From Rate Constraints to Prediction Bounds
A rate describes how quickly a quantity changes. In a model, however, the rate may not be known exactly. Measurements or physical limits might tell us only that the rate remains between a lower and an upper value. Even without knowing the precise rate at every moment, we can still bound the total change over an interval.
This is useful when assessing a prediction. If the predicted change is outside every value permitted by the rate constraints, the prediction cannot satisfy those constraints. If it falls within the permitted range, the constraints alone do not establish that the prediction is correct; they only fail to rule it out.
Let \(y(t)\) be a quantity whose rate of change is \(r(t)=y'(t)\), for \(a\leq t\leq b\). The fundamental theorem of calculus gives the accumulated change as the integral of the rate. The bounds developed here follow by applying that relationship to lower and upper limits on \(r(t)\).
Bounds from Constant Rate Limits
The interval length \(b-a\) is nonnegative, so multiplying the rate inequalities by it preserves their direction. The lower and upper limits therefore bound the change: the quantity cannot increase by less than the lower rate sustained over the whole interval, or by more than the upper rate sustained over the whole interval.
Here is the integral justification in full. Since \(m\leq r(t)\leq M\), the functions \(r(t)-m\) and \(M-r(t)\) are nonnegative. Their integrals are nonnegative as well. Using \(y(b)-y(a)=\int_a^b r(t)\,dt\), we obtain:
Combining these inequalities proves the theorem. The result also applies when the rate bounds hold except at finitely many points and the rate is integrable, because changing values at finitely many points does not change the integral.
The units provide a useful check: rate units multiplied by time units must give the units of change in \(y\). If \(y\) is measured in litres and \(t\) in minutes, then \(r\) must be measured in litres per minute.
Worked Example: A Constrained Filling Rate
A tank contains \(40\) litres at time \(t=0\). Over the next \(12\) minutes, its filling rate \(r(t)\) is known to remain between \(2.4\) and \(3.1\) litres per minute. Bound the volume after \(12\) minutes and test a proposed prediction of \(80\) litres.
The interval length is \(12-0=12\) minutes. The minimum possible change is \(2.4\text{ L/min}\times12\text{ min}=28.8\text{ L}\). The maximum possible change is \(3.1\text{ L/min}\times12\text{ min}=37.2\text{ L}\). Thus the ending volume must satisfy
The proposed \(80\) litres is above the upper bound of \(77.2\) litres, so it is impossible under the stated rate constraint. As a direct check, it would require a change of \(80-40=40\) litres and hence an average rate of \(40\text{ L}/12\text{ min}=3.\overline{3}\text{ L/min}\), which exceeds \(3.1\text{ L/min}\).
Combining Bounds Across Time Intervals
Rate limits may differ from one interval to the next. Bound the change separately on each interval, then add the resulting changes. This works because total change across consecutive intervals is the sum of the changes on those intervals.
Use the times at which the stated rate limits change.
Multiply each interval's lower and upper rate limits by that interval's duration.
Sum all lower bounds for the total lower bound, and all upper bounds for the total upper bound.
The resulting bounds on total change give bounds on the ending quantity.
Worked Example: Rate Limits That Change Over Time
A population is \(120\) at \(t=0\), with time measured in hours. For the first \(2\) hours, its rate is between \(4\) and \(6\) individuals per hour. For the next \(3\) hours, its rate is between \(-1\) and \(2\) individuals per hour. Bound the population at \(t=5\), and test a proposed value of \(140\).
On \([0,2]\), the change is between \(4\times2=8\) and \(6\times2=12\) individuals. On \([2,5]\), it is between \((-1)\times3=-3\) and \(2\times3=6\) individuals. Adding the lower bounds gives \(8+(-3)=5\); adding the upper bounds gives \(12+6=18\). Therefore the total change is between \(5\) and \(18\), and
The proposed value \(140\) is impossible under these constraints because it exceeds \(138\). For a direct check, reaching \(140\) would require a total change of \(20\), but the maximum permitted total change is \(18\).
A negative lower rate bound is not an error: it allows the quantity to decrease. In the second interval above, the population could fall by as much as \(3\) individuals. The lower total bound correctly includes that possible decrease rather than treating every rate as positive.
Time-Varying Rate Bounds
Sometimes the limits themselves vary with time. If integrable functions \(L(t)\) and \(U(t)\) satisfy \(L(t)\leq r(t)\leq U(t)\) on \([a,b]\), then integrating these inequalities gives the corresponding bounds on change:
The justification is the same monotonicity of integration used above: both \(r(t)-L(t)\) and \(U(t)-r(t)\) are nonnegative, so their integrals are nonnegative. When \(y(a)=y_0\), add \(y_0\) to both bounds to obtain the range of possible ending values.
Worked Example: Bounds That Increase with Time
Suppose a quantity starts at \(10\) when \(t=0\), and its rate satisfies \(1+t\leq r(t)\leq3+t\) for \(0\leq t\leq2\). Find the endpoint range and determine whether an ending value of \(17\) is compatible with the constraints.
Integrating the lower rate bound gives \(\int_0^2(1+t)\,dt=[t+t^2/2]_0^2=2+2=4\). Integrating the upper bound gives \(\int_0^2(3+t)\,dt=[3t+t^2/2]_0^2=6+2=8\). Thus the change lies between \(4\) and \(8\), so the ending value lies between \(10+4=14\) and \(10+8=18\).
The value \(17\) is within that range, but being within the range alone does not prove that it can occur. To check compatibility, consider \(r(t)=2.5+t\). Its distance above the lower bound is \((2.5+t)-(1+t)=1.5\), and its distance below the upper bound is \((3+t)-(2.5+t)=0.5\), so it satisfies both constraints at every time. Its accumulated change is
Starting from \(10\), this rate gives \(10+7=17\). Substitution checks the endpoint, and the pointwise comparisons check that the chosen rate respects the constraints. Therefore \(17\) is compatible with the stated bounds.
Signed Rates and Average-Rate Tests
When rates can be positive or negative, use signed accumulated change. A positive rate adds to the quantity, while a negative rate subtracts from it. Replacing a negative rate by its absolute value would change the model and can produce an invalid bound.
Another useful check follows by dividing the constant-bound theorem by the positive interval length \(b-a\). The average rate, defined as \((y(b)-y(a))/(b-a)\), must lie between \(m\) and \(M\). Thus an endpoint prediction directly implies an average rate that can be compared with the allowed limits.
Worked Example: A Limit on the Magnitude of a Decrease
A stock begins at \(30\) units and is observed over \(5\) hours. Its rate satisfies \(|r(t)|\leq0.8\) units per hour. Test a proposed ending stock of \(35\) units.
The absolute-value constraint is equivalent to \(-0.8\leq r(t)\leq0.8\). The total change is therefore between \((-0.8)(5)=-4\) and \((0.8)(5)=4\) units. The ending stock must satisfy
The proposed \(35\) units is outside the range, so it is impossible under the rate constraint. Indeed, it requires a change of \(35-30=5\) units and an average rate of \(5/5=1\) unit per hour, greater than the maximum allowed \(0.8\) units per hour.
What a Bound Can—and Cannot—Establish
A bound is a necessary condition on the prediction, not a guarantee that every value between the endpoints is achievable. For example, additional restrictions on how the rate can vary might exclude some values inside a calculated range. To establish that a particular endpoint is achievable, it is enough to provide a rate function that obeys the constraints and whose integral produces that endpoint, as in the time-varying example.
Conversely, if a proposed endpoint lies outside the bound, no rate satisfying the stated constraints can produce it. This is a mathematical incompatibility, conditional on the given initial value, time interval, and rate limits. It does not by itself identify which input or modelling assumption should be changed.
A frequent mistake is to compare a proposed total change with a rate limit without multiplying the rate by the elapsed time. For instance, if a rate is at most \(3\) litres per minute for \(10\) minutes, the maximum change is not \(3\) litres; it is \(3\text{ L/min}\times10\text{ min}=30\) litres. The units reveal the correction: a rate alone is not a quantity of accumulated change.
Check Your Understanding
Use the rate constraints and the relevant interval lengths in each question. Treat bounds as inclusive.
- A quantity starts at \(18\) units. For \(4\) hours its rate is between \(1.5\) and \(2\) units per hour. What range of ending values is permitted?
- A stock starts at \(50\). For \(2\) days its rate is between \(-3\) and \(-1\) units per day. What are the lower and upper bounds on the ending stock?
- For \(0\leq t\leq1\), a quantity has \(2\leq r(t)\leq4\). Can a proposed change of \(5\) occur under these constraints? Explain using the average rate.
- Suppose the rate is between \(t\) and \(2t\) for \(0\leq t\leq2\). What bounds do these constraints place on total change?
- If an endpoint lies inside a bound obtained from rate limits, does that alone prove the endpoint is achievable? State what additional check could establish compatibility.
