uniqueness of IRR

minivan15 · #1 $Old$ October 31st, 2009, 06:14 PM

There was a two part question assigned to my class for homework. It's already been turned in, but I couldn't get the 2nd part, and am curious how it is done:

part a:

suppose there is a k between 0 and n such that either:

(i) $C_0, C_1, ..., C_k <= 0$ and $C_(k+1), C_(k+2),..., C_(n)>= 0$
or

(ii) $C_0, C_1, ..., C_k >=0$ and $C_(k+1), C_(k+2),..., C_(n) <= 0$
show there is a unique i > -1 for which the net present value of this transaction is 0.

this is the part I got. I don't need help on it, but am just introducing it, because I'm sure it somehow is used to prove the next part.

part b:

Let $C_0, C_1, ..., C_n$ be an arbitrary sequence of net cashflows, and let

$F_0 = C_0, F_1 = C_0 + C_1, . . . F_n = C_0 + C_1 + ... + C_n$

Suppose both $F_0$ and $F_n$ are non-zero, and that the sequence ${F_0, F_1, ..., F_n}$ has exactly one change of sign.

Show there is a unique i > 0 such that the net present value of these cash flows is 0 (although there may be one or more negative roots).

This is the part I need help on. I can show what I've done so far, but it really isn't anything except showing there is a root i > 0. I haven't shown that it is unique.

my work:

we need to show that the equation

has a unique solution 0 < v < 1.

if v = 0,

if v = 1,

we are given F_0 and F_n are non-zero and opposite signs of each other, so by the intermediate value theorem there exists a root v, 0 < v < 1.

This is all I have come up with, I don't know if that is a good starting point for the rest of the solution or not. any help would be appreciated! Thanks in advance.

TKHunny · #2 $Old$ October 31st, 2009, 10:08 PM

Yikes!

$\epsilon \neq 0$

If IRR is not unique, then both $(i)$ and $(i+\epsilon)$ produce the same result.

minivan15 · #3 $Old$ October 31st, 2009, 10:46 PM

could you explain what you mean by this a bit more please?

Thank you!

TKHunny · #4 $Old$ October 31st, 2009, 11:22 PM

If IRR is not unique, then there must be some $\epsilon \neq 0$ such that $CashFlows\;Discounted\;@\;i\;=\;CashFlows\;Discounted\;@\;(i\;+\;\epsilon)$

For the simplest case, one cash flow, C, over one year, we have:

For $v = \frac{1}{1+i}$ and $w = \frac{1}{1+i+\epsilon}$

$Cv = Cw \implies C(1+i) = C(1+i+\epsilon) \implies \epsilon = 0$

Do we have to say more than that?

minivan15 · #5 $Old$ November 1st, 2009, 12:39 AM

I'm sorry, but I'm not sure your answer is valid. I agree that it works with you for the simple case you illustrated, but what about for a more complicated case?

i.e.

$C_0+C_1*v+C_2*v^2+...+C_n*v^n=C_0+C_1*w+C_2*w^2+...+C_n*w^n$

in this case, v = w is definately ONE solution, but is there a way you were alluding to that says it is the ONLY solution?

* edit*

for example, consider (and solve) the equation

$2x+3x^2=2y+3y^2$

$2(x-y)+3(x-y)(x+y)=0$

$(x-y)[2+3(x+y)]=0$

$x=y$ or $y=-2/3 -x$

so in this case the solution x = y was not unique.

TKHunny · #6 $Old$ November 1st, 2009, 03:01 PM

Hmpf! For a minute there, I thought I had something because your x and y are independent and my (i) and (i+ $\epsilon$ ) clearly are not. However, full recognition of the relationship leads to the same dilemma.

I considered making $\epsilon > 0$ , but that is insufficiently general.

Perhaps some other kind soul will put me out of my misery while I am thinking about it.

Wilmer · #7 $Old$ November 2nd, 2009, 10:57 AM

Quote:

Originally Posted by minivan15 $View Post$

for example, consider (and solve) the equation
$2x+3x^2=2y+3y^2$
$2(x-y)+3(x-y)(x+y)=0$
$(x-y)[2+3(x+y)]=0$
$x=y$ or $y=-2/3 -x$
so in this case the solution x = y was not unique.

That's really a quadratic equation,
with solution x = y or x = -(y + 2/3);
and (I think) that due to the "nature" of IRR,
x = -(y + 2/3) is not valid.

Anyhow, not being sure, plus confused to no end with those
scary(!) flows of yours, I asked someone this "fresh" question:
.................................................. ..................................................
x^2 + 3x = k^2 + 3k ; x obviously = k, right?

x^2 + 3x - k^2 - 3k = 0

x = {-3 +- sqrt[(2k + 3)^2]} / 2

x = (-3 + 2k + 3)/2 or x = (-3 - 2k - 3)/2

x = k or x = -k-3

WHY?
.................................................. .................................................
Answer was:
.................................................. .................................................
You should ask that question about the first line of your post. Clearly if x = k
then that makes the first equation true. But why do you think that's the
only way the first equation can be true?

Compare to: "x2 = k2 ; obviously x = k right? er but 32 = (-3)2!"

Perhaps it would be instructive to substitute in x = -k-3 to the equation
x2+3x = k2+3k and see for yourself that both sides really do match up.
(You just need to expand the LHS and collect terms.)

Also even if x = k was the only solution of your first equation, the fact
you arrived at the end at x = k or x = -k-3 wouldn't be a problem unless
you also noted that every step was "if and only if". Otherwise you can
quite easily get surplus solutions at the end. Compare to...

"Let x = y. Therefore x2 = y2. This has two solutions: x = y or x = -y. Er what gives?"

What gives is that some of the steps of this argument are only ==> not <==>.
Hence "x = y or x = -y" is an *implication* of the premise "x = y" but is
not equivalent. This is why at the end of a "==>" argument you have to
check that all your solutions are valid and throw out any spurious "solutions".
.................................................. .................................................

Amen

TKHunny · #8 $Old$ November 2nd, 2009, 05:07 PM

I was more done than I thought? Very unsatisfying.

Of course, this becomes monumentally more difficult as the number of cash flows increases. I'm trying the get DesCartes' Rule of Signes to help us out, but it isn't quite making sense. I can't control the direction of the Cash Flows.

Wilmer · #9 $Old$ November 2nd, 2009, 09:08 PM

Well, as far as I know, the definition of IRR is:
"The internal rate of return on an investment or potential investment is the annualized
effective compounded return rate that can be earned on the invested capital."

A simple example:
3 annual cash flows of 2000, 3000 and 5000 at 12%:
0[-7737] 1[2000] 2[3000] 3[5000]
This is from (rounded):
2000/1.12^1 + 3000/1.12^2 + 5000/1.12^3 = 1786 + 2392 + 3559 = 7737

Not knowing that the IRR is 12%, we then have:
0[-7737] 1[2000] 2[3000] 3[5000]
and need to calculate the resulting IRR (IRR = r):
2000/(1+r)^1 + 3000/(1+r)^2 + 5000/(1+r)^3 = 7737

We can solve for r, but only by using iteration.
And we know there is only one possible r.
So why try and prove it?

TKHunny · #10 $Old$ November 3rd, 2009, 06:04 AM

Well, somewhere we should generate a sound theoretical basis for our actions. We certainly should make undergraduates work on such proofs.

Wilmer · #11 $Old$ November 3rd, 2009, 06:13 AM

Quote:

Originally Posted by TKHunny $View Post$

Well, somewhere we should generate a sound theoretical basis for our actions.

Best way to generate a sound is eat beans

flamumhum · #12 $Old$ November 3rd, 2009, 08:14 PM

I accept with information:
Suppose both

and

are non-zero, and that the sequence

has exactly one change of sign.

Show there is a unique i > 0 such that the net present value of these cash flows is 0 (although there may be one or more negative roots).

This is the part I need help on. I can show what I've done so far, but it really isn't anything except showing there is a root i > 0. I haven't shown that it is unique.

my work:

we need to show that the equation

has a unique solution 0 < v < 1.

if v = 0,

if v = 1,

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TKHunny · #13 $Old$ November 3rd, 2009, 09:29 PM

1 sign change? See, that is the direction.

I think it obvious that if all the cash flows are positive and all the discounts are integer multiples of 2, we're done.