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Is there a standardised formula for continuously increasing interest rate of an investment? Given a compound interest formula of A = P(1 + (r/n)^nt), is there a way to combine daily compounding with a continuously increasing interest rate that starts each year at x and ends it an increase of y%?

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  • Can you explain how is this a personal money matter? It sounds a lot more like accounting homework. Oct 28, 2023 at 19:32
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    Instinct says you're trying to integrate under an exponential curve, but it has been much too long. And the thought experiment of smooth increase in rate of increase seems unlikely to match any real world case...
    – keshlam
    Oct 28, 2023 at 20:34

2 Answers 2

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Although a continuously increasing rate would be an unlikely practical scenario the question is answered here for mathematical fun.

Here is a simple example, compounding z over 4 years with interest rate x increasing by y percent annually.

z = 100
x = 0.05
y = 0.10

start amount   a = z = 100
after year 1   a = a (1 + x (1 + y)^0) = 105
after year 2   a = a (1 + x (1 + y)^1) = 110.775
after year 3   a = a (1 + x (1 + y)^2) = 117.477
after year 4   a = a (1 + x (1 + y)^3) = 125.295

Expressed as a formula, with m = 4

eq

The expression does not simplify to a closed-form.

Excel equivalent

=100*PRODUCT(1 + 0.05*(1 + 0.1)^(SEQUENCE(4)-1))

125.295


The same method can be used with monthly or daily rates in which case the interest rate will increase monthly or daily respectively, suitable for monthly or daily compounding.

For example, with starting monthly interest rate x increasing every month by y (0.8%)

z = 100
x = 0.004
y = 0.008

year 1 month 1 rate = 0.004                = 0.400000 %
year 1 month 2 rate = 0.004*(1 + 0.008)    = 0.403200 %
year 1 month 3 rate = 0.004*(1 + 0.008)^2  = 0.406426 %
year 1 month 4 rate = 0.004*(1 + 0.008)^3  = 0.409677 %
...
year 2 month 1 rate = 0.004*(1 + 0.008)^12 = 0.440135 %
etc.

so in this case the monthly rate will increase by 10.034% year to year, e.g.

0.440135 % / 0.4 % - 1 = 10.034 %

Compounding the amount z for 48 months with these rates

=100*PRODUCT(1 + 0.004*(1 + 0.008)^(SEQUENCE(48)-1))

126.16

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  • Is y the starting interest rate?
    – RonJohn
    Oct 30, 2023 at 2:01
  • x is the starting rate: 5% (in the annual example), which increases by y (10%) every year: so 5%, then 5.5%, then 6.05%, then 6.655% for the years 1 to 4. Apr 28 at 12:03
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As stated in the comment for the answer this calls for an exponential function. For an annual rate of y% (say 0.05) take z = ln(1+y) = 0.04879... as the continuous rate of interest. Now you can take exp(t * z) and for any time t (years, can be any fraction) you have the continuous compounded result.

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