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Calculus-Based Model Construction · Tutorial 1492 of 1495

Capacity-Limited Accumulation Models

Learn to choose a capacity-limited rate rule, solve representative models, and check whether predicted stock remains within its allowable range.

Advanced 12 min read

What You'll Learn

  • Distinguish a hard capacity stop from a rate that slows as capacity is approached.
  • Define a proportional-to-remaining-capacity rate and derive its solution.
  • Prove that the solution to this initial-value model is unique.
  • Construct a piecewise model with a maximum rate and a capacity-dependent rate.
  • Check a model’s units, initial value, rate law, and capacity bounds.
  • Recognize why a constant positive rate cannot continue unchanged at a finite capacity.

Why Capacity Changes an Accumulation Model

A stock is a quantity that accumulates over time: water in a tank, material in a storage system, or participants in a program. If the system has a finite capacity, an accumulation model must represent what happens as the stock approaches that limit. A constant positive rate cannot continue indefinitely without eventually predicting more stock than the system can hold.

There are two common ways to represent capacity. A hard stop allows a positive rate below capacity and sets the rate to zero once capacity is reached. A soft limit reduces the rate as the remaining capacity shrinks. These are different assumptions about the system, so choosing between them is part of building the model.

Definition: Let \(S(t)\) denote a stock at time \(t\), and let \(K>0\) be its capacity. A capacity-respecting prediction satisfies \(0\leq S(t)\leq K\) throughout the time interval being modeled. The accumulation rate is \(S'(t)\); a stock-dependent rate specifies this rate as a function of the current stock.

The rate has units of stock per unit time, while \(K\) and \(S(t)\) have units of stock. In a stock-dependent rule, the rate should also have a physically meaningful value at the capacity. If a system cannot accept additional stock, a natural requirement is that its net accumulation rate be zero at \(S=K\).

A Rate Proportional to Remaining Capacity

Suppose the system accumulates more quickly when there is more unused capacity, and that its rate is proportional to the gap between its current stock and its capacity. With a constant \(k>0\), the model is

$$ S'(t)=k\bigl(K-S(t)\bigr), \qquad S(0)=S_0. $$

Here \(k\) has units of inverse time, so that \(k(K-S)\) has units of stock per time. Assume \(0\leq S_0\leq K\). The rate is positive when \(S<K\), zero when \(S=K\), and negative if \(S>K\). The last feature means that the equation would describe movement back toward capacity even from an over-capacity starting value; for a model of a nonnegative, capacity-limited stock, we use initial values within the allowed range.

Formula: For \(k>0\), \(K\) constant, and initial value \(S(0)=S_0\), the solution of \(S'=k(K-S)\) is $$ S(t)=K-(K-S_0)e^{-kt}. $$

This formula gives a smooth approach to capacity. It does not predict that the stock reaches \(K\) at a finite time when it starts below \(K\). The distinction matters: the soft-limit model slows the rate progressively, whereas a hard-stop model can reach capacity and remain there.

Worked Example: A Tank Approaching Its Capacity

A tank has capacity \(K=1000\) litres and initially contains \(S_0=200\) litres. Suppose its net accumulation rate is \(0.3(1000-S)\) litres per day, with \(S\) measured in litres. Find the predicted stock after four days and check that it respects the capacity.

The rate constant is \(k=0.3\) per day. Substituting \(t=4\), \(K=1000\), and \(S_0=200\) into the formula gives

$$ S(4)=1000-(1000-200)e^{-0.3(4)} =1000-800e^{-1.2}. $$

Using \(e^{-1.2}\approx 0.3011942\), the remaining gap is approximately

$$ 800(0.3011942)=240.9554\text{ litres}, \qquad S(4)\approx1000-240.9554=759.0446\text{ litres}. $$

Rounded to one decimal place, the prediction is \(759.0\) litres. It is between \(0\) and \(1000\) litres. To check the model itself, differentiate the formula: \(S'(t)=0.3(1000-200)e^{-0.3t}=240e^{-0.3t}\) litres per day. Also, \(0.3(1000-S(t))=0.3(800e^{-0.3t})=240e^{-0.3t}\) litres per day. The initial value is \(S(0)=1000-800=200\) litres, as required.

Why the Formula Is the Unique Solution

Checking that a proposed formula satisfies a differential equation and an initial value verifies that it is a solution. To establish that it is the solution, we must also show that every solution with that initial value has the same form. The integrating-factor argument below does both by determining the form of an arbitrary solution.

1
Measure the gap.
For any solution \(S\), define \(G(t)=K-S(t)\). Since \(K\) is constant, \(G'(t)=-S'(t)\), and the differential equation gives \(G'=-kG\).
2
Use an integrating factor.
Multiply \(G'+kG=0\) by \(e^{kt}\). The product rule gives \((e^{kt}G(t))'=e^{kt}(G'+kG)=0\).
3
Apply the initial value.
A differentiable function with zero derivative on the interval is constant there. Thus \(e^{kt}G(t)=G(0)=K-S_0\), so every solution must have \(G(t)=(K-S_0)e^{-kt}\).
4
Recover the stock.
Since \(S=K-G\), every solution is \(S(t)=K-(K-S_0)e^{-kt}\). This establishes uniqueness as well as the formula.

The same expression makes the capacity check direct. For \(t\geq0\), \(0<e^{-kt}\leq1\). Rewriting the solution gives

$$ S(t)=S_0e^{-kt}+K(1-e^{-kt}). $$

Both coefficients are nonnegative and sum to one. When \(0\leq S_0\leq K\), this expression is a weighted average of \(S_0\) and \(K\), so \(0\leq S(t)\leq K\). If \(S_0<K\), then \(K-S(t)=(K-S_0)e^{-kt}>0\) at every finite \(t\), so the model approaches capacity without reaching it. Also, \(S'(t)=k(K-S_0)e^{-kt}>0\), so the stock increases throughout that case.

A Hard Stop at Capacity

A hard-stop rule is useful when the system accepts stock at a specified constant rate until it is full, and accepts none afterward. Let the rate below capacity be \(r>0\), and let \(0\leq S_0<K\). Then the stock follows \(S'=r\) while \(S<K\), and \(S'=0\) once it is full. The time to reach capacity is found from \(S_0+rt=K\), giving \(t_*=(K-S_0)/r\). The resulting stock prediction is

$$ S(t)= \begin{cases} S_0+rt, & 0\leq t\leq t_*,\\ K, & t\geq t_*. \end{cases} $$

The two expressions agree at \(t=t_*\), because \(S_0+rt_*=K\). The stock is continuous there, but its rate changes from \(r\) to \(0\). Thus the derivative of this piecewise model need not exist at the switching time; the rate rule applies on either side of the switch.

Worked Example: Filling a Storage Unit with a Hard Stop

A storage unit holds at most \(50\) kilograms. It contains \(38\) kilograms initially and receives material at \(12\) kilograms per day until it is full. Find when the capacity is reached and the stock after four days.

The unused capacity is \(50-38=12\) kilograms. At \(12\) kilograms per day, the time to fill is

$$ t_*=\frac{50-38\text{ kg}}{12\text{ kg/day}} =\frac{12}{12}\text{ day}=1\text{ day}. $$

Since four days is after the switching time, the piecewise rule gives \(S(4)=50\) kilograms. Check the boundary: before filling, \(S(1)=38+(12)(1)=50\) kilograms; after filling, the model holds the stock at \(50\) kilograms. Therefore it never predicts more than the \(50\)-kilogram capacity. The predicted rate is \(12\) kilograms per day before the unit is full and zero afterward.

A Maximum Rate Followed by a Soft Limit

A system may have a maximum operating rate at low stock, while also slowing as it approaches capacity. One model uses the smaller of a fixed maximum rate \(r_{\max}\) and a capacity-dependent rate \(k(K-S)\):

$$ S'(t)=\min\{r_{\max},\,k(K-S(t))\}, \qquad S(0)=S_0. $$

For \(S\leq K-r_{\max}/k\), the capacity-dependent rate is at least \(r_{\max}\), so the rate is capped at \(r_{\max}\). Above that threshold, the rate is \(k(K-S)\). The transition stock is \(K-r_{\max}/k\), provided it lies between the initial stock and capacity. Once the system enters the second regime, the proportional-gap formula applies with the transition stock as its new initial value.

Worked Example: A Capped Rate That Slows Near Capacity

Let a system have capacity \(K=100\) units, initial stock \(S_0=0\), maximum rate \(r_{\max}=10\) units per day, and \(k=0.2\) per day. Use the smaller of \(10\) and \(0.2(100-S)\) as its rate. Find its stock after ten days.

The transition occurs where the two candidate rates are equal:

$$ 0.2(100-S)=10 \quad\Longrightarrow\quad 100-S=50 \quad\Longrightarrow\quad S=50\text{ units}. $$

At \(10\) units per day, the time to accumulate the first \(50\) units is \(50/10=5\) days. Thus \(S(t)=10t\) units for \(0\leq t\leq5\), and \(S(5)=50\) units. For \(t\geq5\), the stock follows the proportional-gap rule with initial stock \(50\) at time \(5\):

$$ S(t)=100-(100-50)e^{-0.2(t-5)} =100-50e^{-0.2(t-5)}. $$

At \(t=10\), \(S(10)=100-50e^{-1}\). Since \(e^{-1}\approx0.3678794\), this is \(100-18.3940=81.6060\) units, or \(81.61\) units rounded to two decimal places. The rate in the second regime at that time is \(0.2(100-81.6060)=3.6788\) units per day, which is below the \(10\)-unit-per-day maximum. The transition also checks: both regimes give \(50\) units at day five, and both give a rate of \(10\) units per day there. The predicted stock stays below the \(100\)-unit capacity.

Common Mistake: Continuing a Constant Rate Past Capacity

Suppose a container has capacity \(100\) litres, currently holds \(90\) litres, and receives liquid at a constant \(5\) litres per hour. A mistaken calculation applies that rate without change for three hours:

$$ 90+(5)(3)=105\text{ litres}. $$

The arithmetic is correct, but the prediction exceeds capacity. The problem is the assumption that the full rate continues after the container fills. Under a hard-stop rule, the time to fill is \((100-90)/5=2\) hours, and the stock remains \(100\) litres during the third hour. Under a soft-limit rule, the rate must instead decrease as the stock approaches \(100\) litres. The model’s rate assumption, not merely its final arithmetic, determines whether the prediction respects the limit.

Choosing and Checking a Capacity Rule

A capacity value alone does not determine the rate law. A hard stop is appropriate when the rate is effectively unchanged below capacity and then switches off. A soft limit is appropriate when the rate declines as available space disappears. A capped soft-limit rule can represent a maximum rate at low stock together with gradual slowing near capacity.

Before using a prediction, check the model in a consistent order:

  • Confirm that \(S\) and \(K\) have the same stock units and that the rate has units of stock per time.
  • Check the initial value and the rate at meaningful boundary values, especially \(S=K\).
  • For a piecewise rule, calculate the switching stock or time and verify that the formulas agree at the switch.
  • Check the predicted stock against \(0\leq S(t)\leq K\) over the modeled interval, not just at one convenient time.
  • Ask whether the chosen rule describes the actual mechanism: reaching a stop, slowing continuously, or both.

These checks connect the calculus to the intended system. A formula can be algebraically correct yet unsuitable if it assumes a rate that continues beyond the physical capacity. A capacity-limited model should encode the limit in its rate rule, and its predictions should be checked against the range the system can actually occupy.

Check Your Understanding

For each question, identify the rate assumption and use it to reason about the capacity.

  1. In the model \(S'=k(K-S)\), what does \(k\) measure in terms of units, and what is the rate when \(S=K\)?
  2. If \(0\leq S_0<K\) and \(k>0\), why does the proportional-gap model remain below capacity at every finite time?
  3. A stock of \(24\) kilograms fills a \(60\)-kilogram store at \(9\) kilograms per day until full. Under a hard-stop rule, how long does filling take, and what is the stock after five days?
  4. For the rule \(S'=\min\{8,0.4(80-S)\}\), at what stock does the model change from the maximum-rate regime to the capacity-dependent regime?
  5. Give one reason a constant positive rate applied indefinitely may fail as a capacity-limited model.

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