Straight line depreciation - the value reduces by the same amount every time period.
Reducing balance depreciation - the value reduces by a percentage of the previous value every time period.

Ideas

Straight line depreciation
Reducing balance depreciation
Depreciation as a geometric sequence

Straight line depreciation

Straight line depreciation is a bit like simple interest in reverse because the principal is reduced by the same amount every time period. The graph showing the value at regular intervals will appear as a downward-sloping straight line. The slope of the line reflects the fixed quantity lost from the value in each period. Faster depreciation means a graph with a steeper slope.

The straight line method assumes the value of depreciation is constant per period.

Straight line depreciation

\text{Number of periods (years)}

5000

10000

15000

20000

\text{Value } (\$)

This straight line graph shows a \$20\,000 initial value with an annual depreciation of \$3000 p.a. The item reduces to a worth of \$0 after approximately 6.7 years.

Straight line depreciation can also be modelled using an arithmetic sequence. The recurrence relation for the graph above would be V_{n+1}=V_n-3000 where V_0=20\,000 and V_n represents the value at the end of the nth year.

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.

Example 2

A car is initially purchased for \$24\,000 depreciates by \$1700 each year.

Write a recurrence relation, V_n, that gives the value of the car, in dollars, after n years. Write both parts including V_0.

Worked Solution

Create a strategy

Use the recurrence form: V_{n}=V_{n-1}-d where V_0=P.

Apply the idea

The initial value of the car is P=24\,000, and the yearly depreciation is d=1700. So the recurrence relation is:

\displaystyle V_n

\displaystyle =

\displaystyle V_{n-1}-1700,

where V_0=24\,000

Use the sequence facility of your calculator to determine the value of the car after 7 years.

Worked Solution

Create a strategy

Enter the recurrence relation you found in part (a) along with N=7 into your calculator.

Apply the idea

You should get: V_7= 12\,100.

The value of the car after 7 years will be \$12\,100.

After how many years will the value of the car first fall below \$10\,100? Your answer should be a whole number.

Worked Solution

Create a strategy

Input the recurrence relation given in the question into your calculator. Then scroll through the table of values until you find the first value of n for which the value is less than \$10\,100.

Apply the idea

In the list on your calculator, the first value less than 10\,100, should be 8700. The corresponding value of n should be n=9.

After 9 years, the value of the car will fall below \$10\,100.

Idea summary

For straight line depreciation, the principal, P, is reduced by the same amount, d, every time period. The graph showing the value will appear as a downward-sloping straight line.

Straight line depreciation can be modelled using a recurrence relation of the form V_{n+1}=V_n-d, \, V_0=P.

Reducing balance depreciation

In a similar way to how investments with compound interest increase by a percentage of the value at the start of a time period, assets that are subject to reducing-balance depreciation decrease in value by a percentage of the value at the start of each time period.

This is the more common form of depreciation. We will calculate this kind of depreciation using two methods:

Using the reducing balance depreciation formula, similar to calculating compound interest.
Using a geometric sequence to model the situation.

The formula is just slightly different from the compound interest formula. The difference is that we are reducing the value so we must multiply the principal by a number less than 1 each time. For example, reducing by 5\% is the same as multiplying by 100\%-5\% or 95\% or 0.95. Therefore the formula has 1-r in the bracket instead of 1+r. (Or we could in fact consider it the same formula with a negative rate).

The depreciation formula is A=P(1+r)^n where P is the principal (or initial) amount, r is the depreciation rate per time period, n is the number of time periods, A is the value of the item being depreciated.

Note: The amount an item is worth after depreciation is also called the expected value, book value or residual value.

Examples

Example 3

Han's share portfolio of \$83\,000 fell 14\% per month for the first 4 months of the Global Financial Crisis and then 3\% per month for the 5 months after that.

What was the value of his portfolio after 9 months?

Worked Solution

Create a strategy

Use the depreciation formula: A=P(1+r)^n with different rates of depreciation during the different time periods.

Apply the idea

For the first 4 months, P=83\,000, r=14\%=0.14 and n=4. Both n and r are already in months.

\displaystyle A	\displaystyle =	\displaystyle 83\,000 \times (1-0.14)^4	Substitute P, \, r, and n
	\displaystyle =	\displaystyle \$45\,401.677\,28	Evaluate and round

For the 5 months after that, P=45\,401.677\,28, r=3\%=0.03 and n=5. Both n and r are already in months.

\displaystyle A	\displaystyle =	\displaystyle 45\,401.677\,28 \times (1-0.03)^5	Substitute P, \, r, and n
	\displaystyle =	\displaystyle \$38\,997.97	Evaluate and round

The value is \$38\,997.97 after the full 9 months.

Idea summary

Depreciation formula:

\displaystyle A=P(1-r)^n

\bm{P}

is the principal

\bm{r}

is the depreciation rate per time period

\bm{n}

is the number of time periods

\bm{A }

is the depreciated value of the item

Depreciation as a geometric sequence

Reducing balance depreciation is where the value at the start of each year is multiplied by a constant rate of depreciation. Therefore we can solve depreciation problems using the geometric sequence forms, where the common ratio will always be less than 1. Notice the explicit rule is the same as the depreciation formula.

For an item with initial value, P, at a depreciation rate of r per period, the sequence of the value of the item over time forms a geometric sequence with a starting value of P and a common ratio of (1-r).

The sequence which generates the value, V_n, of the item at the end of each depreciation period is:

Recursive form: V_n=V_{n-1} \times (1-r), where V_0=P
Explicit form: V_n=P(1-r)^n

Examples

Example 4

A brand new car depreciates in value each year and its value is modelled by:

V_n=0.89V_{n-1},\,V_0=21\,000 where V_n is the value, in dollars, of the car after n years.

How much was the car purchased for?

Worked Solution

Create a strategy

Use the value of V_0.

Apply the idea

The car was purchased for \$21\,000.

As a percentage, what is the annual depreciation rate?

Worked Solution

Create a strategy

Use the recursive form: V_n=V_{n-1} \times (1-r), where V_0=P

Apply the idea

If we compare our given rule to the above recursive form we can see that 0.89=1-r. So we need to solve this equation to find r.

\displaystyle 1-r	\displaystyle =	\displaystyle 0.89	Write the equation
\displaystyle -r	\displaystyle =	\displaystyle 0.89-1	Subtract 1 from both sides
\displaystyle -r	\displaystyle =	\displaystyle -0.11	Evaluate
\displaystyle r	\displaystyle =	\displaystyle 0.11	Multiply both sides by -1
\displaystyle r	\displaystyle =	\displaystyle 11\%	Write as a percentage

Use the sequence facility of your calculator to determine the value of the car after 9 years.

Worked Solution

Create a strategy

Input the recurrence relation given in the question into your calculator along with n=9.

Apply the idea

You should get the following value:V_9=\$7357.48The value of the car after 9 years is \$7357.48.

When a car is worth less than \$500 it is deemed only useful for parts. At the end of which year is the car only useful for parts?

Worked Solution

Create a strategy

Input the recurrence relation given in the question into your calculator. Find the first value of V_n that is less than 500.

Apply the idea

By looking through the list generated by your calculator, should should find that the first value less than \$500 is \$448.84 which occurs when n=33.

At the end of the 33rd year, the car will only be useful for parts.

Idea summary

The sequence which generates the value, V_n, of the item at the end of each depreciation period is:

Recursive form:

\displaystyle V_n

\displaystyle =

\displaystyle V_{n-1} \times (1.r),

where V_0=P

Explicit form:

\displaystyle V_n

\displaystyle =

\displaystyle P(1-r)^n

Outcomes

ACMGM070

use arithmetic sequences to model and analyse practical situations involving linear growth or decay; for example, analysing a simple interest loan or investment, calculating a taxi fare based on the flag fall and the charge per kilometre, or calculating the value of an office photocopier at the end of each year using the straight-line method or the unit cost method of depreciation

ACMGM074

use geometric sequences to model and analyse (numerically, or graphically only) practical problems involving geometric growth and decay; for example, analysing a compound interest loan or investment, the growth of a bacterial population that doubles in size each hour, the decreasing height of the bounce of a ball at each bounce; or calculating the value of office furniture at the end of each year using the declining (reducing) balance method to depreciate

ACMGM096

with the aid of a calculator or computer-based financial software, solve problems involving compound interest loans or investments; for example, determining the future value of a loan, the number of compounding periods for an investment to exceed a given value, the interest rate needed for an investment to exceed a given value

6.05 Analysing depreciation

Introduction

Ideas

Straight line depreciation

Examples

Example 1

Example 2

Reducing balance depreciation

Examples

Example 3

Depreciation as a geometric sequence

Examples

Example 4

Outcomes

ACMGM070

ACMGM074

ACMGM096

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