PT - Depreciation Lesson

Depreciation

Linear Depreciation

A car loses its value over time due to normal wear and tear. If you buy your car for one price, chances are you won’t be able to sell it for that same price even if it’s only a year later! The loss in value is called depreciation.

Let’s splurge in this example and buy a Lamborghini for $274,390.00 (wow). With linear depreciation, the car depreciates the same amount each year.

Car’s Value = Original Cost - Depreciation x Years

We can shorten this to the formula: LaTeX: V = C - dt $V = C - dt$

Example 1: What is the value of the Lamborghini after 5 years with a depreciation loss of $10,000.00 per year?

LaTeX: V = C - dt $V = C - dt$

V = $274,390.00 - $10,000.00(5)

V = $224,390.00

The car lost $50,000.00 in value in 5 years.

Example 2: Try this one on your own and then check your answer below. What is the value of a sedan that was purchased for $18,575.00 after 3 years with a depreciation loss of $1,500.00 per year?

LaTeX: V = C - dt $V = C - dt$

V = $18,575.00 - $1,500.00(3)

V = $14,075.00

This formula can also be used to find the depreciation rate if you know the purchase price and what the car is worth years later.

Example: What is the yearly depreciation loss if a coupe that was purchased for $7,000.00 is now worth $4,075.00 after 4 years?

LaTeX: V = C - dt $V = C - dt$

$4,075.00 = $7,000.00 - 3d Subtract $7,000 from each side.

-$2,925.00 = -3d Divide each side by -3.

$957.00 = d

Exponential Depreciation

With exponential depreciation, the car depreciates the same percentage each year.

LaTeX: V=C(1-r)^t $V=C(1-r)^t$

V = the car’s new value

C = the original cost

r = rate of depreciation

t = time in years

Example 1: Let's use the same Lamborghini example that sells for $274,390.00. What is the value of the Lamborghini after 5 years with a depreciation loss of 10% per year?

LaTeX: V=C(1-r)^t $V=C(1-r)^t$

V = $274,390.00(1 – 0.10)⁵

V = $162,024.55

The car lost $112,365.45 in value in 5 years!

Example 2: Try this one on your own and then check your answer below. What is the value of a sedan that was purchased for $18,575.00 after 3 years with a depreciation loss of 15% per year?

LaTeX: V=C(1-r)^t $V=C(1-r)^t$

V = $18,575.00(1 – 0.15)³

V = $11,407.37

Just like the other formula, this can also be used to find other missing values.

Example: What was the original cost of a coupe that is now worth $9,200.00 after 4 years at a depreciation rate of 12%?

LaTeX: V=C(1-r)^t $V=C(1-r)^t$

$9,200.00 = C(1 – 0.12)⁴Simplify (1 – 0.12)⁴

$9,200.00 = C(0.5997) Divide each side by 0.5997.

$15,341 = C

Comparing Depreciation

The graph below shows the linear depreciation and exponential depreciation of the Lamborghini in the other two examples.

Linear Equation: y = $274,390.00 – $10,000.00x

Exponential Equation: y = $274,390.00(1 – 0.10)^x

By comparing the graphs, we can determine a few things that are noteworthy:

Exponential depreciation loses value much faster than linear depreciation.
Even so, they both end up losing the same amount of value after about 25.5 years. Do you see this intersection on the graph?
The car will eventually be worth $0.00 with the linear depreciation but never quite gets to $0.00 using exponential depreciation.
No matter what, you will lose a lot of money trying to resell your Lamborghini!

Depreciation Practice

Answer the questions based on the following graph.

The functions on the graph are:

$y=5000-100x$
$y=5000(1-0.10)^x$

1. Which line represents linear depreciation? Which represents exponential depreciation?

2. Which equation represents linear depreciation? Which represents exponential depreciation?

3. Explain what each equation means in the real-world scenario. What does each number represent?

4. How much will this car be worth in 5 years using linear depreciation? How about exponential depreciation?

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