Financial Interest Calculation

The first formula is for the amount after one year of simple interest, while the second is the amount after m years of compound interest. They will be equal if i = r and m = 1.

Why do you ask?
 
Dr.Peterson said:
The first formula is for the amount after one year of simple interest, while the second is the amount after m years of compound interest. They will be equal if i = r and m = 1.

Why do you ask?
Click to expand...

There is no particular reason for asking. I just saw this in a textbook, but I could not determine the solution to it.

Will m always need to equal 1 for this equation to be equal?
 
Dr.Peterson said:
The first formula is for the amount after one year of simple interest, while the second is the amount after m years of compound interest. They will be equal if i = r and m = 1.

Why do you ask?
Click to expand...
Because the formula

[MATH]P * \left (1 + \dfrac{r}{100c} \right )^{(c * n)}[/MATH]
is the formula for calculating ending principle on initial principal P invested for n periods at an annual rate of r percent and compounded c periods per year, I'd prefer to leave simple interest out of a discussion that asks to distinguish between (1 + r/100) and (1 + r/100)^n. I'd say the first is the formula for the ending value of one monetary unit invested for one year at r% annually and compounded annually whereas the second is the formula for the ending monetary value of one monetary unit invested for n years at r% annually and compounded annually.

The simple interest formula is quite different, with multiplication instead of exponentiation

[MATH]P * \left (1 + \dfrac{nr}{100c} \right ).[/MATH]
Obviously, you are correct that both formulas reduce mathematically to the same

[MATH]P * \left (1 + \dfrac{r}{100} \right ) \text { if } c = 1 \text { and } n = 1.[/MATH]
 
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Explain this! said:
There is no particular reason for asking. I just saw this in a textbook, but I could not determine the solution to it.

Will m always need to equal 1 for this equation to be equal?
Click to expand...
[MATH]a^1 = a \text { always.}[/MATH]
But no

[MATH]a = a^m \implies m = 1 \text { UNLESS } a = 0 \text { or } a = 1.[/MATH]
So there are two exceptions to the rule you gave. It is not universal. The rule I gave is universal.
 
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Explain this! said:
There is no particular reason for asking. I just saw this in a textbook, but I could not determine the solution to it.

Will m always need to equal 1 for this equation to be equal?
Click to expand...
Please quote exactly what the book said. Did it ask this question, or did it just use the two different formulas in different contexts, and you wondered how they are related?

If I were given the equation P(1 + r) = P(1 + i)m to solve, I would say that there are infinitely many combinations of r, i, and m that would make it true, and P can be any number at all. Give values for r and i, we could say that m = log(1+r)/log(1+i).

My quick answer was not intended as a solution of that equation, just an obvious observation that it would be true if (but not only if) the variables were matched up. My main point was that the two formulas as stated (which even have different variables for the rate, suggesting that they have different meanings) are commonly used in different contexts. They will happen to be equal under certain conditions, but that is meaningless, really.

In any case, I doubt that the book would intend any of this, which is why I want to see the source.
 
Dr.Peterson said:
Please quote exactly what the book said. Did it ask this question, or did it just use the two different formulas in different contexts, and you wondered how they are related?

If I were given the equation P(1 + r) = P(1 + i)m to solve, I would say that there are infinitely many combinations of r, i, and m that would make it true, and P can be any number at all. Give values for r and i, we could say that m = log(1+r)/log(1+i).

My quick answer was not intended as a solution of that equation, just an obvious observation that it would be true if (but not only if) the variables were matched up. My main point was that the two formulas as stated (which even have different variables for the rate, suggesting that they have different meanings) are commonly used in different contexts. They will happen to be equal under certain conditions, but that is meaningless, really.

In any case, I doubt that the book would intend any of this, which is why I want to see the source.
Click to expand...

The information is too large for me to copy or to attach a file. The server is unable to process any file that I try to attach. The book is entitled Mathematics of Finance by Simpson and others fourth edition.
 
You cant just take a picture or screenshot of one page? I don't need the whole book, just the bit that prompted your question.
 
Thanks. That's much better. Now we have the actual statement you are asking about, in context. (There seems to be a printing error that amounts to a fold line that makes some bits hard to read, and it's hard to tell m from n, but I think I've got it after enlarging.)

FMH117556.png

The relevant part says this:

Let r denote the effective rate equivalent to a given nominal rate j converted m times a year. We shall use i exclusively for rate per conversion period. At the rate i = j/m, P amounts in 1 year to P(1 + i)^m. At the effective rate r, P amounts in 1 year to P(1 + r), Since equivalent rates yield equal amounts in the same length of time,​
P(1 + r) = P(1 + i)^m,​
or​
1 + r = (1 + i)^m,​
whence​
r = (1 + i)^m - 1. ......... (2)​
By use of (2) we can find the effective rate r equivalent to the nominal rate j = im converted m times a year, that is, equivalent to the rate i = j/m per conversion period.​

This is one of the possible contexts I had imagined. As I said, the different rates have different meanings, and the context defines them:

r = effective interest rate (that is, the simple rate that results in the same amount after one year)​
i = rate per conversion period (is that the same thing as compounding?)​
m = conversion periods per year​

The key idea is that they get the same amount by thinking of the interest in two different ways, and then solve to find how the two different rates are related.

I'll let the financial experts fill in any details.
 
Dr. Peterson is absolutely correct.

I find the book's vocabulary a bit odd; at least it is not standard in U.S. banking, but the profession's vocabulary is not wholly consistent in any case.

What your book calls "conversion period" is more usually called "compounding period" or "payment period" in the U.S. It reflects when interest is paid or else added to principal.

The amount of interest due for the period t is

[MATH]S_t * \dfrac{i}{100m}, \text { where}[/MATH]
[MATH]S_t = \text {principal at the START of period number t;}[/MATH]
[MATH]i = \text {the CONTRACT interest rate expressed as an annual percentage; and}[/MATH]
[MATH]m = \text {the number of compounding periods in a year.}[/MATH]
When interest is compounded, the amount of interest earned during the period is added to starting principal for that period to calculate ending principal for that period.

[MATH]\therefore E_t = S_t + \left (S_t * \dfrac{i}{100m} \right ) = S_t * \left ( 1 + \dfrac{i}{100m} \right ) \text {where}[/MATH]
[MATH]E_t = \text {principal at the END of period number t.}[/MATH]
[MATH]\text {And obviously } S_{t+1} = E_t.[/MATH]
Now you can prove that the principal at the end of m compounding periods, which means at the end of 1 year, is

[MATH]_ = S_1 * \left ( 1 + \dfrac{i}{100m} \right )^m.[/MATH]
I am not going to prove that mathematically, but merely demonstrate it with an example. The Federal Reserve invests $100 billion with the European Central Bank at an annual rate of 2% compounded quarterly. How much does ECB owe the Fed at the end of a year?

[MATH]\text {principal at start of first period} = 100,000,000,000.[/MATH]
[MATH]\text {interest earned in first period} = 100,000,000,000 * \dfrac{2}{100 * 4} = 500,000,000.[/MATH]
[MATH]\text {principal at end of first period} = 100,000,000,000 + 500,000,000 = 100,500,000,000 \implies[/MATH]
[MATH]\text {principal at start of second period} = 100,500,000,000.[/MATH]
[MATH]\text {interest earned in second period} = 100,500,000,000 * \dfrac{2}{100 * 4} = 502,500,000.[/MATH]
[MATH]\text {principal at end of second period} = 100,500,000 + 502,500,000 = 101,002,500,000 \implies[/MATH]
[MATH]\text {principal at start of third period} = 101,002,500,000[/MATH]
[MATH]\text {interest earned in third period} = 101,002,500,000 * \dfrac{2}{100 * 4} = 505,012,500.[/MATH]
[MATH]\text {principal at end of third period} = 101,002,500,000 + 505,012,500 = 101,507,512,500 \implies[/MATH]
[MATH]\text {principal at start of fourth period} = 101,507,512,500.[/MATH]
[MATH]\text {interest earned in fourth period} = 101,507,512,500 * \dfrac{3}{100 * 4} = 507,537,562.50.[/MATH]
[MATH]\text {principal at end of fourth period} = 101,507,512,500 + 507,537,562.50 = 102,015,050,062.50.[/MATH]
Now, that is a lot of messy arithmetic, and, if we were dealing with smaller numbers, we would run into rounding issues.

It is a lot easier to get out the calculator and compute

[MATH]100,000,000,000 * \left ( 1 + \dfrac{2}{100 * 4} \right )^4 = 102,015,050,062.50[/MATH]
Now notice that the Fed would have got exactly the same amount had it invested with annual compounding instead of quarterly compounding at a contract rate of

[MATH]2.015,050,062,500 \% \approx 2.015\%.[/MATH]
That is what your book is calling the "equivalent" rate, which is not a standard term in US banking. It might be called the "effective annual rate" or the "annual yield."
 
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