# Formula for continuously increasing interest rate

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%?

• Can you explain how is this a personal money matter? It sounds a lot more like accounting homework. Oct 28, 2023 at 19:32
• 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... Oct 28, 2023 at 20:34

## 2 Answers

A simple case, compounding `z` over 4 years with interest rate `x` increasing by `y` percent annually.

``````z = 100
x = 0.05
y = 0.10

a = z = 100
a = a (1 + x (1 + y)^0) = 105
a = a (1 + x (1 + y)^1) = 110.775
a = a (1 + x (1 + y)^2) = 117.477
a = a (1 + x (1 + y)^3) = 125.295
``````

With m = 4

$a=z\prod_{k=0}^{m-1}(1+x(1+y)^k)=125.295$

The expression does not simplify to a closed-form.

Such a formula could only be used for whole years. Any remaining partial year could be added on. So calculate the whole years with an annual rate equivalent to the daily-compounded rate, then add on any daily interest. To convert between rates. . .

``````r  = annual nominal rate compounded annually
rd = annual nominal rate compounded daily

r  = (1 + rd/365)^365 - 1
rd = 365 ((1 + r)^(1/365) - 1)
``````
• Is `y` the starting interest rate? Oct 30, 2023 at 2:01
• @RonJohn I added clarification in the answer. Oct 30, 2023 at 8:30

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.