Depreciation is the term used for the loss of value of an item over time. The loss of value is often due to the fact that the item has been used regularly and some of its utility has been lost. Cars, for example, tend to wear out with use and do not run as well or look as good as they did when new. Also, an item can depreciate in value because fashions change and older things sometimes become less desirable.

Flat rate and unit cost depreciation

If the amount of depreciation is a fixed amount per period, or a fixed percentage of the initial value, it is called flat rate depreciation. If the amount of depreciation is based on the unit output of the asset, such as kilometres travelled for a car, it is called unit cost depreciation. In each case, the rate of depreciation remains the same across the life of the asset, and so the asset's value shows linear decay over time, hence, this type of depreciation is often referred to as straight line depreciation.

Flat rate and unit cost depreciation can be modelled using the following recurrence relation:V_{n+1}=V_n-d, \, V_0=aWhere V_{n+1} is the value of the asset after n+1 time periods, d is the amount of depreciation per time period, and a is the initial value of the asset.

Examples

Example 1

The graph shows the depreciation of a car's value over 4 years.

\text{Age (years)}

9000

18000

27000

36000

\text{Value } (\$)

What is the initial value of the car?

Worked Solution

Create a strategy

Find the value at year 0.

Apply the idea

The y-intercept on the graph is (0,36\,000). So at year 0 the value was \$36\,000.

The initial value is \$36\,000.

By how much did the car depreciate each year?

Worked Solution

Create a strategy

Subtract the value of the car after one year from its initial value.

Apply the idea

First we need to know how much each interval on the vertical axis is increasing by.

\text{Age (years)}

9000

18000

27000

36000

\text{Value } (\$)

Each \$9000 on the vertical axis is divided into 5 intervals. So each interval represents 9000\div 5=\$1800.

After 1 year, the value of the car is 1 interval above \$27\,000. So the value is 27\,000+1800=\$28\,800.

Now we are ready to subtract the value of the car after one year from its initial value.

\displaystyle \text{Depreciation}	\displaystyle =	\displaystyle 36\,000 - 28\,800	Subtract the values
	\displaystyle =	\displaystyle \$7200	Evaluate

After how many years will the car be worth \$14\,400?

Worked Solution

Create a strategy

Divide the amount of depreciation that we want by the depreciation rate per year.

Apply the idea

The depreciation amount can be found by subtracting \$14\,400 from the initial value:

\displaystyle \text{Depreciation}	\displaystyle =	\displaystyle 36\,000-14\,400	Subtract 14\,400 from the initial value
	\displaystyle =	\displaystyle \$21\,600	Evaluate

To find how many years this will take, divide the depreciation by the depreciation rate of \$7200 per year.

\displaystyle \text{Years}	\displaystyle =	\displaystyle \dfrac{21\,600}{7200}	Divide by the depreciation per year
	\displaystyle =	\displaystyle 3	Evaluate

The car will be worth \$14\,400 after 3 years.

What is the value of the car after 4 years ?

Worked Solution

Create a strategy

In the graph, find the y-value of the point in the depreciation line which corresponds to 4 years.

Apply the idea

\text{Age (years)}

9000

18000

27000

36000

\text{Value } (\$)

The value that corresponds to the 4 years hits the vertical axis 1 interval below 9000.

We learnt from part (b) that each interval represents \$1800.

So the corresponding value is 9000-1800=\$7200.

The car will be worth \$7200 after 4 years.

Idea summary

Flat rate and unit cost depreciation can be modelled using the following recurrence relation:

\displaystyle V_{n+1}=V_n-d, \, V_0=a

\bm{V_{n+1}}

is the value of the asset after after n+1 time periods

\bm{d}

is the amount of depreciation per time period

\bm{a}

is the initial value of the asset

Reducing balance depreciation

Whereas flat rate and unit cost have a constant amount of depreciation per time period, a reducing balance depreciation has a constant depreciation ratio or percentage. For example, an asset may depreciate at a constant rate of 10\% per year. If a graph of this is plotted, it will be non-linear.

We can use the following recurrence relation:V_{n+1}=RV_n, \, V_0=aWhere V_{n+1} is the value of the asset after n+1 time periods, R is the remaining proportion of the asset after each time period (usually expressed as a decimal), and a is the initial value of the asset.

Unlike flat rate and unit cost, reducing balance depreciation is an example of a geometric decay.

It is helpful to compare the graphs of flat rate and unit cost depreciation with that of reducing balance depreciation to see the differences.

Consider an example with an initial investment of \$1000. Below, the first graph with the blue dots represent a constant drop in value of \$100 each year. The second graph with the green dots represent a drop in value of 10\% each year.

500

600

700

800

900

1000

1100

Flat rate

500

600

700

800

900

1000

1100

Reducing balance

As the graphs show, a constant or flat rate of depreciation ends up being modelled by a linear function or an arithmetic sequence. A percentage rate of depreciation is best modelled by an exponential function or a geometric sequence.

We can use the sequence facility of our CAS calculator to analyse situations involving depreciation. Identify the type of depreciation involved and create a table of values or graph to display the value over time.

Examples

Example 2

After one year, the value of a company’s machinery had decreased by \$16\,020 from \$89000. The value of the machinery depreciates by a constant percentage each year.

At what rate did the machinery depreciate in the first year?

Worked Solution

Create a strategy

Divide the given depreciation amount by the initial value and write as a percentage.

Apply the idea

\displaystyle \text{Depreciation rate}	\displaystyle =	\displaystyle \dfrac{16\,020}{89\,000}	Divide 16\,020 by 89\,000
	\displaystyle =	\displaystyle 0.18	Evaluate
	\displaystyle =	\displaystyle 18\%	Write as a percentage

What will the machinery be worth at the end of the second year?

Worked Solution

Create a strategy

Subtract the depreciation amount in the second year from the value at the end of the first year.

Apply the idea

The value of the machinery at the end of the first year can be found by subtracting the depreciation amount from the initial value.

\displaystyle \text{Value}	\displaystyle =	\displaystyle 89\,000-16\,020	Subtract 16\,020 from the 89\,000
	\displaystyle =	\displaystyle \$72\,980	Evaluate

The depreciation amount in the second year is 18\% of the depreciated value at the end of the first year. So the value of the machinery at the end of the second year is:

\displaystyle \text{Value}	\displaystyle =	\displaystyle 72\,980 - 0.18 \times 72\,980	Write the values
	\displaystyle =	\displaystyle \$59\,843.60	Evaluate

Write a recurrence relation, V_{n+1}, that gives the value of the machinery at the end of year n.

Worked Solution

Create a strategy

Use the recurrence form: V_{n+1}=RV_n where V_0=a.

Apply the idea

The initial value of the machinery is V_0=89\,000. The value of the machinery is decreased by 18\% each year. So:

\displaystyle R	\displaystyle =	\displaystyle 100 - 18	Subtract 18\% from 100\%
	\displaystyle =	\displaystyle 81\%	Evaluate
	\displaystyle =	\displaystyle 0.82	Write as decimal

So the recurrence relation is: V_{n+1}=0.82V_n, \, V_0=89\,000

The company bought this machinery at the end of 2011. When the value of the machinery falls below \$3000. they will invest in new machinery. In which year will this occur?

Worked Solution

Create a strategy

Input the recurrence relation given in the question into your calculator. Find the first value of n that is less than 3000.

Add the value of n to 2011.

Apply the idea

By looking through the list generated by your calculator, you should find that the first value less than \$3000 is \$2500.57 which occurs when n=18.

So the year in which this will occur is:

\displaystyle \text{Year}	\displaystyle =	\displaystyle 2011 + 18	Add n=18 to 2011
	\displaystyle =	\displaystyle 2029	Evaluate

Idea summary

In reducing balance depreciation, we can use the following recurrence relation:

\displaystyle V_{n+1}=RV_n, \, V_0=a

\bm{V_{n+1}}

is the value of the asset after after n+1 time periods

\bm{R}

is the remaining proportion of the asset after each time period (usually expressed as a decimal)

\bm{a}

is the initial value of the asset

Outcomes

U3.AoS2.2

the use of first-order linear recurrence relations to model flat rate and unit cost, and reduce balance depreciation of an asset over time, including the rule for the future value of the asset after 𝑛 depreciation periods

U3.AoS2.6

demonstrate the use of a recurrence relation to determine the depreciating value of an asset or the future value of an investment or a loan after 𝑛 time periods for the initial sequence

U3.AoS2.4

the use of first-order linear recurrence relations to model compound interest investments and loans, and the flat rate, unit cost and reducing balance methods for depreciating assets, reducing balance loans, annuities, perpetuities and annuity investments

5.03 Depreciation

Introduction

Ideas

Flat rate and unit cost depreciation

Examples

Example 1

Reducing balance depreciation

Examples

Example 2

Outcomes

U3.AoS2.2

U3.AoS2.6

U3.AoS2.4

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