Value Over Time and Depreciation

Over time the value of items can change due to various factors such as inflation, market demand, and the nature of the item itself.

These changes in value are referred to as:

  • Appreciation: An increase in value over time. This is often observed in assets like real estate and collectibles, where the value tends to rise as demand increases.
  • Depreciation: A decrease in value over time. This is common in items such as cars, electronic devices, or machinery, which lose value as they age or as new models replace them.

Appreciation is more challenging to predict and quantify, as it depends on various unpredictable factors like market trends, demand fluctuations, and external economic conditions. In contrast, depreciation is often more straightforward to calculate, as it can typically be estimated using the principal value of the item and its expected lifespan.

Use this page to revise the following concepts of value over time and depreciation:

Flat rate depreciation

Flat rate depreciation refers to the decrease in value of an item by a fixed amount over time. This is calculated similarly to simple interest, but instead of adding value, a fixed amount is subtracted each time period .

Note

Flat rate depreciation can also be called Straight line depreciation. This term is especially common in the accounting field.

Flat rate depreciation recurrence relation

Flat rate depreciation decreases the value of an item by a fixed amount over time. The depreciation amount per time period is calculated as:

\[D = \frac{r}{100}\ \times\ V_{0}\]

Where:

  • \(D\) is the depreciation amount per time period
  • \(r\) is the interest rate per time period (as a percentage)
  • \(V_0\) is the initial value or principal amount

A recurrence relation can be used to describe incremental changes to the value of an item over number of time periods. For flat rate depreciation:

\[V_{0} = a, \quad{V}_{n + 1} = V_{n} - D\]

Where:

  • \(V_{n + 1}\) is the future value after \(n + 1\) time periods
  • \(V_n\) is the value at the nth time period
  • \(a\) is the constant (representing the principal value)
  • \(n\) is the number of time periods

​

Worked Example

New office furniture is purchased for \(\$10,000\) and is expected to depreciate at a flat rate of \(8.4\%\) per year. Calculate the value of the office furniture at the end of each year for the next three years.

Determine the given variables

\[\begin{align}&V_0 = 10000 \\ &r = 8.4 \\ &n = 3\end{align}\]

Calculate the flat rate of depreciation amount each year

\[\begin{align}D &= \frac{8.4}{100}\ \times\ 10000 \\ &= 840\end{align}\]

Construct the recurrence relation

\[V_0 = 10000, \quad{V}_{n + 1} = V_n - 840\]

Calculate the values each year

\[\begin{align}V_1 &= V_0 - 840 = 10000 - 840 = 9160 \\ V_2 &= V_1 - 840 = 9160 - 840 = 8320 \\ V_3 &= V_2 - 840 = 8320  - 840 = 7480\end{align}\]

The office furniture has a value of \(\$9,160\) at the end of the first year, \(\$8,320\) at the end of the second year, and \(\$7,480\) at the end of the third year.


Rule for flat rate depreciation recurrence relation

A recurrence relation requires knowledge of the value from the previous time period to calculate the value for the current time period. This can become tedious as the number of time periods increases. Since the amount subtracted is directly proportional to the number of years, the value at any time period can be simplified using the formula:

\[V_n = V_0 - nD\]

Worked Example

New office furniture is purchased for \(\$10\ 000\) and is expected to depreciate at a flat rate of \(8.4\%\) per year. Calculate the value of the office furniture after \(8\) years.

Determine the given variables

\[\begin{align}V_0 &= 10000 \\ r &= 8.4 \\ n &= 8\end{align}\]

Calculate the flat rate of depreciation amount each year

\[\begin{align}D &= \frac{8.4}{100}\ \times\ 10000 \\ &=840\end{align}\]

Substitute into the rule

\[\begin{align}V_8 &= V_0 - 8 \times D \\ &= 10000 - 8 \times 840 \\ &=3280 \end{align}\]

The office furniture has a value of \(\$3280\) after \(8\) years.


Use these tabs to view the worked solution for the question above.

The value after \(n\) years can be determined using this formula:

\[V_n = V_0 - n\left(\frac{r}{100}\ \times\ V_0\right)\]

where \(r\) is the rate of depreciation.

\[\begin{align} &n = 15\\ &r = 2.5\\ &V_0 = \$120\ 000\\ &V_{15} = 120000 - 15\left(\frac{2.5}{100}\ \times\ 120000\right) =1 20000 - 15\left(3000\right) = \$75\ 000 \end{align}\]

Unit cost depreciation

Unit cost depreciation is calculated similarly to flat rate depreciation. Instead of calculating the fixed amount of depreciation from a depreciation rate, the unit cost is used instead. This is typically used when the depreciation is dependent on the usage of item.

Recursion relation:

\[V_0 = a, \quad{V}_{n + 1} = V_n - D\]

Where:

  • \(V_{n+1}\) is the future value after \(n + 1\) time periods
  • \(V_n\) is the value at the nth time period
  • \(a\) is a constant representing the principal value
  • \(n\) is the total quantity of usage units
  • \(D\) is the depreciation per time period

Rule:

\[V_n = V_0 - nD\]

​

Worked Example

A commercial washing machine is purchased for \(\$18\ 000\). It is expected to depreciate by \(\$900\) for every \(3000\) wash cycles. If the machine performs \(4000\) wash cycles per year, calculate its value after \(8\) years using the unit cost depreciation method.

Determine the given variables
\[\begin{align}&V_0 = 18000 \\ &n=8\ \times\ 4000 = 32000\end{align}\]

Calculate the unit cost from the given information

\({Total cost} = 900\)

\(\text{Number of units} = 3000\)

\[\begin{align}D = \text{Unit cost} &= \frac{\text{Total cost}}{\text{Number of units}} \\ &= \frac{900}{3000} = 0.3\end{align}\]

Substitute into the rule

\[\begin{align}V_{32000} &= 18000 - 32000(0.3) \\ &= 8400\end{align}\]

The washing machine has a value of \(\$8400\) after \(8\) years.


Use these tabs to view the worked solution for the question above.

The remaining value after \(100\ 000\) units per year for \(4\) years can be determined using this formula:

\[V_n = V_0 - nD\]

Where

  • \(n\) = total number of units
  • \(D\) = unit cost

\[\begin{align} &n = 100000\ \times\ 4 = 400000\\ &D = \frac{350000}{1000000} = \$0.35\\ &V_{400000} = 350000 - 400000\left(0.35\right) = \$210000 \end{align}\]

Reducing balance depreciation

Reducing balance depreciation is similar to compound interest calculations, where the depreciation amount is based on the principal amount at each increment. Therefore, it is calculated in the same manner, by finding the depreciation factor:

\[R = 1 - \frac{r}{100}\]

Reducing balance depreciation

Recurrence Relation And using the recurrence relation:

\[V_0 = a,\quad V_{n + 1} = {RV}_n\]

Where:

  • \(R\) is the depreciation factor

​

Worked Example

A car is purchased for \(\$26\ 000\) and depreciates at a reducing balance rate of \(15\%\text{ p.a.}\) Determine the value of the car at the end of each year for the first \(3\) years.

Determine the given variables

\[\begin{align}&V_0 = 26000 \\ &r = 15 \\ &n = 3\end{align}\]

Calculate the depreciation factor

\[
\begin{align}
R &= 1 - \frac{15}{100} \\
&= 0.85
\end{align}
\]

Construct the recurrence relation

\[ V_0 = 26000,\quad\quad V_{n + 1} = 0.85V_n\]

Calculate the values each year

\[
\begin{align}
V_1 &= RV_0 = 0.85\left(26000\right) = 22100 \\
V_2 &= RV_1 = 0.85\left(22100\right) = 18785 \\
V_3 &= RV_2 = 0.85\left(18785\right) = 15967.25
\end{align}
\]

The car has a value of \(\$22, 100\) at the end of the first year, \(\$18, 785\) at the end of the second year, and \(\$15, 967.25\) at the end of the third year.


Use these tabs to view the worked solutions for the question above.

The value can be found using the recurrence relation

\[V_0 = a,\ V_{n + 1} = {RV}_n\]

where \(R\) = depreciation factor.

First, to determine the depreciation factor,

\[R = 1 - \frac{r}{100} = 1 - \frac{25}{100} = 0.75\]

This can be used to calculate the value based on the previous year’s value:

\[\begin{align}&V_1 = 0.75\ (V_0) = 0.75\left(6750\right) = \$5062.50 \\ &V_2 = 0.75\ (V_1)  =0.75\left(5062.50\right) = \$3796.88\end{align}\]

Rule for Reducing Balance Depreciation Recurrence Relation

The tedious nature of computing a recurrence relationship for a greater number of periods also exist for reducing balance depreciation calculations. This can be simplified using the rule:

\[V_n = R^nV_0\]

​

Worked Example

A car is purchased for \(\$26\ 000\) and depreciates at a reducing balance rate of \(15\%\text{ p.a.}\) Determine the value of the car after 8 years.
Determine the given variables

\[\begin{align}&V_0 = 26000 \\ &r = 15 \\ &n = 8\end{align}\]

Calculate the depreciation factor

\[
\begin{align}
R &= 1 - \frac{15}{100} \\
&= 0.85
\end{align}
\]

Substitute into the rule

\[V_8 = \left(0.85\right)^{8}\left(26000\right) = 7084.753\ \ldots\ = 7084.75\]

The car has a value of \(\$7084.75\) after \(8\) years.