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Could someone please show me how to calculate the payback period with compounded interest? So if I had a initial investment of $100, paying 1.5% interest annually that is reinvested, how do I calculate how long it would take to recover my initial investment?

Many thanks

  • what is the investment in? A CD?, a bond?, a savings account? a share of a company or mutual fund? – mhoran_psprep Mar 29 '19 at 10:42
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    What do you mean "recover" your initial investment? You haven't lost anything to recover. Do you mean how long it will take to double? – D Stanley Mar 29 '19 at 13:26
  • Also if it's reinvested you're not recovering anything.... the interest is just as lost as the initial investment. After a year you'd have 101.50 invested with nothing "recovered" – xyious Apr 2 '19 at 18:45
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Suppose your investment had a 10% fee and you want to know when it reaches $100

with s = initial investment
     a = target amount
     r = annual effective rate
     n = number of years

s = 90
a = 100
r = 0.015

n =  log(a/s)/log(1 + r) = 7.07658

So it would reach 100 in just over 7 years, i.e.

After 7 years

a = s (1 + r)^7 = 99.89

After 8 years

a = s (1 + r)^8 = 101.38
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An investment earning 1.5% interest will double in 48 years according to: 72 / 1.5 . That's an approximate method.

An investment earning 1.5% interest will double in 46.56 years according to: ln(2) / ln(1.015) . That's yearly compounding. The value 1.015 relates to 1 + (1.5 / 1 year / 100).

An investment earning 1.5% interest will double in 554.86 months according to: ln(2) / ln(1.00125) . That's monthly compounding. The value 1.00125 relates to 1 + (1.5 / 12 months / 100).

An investment earning 1.5% interest and compounded monthly will slightly more than double in 555 months according to a run of this computer program:

Begin

c:= 0;

a:= 100;

While a < 200 Do

Begin

b:= a * 0.015 / 12;

a:= a + b;

c:= c + 1;

End;

WriteLn(a:10:2);

WriteLn(c:10);

ReadLn;

End.

  • If you use b:= a * 0.015 / 12 that is a nominal rate of 1.5% compounded monthly, which is not the same as an effective annual rate of 1.5%. (See link.) For an equivalent calculation you should use b:= a * x / 12 where x = 12 ((1 + 0.015)^(1/12) - 1) = 0.0148979. That is, x is the equivalent nominal annual rate compounded monthly. This figure will give you the same result as the annual one, but in terms of months. – Chris Degnen Mar 31 '19 at 20:16
  • Using 0.0148979 in the computer program results in 559 months as 46.58 years. – S Spring Apr 1 '19 at 4:45
  • Yes. Your program counts whole months and slightly overshoots the 200 target. The same result is obtained by 12 * ln(2) / ln(1.015) = 558.67 months which rounds up to 559 whole months. – Chris Degnen Apr 1 '19 at 9:05
  • I included ln(2) / ln(1.015) to begin with. A conversion from years to months is not worth mentioning. – S Spring Apr 1 '19 at 16:04
  • Oh, for monthly compounding it's ln(2) / ln(1.00125) and results in 554.86 months. The computer program which compounds monthly results in 555 months when using 1.5% as 0.015 . – S Spring Apr 1 '19 at 16:47

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