Varying interest rate question with different payout times

Question

Just wondering if anyone could help me with the most efficient solution to working this problem out. The only way I can think of right now is to work backwards but I know this is a tedious and long winded answer. Any help to find the most efficient method would be appreciated.

A bank offers 2.7% p.a. payable monthly on amounts less than 30,000, 4.2% p.a. payable quarterly on amounts between 30,000 and 50,000, and 6.1% p.a. payable half-yearly on amounts of 50,000 or more. Interest is payable at the lowest monthly rate until the end of the month when the balance (including accumulated interest) first reaches 30,000 or more. Interest is payable at the lower quarterly rate until the end of the quarter when the balance (including accumulated interest) first reaches 50,000 or more. Thereafter the highest half-yearly rate applies. What is the amount required now so that there is 60,000 in the account in twenty-two years’ time? How many years does it take to obtain the highest half-yearly rate?

Answer 1

Work backwards from the target, which allows you to figure out the applicable interest rates. You have $60,000$ at the end of $22$ years. Assume the balance at $21.5$ years is at least $50,000$, so the interest is at $6.1\%$. The balance at the end of $21.5$ years is then $\frac {60,000}{1.0305}\approx 58,224.16$ validating the assumption. Keep going backwards until the balance falls below $50,000$ and impose the next lower interest rate. A spreadsheet will make it easy.

Answer 2

Here's a MATLAB solution. Phase 1, 2, and 3 correspond to the three interest rates.

left = 22*12;              % months left
x3 = 6e4;                  % amount at the end of phase 3
%% Semesters spent in phase 3
s3 = floor(log(x3/5e4) / log(1+0.061/2));
left = left - 6*s3;
x2 = x3 / (1+0.061/2)^s3;  % amount at the end of phase 2
%% Quarters spent in phase 2
q2 = floor(log(x2/3e4) / log(1+0.042/4));
left = left - 3*q2;
x1 = x2 / (1+0.042/4).^q2; % amount at the end of phase 1
%% Amount required at the beginning of phase 1
x0 = x1 / (1+0.027/12).^left;

The amounts at the beginning of each phase are $x_0 = 25,034$, $x_1 = 30,032$, and $x_2 = 50,103$.

It takes $81$ months to pass the $30,000$ threshold, $49$ quarters to then cross the $50,000$ threshold, and $6$ semesters to reach the $60,000$ goal.

Answer 3

I know no "simple ways," so in some sense I'm not answering your question. However my approach differs a bit from the other answers so I'll add it in case it's of any interest:

I'll use the standard compound interest formula, $A=P(1+r)^n$ many times. $A$ is the amount of money at the end, $P$ the amount at the beginning, $r$ the rate per period of compounding, and $n$ the number of compounding periods.

$60000=50000(1.0305)^n$ gives $n=6.0684...$

So it takes about 3 years (6 half years) to grow that last bit.

$P_1\cdot (1.0305)^6=60000$ gives $P_1=\$50102.95$ as about the balance you want at the 19th year for those final three years.

Doing the same thing again for the growth from 30 to 50 thousand:

$50102.95=30000(1.0105)^n$ gives 49.1 periods or 12 years and 3 months, giving $50102.95=P_2(1.0105)^{49}$ gives $P_2=\$30031.96$ or so for the balance at the beginning of those 49 quarters.

To get to 30,000, we have most likely 22 years minus 3 years minus 12 years 3 months, which is 6 years nine months or 81 months. We can adjust for any problems.

$30031.96=P_3(1+.027/12)^{81}$ Giving $P_3=\$25033.55$ as the amount you want at the beginning of the 22 years.

To check the answer, put it into the formula below and tweak, making sure the necessary quantities are reached along the way:

$A=P(1+\frac {.027} {12})^{81}(1.0105)^{49}(1.0305)^6$

$25033.55(1+\frac {.027} {12})^{81}(1.0105)^{49}(1.0305)^6=60000.000123219$

It seems to have come out ok (barring any error on my part), so no tweaks seem necessary.