How do I find the amount of the monthly payments?

Question

We create a fund earning an annual effective rate of 6% with the aim of accumulating $ 12,000 after 20 years with 20 equal payments 𝑋 made at the end of the year.

Immediately after making the ninth payment, the fund's interest rate drops to $𝑖^{(12)}$ = 3%. In order to accumulate the desired $12,000, we then decide to replace the future annual payments of 𝑋 with new monthly payments of the amount 𝑅, with the first payment made in exactly one month. What is the size of the monthly payment 𝑅?

Here's what I did:

I found the amount accumulated after 9 years (using the formula for accumulated value of an annuity) which was $$[\frac{(1.06)^{9}-1}{0.06}]X=11.49X$$ Then I had done the same thing for the next 11 years with the monthly R payments and the new monthly interest rate, but I realized that I forgot to take in account the amount accumulated over the first 9 years so I don't actually know what to do at this point.

Thank you !!

Answer 1

As usual, write out the cash flows. In this case, we set up two equations; the first for the accumulated value at the end of the $20$-year term under the assumption that level annual payments of $X$ of an annuity-immediate at effective annual interest rate $i = 0.06$ will have accumulated value $12000$:

$$\require{enclose} 12000 = X (1+i)^{19} + X (1+i)^{18} + \cdots + X (1+i)^0 = X s_{\enclose{actuarial}{20} i} = X \frac{(1+i)^{20} - 1}{i} \approx 36.7856X,$$ hence $X \approx 326.215$ is the level annual payment originally required.

But if the interest rate changes immediately after the ninth annual payment of $X$ to $i^{(12)} = 0.03$, and the schedule of payments is changed to monthly, then the effective monthly interest rate is $j = i^{(12)}/12 = 0.0025$, and the cash flow becomes $$12000 = X(1+i)^{19} + \cdots + X(1+i)^{11} + R(1+j)^{131} + R(1+j)^{130} + \cdots + R(1+j)^0.$$ Notice that this equation of value mixes annual payments and monthly payments, as there is no need to convert the first nine payments of $X$ into annual payments based on a monthly interest rate--those payments have already been made. Since monthly payments are made at the end of each month, and there are $11$ years of such payments, the first monthly payments has $12(11)-1 = 131$ months to accrue interest at effectively monthly rate $j$. We may now write this in actuarial notation as $$12000 = X(1+i)^{11} s_{\enclose{actuarial}{9}i} + R s_{\enclose{actuarial}{132}j} = (326.215)(1.06)^{11} \frac{(1.06)^9 - 1}{0.06} + R \frac{(1.0025)^{132} - 1}{0.0025}.$$ Solving this equation yields $R \approx 31.2758$.

Answer 2

EDIT
I am assuming the first deposit is made when you start the savings and at the end of the year thereafter. If the first payment is made at the end of the first year, we will have to adjust accordingly as all deposits will earn interest for one year less. I have given answers for that case too. Before I go to the details, here are different values of $R$ with different assumptions that I made -

A) No compounding -
i) Deposits starting Year 0 and for 20 years thereafter (which of course changes to monthly after 9th deposit).
R = \$31.60

ii) Deposits starting at the end of first year and for 20 years thereafter (which of course changes to monthly after 9th deposit).
R = \$34.04

B) With compounding -
i) Deposits starting Year 0 and for 20 years thereafter (which of course changes to monthly after 9th deposit).
R = \$40.54

ii) Deposits starting at the end of first year and for 20 years thereafter (which of course changes to monthly after 9th deposit).
R = \$43.71

Details -
A) Without further compounding as in some recurring deposits
$A(T) = nX + nX \times I + (n-1)X \times I + ..+ X \times I = nX + \frac{n(n+1)}{2}XI$ ...(i)

$A(T)$ is accumulated amount including interest after $T$ time period.

Using (i), find $X$ when $A = \$12000, n = 20, I = 6\%$.

Then using (i), find $A(8)$ where $n = 8, I = 6\%$ and you know $X$ now. $A(8)$ as you are at the end of the $8th$ year or at the end of 9th year if first payment was at the end of the first year.

Now you make the $9th$ payment and the interest rate immediately drops. You would have been left with $11$ annual payments. Now you will make $R$ monthly payment for $143$ months (if we go by $20$ payments were being made starting from year $0$) starting the end of first month of $9th$ year. If the first payment was at the end of year one and the last payment was to be at the end of year 20, then we would make $132$ payments.

$A(12) = \displaystyle 12 \times 9X \times I + 143R + \frac{143 \times 144}{2 \times 12}RI = 12000 - A(8) - X$

Please note this time $I = 3\%$

B) With compounding - please note that as annual effective rate takes care of any compounding within the year, we should compound only annually instead of monthly -

$A(T) = \sum \limits_{k=1}^{n} X(1 + I)^k = X\frac{(1+I)^{n+1}-I-1}{I}$...(ii)

For payments only starting at the end of first year

$A(T) = \sum \limits_{k=0}^{n-1} X(1 + I)^k = X\frac{(1+I)^n-1}{I}$...(iii)

When it switches to monthly payments -

Annual sum $Y = 12R + \frac{I}{12} (11R + 10R+...R) = 12R + \frac{11 \times 12 \times 0.03}{12 \times 2}R = 12.165R$.

$Y$ should be replaced in (ii) or (iii) for compounding.