Is My Textbook Incorrect On Explaining How To Solve Annual Interest Compounded Monthly?

Question

I doubt that the textbook solution is correct. If I have $\$100$ and put it into a bank with annual interest compounded monthly of $6\%$, how much money $y$ would I have after $t$ years? The equation that the textbook provides is $y = 100(1+x)^{12t}$. The textbook states that for annual interest compounded monthly, the rate of $x$ equals $6\%/12$, which would equal $0.005$. But putting this into the equation with $t$ equaling $1$ for $1$ year, I get $106.1677812$ instead of the expected answer of simply $106$. Anyone know how to calculate $x$ to just get the correct exact answer of $106$? Is the equation incorrect as well?

Answer 1

This is what "compounded monthly" means: at the end of the first month, the bank calculates how much interest you should have earned for the first month. By convention (and perhaps law), the bank treats each month as if it were exactly $1/12$ of a year and credits you for $1/12$ of the annual interest. The annual interest on $100$ would be $6,$ so you are credited $6 \times 1/12 = 0.5.$

Or you can more simply say that the bank increases your balance by $6\% \times 1/12 = 0.5\%.$ Either way, the first month's interest is $0.5.%

But at the end of the second month, you are credited with interest on all the money you have on deposit, including the interest you earned in the first month, not just the original principal. Another way to put this is, you earn interest on the original deposit plus interest on the interest you have already earned.

And in every month after that the bank again credits you with interest on the original principal plus interest on all the interest earned in previous months.

The extra $0.1677812$ is the accumulation of interest earned on previous months' interest.

In the real world, banks do not usually keep track of balances with such precision; they might only keep track to the nearest $0.01$ of the unit of currency. So sometimes the interest will be rounded down, and sometimes it will be rounded up, and the end result may not be precisely what the formula says. But the result is usually quite close, and much more accurate than assuming you have simple interest (paid on the original deposit only, not on any previously earned interest) when you actually have compound interest.

Answer 2

The nominal interest rate compounded monthly is $i^{(12)}=6\%$. So the effective annual interest rate $i$ is $$ i=\left(1+\frac{i^{(12)}}{12}\right)^{12}-1\approx 6.168\% $$ So after $t$ years you have the amount $S(t)$ $$ S=S_0(1+i)^t=S_0\left(1+\frac{i^{(12)}}{12}\right)^{12\,t} $$ where $S_0$ is the initial amount of money.

In general, for a nominal interest rate $i^{(m)}$ compounded $m$ times in a year, you have an effective annual interest rate $$ i=\left(1+\frac{i^{(m)}}{m}\right)^{m}-1 $$ The quantity $i_m=\frac{i^{(m)}}{m}$ is called interest rate per conversion period. Compounded monthy means $m=12$, quarterly $m=4$, weekly $m=52$ and so on.

Answer 3

An annual $6\%$ interest compounded monthly can mean either $0.005\%$ applied every month, or the interest is added monthly for a year's total of $6\%$.

In the second case (for $1$ year), we have the equation

$$(1+x)^{12}=1.06$$

In order to solve this, we use logarithms. Any base will do - I have used base $e$, the natural base.

Proceeding,

$$(1+x)^{12}=1.06$$

$$\ln\left((1+x)^{12}\right)=\ln(1.06)$$

$$12\ln(1+x)=0.05826890812397577552571835111851$$

$$\ln(1+x)=0.00485574234366464796047652925988$$

$$1+x=1.0048675505653430375411989455875$$

$$x=0.0048675505653430375411989455875$$

$x$ is the amount of interest added a month, to be written in the form of the question it needs to multiplied by $12$, giving an annual rate of $\approx 5.84\%$.

Another method is to take the $12^{th}$ root of both sides and subtract $1$ - this gives the same value for $x$.