Why Calibrate a Rate Model?
A proposed rate function describes how quickly a quantity changes over time. Its form may be chosen from the setting, but it often contains unknown parameters. Calibration means selecting parameter values so the function agrees with measured rates.
For example, a model might say that a flow rate increases linearly, or that a reaction rate decreases exponentially. Measurements can determine the model's intercept, slope, initial rate, or decay constant. A fitted model is not automatically a good model: its units must be consistent, its parameters must make sense in context, and its predictions should be compared with the observed rates.
If the measured rates are \(y_i\) at times \(t_i\), the model predicts \(r(t_i;\boldsymbol{\theta})\). The difference between an observed rate and its prediction is the residual:
A residual of zero means that the model matches that particular measurement exactly. A nonzero residual records the size and direction of the mismatch: positive means the observation is above the prediction, while negative means it is below.
From Measurements to Parameter Equations
For a model with one unknown parameter, one suitable measurement may determine that parameter. For a model with several unknown parameters, more information is usually needed. A basic strategy is to substitute each observed time and rate into the proposed function, creating equations for the unknowns.
When there are exactly as many independent measurements as parameters, these equations may determine a parameter choice that matches every measurement exactly. But exact matching is not guaranteed: the equations might be incompatible, or multiple parameter choices might give the same measurements. With more measurements than parameters, a model will generally not pass through every observed rate. Then a fitting criterion is needed to choose a compromise.
Identify the unknown parameters and the units each parameter must have.
Substitute each observation's time and measured rate into the model.
Use the equations directly when they can determine the parameters; for extra, imperfect measurements, a criterion such as least squares can select a fit.
Verify the fitted predictions, inspect remaining residuals, and assess the units and plausible meaning of the parameters.
Worked Example: Calibrating a Linear Rate
Worked Example: Calibrating a Linear Rate
A pump's flow rate is modeled by \(q(t)=a+bt\), where \(t\) is measured in minutes and \(q\) in litres per minute. Measurements give \(q(0)=6\) and \(q(4)=14\). Find the parameters, then compare the model with a separate measurement of \(q(6)=17\).
At \(t=0\), substitution gives \(q(0)=a+b(0)=a\). Therefore \(a=6\) litres per minute.
At \(t=4\), the model gives \(14=a+4b\). Substituting \(a=6\) gives \(14=6+4b\), so \(8=4b\) and \(b=2\). Since \(bt\) must have units of litres per minute and \(t\) has units of minutes, \(b\) has units of litres per minute squared.
Check the two calibration measurements: \(q(0)=6+2(0)=6\), and \(q(4)=6+2(4)=14\). At \(t=6\), the prediction is \(q(6)=6+12=18\) litres per minute. The separate observed rate is \(17\), so the residual, observation minus prediction, is \(17-18=-1\) litre per minute. The model overpredicts that rate by one litre per minute.
The parameter \(a\) is the modeled rate at time zero. The positive slope \(b\) says the modeled flow increases by two litres per minute for each additional minute. The mismatch at six minutes is a reminder that fitting two measurements exactly does not guarantee exact agreement with other measurements.
Worked Example: Calibrating an Exponential Rate
Worked Example: Calibrating an Exponential Rate
A reaction rate is modeled by \(r(t)=Ae^{-kt}\), where \(t\) is in days and \(r\) is in milligrams per day. A measurement at day zero is \(12\) milligrams per day, and a measurement at day three is \(6\) milligrams per day. Find \(A\) and \(k\), then compare with a measured rate of \(2.8\) milligrams per day at day six.
At day zero, \(12=Ae^{-k(0)}=A\), so \(A=12\) milligrams per day. At day three, substitute \(r(3)=6\):
Taking natural logarithms gives \(\ln(1/2)=-3k\). Since \(\ln(1/2)=-\ln 2\), it follows that \(k=\frac{\ln 2}{3}\) per day. The units are inverse days because \(kt\) must be dimensionless.
Check the fitted values. At \(t=0\), the exponential factor is \(1\), so \(r(0)=12\). At \(t=3\), the exponent is \(-\ln 2\), so \(r(3)=12e^{-\ln 2}=12/2=6\). At \(t=6\), the exponent is \(-2\ln 2\), so \(r(6)=12e^{-2\ln 2}=12/4=3\) milligrams per day. Compared with the observation \(2.8\), the residual is \(2.8-3=-0.2\) milligrams per day.
The positive value of \(k\) describes decay. In this model the rate halves every three days: increasing \(t\) by three multiplies the exponential factor by \(e^{-3k}=1/2\). The day-six discrepancy shows that this interpretation and the chosen exponential form do not give an exact match to every observation.
When Measurements Do Not Match Exactly
Measured rates may differ from the proposed function because of measurement error or because the function is only an approximation to the real process. When there are more observations than parameters, it is generally unreasonable to demand that every residual be zero. One common choice is least squares: select parameters that minimize the sum of the squared residuals.
Squaring prevents positive and negative residuals from cancelling and gives a nonnegative measure of total mismatch. It also makes the fitting criterion a function of the unknown parameters that can often be minimized using calculus. The resulting fit is a compromise; it is not a claim that the observations are all exact or that the model is true.
Why the Least-Squares Line Has Its Fitted Values
For a linear rate model \(r(t)=a+bt\), suppose there are measurements \((t_i,y_i)\). The least-squares objective is \(S(a,b)=\sum_i(y_i-a-bt_i)^2\). At an interior minimum, the partial derivatives with respect to \(a\) and \(b\) must be zero. Differentiating gives:
Thus the fitted residuals satisfy \(\sum_i e_i=0\) and \(\sum_i t_i e_i=0\). These conditions follow from the choice of the least-squares objective; they do not mean that each residual is zero. If the observation times are not all identical, the quadratic objective has a unique minimum in \(a\) and \(b\): its curvature in any nonzero parameter change is proportional to \(\sum_i(\Delta a+t_i\Delta b)^2\), which is positive unless that change gives zero at every distinct time. This explains why least squares yields a well-defined straight-line fit when the data include at least two distinct times.
Worked Example: A Least-Squares Linear Rate
Worked Example: A Least-Squares Linear Rate
Suppose measured production rates, in kilograms per hour, are \(2\), \(4\), and \(5\) at times \(t=0\), \(1\), and \(2\) hours, respectively. Fit \(r(t)=a+bt\) by least squares.
The objective is \(S(a,b)=(2-a)^2+(4-a-b)^2+(5-a-2b)^2\). The zero-partial-derivative conditions are \(\sum e_i=0\) and \(\sum t_i e_i=0\). Writing them using the observations gives \(3a+3b=11\) and \(3a+5b=14\): the first equation comes from \(3a+b(0+1+2)=2+4+5\), and the second from \(a(0+1+2)+b(0^2+1^2+2^2)=0(2)+1(4)+2(5)\).
Subtracting the first equation from the second gives \(2b=3\), so \(b=3/2\). Substituting into \(3a+3b=11\) gives \(3a+9/2=11\), hence \(3a=13/2\) and \(a=13/6\). Therefore the fitted model is:
The intercept has units of kilograms per hour. Since \(bt\) must also have those units, \(b\) has units of kilograms per hour squared. The predictions at \(t=0,1,2\) are \(13/6\), \(13/6+3/2=11/3\), and \(13/6+3=31/6\). The residuals, observed minus predicted, are \(2-13/6=-1/6\), \(4-11/3=1/3\), and \(5-31/6=-1/6\), each in kilograms per hour.
As a check on the least-squares conditions, the residuals sum to \(-1/6+1/3-1/6=0\), and their time-weighted sum is \(0(-1/6)+1(1/3)+2(-1/6)=0\). The squared-error total is \(1/36+1/9+1/36=6/36=1/6\) in squared rate units. None of the three rates is matched exactly, but the chosen line minimizes that total among linear models.
Interpret the Fit, Not Just the Numbers
A parameter value is useful only when its meaning is understood in the model and the setting. Check that each term in the rate function has the units of a rate. For a linear model \(a+bt\), \(a\) has rate units and \(b\) has rate-per-time units. For an exponential model \(Ae^{-kt}\), \(A\) has rate units and \(k\) has inverse-time units. A unit mismatch often reveals an incorrectly specified or incorrectly interpreted parameter.
Also consider whether the parameter values are plausible. A negative decay constant in a model intended to describe decay instead produces growth. A negative rate prediction may be impossible for a particular application, even if it follows algebraically from a fitted line. Such a result does not by itself identify the remedy, but it is a reason to question whether the chosen form or range of use is appropriate.
Finally, inspect the remaining discrepancies in the same units as the measured rates. Report predictions and residuals rather than saying only that the model “fits.” A small residual may be acceptable relative to the measurement scale; a large one may matter. Calibration uses the available rate observations, but it cannot establish that the model will behave well at times or conditions not represented by those observations.
Common Pitfall: Treating a Rate Parameter as a Rate
In \(r(t)=Ae^{-kt}\), a common error is to assign \(k\) the same units as \(r\). That would make the exponent \(kt\) have units of rate multiplied by time, so it would not be dimensionless as an exponential's exponent must be. The correction is to give \(k\) units of inverse time, making \(kt\) dimensionless. Meanwhile \(A\), the value of the model at \(t=0\), has the same units as the rate.
Another mistake is to regard parameters obtained from two measurements as a guarantee of accuracy everywhere. Exact calibration only confirms agreement at the measurements used to determine the parameters. Other rates still need to be compared with the model's predictions.
Check Your Understanding
Use these questions to test whether you can set up and assess a rate calibration.
- For \(r(t)=a+bt\), what units must \(a\) and \(b\) have if \(r\) is measured in litres per minute and \(t\) in minutes?
- In \(r(t)=Ae^{-kt}\), why must \(kt\) be dimensionless, and what units does that require for \(k\)?
- If an observed rate is \(7\) and the model predicts \(8\), what is the residual using observation minus prediction, and what does its sign indicate?
- Why might a least-squares fit with three measurements and two parameters fail to match every measurement exactly?
- What does exact agreement at the measurements used for calibration establish, and what does it not establish?
