### Author Topic: mathematically inept seeks someon to explain compound formula s l o w l y...  (Read 5470 times)

#### MadisonStreet

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« on: May 30, 2013, 12:40:08 PM »
I've been back and forth on posting this question for fear of revealing just how mathematically inept I am.  It just always seems to take me longer than most to wrap my head around numbers, especially so in terms of money and finance.

So, would someone mind explaining, step by step, and in very layman's terms, how the compounding formula works?

I've only recently discovered MMM and I'm scrambling to get caught up with all the posts, but so far they've shed some serious sunlight.  MMMism also confirms so many things I think I inately already knew so it's not hard to embrace the message, however, everytime I see how a cup of coffee can become a bucket o' cash over 10 years, I develop a brain cloud.  It's not the math, it's me.

Thanks in advance all,
Growing a mustache in Spain

#### mpbaker22

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« Reply #1 on: May 30, 2013, 12:47:53 PM »
I think you mean compound interest, in terms of investing or paying off debt.
So, let's take a simple example.  You take out a loan of \$100 at 5% interest compounded each year.  Let's assume you don't pay anything.  This \$100 becomes 100*1.05 (5% interest is 1.05).  \$100 becomes \$105 after one year.
After two years, it's \$105*1.05, or \$110.25.
In this scenario, where you aren't paying off the loan monthly/yearly, the amount you owe at the end of the n-th year is 100*1.05^n (after 9 years, it's 100*1.05^9 = \$155.1328, for example).
The point is, your loan doesn't increase by \$5 every year.  It increases by \$5 the first year, \$5.25 the second year, \$5.51 the 3rd year, etc.
With a higher interest rate, like a 20% credit card, \$100 becomes \$515.978 after 10 years (as opposed to \$300 at "simple" interest).

This formula gets more complicated if you pay off a set amount and add some every month (like a credit card at minimum payments), but that's the gist.

#### MadisonStreet

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« Reply #2 on: May 30, 2013, 12:59:07 PM »
Thanks for your reply!!  So, vice-versa, this also works for calculating how much X expense could be worth in X years, if invested/saved gaining X compounded interest, correct?  I see that the median 7% compounded interest in 10 years is the default on MMM (the 752 weekly/173 monthly)...

#### Self-employed-swami

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« Reply #3 on: May 30, 2013, 01:02:40 PM »
Quote
M = P( 1 + i )^n

M is the final amount including the principal.

P is the principal amount.

i is the rate of interest per year.

n is the number of years invested.

Stolen from http://math.about.com/od/formulas/a/compound.htm

So, if we take the example of someone who invests \$1000, at 5%, and it compounds annually.

In year 1, they would have
M = P( 1 + i )^n
M = 1000(1.05)^1 = 1050

In year 2, there are 2 ways to calculate it.
Using the formula above
M = P( 1 + i )^n
M = 1000(1.05)^2 = 1000(1.1025) = 1102.5

Or we can use the 1050 from the first year, as the new P number
M = P( 1 + i )^n
M = 1050(1.05)^1 = 1102.5

So, if you can see what is happening here, the interest earned at the end of the first compounding period (here it is a year), becomes the input as the principal in the second year.  That rolling over of the interest, into the principal for the next compounding period, is how compounding interest works.

Let's do another example to illustrate just how much compounding can make a difference!

Say you have 1000 to invest, and it will earn 5% annually (from an very stable fund, for example), and it will pay out that interest rate for 25 years.  You are given the option to let that interest compound (roll over) or you can take it out annually, whichever you prefer.

If you take the interest out every year, the formula doesn't need to compound (simple interest), so it will look like this:

M = P*i*n + P
M = (1000 x 0.05 x 25) + 1000 = 1250 + 1000
M = 2250

But, in this case, you need to remember, that you were taking the interest out every year, so you would have collected \$50/year for 25 years, meaning you earned \$1250 in interest, plus you have your original \$1000 left at the end.

But, if you let the interest compound, it would look like this:

M = P( 1 + i )^n
M = 1000(1.05)^25 = 3386.35

So, if the interest compounds, you wind up with \$1136.35 of EXTRA growth at the end of the 25th year.

If that has helped at all, (or not) please let me know.

Other things that are important, that will make your money grow faster are:
-higher interest rates (even 0.1% can make a big difference, after a large number of compounding cycles)
-shorter compounding cycles
« Last Edit: May 30, 2013, 01:35:23 PM by Self-employed-swami »

#### MadisonStreet

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« Reply #4 on: May 30, 2013, 01:34:29 PM »
Hi, yes this (and the previous posting) did help considerably - more often than not, I just to see it written out.  Just to be clear though, ^n is simply the length of time (could be years, months...?), and therefore, how many times to repeat the cycle/formula to arrive at your final M?

#### matchewed

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« Reply #5 on: May 30, 2013, 01:38:49 PM »
Regarding the years/months. The compounding frequency will have to be defined. Do you compound monthly? If yes then do months. Annually? Then years. So n=time. You just have to define the time frame.

#### Self-employed-swami

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« Reply #6 on: May 30, 2013, 01:45:14 PM »
Yes!

N is the number of compounding cycles.  In the examples I used, n was a year, so n=25 is 25 years.

However, lots of debts and savings accounts compound interest monthly, or even daily.

If, for example, my yearly savings account gives interest of 5% annually, but compounds daily, the formula needs to be changed slightly, since our interest (i) and compounding cycles (n) are using different time references (annually for the interest, daily for the compounding cycle).  So, we need to adjust the interest, to daily.  We can do that, by taking the interest in decimal form (0.05) and dividing it by the number of days, in a year (365).

So then, our formula looks like this:

M = P( 1 + i /365)^n
M = 1000(1+0.05/365)^365 = 1000 X 1.0512 = \$1051.27

If we compare this to above, where one compounding cycle (1 year) was used in the same timeframe (1 year) you see that you earned an extra \$1.27, with having daily compounding, over the yearly compounding.

Does that help at all?

#### mpbaker22

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« Reply #7 on: May 30, 2013, 01:55:41 PM »
What Swami is getting at is that 5% interest, compounded daily doesn't mean 5% interest is added daily.  It means the daily allotment of 5% over the course of the year (that is 5% divided by 365) is added daily.
so 5% compounded daily is actually (1+(.05/365))^365=5.12% per year.
for an easier view of the equation - http://www.wolframalpha.com/input/?i=%281%2B.05%2F365%29^365

Does anyone know - if something advertises as 5% APR, does that mean their daily compounding over a year would be 5%?

#### Self-employed-swami

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« Reply #8 on: May 30, 2013, 02:02:14 PM »
I think that depends on the advertising rules, but I am not sure.

#### DoubleDown

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« Reply #9 on: May 31, 2013, 08:54:24 AM »
And generally you can ignore the frequency of compounding when considering investment returns. The difference between monthly, daily, or "continual" compounding is small enough to be ignored, and certainly will be overshadowed by fluctuations in the returns themselves. If math is a challenging subject, then I would just stick with the easy annual examples above, such as using "5% each year" and ignoring any of the more exotic formulas with higher frequency of compounding.

#### happy

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« Reply #10 on: May 31, 2013, 09:23:23 AM »
There is a useful compound interest calculator here. http://jaw.iinet.net.au/stuff/interest.html  Once you understand the concept you can just use the calculator instead of worrying about the equation.

#### No Name Guy

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« Reply #11 on: May 31, 2013, 03:02:01 PM »
The exact formula for continuous compounding is:

A = Ao * e ^ (r*t)

Spoken, it would be "A" equals "A" naught times "E" raised to the power of "R" times "T".

A is the final value
Ao is the starting value, or initial principal
e is the natural exponent (the other way of the natural logarithm) - approximately 2.718.  See http://en.wikipedia.org/wiki/E_(mathematical_constant) which also has a section on compound interest.
r is the interest rate - in the same time units as the time used (e.g. use annual rate for years, monthly rate for months, etc)
t is time.

So, if you're working in terms of years, the rate is the annual rate, and t is in years.

Let's put some numbers to an example:

Initial principal = \$1,000
Rate = 3.4% per annum
Time = 5 1/2 years.

What will I have after the above?

A = \$1,000 * e ^ (0.034 * 5.5) = \$1000* e^ (0.187) = \$1,205.63

As an Excel Function type the following:
= 1000*exp(0.034*5.5)

A variation on this is the "Rule of 70" or "Rule of 72" or "Rule of 69".

http://en.wikipedia.org/wiki/Rule_of_72

The short version is that these are approximations for determining how long it takes for an investment to double.  Divide 69, 70 or 72 by the annual interest rate and the result is approximately how long it will take to double your money.

Example:
8% rate of return.  72 divided by 8 = 9, so it'll take about 9 years to double your money.  The exact answer is 8.66 years, but 9 is a good first order approximation.

Example:  5% rate of return.  70 divided by 5 = 14, so it'll take about 14 years to double your money.  The exact answer is 13.86 years - again, the rule returns a good first order approximation.

(Formula for exact is years = Ln 2 / annual interest rate as a decimal)

or in Excel type in a cell
= ln(2) / (annual interest rate as a decimal)

where annual interest rate as a decimal corresponds to:
2% = 0.02
5% = 0.05
etc.