
Write each start time, end time, and applicable rate. Confirm that rates and times use compatible units.
2
Calculate the net rate.
For a treated interval, subtract the killing rate from the growth rate: \(g=r-k\). For an untreated interval, \(g=r\).
3
Accumulate the exponent.
Multiply each net rate by its interval duration and add those products, retaining sufficient precision.
4
Calculate and check.
Apply the exponential to the starting volume. Check the direction of change against the sign of each net rate and verify continuity at every switch.

Worked Example: Tumor Volume During a Treatment Interval

Suppose a hypothetical tumor has volume \(1.20\ \mathrm{cm}^3\) at day 4. Its untreated growth rate is \(r=0.025\ \mathrm{day}^{-1}\), and treatment begins on day 4 with a constant killing rate \(k=0.040\ \mathrm{day}^{-1}\). Calculate the predicted volume on day 12.

The treated interval lasts \(12-4=8\) days. Its net rate is \(g=r-k=0.025-0.040=-0.015\ \mathrm{day}^{-1}\). The product of this rate and the interval length is dimensionless: \((-0.015\ \mathrm{day}^{-1})(8\ \mathrm{day})=-0.120\).

$$ V(12)=1.20e^{(-0.015)(8)} =1.20e^{-0.120} \approx 1.064\ \mathrm{cm}^3. $$

The model predicts a decrease because the net rate is negative. No additional factor is applied at treatment initiation: the volume at day 4 is \(1.20\ \mathrm{cm}^3\) on both sides of that switch under the stated assumptions.

Worked Example: Carrying Volume Across Three Intervals

Consider a hypothetical tumor with \(V(0)=2.00\ \mathrm{cm}^3\). Specify a net rate of \(+0.030\ \mathrm{day}^{-1}\) from day 0 to day 7, \(-0.020\ \mathrm{day}^{-1}\) from day 7 to day 21 during treatment, and \(+0.015\ \mathrm{day}^{-1}\) from day 21 to day 35 after treatment. For the final untreated interval, the stated \(+0.015\ \mathrm{day}^{-1}\) is the assumed untreated growth rate. Calculate volumes at days 7, 21, and 35.

At day 7, the first interval contributes \((0.030)(7)=0.210\) to the exponent:

$$ V(7)=2.00e^{0.210}\approx 2.467\ \mathrm{cm}^3. $$

For day 21, use the original volume and add the second interval's exponent, \((-0.020)(14)=-0.280\), to the first interval's \(0.210\). Thus the total exponent is \(-0.070\):

$$ V(21)=2.00e^{0.210-0.280} =2.00e^{-0.070} \approx 1.865\ \mathrm{cm}^3. $$

The final interval contributes \((0.015)(14)=0.210\). Adding all three contributions gives a total exponent of \(0.140\):

$$ V(35)=2.00e^{0.210-0.280+0.210} =2.00e^{0.140} \approx 2.301\ \mathrm{cm}^3. $$

The volume decreases during the second interval and increases during the third, as indicated by the signs of the rates. The calculations use the unrounded exponents and original starting volume; rounded intermediate volumes are not reused. At days 7 and 21 the volume is continuous, while the rate governing its subsequent change switches.

Thresholds and Limiting Behavior

On treatment, the sign of \(r-k\) determines the direction of change. If \(k>r\), the model predicts shrinkage; if \(k=r\), it predicts a constant volume; and if \(k<r\), it predicts continued growth despite treatment. This comparison concerns the two model rates, not a direct guarantee of a clinical response.

Worked Example: Time to a Target Volume

Suppose, hypothetically, a tumor begins treatment at \(4.00\ \mathrm{cm}^3\), with \(r=0.030\ \mathrm{day}^{-1}\) and \(k=0.050\ \mathrm{day}^{-1}\) held constant. How long does the model predict it will take to reach \(2.00\ \mathrm{cm}^3\)?

The net rate is \(g=0.030-0.050=-0.020\ \mathrm{day}^{-1}\). Set the interval solution equal to the target and solve for elapsed time:

$$ 2.00=4.00e^{-0.020\Delta t} \quad\Longrightarrow\quad \Delta t=\frac{\ln(2.00/4.00)}{-0.020} =\frac{\ln(0.5)}{-0.020} \approx 34.66\ \mathrm{days}. $$

Thus the predicted time to halve from \(4.00\) to \(2.00\ \mathrm{cm}^3\) is about 34.66 days under these constant-rate assumptions. If the same negative net rate were maintained indefinitely, \(V(t)=V(0)e^{gt}\) would approach zero as time increases, but it would not reach zero at a finite time. That mathematical limit should not be interpreted as a realistic prediction of arbitrarily small measurable tumor volume.

More generally, under a constant net rate, a positive rate leads to unbounded exponential growth in the mathematical model, a zero rate gives a constant volume, and a negative rate leads asymptotically toward zero. These limiting behaviors help identify what the equations imply if their assumptions are extended indefinitely. They also highlight why a constant-rate model may be suitable over a specified interval but inadequate for long-term prediction.

Diagnostics, Alternatives, and Limitations

A hand solution is most useful when its inputs and intermediate steps can be checked. Confirm that time and rate units match; a rate in \(\mathrm{day}^{-1}\) must be paired with an interval in days. Confirm the sign convention: killing is subtracted from growth in \(r-k\). At each switch, verify that the ending volume of the preceding interval is the starting volume of the next. Finally, check that the direction of each predicted change matches the net-rate sign.

The hand solution is exact for the stated piecewise-constant-rate model. If rates vary continuously with time, the corresponding exponential solution depends on the accumulated rate over time rather than one constant rate times an interval duration. Numerical methods can be useful when a model has time-varying rates, interacting compartments, or other equations without a simple closed-form solution. They do not remove the need to check units, assumptions, and behavior at treatment changes.

A common mistake is to round each interval's tumor volume and then use that rounded value as the next interval's starting value. Repeated rounding can shift the final result. The correction is to carry the unrounded value, or equivalently add all rate-times-duration contributions in the exponent and round only the reported final volume. Another error is to treat a treatment switch as an instantaneous reduction in volume even though the model changes only the rate. A rate switch changes the subsequent slope; a volume jump requires a separate, explicit model rule.

The model's limitations are structural: constant per-capita rates imply exponential trajectories within intervals; treatment killing is represented as a fixed rate rather than a changing drug effect; and the equation does not itself represent measurement error, tumor-size constraints, or other biological mechanisms. A good fit over one interval does not establish that those assumptions remain reasonable over another.

Check Your Understanding

Answer each question using the piecewise exponential model. Solutions follow so that the arithmetic and interpretation can be checked.

  1. A tumor has \(V(0)=3.00\ \mathrm{cm}^3\) and net rate \(0.010\ \mathrm{day}^{-1}\) for 5 days. What is \(V(5)\), and should it increase or decrease?
  2. Growth and killing rates are \(r=0.040\ \mathrm{day}^{-1}\) and \(k=0.040\ \mathrm{day}^{-1}\). What does the model predict during treatment?
  3. Why must the starting volume for the second interval equal the ending volume for the first interval when there is no instantaneous volume change?
  4. A tumor has volume \(5.00\ \mathrm{cm}^3\) at the start of an interval. It shrinks under a constant net rate of \(-0.025\ \mathrm{day}^{-1}\). How long does the model take to reach \(2.50\ \mathrm{cm}^3\)?

Solutions. (1) \(V(5)=3.00e^{(0.010)(5)}=3.00e^{0.050}\approx3.154\ \mathrm{cm}^3\); it increases because the net rate is positive. (2) The net rate is zero, so \(V(t)=V(0)e^0=V(0)\): volume remains constant during that interval. (3) The rate changes at the switch but the state variable is continuous under the stated assumption; hence the value reached at the boundary is carried forward, not reset. (4) Solve \(2.50=5.00e^{-0.025\Delta t}\). Then \(\Delta t=\ln(0.5)/(-0.025)\approx27.73\ \mathrm{days}\).

Clinical Biostats