Formula: actuarial accumulated value of an annuity paid at the end of the year
Example 1:
Investor A just turned 40 years old and wants to be a millionaire by age 50. He wants to know how much money to invest at the end of each year assuming his investments will earn an interest rate of 7%.
Solution 1:
In order to solve this equation we will need to introduce a new variable, A, where A is the amount paid (invested) at the end of each year. We will also set the equation equal to the final accumulated value of $1,000,000.
$1,000,000 = A *

So i = .07 and n = 10 for this example.
Solving the equation for A we get A = $72,377.50.
In order for this investor to be a millionaire by his 50th birthday he will need to invest $72,377.50 at the end of each year earning 7%.
Example 2:
Investor B at age 24 determines he will have $12,000 to invest at the end of each year. Assuming a return of 7% per year he wants to know how many years it will take to become a millionaire.
Solution 2:
We know A = $12,000 so
$1,000,000 = $12,000 *

Substituting for i = .07 and solving for n gives us:
n = 26.07
So it will take investor B a little over 26 years if he invests $12,000 at the end of each year that earns 7% interest. He will be 50 years old.
Conclusion:
Investor A needs to put away a lot more money each year than Investor B. This is due to Investor B having an additional 16 years of compounding before he reaches 50.
I wanted to give a fairly simple formula that anyone could use to come up with similar solutions. Things can get a lot more complicated with varying interest rates, increasing payments and monthly contributions.

Compound interest plus time is a magical thing. It's so much easier to build wealth if you start really early.
ReplyDeleteHey PIP,
DeleteI agree. That's why it's so important to start early so you aren't playing catch-up.
Thanks for stopping by!
In both of these examples, you assume the investor starts with $0 accumulated, right?
ReplyDeleteHow could you tweak the formula to account for money already invested? For example, someone has $500K investments earning X rate and wants to know how many additional years it would take for that to reach a certain value given regular additional contributions?
Hi Executioner,
DeleteGood question. Yes, in both examples the starting investment is $0 with the first investment made at the end of the first period, one year.
If someone already had a $500k investment and say you wanted to get to a million then you would do the same as above except at the end of the equation you would add 500,000(1+i)^n since the 500k is also being compounded at the same interest rate for n years.
If I get a chance when I get home I'll try and do the calculations.
Starting with 500k , contributing 12k at the end of each year at 7% interest I came up with 8.23 years to reach 1,000,000. If you want me to explain how I got that let me know.
DeleteI was hoping I could ignore the fact that I won't be able to reach my 1 million bucks before I get 50... I am now trying to convince myself to be able to generate income with less than that and still retire early, otherwise good bye early retirement.
ReplyDeleteHi Martin,
DeleteIt may come down living below your means. I know I'm spending more now than I plan to in retirement so I need to cut down spending also. If you can get rid of a few things that aren't necessities it will increase your savings rate and speed up your retirement. I'm stil working on that as well.
Take care!
What about kids? hehe, can I cut down spending :))) I am trying but not sure I could find more. Well my next step would actually be eliminating debt. It is a quite big leak paying interest. Then I will see.
DeleteThat is a simple equation. Sure makes me wish I had paid more attention to algebra in Junior High. I'll play with those.
ReplyDeleteHi Payoffmyrentals,
DeleteThe first two examples are solved using natural logs. If you have a starting amount like executioner mentioned you won't needs logs.
I enjoy mathematics so I don't minding helping if you get stuck. I may do a whole series on theory of interest with lots of different helpful equations.
Thanks for stopping by!