Hi All..

I just needed some help solving a question to find the internal rate of return (IRR). I have attached the problem.

How do I find IRR from the equation?

I have the solution of 13.45%

Thank you

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- Apr 5th 2009, 09:19 AMdadonRearranging to find IRR
Hi All..

I just needed some help solving a question to find the internal rate of return (IRR). I have attached the problem.

How do I find IRR from the equation?

I have the solution of 13.45%

Thank you - Apr 5th 2009, 10:11 AMJhevon
first combine the fractions on the right and multiply the left by $\displaystyle \frac {(1 + IRR)^5}{(1 + IRR)^5}$, we get:

$\displaystyle \frac {10000(1 + IRR)^5}{(1 + IRR)^5} = $ $\displaystyle \frac {2000(1 + IRR)^4 + 2500(1 + IRR)^3 + 3000(1 + IRR)^2 + 3500(1 + IRR) + 4000}{(1 + IRR)^5}$

now, since the denominators are the same on both sides, we can just equate the numerators, to get:

$\displaystyle 10000(1 + IRR)^5 = $ $\displaystyle 2000(1 + IRR)^4 + 2500(1 + IRR)^3 + 3000(1 + IRR)^2 + 3500(1 + IRR) + 4000$

haha, we could have gotten this by multiplying through by $\displaystyle (1 + IRR)^5$ :p. an easier way. can you finish? - Apr 5th 2009, 12:54 PMdadon
Hey

Thanks for your reply

Ok I expanded the brackets dividing the whole thing by 1000 first

...very tedious method used Pascals triangle

Then I collected the like terms together this is what I got.. (I hope I didn’t make a mistake!!)

$\displaystyle 25IRR + 7.51RR^2 + 89.5IRR^3 + 48IRR^4 + 10IRR^5 – 4001 = 0$

Then divided the whole thing by 0.5

$\displaystyle 50(IRR) + 155(IRR)^2 + 179(IRR)^3 + 96(IRR)^4 + 20(IRR)^5 - 8002 = 0$

LOL after all that I’m still nowhere close to the answer 13.45%...

I’m I going the right way about this.. surely there is a simpler method? - Apr 5th 2009, 01:58 PMdadon
Hey All..

Sorry I think the example I gave before was abit tedious on here.. I have found a better one which I have attached.. :)

The question is still the same how do I find IRR?

Thank you - Apr 5th 2009, 02:56 PMJhevon
- Apr 6th 2009, 05:08 AMdadon
- Apr 6th 2009, 06:23 AMdadon
Hey me again :P

Ok I get the quadratic stage.. but normally the powers are high like the first example posted.. how do I solve for higher powers..Is there like a way to do it on calculator?

are there more formulas?

Thank you - Apr 6th 2009, 06:43 PMJhevon
that is not right. dividing through by 100 and expanding everything, we get:

$\displaystyle 100(IRR)^5 + 480(IRR)^4 + 895(IRR)^3 + 775(IRR)^2 + 250(IRR) - 50 = 0$

i am sorry, but i do not see a nice way to solve this. there is no general formula for solving general polynomials of degree 5 or above. and this guy does not look like a nice guy to factor.

so yeah, there is no general formula for solving quintic polynomials. if you can use technology for this, do so. otherwise, this question is just mean. - Apr 7th 2009, 04:12 AMdadon