... after 7 years?
Answers (3)
My calculations are as follows with a depreciation of 14% annually, calculated each year after previous depreciation:
Year 1- $15.750 new, less $2.205 @ 14% = $13.545
Year 2- $13.545 less $1.896 @ 14% = $11.649
Year 3- $ 11.649 less $1.631 @ 14% = $10.018
Year 4- $ 10.018 less $ 1.403 @ 14% = $8.615
Year 5 - $8.615 less $ 1.206 @ 14% = $7.409
Year 6 - $7.409 less $ 1.037 @ 14% = $6.372
Year 7 - $6.372 less $ 0.892 @ 14% = $5.480
Year 8 - $5.480 less $0.767 @ 14% = $4.713
I hope this helps.
At the time of purchase the tractor has a value v0 = $15750
The annual rate of depreciation is a constant r = 0.14.
It is an example of geometric degressive depreciation
What is the remaining value after 7 years?
Well, after one year v1 = v0 * r
After 2 years v2 = v1 * r
...
After 7 years v7 = v6 * r.
Walking backward, you get:
v7 = v5 * r * r
v7 = v4 * r * r * r
...
until you finally get:
v7 = v0 * r * r * r * r * r * r * r = v0 * r^7
Sometimes it helps to think in recursions.
Something went DEFINITELY WRONG last night.
I wrote r=0.14 but tacitly calculated with (1-r)=0.86.
If
v0 = 15750
r = 0.14
R = (1 - r) = 0.86
then
v1 = v0 - (v0 * r) = v0 * (1 - r) = v0*R
v2 = v1*R
and as v1 = v0*R -> v2 = v0*R*R
...
v7 = v6*R = v5*R*R = ... = v0*R^7
After 7 years your tractor will have a value of $5497.86.