# Rearranging to find IRR

• April 5th 2009, 09:19 AM
Rearranging 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
• April 5th 2009, 10:11 AM
Jhevon
Quote:

Originally Posted by dadon
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

first combine the fractions on the right and multiply the left by $\frac {(1 + IRR)^5}{(1 + IRR)^5}$, we get:

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

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

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

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

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!!)

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

Then divided the whole thing by 0.5
$50(IRR) + 155(IRR)^2 + 179(IRR)^3 + 96(IRR)^4 + 20(IRR)^5 - 8002 = 0$

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?
• April 5th 2009, 01:58 PM
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
• April 5th 2009, 02:56 PM
Jhevon
Quote:

Originally Posted by dadon
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

the same approach as above will work, except you end up with a quadratic equation, which is easier to manage
• April 6th 2009, 05:08 AM
Quote:

Originally Posted by Jhevon
the same approach as above will work, except you end up with a quadratic equation, which is easier to manage

Hey :)

Ok I tried what you said

It gives me this equation:
6000(IRR) + 4000(IRR)^2 - 2000 = 0
3(IRR) + 2(IRR)^2 - 1 = 0

and I get the answer 0.2808 and -1.7808

Thank you
• April 6th 2009, 06:23 AM
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
• April 6th 2009, 06:43 PM
Jhevon
Quote:

Originally Posted by dadon
Hey

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!!)

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

Then divided the whole thing by 0.5
$50(IRR) + 155(IRR)^2 + 179(IRR)^3 + 96(IRR)^4 + 20(IRR)^5 - 8002 = 0$

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?

that is not right. dividing through by 100 and expanding everything, we get:

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

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.

Quote:

Originally Posted by dadon
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

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.
• April 7th 2009, 04:12 AM
$100(IRR)^5 + 480(IRR)^4 + 895(IRR)^3 + 775(IRR)^2 + 250(IRR) - 50 = 0$