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How to calculate the interest forumula

by Banaticus
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Banaticus
#1
Dec14-11, 03:52 PM
P: 30
1. The problem statement, all variables and given/known data
If you deposit $deposit into a bank account each year and get y% annual interest, the simple formula is
[tex]\frac{deposit((1+y)^{years}-1)}{y}[/tex]
How is this formula calculated? What's going on?

Let's look at $50, 6% interest, two years.
Maybe I'm supposed to be dropping $50 in just as the first year ends, so I don't build any interest on it the first year. I then build interest on it the second year, then drop in another $50 just as the year ends:
[tex]=(0*1.06+50)*1.06+50[/tex] no interest the first year + 50, then interest on that and another 50
=50*1.06+50
=103
That squares with the formula given earlier. So, let's look at three years
[tex]=(0*1.06+50)*1.06*1.06+50*1.06+50[/tex]
That could be rewritten as
[tex]=50x^2+50x+50[/tex] where x is 1+interest and the variable is (years-1, years-2... 1)
That comes to the same answer as the original formula, $159.18, so things are working out.

Let's look at 7 years. How does:
[tex]50*1.06^6+50*1.06^5+50*1.06^4+50*1.06^3+50*1.06^2+50*1.06+50[/tex]
get turned into:
[tex]\frac{50(1.06^7-1)}{.06}[/tex]

Let's play around with it.
[tex]50(1.06^6+1.06^5+1.06^4+1.06^3+1.06^2+1.06+1)[/tex]
So
[tex]1.06^6+1.06^5+1.06^4+1.06^3+1.06^2+1.06+1=\frac{1.06^7-1}{.06}[/tex]
right? Maybe if I explicitly state the interest...
[tex](1+.06)^6+(1+.06)^5+(1+.06)^4+(1+.06)^3+(1+.06)^2+(1+.06)+1[/tex]
[tex](1^6+.06^6)+(1^5+.06^5)+(1^4+.06^4)+(1^3+.06^3)+(1^2+.06^2)+(1+.06)+1[/tex]
[tex](1+.06^6)+(1+.06^5)+(1+.06^4)+(1+.06^3)+(1+.06^2)+(1+.06)+1[/tex]
[tex]1+.06^6+1+.06^5+1+.06^4+1+.06^3+1+.06^2+1+.06+1[/tex]
[tex].06^6+.06^5+.06^4+.06^3+.06^2+.06+7[/tex]
This isn't helping, it's not taking me in the direction that I want to go. How does:
[tex]50*1.06^6+50*1.06^5+50*1.06^4+50*1.06^3+50*1.06^2+50*1.06+50[/tex]
which is 419.6918825 get turned into:
[tex]\frac{50(1.06^7-1)}{.06}[/tex]
which is 419.6918825? How is that original equation calculated?
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Simon Bridge
#2
Dec14-11, 05:43 PM
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If T(i) is the total after year i:

you start off with D - that would be T(0)=D.
after one year you have the original deposit, plus the interest, which is rD, where r=y/100 (since y is a percentage)
so in year 1 you have T(1)=(1+r)D

year two you have the amount in T(1) plus interest on that:
T(2)=(1+r)T(1) = D(1+r)2

which would kinda suggest that [itex]T(n)=D(1+r)^n[/itex]

The amount of interest that has been earned must be the total in your bank account less the initial deposit: I = T(n)-D

[tex]I = D \big [ (1+r)^n - 1 \big ][/tex]

Which would be the usual formula for finding the interest earned after n compounding periods (in this case, annually).
This has a similar shape to the formula you are examining.
What was that formula supposed to find out, exactly?
Ray Vickson
#3
Dec14-11, 07:20 PM
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Quote Quote by Simon Bridge View Post
If T(i) is the total after year i:

you start off with D - that would be T(0)=D.
after one year you have the original deposit, plus the interest, which is rD, where r=y/100 (since y is a percentage)
so in year 1 you have T(1)=(1+r)D

year two you have the amount in T(1) plus interest on that:
T(2)=(1+r)T(1) = D(1+r)2

which would kinda suggest that [itex]T(n)=D(1+r)^n[/itex]

The amount of interest that has been earned must be the total in your bank account less the initial deposit: I = T(n)-D

[tex]I = D \big [ (1+r)^n - 1 \big ][/tex]

Which would be the usual formula for finding the interest earned after n compounding periods (in this case, annually).
This has a similar shape to the formula you are examining.
What was that formula supposed to find out, exactly?
According to the description given by the OP, D gets deposited each year, so at the end of year 2 the account contains [itex] T(2) = D(1+r) + D(1+r)^2, [/itex] etc. This leads to [tex] T(n) = D(x + x^2 + ... + x^n) = D\frac{x^{n+1}-1}{x-1},[/tex] where x = r+1. The value of T(n-1) [= amount at start of year n] would match the formula given by the OP.

RGV

Simon Bridge
#4
Dec14-11, 07:54 PM
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How to calculate the interest forumula

Ah thank you - misread - except for [itex]years = n+1[/itex] (the exponent).
But that would match the comment about missing a year's interest I guess.
Would also make y the interest rate represented as a fraction rather than as a percentage ... which was puzzling me. Both descriptions are used but I see that it is common to refer to a "percentage" and write a fraction (and call it the decimal form of the percentage).
Banaticus
#5
Dec15-11, 12:10 AM
P: 30
Thanks. I got the original formula from http://www.ajdesigner.com/phpinteres...deposits_p.php and it seems to be the same formula that my teacher was using (he didn't mandated anything, he just suggested that we use massive Excel tables to calculate it, so I went out looking for a multiple payment interest generating formula that gave the same answer that my tables were giving), but I just can't see how it was derived, how it was put together. Well, the formula as presented on that webpage was actually:
[tex]P=M((1+\frac{i}{q})^{nq}-1)(\frac{q}{i})[/tex] where q was the number of periods in the year, etc., but since I was working with only one interest generating period (simple annual interest) and only one deposit during that time period, I simplified the formula by putting in 1 for q and then removing it from the equation.
Quote Quote by Ray Vickson View Post
This leads to [tex] T(n) = D(x + x^2 + ... + x^n) = D\frac{x^{n+1}-1}{x-1},[/tex] where x = r+1. The value of T(n-1) [= amount at start of year n] would match the formula given by the OP.
Thanks. :) I don't see how the second formula was calculated from the first though.
Ray Vickson
#6
Dec15-11, 10:22 AM
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Quote Quote by Banaticus View Post
Thanks. I got the original formula from http://www.ajdesigner.com/phpinteres...deposits_p.php and it seems to be the same formula that my teacher was using (he didn't mandated anything, he just suggested that we use massive Excel tables to calculate it, so I went out looking for a multiple payment interest generating formula that gave the same answer that my tables were giving), but I just can't see how it was derived, how it was put together. Well, the formula as presented on that webpage was actually:
[tex]P=M((1+\frac{i}{q})^{nq}-1)(\frac{q}{i})[/tex] where q was the number of periods in the year, etc., but since I was working with only one interest generating period (simple annual interest) and only one deposit during that time period, I simplified the formula by putting in 1 for q and then removing it from the equation.

Thanks. :) I don't see how the second formula was calculated from the first though.
I don't know what your "second" and "first" refer to, since you cite several posts, each having some formulas.

RGV
Banaticus
#7
Dec15-11, 11:33 AM
P: 30
Quote Quote by Ray Vickson View Post
I don't know what your "second" and "first" refer to
They refer to the portion of your post quoted immediately above the words "second" and "first" in the post you quoted. The inability of this forum to display a quote in a quoted post can swiftly become confusing, so let me just repost it:
Quote Quote by Ray Vickson View Post
[tex]D(x + x^2 + ... + x^n) = D\frac{x^{n+1}-1}{x-1},[/tex]
Where does the x^{n+1}-1, etc., come from, how was that formula generated?
Simon Bridge
#8
Dec15-11, 06:19 PM
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let [itex]s=x+x^2+\ldots +x^{n-1}+x^n[/itex]

then [itex]xs=x^2+x^3+\ldots + x^n + x^{n+1}[/itex]

then [itex]xs-s = x^{n+1}-x[/itex] or:

[tex]s=\frac{x^{n+1}-x}{x-1}[/tex]

hmmm ... did I miss something again?
(That extra factor of x would be the interest in the (n+1)th year, so it doesn't count?)

anyway: think "geometric series".


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