Formula for decreasing annuity by k% every year
Lets say you have an annuity where you make annual deposits for n years, with the first payment being x. Then each subsequent deposits is k% smaller than the previous. And you are given the effective annual interest rate i%. How do you find the accumulated value for this annuity after the last deposit?
I know how to find present value;
$PV=x\left(\frac{\left(1-\left(\frac{\left(1-k\right)}{1+i}\right)^n\right)}{1-\left(\frac{\left(1-k\right)}{1+i}\right)}\right)$
but how do you find the accumulated value??
$\endgroup$ 31 Answer
$\begingroup$Whenever periodic payments are made in geometric progression, i.e., there is a constant ratio between successive payments, the resulting cash flow is equivalent to a level payment annuity with a modified interest rate.
For instance, suppose we have an $n$-year annuity-immediate with first payment of $1$, and effective annual interest rate of $i$. Each successive payment is $r$ times the previous payment. So the equation of value for the present value is $$PV = v + rv^2 + r^2 v^3 + \cdots + r^{n-1} v^n$$ where $v = 1/(1+i)$ is the effective annual present value discount factor. We observe that if $v'$ is a modified discount factor that satisfies $$v' = rv,$$ then we may write the above as $$\require{enclose} PV = \frac{1}{r}\left( rv + (rv)^2 + (rv)^3 + \cdots + (rv)^n \right) = \frac{1}{r}\left(v' + (v')^2 + \cdots + (v')^n\right) = \frac{1}{r} a_{\enclose{actuarial}{n} j},$$ where $j$ is some modified interest rate for which $v' = rv$ holds; i.e., $$\frac{1}{1+j} = v' = rv = \frac{r}{1+i}.$$ Thus $$j = \frac{1+i}{r} - 1.$$ The accumulated value may be written $$AV = \frac{1}{r} (1+i)^n a_{\enclose{actuarial}{n}j};$$ however, we may also write $$\begin{align} AV &= (1+i)^{n-1} + r(1+i)^{n-2} + r^2 (1+i)^{n-3} + \cdots + r^{n-1} \\ &= r^{n-1} \left( \frac{(1+i)^{n-1}}{r^{n-1}} + \frac{(1+i)^{n-2}}{r^{n-2}} + \frac{(1+i)^{n-3}}{r^{n-3}} + \cdots + 1 \right) \\ &= r^{n-1} \left((1+k)^{n-1} + (1+k)^{n-2} + (1+k)^{n-3} + \cdots + 1 \right), \end{align}$$where this time we require $$1+k = \frac{1+i}{r},$$ or $k = \frac{1+i}{r} - 1 = j$. These modified rates are the same in both cases, hence $$AV = r^{n-1} s_{\enclose{actuarial}{n} j}.$$
Now, speaking to your original question, in your situation the ratio is $$r = 1-k,$$ where $k > 0$ is the percentage by which each payment decreases; e.g., if $k = 0.05$ and the initial payment is $x = 100$, then the second payment is $100(1-0.05) = 95$, the third is $95(1-0.05) = 90.25$, etc. Then we have for the present value $$PV = \frac{x}{1-k} a_{\enclose{actuarial}{n}j} = \frac{x}{1-k} \frac{1 - \left(\frac{1-k}{1+i}\right)^n}{\frac{1+i}{1-k} - 1} = x \frac{1 - \left(\frac{1-k}{1+i}\right)^n}{i+k}.$$ The accumulated value is $$AV = x (1-k)^{n-1} s_{\enclose{actuarial}{n}j} = x (1-k)^{n-1} \frac{\left(\frac{1+i}{1-k}\right)^n - 1}{\frac{1+i}{1-k} - 1} = x (1-k)^n \frac{\left(\frac{1+i}{1-k}\right)^n - 1}{i+k}.$$And now we can see in both formulas that they are closely related if we rewrite the first as $$PV = x \frac{1}{(1+i)^n} \frac{(1+i)^n - (1-k)^n}{i+k}$$ and the second as $$AV = x \frac{(1+i)^n - (1-k)^n}{i+k}.$$ It becomes immediately obvious that $$AV = (1+i)^n PV.$$As a final note, similar formulas apply for annuities-due but I leave these as an exercise for the reader.
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