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I have to calculate the monthly value I have to save with 4.5 % interest to get € 529411 in 35 years.

As I know, it is about € 520 but I need a formula to integrate it into my software.

Thank you

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yes I am sure, if you save evety month about 520 € für 35 years with 4.5 % interest you will get something about 530.000 - i check it on other sits :) –  Fincha Nov 12 '11 at 12:46
    
there is a simple formula $$V=V_0(1+\frac{p}{100})^{420}$$ , but it gives that monthly interest in percents is $p=1.66$ % not $4.5$ –  pedja Nov 12 '11 at 13:12
    
Is the interest only compounded annually? Or monthly? –  Raskolnikov Nov 12 '11 at 13:13
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2 Answers 2

up vote 3 down vote accepted

The accumulated future worth $F$ at the end of $n=35\times 12$ equal deposits$^1$ $A$ compounded monthly at an interest rate per month $i=0.045/12$ is the sum

$$F=\sum_{k=1}^{n}A(1+i)^{n-k}=A\frac{(1+i)^{n}-1}{i}.$$

So

$$A=F\dfrac{i}{(1+i)^{n}-1}=529411\dfrac{0.045/12}{(1+\dfrac{0.045}{12} )^{35\times 12}-1}=520.18.$$

$^1$ Each deposit is assumed to be made at the end of each month.

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This is all on the assumption that payments are made at the end rather than the beginning of each month though, isn't it? For example if just one deposit is made (so $n=1$ in the first formula), then you're left with $F=A$ -- in other words, no interest is added. If the payments are made at the beginning of each month, you need to multiply $F$ above by $(1+i)$. –  Shane O Rourke Nov 12 '11 at 13:52
    
@ShaneORourke: Yes, that was my assumption. –  Américo Tavares Nov 12 '11 at 13:57
    
@ Americo Tavares: But isn't it more natural in the context of savings to take it that the payments are made at the beginning of each month, rather than at the end? In other words, to suppose it's a due rather than an ordinary annuity? The formulae above would seem to apply more naturally in the case of loan repayments. –  Shane O Rourke Nov 12 '11 at 14:19
    
thank you very much! –  Fincha Nov 12 '11 at 14:31
    
@ShaneORourke: I used the assumption of end-of-period compounding by two reasons: 1) it is used in Engineering Economics by Riggs, Bedworth and Randhawa in the derivation of the sinking fund factor $A/F$. 2) It agrees with the numerical value indicated by OP. –  Américo Tavares Nov 12 '11 at 14:42
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The formula is

$$\text{Monthly payment} = \dfrac{\text{Monthly interest}}{(1+\text{Monthly interest})^{\text{Duration}}-1} \times \text{End capital}$$

If you compute the monthly interest as $0.045/12$, you'll get $520.18$ as the end result.

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I think you need to add -1 to the denominator. –  Américo Tavares Nov 12 '11 at 13:24
    
Yeah, forgot it. –  Raskolnikov Nov 12 '11 at 13:25
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