Compound interest questions -- Annual rate of 3.11% which is compounded 3 times each year

pbuk
Gold Member
Check his algebra again. It is right.
Oh nuts, really? See, when it comes to numbers I'm lost without a spreadsheet

pbuk
Gold Member
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Chestermiller
Mentor
The only instantaneous interest rate that makes sense to me is $$100\frac{A'}{A}=100n\ln{\left(1 + 0.01 \frac{r}{n} \right)}$$which is a constant, independent of t.

brake4country and pbuk
pbuk
Gold Member
The only instantaneous interest rate that makes sense to me is $$100\frac{A'}{A}=100n\ln{\left(1 + 0.01 \frac{r}{n} \right)}$$which is a constant, independent of t.
Yes, a constant with dimensions ## T^{-1} ## . If you substitute the original equation 𝐴(𝑡)=𝐼(1+ (0.01×𝑟/n))^𝑛𝑡 into that then you recover the answer to Q1, the rate of growth in \$ per year as a function of t.

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Oops - almost right. You had the term 1+(0.0311)/3 in your original answer which was correct but now you have 3.0311/3 which is not.
I just used common denominator =)

pbuk
WWGD
Gold Member
2019 Award
I know I am right in practice, I do this for a living.
So much for my theory on peanut butter ( pb) ;)

Math_QED and pbuk
pbuk
Gold Member
I just used common denominator =)
Yes of course

etotheipi
Gold Member
2019 Award
I think you mean discrete ## n ##. And it's not a 'very good approximation': at every time ## t_i ## that the discrete model is defined it is equal to the value of the continuous model at ## t_i ##.
No, I mean finite ##n##. If we take the function to be continuous on the interval containing all real ##t##, then ##A(t) = I(1+\frac{r}{n})^{nt}## approaches ##A_{2}(t) = Ie^{rt}## in the limit of large ##n##. However since the former is very close to the latter even for finite ##n##, the latter is a very useful time-saving approximation. The former is still exactly an exponential function, ##A(t) = Ie^{Rt}##, but with a slightly smaller rate constant ##R =n\ln{\left(1 + \frac{r}{n} \right)}##. You can show that ##\lim_{n \rightarrow \infty} n\ln{\left(1 + \frac{r}{n} \right)} = r##. My point is that an analyst would probably approximate discrete compounding with an exponential of unchanged rate constant

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pbuk
Gold Member
No, I mean finite ##n##. If we take the function to be continuous on the interval containing all real ##t##, then ##A(t) = I(1+\frac{r}{n})^{nt}## approaches ##A_{2}(t) = Ie^{rt}## in the limit of large ##n##. However since the former is very close to the latter even for finite ##n##, the latter is a very useful time-saving approximation. The former is still exactly an exponential function, ##A(t) = Ie^{Rt}##, but with a slightly smaller rate constant ##R =n\ln{\left(1 + \frac{r}{n} \right)}##. You can show that ##\lim_{n \rightarrow \infty} n\ln{\left(1 + \frac{r}{n} \right)} = r##. My point is that an analyst would probably approximate discrete compounding with an exponential of unchanged rate constant
I see. However we are not interested in large ## n ##, we are only interested in two values, ## n = 3 ## and ## n = 1 ##. If we use ## n = 3 ## then the value of the investment after 1 year using the headline rate ## r ## is ## I (1 + \frac{r}{3} ) ^ 3 ## and if we use ## n = 0 ## then the value of the investment after 1 year using the equivalent annual rate ## r_a ## is ## I ( 1+ r_a ) ^ 1 ##. From the definition of the equivalent annual rate ## r_a = (1 + \frac{r}{n})^n - 1 ## you can immediately see that these two calculations are exactly equal.

Because we only define the annual rate of return at integral numbers of years, the continuous model is not an approximation, it is exactly equal to the discrete model at every point that the discrete model is defined.

A model where a headline rate is compounded continuously to give an effective annual rate which you have correctly calculated as ## e^r - 1 ## is not useful either in investment theory or practice.

etotheipi
etotheipi
Gold Member
2019 Award
That is all fine, but when I wrote that I was referring to this
And as it happens the discrete model is not how it happens anyway - interest is normally accrued on a daily basis (using some conventional calculation) increasing the value of the deposit (almost) continuously. The accrued amount is not included in the amount that is used to calculate interest.

pbuk
Gold Member
That is all fine, but when I wrote that I was referring to this
Oh I see

Reading that back I could have phrased it better:

Interest is normally calculated on a daily basis (using some conventional calculation - see below) increasing the value of the investment (almost) continuously. However this does not mean that the interest compounds continuously, the calculated amount is added to a separate balance ("accrued interest") from the sum that is used to calculate interest (the "principal"). At each relevant date the balance of accrued interest is transferred to the principal ("rolled up") or paid to the investor.

The daily interest calculation is set by the contract and/or conventions of the particular financial instrument and will follow one of a number of surprisingly arcane procedures some of which are outlined here.

etotheipi
Chestermiller
Mentor
Oh I see

Reading that back I could have phrased it better:

Interest is normally calculated on a daily basis (using some conventional calculation - see below) increasing the value of the investment (almost) continuously. However this does not mean that the interest compounds continuously, the calculated amount is added to a separate balance ("accrued interest") from the sum that is used to calculate interest (the "principal"). At each relevant date the balance of accrued interest is transferred to the principal ("rolled up") or paid to the investor.

The daily interest calculation is set by the contract and/or conventions of the particular financial instrument and will follow one of a number of surprisingly arcane procedures some of which are outlined here.
In line with this and what @etotheipi said in post #11, A(t) should really be a discontinuous function of t, with the discontinuities occurring on every compounding date (or whatever it's called), and A(t) should more properly be expressed with the nt in the exponent being replaced by $$\sum_{m=1}^{\infty}u\left(t-\frac{m}{n}\right)$$where u(x) is the unit step function, equal to 0 if its argument is less than zero and 1 if its argument is greater than zero. So, more properly, A(t) should read $$A(t) = I\left(1+\frac{r}{n}\right)^{\sum_{m=1}^{\infty}u\left(t-\frac{m}{n}\right)}$$

etotheipi