Re: Tough Financial Problem
- From: hagman <google@xxxxxxxxxxxxx>
- Date: Mon, 09 Jul 2007 00:49:36 -0700
On 9 Jul., 09:06, Mark <mnbaya...@xxxxxxxxx> wrote:
On Jul 6, 5:47 pm, Brian VanPelt <bvanp...@xxxxxxxxxx> wrote:This is not exactly what your formula says (and 2.5% inflation
On Fri, 06 Jul 2007 16:44:24 -0700, Mark <mnbaya...@xxxxxxxxx> wrote:
The problem I'm working on is actually much more difficult than this,
but here's the simple version. It's the interest I can't seem to
figure out.
How much money would you need to last you exactly 20 years, if you
spend $3000/mo but earn 5% interest, compounded monthly, on your
initial investment?
Let's assume that the annual percentage rate is 5% so that the monthly
interest rate is .05/12. Suppose A is the amount you need to finance
this 20 years and that you withdraw $3000 at the _beginning_ of each
month. Furthermore, for simplicity, let x = 1 + .05/12. I will do a
few months until a pattern emerges and then get the formula you need.
After 1 month you have
(A - 3000) x
since you withdrew the money at the beginning of the month and the
money left over accrued interest during that month.
After 2 months you have
((A - 3000) x - 3000) x = Ax^2 - 3000x^2 - 3000x
After 3 months you have
(Ax^2 - 3000x^2 - 3000x - 3000) x = Ax^3 - 3000x^3 - 3000x^2 - 3000x
Do you get the pattern?
After n months you have
Ax^n - 3000x^n - 3000x^(n - 1) - ... -3000x
Now, after 20 years, or 240 months, you are suppose to have exactly $0
left, so for n = 240 we have
Ax^240 - 3000x^240 - 3000x^239 - ... - 3000x = 0
Or
Ax^240 = 3000x^240 - 3000x^239 - ... - 3000x
On the right side, factor out the 3000x to get
Ax^240 = 3000 x (x^239 + x^239 + ... + x + 1)
Divide both sides by x^240
A = 3000(x^-239)(x^239 + x^238 + ... + x + 1)
Finally, multiply and divide by x - 1 to get
A = 3000(x^-239)(x^240 - 1) / (x - 1)
Substitute x = 1 + .05/12 to get
A = 456,470.01
Hopefully, you will see how this can be modified for different amounts
to develop a general formula.
Hope this helps,
Brian
Thanks for the detailed reply. I came up with a multi-pass (loop)
formula, I was beginning to think you couldn't do it on one line.
However, I did mention that my problem was actually a bit more
complicated than that. What if the $3000 varies from month to month?
i.e.,
monthly_expenses = e^(month*0.0025)
where e = 2.71828183
and month is how many months since the start. (the first month costs
$3000, the second month costs $3007, the third costs $3015... etc. --
due to inflation)
*monthly* seems a bit pessimistic)
If A[n] is your amount after month n, you have
A[n+1] = (A[n] - initialexpenses*inflationfactor^n)*interestfactor
Let B[n]:=A[n]/inflationfactor^n, then we have
B[n+1]=A[n+1]/inflationfactor^(n+1)
= (A[n]-initialexpenses*inflationfactor^n)*interestfactor/
inflationfactor^(n+1)
= (A[n]/inflationfactor^n - initialexpenses)*interestfactor/
inflationfactor
= (B[n] - initialexpenses)*(interestfactor/inflationfactor)
Thus the problem is the same (A[n]=0 happens at the same time that
B[n]=0)
with the interestfactor replaced by interestfactor/inflationfactor.
With 5/12% paid each month, the interestfactor is 1+5/1200
and with an inflation of 2.5%, the inflation factor is
(1+2.5/100)^(1/12)
which is approximately 1+2.5/1200.
Again just approximately, the quotient (1+5/1200)/(1+2.5/1200)
is about 1+5/1200-2.5/1200 = 1 + 2.5/1200.
Long story short: If the inflation is not detrimentally high
you may simply subtract it from the interest rate for an approximation
that is more precise than your belief in constant inflation.
.
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