Duration of an increasing annuity

Question

I am having the problem computing the duration of an increasing annuity. I know that the annuity is payable for 20 years, the first payment is $24,000$ and each next payment is $1,200$ larger than the previous one. The interest rate is 5%.

So, I know that the the expression for duration is: \begin{equation} \text{duration} = \sum\frac{ t x_t v^t}{x_t v^t} \end{equation} Computing the expression in the denominator is easy. What is more difficult is the expression in the nominator, since we get there: \begin{equation} duration = \frac{24000 (Ia)_{20} + \sum_{n=1}^{20}n(n-1)1200v^n}{24000(a)_{20} + 1200\frac{(a)_{20} - 20v^{20}}{i}} \end{equation} So the problematic expression is the sum in the nominator. Of course I could simply compute each value by hand and sum it up but I think that there is a smarter way to do it. I wouldn't have any problem for computing this if it was an infinite annuity, since then I could just plug these expressions: \begin{equation} \sum_{n=1}^{\infty}n^2x^n = \frac{x(x+1)}{(1-x)^3} \end{equation} \begin{equation} \sum_{n=1}^{\infty}nx^n = \frac{x}{(1-x)^2} \end{equation} However, I cannot come up with the analogous expressions for finite sums. Any help will be appriciated!

Answer 1

You should find that in any extended formula collection. \begin{align} (1-x)\sum_{n=1}^N nx^n &=\sum_{n=1}^N nx^n-\sum_{n=2}^{N+1} (n-1)x^{n}\\ &=\sum_{n=1}^N (n-(n-1))x^n + (1-1)x^1-Nx^{N+1}\\ &=x\frac{1-x^{N}}{1-x}-Nx^{N+1} \end{align}

Similarly for the quadratic coefficients you find \begin{align} (1-x)^2\sum_{n=1}^N n^2x^n &=\sum_{n=1}^N n^2x^n-2\sum_{n=1}^N n^2x^{n+1}+\sum_{n=1}^N n^2x^{n+2}\\ &=\sum_{n=1}^N n^2x^n-2\sum_{n=2}^{N+1} (n-1)^2x^{n}+\sum_{n=3}^{N+2} (n-2)^2x^{n}\\ &=\sum_{n=1}^{N+1}(3)x^n-(N+1)^2x^{N+1}+N^2x^{N+2}-x\\ &=3x\frac{1-x^{N+1}}{1-x}-(N+1)^2x^{N+1}+N^2x^{N+2}-x \end{align}

Answer 2

I don´t know why do you want to calculate the duration. In the exercise it is given. You can calculate the present value of all payments. The formula is

$PV=a\cdot \frac{q^n-1}{i\cdot q^n}+\frac{d}i\cdot \left(\frac{q^n-1}{i\cdot q^n}-n\cdot q^{-n} \right)$

with $q=1+i$, $n:$=years of payments, $i:$=interest rate, $d:$=increasing value of the payments per year

The formula cannot be solved for n. The present value in your exercise is

$PV=24,000\cdot \frac{1.05^{20}-1}{0.05\cdot 1.05^{20}}+\frac{1200}{0.05}\cdot \left(\frac{1.05^{20}-1}{0.05\cdot 1.05^{20}}-20\cdot 1.05^{-20} \right)=417,279.14$

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For $n=3$ it can be tested if the formula works.

$PV=\frac{24000}{1.05}+\frac{25200}{1.05^2}+\frac{26400}{1.05^3}=68519.6$

Using the formula

$PV=24000\cdot \frac{1.05^{3}-1}{0.05\cdot 1.05^{3}}+\frac{1200}{0.05}\cdot \left(\frac{1.05^{3}-1}{0.05\cdot 1.05^{3}}-3\cdot 1.05^{-3} \right)=68519.6$