How to solve a hard annuity question, changing payments

Question

I am having a lot of confusion trying to work out this problem because it seems like a lot is going on all at once. I am looking for someone to please help.

A man would like to buy an annuity of $15$ years, that provides $30$ semi annual payments. The first payment $6$ months after the purchases the annuity. During a year, the payments are the same amount. After the first year however, at the beginning of each year the payments are decreased by $4$ percent.

if $i^{(2)}=0.10$, how much does this annuity cost?

My work:

Well I am quite confused on the wording, but I tried to make a time diagram. At the first 6 month mark, he makes a payment of P. 6 months later a payment of 0.96P, six months later 0.96P, six month later 0.92P, six month later 0.92P,..

6 months before the start of the fifteenth year a payment of 0.44P, six month later a final payment of 0.40P.

But now I don't know how I can use the semiannual rate and annuity formulas because I have different payments.

I tried to do it by discounting to bring everything in terms of value at time 0, but to sum all the inbetween without some sort of trick would be extremely difficult, I also am just overall confused on if that is even the correct method

Any help?

Answer 1

Let $d=0.96, q=1.05$. and $a$ the initial annuity payment. Then the future value is

$\quad S_{30}=a\cdot \left[ q^{29}+q^{28}+dq^{27}+dq^{26}+\ldots+d^{14}q+d^{14} \right] \qquad (1)$

$ \frac{d}{q^2}S_{30} =a\cdot \left[\qquad \qquad \quad dq^{27}+dq^{26}+\ldots . +d^{14}q+d^{14}+\frac{d^{15}}{q}+\frac{d^{15}}{q^2} \right] \qquad (2)$

Substracting (2) from (1)

$S_{30}\cdot \left(1-\frac{d}{q^2}\right)=a\cdot \left(q^{29}+q^{28}-\frac{d^{15}}{q}-\frac{d^{15}}{q^2} \right)$

$S_{30}=a\cdot \frac{q^{29}+q^{28}-\frac{d^{15}}{q}-\frac{d^{15}}{q^2}}{1-\frac{d}{q^2}}$

Expanding the fraction by $q^2$

$S_{30}=a\cdot \frac{q^{31}+q^{30}-d^{15}q-d^{15}}{q^2-d}$

$S_{30}=a\cdot \frac{(1+q)\cdot (q^{30}-d^{15})}{q^2-d}$

And the present value is

$S_{0}=\frac{a}{q^{30}}\cdot \frac{(1+q)\cdot (q^{30}-d^{15})}{q^2-d}$

Remark

After the first year however, at the beginning of each year the payments are decreased by $4$ percent.

My interpetation is that the payments at year $t$ is $a\cdot 0.96^t$. The decrease of $4\%$ is referring to the level of payments of the previous year.

Answer 2

This may help:

v+(1-.04)v^2+(1-.04)^2v^3+...+(1-.04)^14v^15

v[1+(.96/1.1025)^1+(.96/1.1025)^2+...+(.96/1.1025)^14]

Note: You have to use the effective annual rate of (1+.1/2)2=1.1025 in the v factor

=v[(1-(.96/1.1025)^15)/(1-(.96/1.1025))]=3.505

*This method uses the geometric series formula *

Answer 3

Let the amount of the principle remaining in the annuity after $k$ payments have been made be $S_k$, and the initial payment be $P$.

Assume the semi-annual interest accrued is $q = 05$ (the semi-annual compounding of a ten percent annual rate) and the payments decrease at each payment $2i$ by a factor of $\delta = 0.96$.

Then we can see that payment $k$ is always $P\delta^{\lfloor k/2 \rfloor}$ where $\lfloor k/2 \rfloor$ means the greatest integer not exceeding $k/2$. For example, payment $1$ is $P$, and payments $2$ and $3$ are both $P\delta^1$.

We cab cast the problem as follows:

For payments number $2i+1$, $$ S_{2i+1} = (1+\alpha) S_{2i} - P \delta^i $$ For payments number $2i$, $$ S_{2i} = (1+\alpha) S_{2i-1}- P \delta^i $$ And $S_{30}=$. Find $S_0$.

We solve this as follows: First, to get rid of the ugly odd/even structure, we would like to know $S_{2i+2}$ in terms of $S_{2i}$:

$$S_{2i+2} = (1+\alpha) S_{2i+1}- P \delta^{i+1} = (1+\alpha) \left( (1+\alpha) S_{2i} - P \delta^i\right) - P \delta^{i+1}\\ S_{2i+2} = (1+\alpha)^2 S_{2i} - (1+\alpha+\delta) P \delta^i $$

This is a fairly easy series to sum in closed form, if you know a cute starting trick, and know how to sum a geometric series. The cute trick is to let $ S_{2i} = (1+\alpha)^{2i} a_{i}$; then $S_0 = a_0$ and the equation becomes $$ (1+\alpha)^{2(i+1)} a_{i+1} = (1+\alpha)^{2(i+1)} a_i - (1+\alpha+\delta) P \delta^i \mbox{ or}\\ a_{i+1} = a_i - \frac{(1+\alpha+\delta) P}{(1+\alpha)^{2(i+1)}} $$ Since $a_i$ just decreases, each time, by $\frac{(1+\alpha+\delta) P}{(1+\alpha)^{2(i+1)}}$, clearly for all $i>0$ $$ a_i = a_0 - \sum_{k=0}^{i-1} \frac{(1+\alpha+\delta) P}{(1+\alpha)^{2k}}= a_0 - (1+\alpha+\delta) P\sum_{k=0}^{i-1}\frac{1}{(1+\alpha)^{2k}} $$ and we know how to sum that geometric series, giving $$ a_i = a_0 - \frac{1-(1+\alpha)^i}{1-(1+\alpha)} = a_0 - (1+\alpha+\delta)\frac{(1+\alpha)^i-1}{\alpha} \\ S_{2i} = (1+\alpha)^{2i} \left( S_0 - (1+\alpha+\delta)\frac{(1+\alpha)^i-1}{\alpha} \right) $$ If we substitute $i=15, \alpha = 0.05, \delta = 0.95$ into $S_{30}=0$, we get $$ S_{30} =0 = (1.05)^{30} \left( S_0 - 2.01P\frac{(1.05)^{15}-1}{0.5} \right) \\ S_0 = 2.01\frac{(1.05)^{15}-1}{0.5} \approx 21.579 P $$