Inventory Change from Inflow and Outflow Rates

From Rates to Inventory

An inventory changes when quantities enter or leave it. A warehouse receives deliveries and ships orders. A reservoir receives water and releases water. In each case, knowing just the inflow or just the outflow is not enough to determine the change: both affect the amount on hand.

The key is to compare the rates. Inflow adds to the inventory, while outflow subtracts from it. Their difference is the net rate of change. Integrating that net rate over a time interval gives the change in inventory during the interval. Adding that change to the initial amount gives the ending inventory.

Central relationship: Inventory change depends on inflow minus outflow. The ending inventory is the initial inventory plus the accumulated net change over the time interval.

Defining the Rates and Inventory

Let $I(t)$ denote the amount in inventory at time $t$. Let $r_{\text{in}}(t)$ be the inflow rate and $r_{\text{out}}(t)$ the outflow rate. Both rates are expressed in inventory units per unit of time. For example, if inventory is measured in boxes and time in hours, each rate has units of boxes per hour.

Definition: The net inventory rate at time $t$ is the inflow rate minus the outflow rate: $$ r_{\text{net}}(t)=r_{\text{in}}(t)-r_{\text{out}}(t). $$ When the rates describe instantaneous flow, this net rate is the derivative of inventory: $$ I'(t)=r_{\text{in}}(t)-r_{\text{out}}(t). $$

If $r_{\text{net}}(t)>0$, the inventory is increasing at that time. If $r_{\text{net}}(t)<0$, it is decreasing. If the two rates are equal, the inventory has zero instantaneous rate of change. These statements concern the direction of change, not whether the inventory itself is positive or negative.

Suppose the rates are continuous or piecewise continuous on an interval from $t=a$ to $t=b$, and the inventory at the beginning is $I(a)$. The Fundamental Theorem of Calculus gives the change over the interval by integrating the derivative.

Theorem: Under these conditions, the change in inventory and the ending inventory are $$ I(b)-I(a)=\int_a^b \bigl(r_{\text{in}}(t)-r_{\text{out}}(t)\bigr)\,dt $$ and $$ I(b)=I(a)+\int_a^b \bigl(r_{\text{in}}(t)-r_{\text{out}}(t)\bigr)\,dt. $$

The integral has units of inventory: multiplying a rate in boxes per hour by time in hours produces boxes. The integral is a net change. It can be positive, negative, or zero; it is not necessarily the total amount that passed into or out of the inventory.

Why Subtracting the Rates Works

Over a short time interval of length $\Delta t$, the amount entering is approximately $r_{\text{in}}(t)\Delta t$, and the amount leaving is approximately $r_{\text{out}}(t)\Delta t$. The inventory change over that short interval is therefore approximately the amount entering minus the amount leaving.

\Delta I\approx r_{\text{in}}(t)\Delta t-r_{\text{out}}(t)\Delta t =\bigl(r_{\text{in}}(t)-r_{\text{out}}(t)\bigr)\Delta t.

Divide the interval from $a$ to $b$ into smaller intervals. Add the approximate changes from all the small intervals. As the intervals become arbitrarily short, this sum approaches the definite integral of the net rate. The individual changes add to the total change, so the initial inventory plus that total change is the ending inventory.

Set the sign convention.
Count inflow as positive and outflow as negative in the balance.

Find the net rate.
Subtract the outflow rate from the inflow rate at each time.

Accumulate the net change.
Integrate the net rate over the interval of interest.

Apply the initial amount.
Add the net change to the inventory at the starting time.

Worked Examples

Worked Example: Variable Inflow and Outflow Rates

A warehouse starts with $120$ boxes. For $0\le t\le 4$ hours, deliveries arrive at $r_{\text{in}}(t)=30+2t$ boxes per hour, and shipments leave at $r_{\text{out}}(t)=10+t$ boxes per hour. Find the inventory after four hours.

First find the net rate:

r_{\text{net}}(t)=(30+2t)-(10+t)=20+t \quad\text{boxes per hour}.

Integrate from $t=0$ to $t=4$ to find the net change:

\begin{aligned} \Delta I &=\int_0^4(20+t)\,dt\\ &=\left[20t+\frac{t^2}{2}\right]_0^4\\ &=\left(20(4)+\frac{4^2}{2}\right)-\left(20(0)+\frac{0^2}{2}\right)\\ &=80+8\\ &=88\text{ boxes}. \end{aligned}

Add the change to the initial inventory:

I(4)=120+88=208\text{ boxes}.

Check: An antiderivative of $20+t$ is $20t+\frac{t^2}{2}$, since its derivative is $20+t$. The net rate is positive throughout the interval, so an ending amount greater than the initial $120$ boxes is consistent with the rates.

Worked Example: Rates That Change at a Specific Time

A storage area begins with $50$ crates. During the first three hours, crates enter at $18$ crates per hour and leave at $12$ crates per hour. During the following two hours, crates enter at $10$ crates per hour and leave at $14$ crates per hour. Find the inventory after five hours.

On $0\le t<3$, the net rate is $18-12=6$ crates per hour. On $3\le t\le5$, it is $10-14=-4$ crates per hour. Integrate each rate on its own interval:

\begin{aligned} \Delta I &=\int_0^3 6\,dt+\int_3^5(-4)\,dt\\ &=6(3-0)+(-4)(5-3)\\ &=18-8\\ &=10\text{ crates}. \end{aligned}

Thus the ending inventory is

I(5)=50+10=60\text{ crates}.

Check: The inventory increases by $18$ crates during the first three hours, reaching $50+18=68$. It then decreases by $8$ crates during the next two hours, reaching $68-8=60$. This agrees with the integral calculation, and the inventory remains positive during both intervals.

Worked Example: Net Rate Changes Sign

A tank contains $40$ liters at time $t=0$. For $0\le t\le5$ minutes, liquid enters at $r_{\text{in}}(t)=12+4t$ liters per minute and leaves at $r_{\text{out}}(t)=20$ liters per minute. Find the amount in the tank after five minutes.

The net rate is

r_{\text{net}}(t)=(12+4t)-20=-8+4t \quad\text{liters per minute}.

Integrate to find the net change:

\begin{aligned} \Delta I &=\int_0^5(-8+4t)\,dt\\ &=\left[-8t+2t^2\right]_0^5\\ &=\bigl(-8(5)+2(5^2)\bigr)-\bigl(-8(0)+2(0^2)\bigr)\\ &=-40+50\\ &=10\text{ liters}. \end{aligned}

Therefore, the amount after five minutes is

I(5)=40+10=50\text{ liters}.

Check: The antiderivative $-8t+2t^2$ differentiates to $-8+4t$. The net rate is negative for $0\le t<2$ and positive for $2<t\le5$. In fact, the inventory at time $t$ is $I(t)=40-8t+2t^2$; at $t=2$, $I(2)=40-16+8=32$ liters. The inventory first decreases and then increases, so a positive total change of $10$ liters over the full interval is plausible.

Common Pitfall: Reversing the Subtraction

A frequent error is to subtract inflow from outflow. In the warehouse example, reversing the order would give

r_{\text{wrong}}(t)=(10+t)-(30+2t)=-20-t.

Integrating this incorrect rate from $0$ to $4$ gives

\int_0^4(-20-t)\,dt=-88.

That would predict an ending inventory of $120-88=32$ boxes. But at every time in the interval, the delivery rate $30+2t$ exceeds the shipment rate $10+t$. The inventory must be increasing, not decreasing. The correction is to define the balance consistently as inflow minus outflow, giving a net change of $+88$ boxes and an ending inventory of $208$ boxes.

Interpreting the Result Carefully

A negative net change does not mean that the inventory is negative. It means the ending amount is less than the starting amount. For instance, a change of $-8$ crates applied to an initial inventory of $50$ crates gives an ending amount of $42$, not $-8$.

Likewise, a positive net change does not mean that no outflow occurred. It means the total amount entering over the interval exceeded the total amount leaving. The integral combines both effects with their signs. If the inventory represents a physical amount, a proposed model should also be checked against the context: an inventory cannot be below zero, and a model predicting that it is may indicate that the assumptions or time interval need attention.

To keep an accumulation problem organized, write the units alongside the rates and the integral. Check that the integration limits match the requested time interval, and apply the initial inventory only after finding the net change. This separates the rate calculation from the final initial-value calculation.

Check Your Understanding

For each question, keep the direction of flow and the time interval explicit.

An inventory receives $25$ units per hour and releases $19$ units per hour. What is its net rate of change?
A tank starts with $70$ liters. Its net rate is $3$ liters per minute for $4$ minutes. What is its ending amount?
In a $6$-hour interval, the inflow rate is $14$ boxes per hour and the outflow rate is $17$ boxes per hour. What is the net change?
Explain why a positive value of $\int_a^b(r_{\text{in}}(t)-r_{\text{out}}(t))\,dt$ does not imply that outflow was zero.
If inflow and outflow rates are given by different formulas before and after a specified time, how should the net change be calculated?

Inventory Change from Inflow and Outflow Rates

What You'll Learn

From Rates to Inventory

Defining the Rates and Inventory

Why Subtracting the Rates Works

Worked Examples

Worked Example: Variable Inflow and Outflow Rates

Worked Example: Rates That Change at a Specific Time

Worked Example: Net Rate Changes Sign

Common Pitfall: Reversing the Subtraction

Interpreting the Result Carefully

Check Your Understanding

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Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

Inventory Change from Inflow and Outflow Rates

What You'll Learn

From Rates to Inventory

Defining the Rates and Inventory

Why Subtracting the Rates Works

Worked Examples

Worked Example: Variable Inflow and Outflow Rates

Worked Example: Rates That Change at a Specific Time

Worked Example: Net Rate Changes Sign

Common Pitfall: Reversing the Subtraction

Interpreting the Result Carefully

Check Your Understanding

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

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