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My question is; How can I approach a common interest rate? (See example below for an explanation)

Consider the example in the table below 

Addition       Interest rate        End of the year
100            1.05                 105        [100*1.05]
15             1.06                 127.2      [(105+15)*1.06]
10             1.08                 148.176    [(127.2+10)*1.08]

Hence my question; How can I find the 'common' interest rate, such that with 1 single interest rate, the End of year three amount stays the same? (See example below)

Addition       Interest rate        End of the year
100            x                    y1         [100*x]
15             x                    y2         [(y1+15)*x]
10             x                    148.176    [(y2+10)*x]

I was thinking the following, but it did not result in the correct interest rate (Consider the first example, I did not round anything in my calculations.):

100 is done times 1.05, 1.06 and 1.08  = 120.204
15 is done times 1.06 and 1.08         = 17.172
10 is done times 1.08                  = 10.8
                                         148.176 (Sum)

My thought on the interest rate was the following: 
1.05   -   120.204
1.06   -   137.376      (120.204 + 17.172)
1.08   -   148.176      (120.204 + 17.172 + 10.8)
           405.756 (Sum)

Interest rate x = [(120.205 / 405.756) * 1.05] + [(137.376 / 405.756) * 1.06] + [(148.176 / 405.756) * 1.08] 
                = 1.064341..

However this method results in an end amount of 148.206 instead of 148.176

Revils
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2 Answers2

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The way you have explained the problem, it is a compound interest one.

The correct equation for it would thus be
$100(1+r)^3 + 15(1+r)^2 + 10(1+r) = 148.176$, which on solving, yields

$r = 0.0642608, i.e. 6.42608\%$

If you check with this interest rate, you will get back 148.176

true blue anil
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  • I just was at that step indeed, I found the same equation. What is the easiest way to solve an equation like that systematically? Since the list of addings can be variable the following could be the equation: 100(1 + r)^5 + 15(1 + r)^4 + 10(1+r)^3 + .... – Revils Nov 18 '15 at 13:44
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    See http://math.stackexchange.com/questions/200617/how-to-solve-an-nth-degree-polynomial-equation – true blue anil Nov 18 '15 at 13:51
  • Well, this example was simplified. The real equations are going to have around 120 different periods with additions and interest rates. Which would give a power of ^120 + ^119 + ^118 + ... + ^1. Would be fun to solve that haha,.. – Revils Nov 18 '15 at 13:57
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If you expand out an equation for your example, you get:

$$100x^3+15x^2+10x=148.176$$

Running the equation through Wolfram Alpha gives a real solution of $1.06426$, so your method of approximating the root is close.

In general, expanding out will get a polynomial, where the roots may not have nice closed forms. This root can be found numerically through interval bisection.

Since you mention this is an interest rate question, it makes sense to assume that all the different interest rates are positive, and there exists some maximum interest rate. Assuming we have no withdrawals (since your question did not state any), we seek to show that there exists a root in the interval $[0,x_{\max}]$, where $x_{\max}$ is our maximum interest rate in the time period.

Substituting $x=0$ would result in the left-hand side of the equation to be $0$. Assuming that there are only deposits, substituting $x=x_{\max}$ would result in the left-hand side being greater than or equal to the right-hand side.

By Intermediate Value Theorem, there is thus a value of $x$ such that the left-hand side would equal to the right hand side, making this interval a viable candidate to begin interval bisection or other numerical methods like interpolation.

Element118
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