Question 1109124: You wish to accumulate $100,000 through monthly payments of $300. If you can earn interest at an annual rate of 4% compounded monthly, how long (to the nearest year) will it take to accomplish your goal?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! using the TI-BA-II financial calculator, you would enter the following:
FV = 100,000
PMT = -300
I/Y = 4% / 12
PV = 0
calculate N to get N = 224.5377205 months.
divide that by 12 to get number of years = 18.71147671.
round to the nearest year to get 19 years.
you can also do this using an online financial calculator.
one such calculator can be found here:
http://www.zenwealth.com/BusinessFinanceOnline/TVM/TVMCalculator.html
using this calculator, you would enter:
PV = 0
PMT = -300
Rate = 4
FV= 100000
select monthly
click on Periods
the calculator will tell you that the number of periods is 224.54.
since you selected monthly, that means the number of month is 224.54.
divide that by 12 to get 18.71166667 years.
rounding to the nearest integers gets you 19 years.
there's a very small difference from what i got before, most likely due to rounding errors between the two.
the rounded answers are the same.
when you use these online calculators, you have to take the sign of the inputs into consideration.
if the future value is positive, then the payment must be negative.
if the future value is negative, then the payment must be positive.
if you make both the future value positive and the payment positive, the calculator will tell you that it can't find the answer.
follow the instructions for whatever calculator you use.
this can also be computed using the following formula and solving for n.
the formula is:
ANNUITY FOR A FUTURE AMOUNT WITH END OF TIME PERIOD PAYMENTS
a = (f*r)/((1+r)^n-1)
a is the annuity.
f is the future amount.
r is the interest rate per time period.
n is the number of time periods.
you would need to solve for n, using this formula.
multiply both sides of the formula by ((1+r)^n-1) and divide both sides of the formual by a to get:
(1+r)^n - 1 = f*r/a
add 1 to both sides of this equation to get:
(1+r)^n = f*r/a + 1
take the log of both sides of this equation to get:
log((1+r)^n) = log(f*r/a + 1)
since log((1+r)^n) = n * log(1+r), the equation becomes:
n * log(1+r) = log(f*r/a + 1)
divide both sides of the equation by log(1+r) to get:
n = log(f*r/a + 1) / log(1+r)
replace f with 100000
replace r with .04/12
replace a with 300.
formula becomes:
n = log(100000 * .04/12 / 300 + 1) / log(1 + .04/12)
solve for n to get:
n = 224.5377205
that's the same answer i got using the TI-BA-II.
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