Here we go another intelligent question at least the guy taking out his provisional licence made me smile!

A single figure can't represent depreciation accurately for an extended period of time though, as I said, because it isn't a linear relationship (between present value and time). You can qualify your statement to make it more accurate:

"Assume a car loses 20% of its year-on-year value for the first 3 years after it's first registered. After that, assume a year-on-year loss of ~5-10%."

Just plug N * (1-D)^t into a calculator to see how innacurate it becomes with time (N = car RRP, D = average depreciation as a decimal, t = period to compound over).

You can change the value of D all you want, but the more 'accurately' it predicts small t (near time), the less accurate it'll be for big t (distant time) and vice versa.

Hahaha. If you want an even more, better, reliabler and acuraterer model then why not use the exponential function?

Something like f(t) = A * exp(-0.216*t) is likely to be more accurate, where f(t) is the final value (FV), A is the RRP and t is the time over which compounding takes place.

You can have a look at a graph of this here: www.desmos.com/calculator/lo2ibgalyq and hover your mouse over the line to view the 'present' value.