## Re: Discount Rate of Multiple Cash Flows

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• ... and n , ... To find the equation you seek requires us to find the root of a high order polynomial. In the example you gave, there is at least one term
Message 1 of 5 , Mar 31, 2008
--- In mathforfun@yahoogroups.com, "settle2117" <acravenho@...> wrote:
>
> Wow you are amazing.
>
> Is it possible to create an equation that solves for "i" every time?
>
> This is what I have so far:
>
> PV= Present Value
> MP= Monthly Payment
> i= nominal interest rate
> n= number of payments
>
> PV= MP*(1-(1/(1+i)^n))/i
>
> I need it to be:
>
> i=
>
> I will always have the following variables given "PV", "MP",
and "n",
> but "i" will never be given.
>

To find the equation you seek requires us to find the root of a high
order polynomial. In the example you gave, there is at least one
term containing i^120, so we're talking a 120-degree polynomial. I'm
not aware of any technique that will yield an algebraic solution.
As far as I know, numerical analyis techniques must be used.

In fact, the problem you pose is exactly what the IRR function in
Microsoft Excel does. (IRR stand for "internal rate of return".)
I'm pretty sure that Microsoft also uses numerical analysis
techniques to do this.

I should say that I'm not the most knowledgable person here in
MathForFun, so maybe somebody else can speak to this, but I'll be
surprised if there is any algebraic solution to your problem.
• Solutions to linear,quadratic and cubic http://www.math.rutgers.edu/~erowland/polynomialequations.html The solution of a quartic
Message 2 of 5 , Apr 1, 2008
http://www.math.rutgers.edu/~erowland/polynomialequations.html

The solution of a quartic
http://mathworld.wolfram.com/QuarticEquation.html
Notice the exponential increase in complexity of the analytic solutions of degree 2,3,4.
Even if there was a general solution for degree n, the steps involved would be forboding asn becomes large. If a formula did exist, in my estimation the chances of making a mistake
for degree 120 is 100%.

Numerical solutions are crisp, clean and quick. The bisection method is no slouch and almost always guarantees convergence. Newton's method is much faster and convergeswell with polynomials. The accuracy doubles with each iteration. However, Newton'smethod can fail in many other functions and is problematic with functions that are difficultto differentiate.
Discussion on quintic.
http://mathworld.wolfram.com/QuarticEquation.htm
References here lay to rest the notion of generality beyond degree 4.

Here is a little gcc root finding routine using the bisection method

Enjoy,
Cino

To: mathforfun@yahoogroups.comFrom: TimKelleyGroups@...: Tue, 1 Apr 2008 01:36:36 +0000Subject: [MATH for FUN] Re: Discount Rate of Multiple Cash Flows

--- In mathforfun@yahoogroups.com, "settle2117" <acravenho@...> wrote:>> Wow you are amazing. > > Is it possible to create an equation that solves for "i" every time?> > This is what I have so far:> > PV= Present Value> MP= Monthly Payment> i= nominal interest rate> n= number of payments> > PV= MP*(1-(1/(1+i)^n))/i> > I need it to be:> > i= > > I will always have the following variables given "PV", "MP", and "n",> but "i" will never be given.> > I appreciate your time.To find the equation you seek requires us to find the root of a high order polynomial. In the example you gave, there is at least one term containing i^120, so we're talking a 120-degree polynomial. I'm not aware of any technique that will yield an algebraic solution. As far as I know, numerical analyis techniques must be used.In fact, the problem you pose is exactly what the IRR function in Microsoft Excel does. (IRR stand for "internal rate of return".) I'm pretty sure that Microsoft also uses numerical analysis techniques to do this.I should say that I'm not the most knowledgable person here in MathForFun, so maybe somebody else can speak to this, but I'll be surprised if there is any algebraic solution to your problem.

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