[Simple Annuities] Calculating deposits needed to reach the FV (Future Value)

[suppressive]

Suppose $646.56 is deposited at the end of every six months into an account earning 6.5% compounded semi-annually. If the balance in the account four years after the last deposit is to be $20000, how many deposits are needed?

Answer: 18 deposits.


I've been stuck on this one for awhile. I'm using the BA II plus calculator to CPT the missing value, but I'm not sure what value the questions is asking for. I've extracted this so far:

PMT (Payment amount) = 646.56
FV (Future value) = 20000
N (# of total payments)= 4 years * 2 times per year = 8
C/Y (Interest compounding per year) = compounded semi-annually = 2
P/Y (Payments per year) = compounded semi-annually = 2
I/Y (Interest per year) = 6.5

Entering these values and pressing CPT PV (present value) gives me = 10993.7741

But it's asking for the number of deposits, so I don't really know where to go from here.

Any help is appreciated

 

[Ishuda]

Suppose you make n deposits at 6 months, 12 months, 18 months, ... n*6 months. How much money would you have? That is, what is the FV of a deposit made every 6 months for n periods at an interest rate of (6.5/2)% per six months?

Suppose you put an amount of money A on deposit for m periods with an interest rate of 6.5% compounded twice a year? Oh, and the m periods is the 4 years after the last deposit so m=8. And the amount A - thats just the FV from the first part.

Put all of that together and you should get something like (1+i)ⁿ = Something. Take logs to get n. Note, I haven't done the computation of n so you might have to round up to insure at least the 20K at the end of the 4 years.

 

[suppressive]

n = number of required deposits (?)
d = 646.56
i = .065 / 2 = .0325
f = 20000

{d[(1 + i)^n - 1] / i} * (1 + i)^8 = f

You need to solve that for n. (using LOGs will be required)
It'll work out to n = 18.0000101624.....

We are not here to learn about BA II calculators.
We told you that before.

Using a calculator, the steps are:
1: calculate PV of 20,000 effective date of last deposit
2: calculate number of deposits required to accumulate to above PV
How you do that on your calculator is for you to find out.

I apologize; the way I've been doing the other questions were using the calculator and then checking my answer by using the formula. Here, I couldn't figure out how to do either/or.

Anyway, I was able to get 18 as the answer using the formula you listed above. Thanks!

Do you mind explaining the "(1+i)^8" part? I know the first part is just the standard Future Value formula of an simple annuity. How come we don't use 8 as the within the other brackets? [(1+i)^n-1]
In other words, why is there 2 different n's?

 

[Ishuda]

...In other words, why is there 2 different n's?

There are two different set of periods [two different n's]. One is when you are depositing on a regular basis (every six months for the n periods) and the other is when you are just letting the money grow without depositing anything (the four years for the m periods)

 

[jonah2.0]

DISCLAIMER: Beer soaked rambling/opinion/observation/reckoning ahead. Read at your own risk. Not to be taken seriously. In no event shall the wandering math knight-errant Sir jonah in his inebriated state be liable to anyone for special, collateral, incidental, or consequential damages in connection with or arising out of the use of his beer (and tequila) powered views.

Do you mind explaining the "(1+i)^8" part? I know the first part is just the standard Future Value formula of an simple annuity. How come we don't use 8 as the within the other brackets? [(1+i)^n-1]
In other words, why is there 2 different n's?

The most efficient way to solve an annuity problem is to make a time diagram, determine the type of annuity, and then apply the proper formula(s).
Draw up a line graph (or a time diagram) from 0 to 26.
Mark point 1 as the end of the 1st semi-annual period, mark point 2 as the end of the 2nd semi-annual period, and so on and so forth until the end of the 18th semi-annual period.
The 1st deposit is at point 1 so you accumulate this to point 18 for 17 semi-annual periods (compound interest formula); the 2nd deposit is at point 2 so you accumulate this to point 18 for 16 semi-annual periods, and so on and so forth.
Thus, at point 18 you would have 646.56(1.0325)^0 + 646.56(1.0325)^1 + . . . +646.56(1.0325)^16+ 646.56(1.0325)^17.
Equivalently, using geometric series, this would be
\(\displaystyle 646.56\frac{(1.0325)^{18}-1}{.0325}\)=15,484.9280058554...(FV of ordinary or end of period payment annuity formula)
Since this amount is to be left untouched for 8 semi-annual periods ( then we would have
\(\displaystyle 646.56\frac{(1.0325)^{18}-1}{.0325}(1.0325)^8\)=19,999.9851480442...(Compound interest formula)
at the end of point 26. Not exactly 20,000 now is it?
This is of course the practical interpretation.
Going by the theoretical equation of value corresponding to this problem, you'd have to set 18 to n and equate the whole thing to 20,000 which is
\(\displaystyle 646.56\frac{(1.0325)^n-1}{.0325}(1.0325)^8=20,000\)
as suggested by my good friends, Sir Denis and Sir Ishuda.
As you may have noticed, there's some slight discrepancy when computing for the theoretical value of n. You can always expect such difference(s) whenever n is the unknown quantity in an annuity problem.

 

[DexterOnline]

Yes Sir Jonah and Sir Wilmer

Look Again,

\(\displaystyle 646.56\frac{(1.0325)^n-1}{.0325}(1.0325)^8=20,000\)

\(\displaystyle (1.0325)^n-1=\frac{20,000(.0325)}{646.56(1.0325)^8}\)

\(\displaystyle (1.0325)^n=1+\frac{20,000(.0325)}{646.56(1.0325)^8}\)

\(\displaystyle n log(1.0325)=log(1+\frac{20,000(.0325)}{646.56(1.0325)^8})\)

\(\displaystyle n=\frac{log(1+\frac{20,000(.0325)}{646.56(1.0325)^8})}{log(1.0325)}\)


\(\displaystyle n=\frac{log(1+\frac{650}{835.08237122121927245946313476563})}{log(1.0325)}\)

\(\displaystyle n=\frac{log(1+0.77836632935915535799548664102498)}{log(1.0325)}\)

\(\displaystyle n=\frac{log(1.77836632935915535799548664102498)}{log(1.0325)}\)

\(\displaystyle n=\frac{0.25002122706837914344884395023443}{0.01389006032843863054539485367879}\)

\(\displaystyle n=18.000010162409698439302075539823\)