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Business Math Increasing interest rate

Wilmer

In Memoriam
Mar 19, 2012
376
Code:
YR    RATE      INTEREST      BALANCE
0                             1000.00
1    .03         30.00        1030.00
2    .033        33.99        1063.99
3    .0363       38.62        1102.61
4    .03993      44.03        1146.64
Above is an example of future value of an amount at an incresing rate:
$1000.00 at rate 3% 1st year, then the rate increasing by .10 each year.
As example, year2 rate = .03 * 1.10 = .033

What is the formula to calculate the future value in such circumstances?
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Code:
YR    RATE      INTEREST      BALANCE
0                             1000.00
1    .03         30.00        1030.00
2    .033        33.99        1063.99
3    .0363       38.62        1102.61
4    .03993      44.03        1146.64
Above is an example of future value of an amount at an incresing rate:
$1000.00 at rate 3% 1st year, then the rate increasing by .10 each year.
As example, year2 rate = .03 * 1.10 = .033

What is the formula to calculate the future value in such circumstances?
I can't see an obvious closed form (rather than a product with one term for each year), but this can be tackled by setting up the differential equation for continuously compounded interest with a linearly increasing interest rate.

The solution is then of the form:

\[ FV(t)=P_0 e^{\frac{r_0*\rho^t}{\log(\rho)}} \]

Where \(P_0,\ r_0\) and \(\rho\) are related to but not quite the principle, the initial interest rate and the annual interest growth factor.

In this case \(P_0\approx 741.228\), \( r_0\approx 0.0281893\) and \( \rho\approx 1.09871\)

CB
 

Wilmer

In Memoriam
Mar 19, 2012
376
Thanks CB.
I thought there was a way, since the rates themselves can be "summed" by formula,
(like in example: .03 + .033 + .0363 + .03993 = .13923; .13923 / 4 = ~.0348)
then use an average...but that doesn't quite work...
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Thanks CB.
I thought there was a way, since the rates themselves can be "summed" by formula,
(like in example: .03 + .033 + .0363 + .03993 = .13923; .13923 / 4 = ~.0348)
then use an average...but that doesn't quite work...
We want:

\[ FV(n)=P_0 \prod_{k=1}^n (1+r_0 \rho^{k-1}) , \ \ n\ge 1\]

where \(P_0\) is the principle, \(r_0\) the initial interest rate and \(\rho\) the annual groth factor for the rate.

Now there may be a way to express the product in a "nice" form but I can't see it.

CB
 

Wilmer

In Memoriam
Mar 19, 2012
376
Agree. In "looper words":
a=1000:r=.03:i=.10:n=4

FOR y = 1 TO n

k = a * r [this period's interest]

a = a + k [this period's resulting principle]

PRINT y,k,a

r = r * i [update rate]

NEXT y