11.05 Depreciation (Enrichment)
Depreciation refers to the situation where items or investments lose value over time. We will consider two types of depreciation in this course:
- 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.
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.
The straight line graph above shows a $\$20000$$20000 initial value with an annual depreciation of $\$3000$$3000 p.a., the item reduces to a worth of $\$0$$0 after approximately $6.7$6.7 years.
Straight line depreciation can also be modelled using an arithmetic sequence. The recursive rule for the graph above would be $V_{n+1}=V_n-3000$Vn+1=Vn−3000 where $V_0=20000$V0=20000 and $V_n$Vnrepresents the value at the end of the $n$nth year.
Worked examples
Example 1
A car purchased for $\$20000$$20000depreciates by $\$2500$$2500 per year. How much is it worth after $6$6 years?
| $\text{Value }$Value | $=$= | $\$20000-\$2500\times6$$20000−$2500×6 |
| $=$= | $\$5000$$5000 |
Example 2
A car purchased for $\$15000$$15000 is worth $\$9000$$9000 after $8$8 years. By how much did it depreciate per year?
| $\text{Depreciation per year}$Depreciation per year | $=$= | $\frac{\text{Value lost}}{\text{Number of years}}$Value lostNumber of years |
| $=$= | $\frac{\$15000-\$9000}{8}$$15000−$90008 | |
| $=$= | $\$750$$750 |
See below to work through an example of using a calculator for problems involving straight line depreciation.
TI Nspire
How to use the TI Nspire to complete the following tasks regarding sequences in a straight line depreciation context.
A car that is initially purchased for $\$36000$$36000 depreciates by $\$2250$$2250 each year.
Using a recursive relationship, generate the value of the vehicle at the end of each year for $5$5 years.
Using an explicit rule, generate the value of the vehicle at the end of each year for $5$5 years.
Determine the value of the car at the end of $8$8 years.
When does this depreciation model predict the car is worthless?
Graph the value of the vehicle from when it was purchased to when it became worthless.
Practice questions
question 1
The graph shows the depreciation of a car's value over 4 years.
What is the initial value of the car?
By how much did the car depreciate each year ?
After how many years will the car be worth $\$14400$$14400 ?
What is the value of the car after 4 years ?
Question 2
A car is initially purchased for $\$24000$$24000 depreciates by $\$1700$$1700 each year.
Write a recurrence relation, $V_n$Vn, that gives the value of the car, in dollars, after $n$n years.
Write both parts (including for $V_0$V0) on the same line, separated by a comma.
Use the sequence facility of your calculator to determine the value of the car after $7$7 years.
After how many years will the value of the car first fall below $\$10100$$10100?
Your answer should be a whole number.
Question 3
A boat, initially purchased for $\$49000$$49000, depreciates at a rate of $\$180$$180 for every $1000$1000 km of use.
Write a recurrence relation, $V_n$Vn, that gives the value of the boat, in dollars, after $n$n thousand kilometres.
Write both parts of the relation (including for $V_0$V0) on the same line, separated by a comma.
Use the sequence facility of your calculator to determine the value of the boat after it has travelled $11000$11000 km.
Use the sequence facility of your calculator to determine after how many kilometres the boat will first fall below $\$47917$$47917 in value.
Your answer should be a multiple of $1000$1000 km.
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.
Reducing balance depreciation formula
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$1 each time. For example, reducing by $5%$5% is the same as multiplying by $100%-5%$100%−5% or $95%$95% or $0.95$0.95. Therefore the formula has $1-r$1−r in the bracket instead of $1+r$1+r. (Or we could in fact consider it the same formula with a negative rate).
$A=P\left(1-r\right)^n$A=P(1−r)n
where: $P$P is the principal (or initial) amount
$r$r is the depreciation rate per time period
$n$n is the number of time periods
$A$A is the value of the item after being depreciated
Note: The amount an item is worth after depreciation is also called the expected value , book value or residual value.
Worked examples
Example 3
Kathleen deposited $\$6500$$6500 into a new superannuation account. This amount decreased by $2%$2% each year for $3$3 consecutive years.
What was the value of her superannuation after $3$3 years? Give your answer to $2$2 decimal places if necessary.
Think: How do we substitute these values into the depreciation formula?
Do:
| $A$A | $=$= | $P\left(1-r\right)^n$P(1−r)n |
| $=$= | $6500\times\left(1-0.02\right)^3$6500×(1−0.02)3 | |
| $=$= | $6500\times0.98^3$6500×0.983 Note: Decreasing by $2%$2% is the same as finding $98%$98% | |
| $=$= | $\$6117.75$$6117.75 |
See below to work through an example of using a calculator for problems involving reduce balance depreciation using the formula and financial application.
TI Nspire
How to use the TI Nspire to complete the following tasks involving reducing balance depreciation using the formula and the inbuilt financial solver.
A new car purchased for $\$28000$$28000 depreciates at a rate $r$r each year.
If the value of the car has reduced to $\$20412$$20412 after $3$3 years, find the value of $r$r as a percentage.
Assuming the rate of depreciation remains constant, how much can the car be sold for after $10$10 years? Give your answer to the nearest dollar.
The owner of the car wants to sell the vehicle before its value falls below $\$12000$$12000. According to the model, by the end of which year should he sell the vehicle?
Practice questions
Question 4
A laptop currently worth $\$1719$$1719 was purchased $4$4 years ago for $\$2000$$2000.
What was the annual depreciation rate, $I$I?
Write your answer as a percentage to two decimal places.
Question 5
Han's share portfolio of $\$83000$$83000 fell $14%$14% per month for the first $4$4 months of the Global Financial Crisis and then $3%$3% per month for the $5$5 months after that.
What was the value of his portfolio after $9$9 months?
Write your answer to the nearest cent.
Question 6
A dishwasher selling for $\$800$$800, depreciates at $11%$11% p.a. .
What is the percentage of the original value that will remain after 1 year?
What is the percentage of the original value that will remain after 2 years?
Write your answer as a percentage to two decimal places.
What is the percentage of the original value that will remain after 3 years?
Write your answer as a percentage to two decimal places.
How many full years will it take for the dishwasher to lose half its original value?
How many years will it take for the dishwasher to lose $90%$90% of its original value?
Depreciation modelled as a geometric sequence
Reducing balance depreciation is where the value at the start of each year is multiplied by a constant rate of depreciation. Therefore we can solve depreciation problems using the geometric sequence forms, where the common ratio will always be less than $1$1. Notice the explicit rule is the same as the depreciation formula.
For an item with initial value, $P$P, at a depreciation rate of $r$r per period, the sequence of the value of the item over time forms a geometric sequence with a starting value of $P$P and a common ratio of $(1-r)$(1−r).
The sequence which generates the value, $V_n$Vn, of the item at the end of each depreciation period is:
- Recursive form:
$V_n=V_{n-1}\times(1-r)$Vn=Vn−1×(1−r), where $V_0=P$V0=P
- Explicit form:
$V_n=P(1-r)^n$Vn=P(1−r)n
Worked example
Example 4
Danielle buys a car for $\$15000$$15000 and is told to expect it to depreciate at a rate of $x%$x% p.a. The progression of the depreciation is shown in the table below.
| Year | Car value at start of year ($) |
Depreciation ($) | Car value at end of year ($) |
|---|---|---|---|
| 1 | $15000$15000 | $2250$2250 | $12750$12750 |
| 2 | $12750$12750 | $1912.50$1912.50 | $10837.50$10837.50 |
| 3 | $10837.50$10837.50 | $1625.63$1625.63 | $9211.88$9211.88 |
(a) Determine $x%$x%, the depreciation rate of the car.
Think: We can use any of the table rows to do this. Use the formula
$\text{Depreciation rate}=\frac{\text{Depreciation in year n}}{\text{Value at start of year n}}\times100%$Depreciation rate=Depreciation in year nValue at start of year n×100%
Do: Using Year 1 we find $rate=\frac{2250}{15000}=0.15=15%$rate=225015000=0.15=15%
Or: Using Year 2 we find $rate=\frac{1912.50}{12750}=0.15=15%$rate=1912.5012750=0.15=15%
Or: Using Year 3 we find $rate=\frac{1625.63}{10837.50}=0.15=15%$rate=1625.6310837.50=0.15=15%
Therefore the depreciation rate is $15%$15% p.a.
(b) Write a recursive rule for the investment.
Think: Each time we are decreasing by $15%$15%, which means multiplying by $0.85$0.85 as ($100%-15%=85%$100%−15%=85%). We want $n=1$n=1 to show the value at the end of year $1$1, therefore let $V_0=$V0= initial value.
Do: Write the rule $V_{n+1}=V_n\times0.85$Vn+1=Vn×0.85, where $V_0=15000$V0=15000
Or: Write the rule $V_n=V_{n-1}\times0.85$Vn=Vn−1×0.85, where $V_0=15000$V0=15000
See below to work through an example of using a calculator for problems involving sequences and reduce balance depreciation.
TI Nspire
How to use the TI Nspire to complete the following tasks involving sequences in a reducing balance depreciation context.
A boat, originally purchased for $\$12000$$12000, depreciates at $15%$15% per annum.
Using a recursive relationship, generate the value of the boat at the end of each year for the first $5$5 years.
Using an explicit rule, generate the value of the boat at the end of each year for the first $5$5 years.
Find the value of the boat at the end of $8$8 years.
At the end of which year does the value of the boat first fall below $\$4000$$4000.
Sketch a graph of the value of the boat over the first ten years.
Practice questions
question 7
A brand new car depreciates in value each year and its value is modelled by
$V_n=0.89V_{n-1}$Vn=0.89Vn−1, $V_0=21000$V0=21000
where $V_n$Vn is the value, in dollars, of the car after $n$n years.
How much was the car purchased for?
As a percentage, what is the annual depreciation rate?
Use the sequence facility of your calculator to determine the value of the car after $9$9 years.
Give your answer to the nearest cent.
When a car is worth less than $\$500$$500 it is deemed only useful for parts. At the end of which year is the car only useful for parts?
question 8
After one year, the value of a company’s machinery had decreased by $\$16020$$16020 from $\$89000$$89000. The value of the machinery depreciates by a constant percentage each year.
At what rate did the machinery depreciate in the first year?
Give your answer as a percentage.
What will the machinery be worth at the end of the second year?
Give your answer to the nearest cent.
Write a recurrence relation, $V_n$Vn, that gives the value of the machinery at the end of year $n$n.
Write both parts of the relation (including for $V_0$V0) on the same line, separated by a comma.
The company bought this machinery at the end of $2011$2011. When the value of the machinery falls below $\$3000$$3000, they will invest in new machinery. In which year will this occur?