# Value of Investment in the Past

An amount of $1000$ is to be accumulated at a compound rate of discount of $9$% per year. (a) Find the present value $3$ years before (b) Find the value of i corresponding to d.

For a) i have done the following: $1000=X(1-0.09)^{-3}$

$X=0.19$

I am not sure on how to go on about (b).

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For (a), did you mean $1000=X(1-0.09)^3$? I'm not sure if I understood it... What are i and d in (b)? – Guest 86 Jan 24 '13 at 23:11
i used (-3) because they said to find the present valye 3 years "before". the before part has confused me – Yenny Chen Jan 24 '13 at 23:17
d=0.09 and i is what we need to find – Yenny Chen Jan 24 '13 at 23:18
Guest86 is asking what the variable $i$ represents, not what its value is. Apparently $d$ is the discount rate, but what is $i$ supposed to represent? – Austin Mohr Jan 24 '13 at 23:33

If you have \$1000 now and you want to know the value three years prior, \$1000 is your initial value. The relevant computation is therefore $$1000(1 - 0.09)^{-3} \approx 1327.01.$$
An alternative approach is to say I have an initial value of $X$ that will be worth \$1000 in three years. In this case, you would solve $$1000 = X(1 - 0.09)^3$$ and find$X \approx 1327.01$. - If you want the value three years ago, it should be less than \$1000, as there has been some interest since then to get to \$1000 – Ross Millikan Jan 24 '13 at 23:54 @RossMillikan I mistook "rate of discount" to be "rate of depreciation". Should we instead be using$1 + 0.09$, or is discount something else entirely? – Austin Mohr Jan 25 '13 at 0:44 I suspect so, but am confused between discount rate and interest rate. – Ross Millikan Jan 25 '13 at 1:18 For (b), it would seem to me that what you seek is a value of interest$i$such that $$1+i = \frac{1}{1-d}$$ where$d\$ is the depreciation. This then means that
$$i=\frac{d}{1-d} = \frac{0.09}{0.91} \approx 0.0989$$