Calculating FV with increasing APR per year.

[miisstina]

Hello all,

So I have $1000 which i put in the bank with an APR of 2.5% which is compounded quarterly. However every year the APR rate increases by 0.05 per year. How would I go about solving this?

 

[Deleted member 4993]

Hello all,

So I have $1000 which i put in the bank with an APR of 2.5% which is compounded quarterly. However every year the APR rate increases by 0.05 per year. How would I go about solving this?

I do not think there is a closed form solution to the problem.

The best way I think is to use a software like Excel and calculate year by year (or quarter by quarter).

 

[Ishuda]

Hello all,

So I have $1000 which i put in the bank with an APR of 2.5% which is compounded quarterly. However every year the APR rate increases by 0.05 per year. How would I go about solving this?

First, we can ignore the $1000 since it is just a straight multiple and we can add that at the end. Next, I assume that 'rate increases by 0.05 per year' actually means the 2nd years rate is 2.55%, the third years rate is 2.26%, etc. That being the case, the sum after at the end of year n years S(n) is given by
S(n) = \(\displaystyle \Sigma_{k=0}^{k=n} (1.025 + 0.0005 k )^4\) = \(\displaystyle \Sigma_{k=0}^{k=n} (0.0005)^4 (2050 + k )^4\)
Letting j = 2050 + k, we have j = 2050 when k = 0 and j = 2050 + n when k = n, so
S(n) = \(\displaystyle 6.25 * 10^{-14} \Sigma_{k=2050}^{k=2050 + n} k^4\)

There is a closed form solution for the series but it is a little messy.

EDIT: Dumb mistakes for actual interest but fix those and still work but slightly different answer.

 

[jonah2.0]

WARNING: Beer soaked rambling/opinion/observation ahead. Read at your own risk. Not to be taken seriously. In no event shall 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.

Courtesy of my good friend Sir DexterOnline2 (protégé of Sir Denis and Sir tkhunny) a.k.a. DexterOnline, AbrahamA, and GhostAccount, we have the following "spectacular" (according to Sir Denis) formula:
\(\displaystyle P (1+i)^n = P(e^{\sum_{k\ =\ 0}^{n-1} ln(1+r(1+g)^k)})\)

See the whole story at the following thread:
http://mathhelpforum.com/business-m...st-non-linear-interest-growth.html#post826419

Thus, P = 1000, r = (1+ .025/4)^4-1, g = .05, etc.
End of 1st year: 1025.235353...
End of 2nd year: 1052.401138...
End of 3rd year: 1081.681018...
.
.
.
End of 10th year: 1366.676383...

 

[Ishuda]

I think he means 2nd year = 2.50 * 1.05 = 2.625, but who the heck knows

The reason it works the way it does (as I showed it) is that 1 and the initial interest divided by the increment is a whole number. If it weren't, then it gets more complicated

 

[stapel]

I have $1000 which i put in the bank with an APR of 2.5% which is compounded quarterly. However every year the APR rate increases by 0.05 per year. How would I go about solving this?

It might help if you specified what, exactly, you're needing to "solve". What is the full and exact text of the exercise? What were the instructions? Are you supposed to count the increase from the beginning of the calendar year, or from some other point? Are you needing to find the value after some period of time? Are you supposed to being finding the effective rate of interest? Or something else?

When you reply, please include a clear listing of your efforts so far, along with responses to the various earlier replies. Thank you!