Annuity Depreciation Problem from Exam FM A manufacturer buys a machine for 20, 000. The manufacturer
estimates that the machine will last 15 years. It will be depreciated using the constant
percentage method with an annual depreciation rate of 20%. At the end of each year, the
manufacturer deposits an amount into a fund that pays 6% annually. Each deposit is equal
to the depreciation expense for that year. How much money will the manufacturer have
accumulated in the fund at the end of 15 years?
The answer is 36,329. I am having trouble seeing why you would use 0.8 in solving this problem. Is the 0.8 the amount of equip remaining after depreciation? In the sol. they multiply by .8 and .2 multiple times, that is throwing me off. Can someone help me solve this problem? 
 A: You use the $0.8$ factor because you depreciate what is left of the value the machine.  For example, the first year you deposit $(0.2) (20,000)$ dollars in the fund, as that is the depreciation in value of the machine.  The machine has value $(0.8)(20000)$ dollars, and the amount deposited into the fund will have value $(0.2) (20000) (1.06)^{14}$ dollars in $15$ years.  So then in year two you deposit $(0.2) (0.8) (20000)$ into the fund.  The machine has value $(0.8)^2 (20000)$ dollars after the second year depreciation and the amount deposited into the fund will have value $(0.2) (20000) (0.8) (1.06)^{13}$ dollars in $14$ years.  And so on.  The result after 15 years in the fund is
$$(0.2) (20000) \sum_{k=0}^{14} (0.8)^k (1.06)^{14-k} = (0.2) (20000)\frac{(1.06)^{15}-(0.8)^{15}}{1.06-0.8}\approx 36328.8$$
ADDENDUM
In general, you can derive the following result.  If $P$ is the present value of the machine, $d$ the depreciation rate, $i$ the interest rate, and $n$ the number of years the machine will last, then the amount in the fund after $n$ years is
$$P d \frac{(1+i)^n-(1-d)^n}{d+i}$$
