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 $\displaystyle \frac {(1 + IRR)^5}{(1 + IRR)^5}$, 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 $\displaystyle (1 + IRR)^5$ . an easier way. can you finish?
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?
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.