A shipping company owns a fleet of heavy trucks if the purchase price of each truck is $245,000 nad its value depreciates by 15% per year, what ist he value of each truck after 4 years?
If you imagine that it loses 15% each year then after no time it is at full value, then after one year it is t(1-0.15) and after two t(1-.15)(1-.15) and so after n years we get:
$\displaystyle T(x) = T(0)(1+x)^n$
T(x) is the value at any given value of n
T(0) is the value at x=0 (ie original value)
x is the factor of depreciation (x>0 for inflation)
n = number, in this case years.
plug in your numbers,
T(0) = 245,000
x = -0.15
n = 4
$\displaystyle T(t) = 245000\cdot(1-0.15)^4 = 127891.53$
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I'll try and explain a bit here:
$\displaystyle T(x) = T(0)(1+x)^n$
the (1+x)^n is because it is gaining/losing a constant amount each year over n years. If was different amounts each time you'd get (1+a)(1+b)(1+c)... and there would be n sets of brackets. However, because it's the same we get (1+x)(1+x)(1+x) n times which is (1+x)^n.
At n=0 for example we get $\displaystyle T(x) = T(0)(1+x)^0$ and since a^0 = 1 we get T(x)=T(0) which is what we'd expect from the definitions. As n increase the importance of the (1+x)^n term will increase and it will rapidly exceed T(0), while it will never reach 0 it will become worthless as it'd drop below a cent.
Hi puzzle,
We can use the depreciation formula $\displaystyle V=a(1-r)^x$, where V will represent the present day value after 4 years (x) at a depreciation rate of 15% (r).
$\displaystyle V=245000(1-.15)^4$
$\displaystyle V=127891.53$
Or, you could just depreciate 15% a year for 4 years and see what the resulting value would be.
[End of Year 1] $\displaystyle 245000 - .15(245000) = 208250$
[End of Year 2] $\displaystyle 208250 - .15(208250) = 177012.50$
[End of Year 3] $\displaystyle 177012,50 - .15(177012.50)=150460.625$
[End of Year 4] $\displaystyle 150460.625-.15(150460.625)=127891.5313$