Car Depreciation

[carbs]

Why do we get different answers when we use the equation to calculate the new value versus taking 78% of each new value each year for the problem below.

A car presently valued at $12,000 depreciates 22% each year. Determine the value of the car 2 years and 6 months from now.

 

[Deleted member 4993]

Why do we get different answers when we use the equation to calculate the new value versus taking 78% of each new value each year for the problem below.

A car presently valued at $12,000 depreciates 22% each year. Determine the value of the car 2 years and 6 months from now.

Please share your work showing the "difference" you are referring here.

 

[carbs]

For example, using the equation V=V₀(1-r)^t you get an answer of $6,448 rounded. The other method we tried was...

Year 0 = 12000
Year 1 = 12000(.78)=9363
Year 2 = 9363(.78)=7300.8
Year 3 = 7300.8(.78)=5694.624

Then did 7300.8-5694.624=1606.176/2=803.088 and then did 7300.8-803.088 to find the new value after 2.5 years. This came out to 6497.712

 

[Deleted member 4993]

For example, using the equation V=V₀(1-r)^t you get an answer of $6,448 rounded. The other method we tried was...

Year 0 = 12000
Year 1 = 12000(.78)=9363
Year 2 = 9363(.78)=7300.8
Year 3 = 7300.8(.78)=5694.624

Then did 7300.8-5694.624=1606.176/2=803.088 and then did 7300.8-803.088 to find the new value after 2.5 years. This came out to 6497.712

You are using t=2.5 for the first-set of calculations, but,

You are using t=3 for the second-set of calculations. Why?

That is the source of difference!

 

[carbs]

well i figured that if i found the difference between year 2 and 3 i could divide that by 2 to account for the .5 after 2 years. Even if you do year 1 and 2 and then use .89 instead of .78 (dividing the depreciation percent by 2) you still get the 6497.712

Im just confused why the two different methods yield different answers

 

[Deleted member 4993]

well i figured that if i found the difference between year 2 and 3 i could divide that by 2 to account for the .5 after 2 years. Even if you do year 1 and 2 and then use .89 instead of .78 (dividing the depreciation percent by 2) you still get the 6497.712

Im just confused why the two different methods yield different answers

In the first method -you have used:

A = 12000 * (1-0.78) * (1 - 0.78) * [1 - 0.78]^(1/2)

and the second method:

A = 12000 * (1-0.78) * (1 - 0.78) * [1 - (1/2) *0.22]

Do you see the difference ?

 

[carbs]

yes! thank you!

 

[Steven G]

Imaging you put $100 in the bank earning interest twice per year and the yearly interest rate is 10%

Now after 6 month you will have earned 5% (1/2 of 10%) on your $100 which will be $5. So after 6 month you will have $105. After another 6 month you will have earned 5% of $105 which is $5.25. Do you see that you earned more the 2nd half of the year? The total interest earned is $10.25. Note that after the 1st 6 month you did NOT earn $10.25/2 = $5.125.

The reason this happened is that you are not computing 5% of the same values ($100 and $105 are different values).

The same happens with depreciation. Suppose that you buy a car for $10,000 that will depreciate 10% every year. Note that the amount the car depreciates less and less each year even though the rate of depreciation is always 10%. If it depreciates different amounts from year to year it is not unreasonable that it depreciates different amounts every 6 month. Is this clear?