IRR
Motivated by e-mail from Dean A.

I always thought I understood IRR.

>Funny! I always thought you didn't know ...
Pay attention. My understanding goes like this:

• You invest \$10K.
• After one year, you withdraw \$5K.
• At the end of 3 years, your portfolio is worth \$7K.
The \$10K should be worth 10(1+x)3 after 3 years. (That's measured in kilo-bucks.)
After 1 year you withdraw \$5K, so that's money that don't earn nothing.
So we'd have 10(1+x)3 - 5(1+x)2 after 3 years.
That's equal to the \$7K final portfolio value ... so to find the annual Rate of Return (IRR) we must solve:
[1]         10(1+x)3 - 5(1+x)2 = 7
The solution turns out to be x = IRR = 0.0896 or 8.96%

>So that 5(1+x)2 represents money without the x% return, right?
That's what I would have said, but consider this:

• You invest \$10K.
• After one year, you withdraw \$5K, so your portfolio is now worth 10(1+x) - 5. (Again, in kilo-bucks.)
The money you withdrew? You bought chocolate bars.
• This portfolio balance of 10(1+x) - 5 is allowed to remain, earning an annual return of x% for another 2 years.
• At the end of 3 years, your portfolio is worth \$7K ... and some chocolate bars.
We'd have:
[2]         { 10(1+x) - 5 }(1+x)2 = 7 (That's the 1-year balance, invested for another 2 years at x% per year.)
Note that it's the same equation as [1].

Aah, but what did you do with the \$5K you withdrew?
If you bought chocolate bars, it vanished (financially speaking) and your return would be the solution to [2], namely 8.96%.

Suppose, instead of buying chocolate bars, you withdrew that \$5K and put it under the pillow.
It then becomes part of your final portfolio, so our equation would now be:
[3]         10(1+x)3 - 5(1+x)2 = 7 + 5
The solution turns out to be x = 0.258 or 25.8%

Indeed, if you actually invested that \$5K at some annual return r% for the remaining 2 years, that'd be part of your final portfolio.
We'd then have:
[4]         10(1+x)3 - 5(1+x)2 = 7 + 5(1+r)2
Note that, if r = 0, it's the same equation as [3]. (Under the pillow, eh?)
If r = 0.04 or 4%, the solution turns out to be x = 0.270 or 27.0%

>Okay, so what's your annual return?

Note that this ritual described above ([1] or [2], to calculate IRR) could also be presented like so:

• You invest A1. In a year it'd be worth A1(1+x).
• Then you withdraw A2, so your portfolio is now worth (after 1 year): P1 = A1(1+x) - A2.
• After 2 years, your portfolio balance is P1(1+x).
• Now you invest another A3.
• Your portfolio balance becomes (after 2 years): P2 = P1(1+x) + A3 = A1(1+x)2 - A2(1+x) + A3.
• In another year, you withdraw A4.
• The portfolio balance is then (after 3 years): P3 = A1(1+x)3 - A2(1+x)2 + A3(1+x) - A4.
• Suppose that, after 4 years, your portfolio is worth P4.
• Then:
 **   A1(1+x)4 - A2(1+x)3 + A3(1+x)2 - A4(1+x) = P4
Yes ... and you gotta solve for x = IRR.

 the infamous Reinvestment Assumption

It seems to be a prevalent understanding (among some financial gurus) that IRR assumes that all money withdrawn from an investment is reinvested at the IRR rate.
I never understood the arguments, but they go something like this (using the above example):

• You invest \$10K.
• You withdraw \$5K at the end of the first year.
• Your portfolio is worth \$7K at the end of 3 years so you have \$5K + \$7K = \$12K in your pocket.
• You started with \$10K so your 3-year gain factor is 12/10 = 1.2.
• That's over 3 years which implies an annual return of 1.21/3 - 1 = 0.0627 or 6.27%.
>So?
So the IRR value is IRR = 8.96%. It overestimates your "actual" annual return (whatever that means) of 6.27%.
>So?
So that means that the higher IRR return implies that you reinvested the \$5K withdrawal at the IRR rate.
That reinvestment gave you that higher IRR return.
>But suppose I put the \$5K under my pillow. What then? It wouldn't be reinvested, right?
That's right ... and you'd have \$12K after 3 years and your annual return would be 6.27%.
>But if I withdrew \$5K then immediately reinvested it at the same rate as the original \$10K, then ...
Then it's like not withdrawing it at all. You'd have an initial \$10K ending up worth 10(1+r)3 ... for some r-value.
>Which equals ... what?
At r = 8.96% (that's IRR) it'd be \$12,936 and at r = 6.27% (that's without reinvesting) it'd be \$12,000.
I conclude that, if you did NOT reinvest the withdrawal (but put the \$5K withdrawal under the pillow), your final portfolio would be worth \$12K.
If you reinvested at the IRR rate of 8.96%, your final portfolio would be worth \$12,936 (not \$7K).
The 3-year gain factor is 1.2936 so the annual gain factor would be 1.29361/3 = 1.0896 which means an annual return of 8.96%.
>Aha! So IRR does imply reinvestment at the IRR rate, right?
But remember, I got that final portfolio of \$12,936 by assuming that the withdrawasl was invested at the IRR rate.
So it's not surprising that we conclude that the withdrawal is invested at the IRR rate.

However, if (after 1 year) we withdrew \$5K then immediately reinvested it at the same rate, we'd get:

• The \$10K is worth 10(1+x) after 1 year.
• We withdraw \$5K then immediately put it back in, giving 10(1+x) - 5 + 5.
• Another 2 years (at the same rate) and we'd have { 10(1+x) - 5 + 5 }(1+x)2    or
[5]         10(1+x)3 = 7.

>Huh? Your \$10K ends up being worth a measly \$7K?
Yes ... and that's a negative annual return of -11.2%.
You started with \$10K and end up with \$7K. That's a loss, eh?

>This is really getting confusing!
Ain't it though?

>But you said that, if you reinvest the \$5K withdrawal at r%, the equation is [4].
Aah, yes, I see the confusion. In [4], put r = x and you don't get [5].
That's because [4] assumes you withdrew \$5K then immediately invested at some rate independent of IRR.
The resultant equation, namely [4], gives x = actual return on investment.
>So it's not IRR?
No, it's whatever is the solution to [4]. Remember that IRR is (by definition) the solution to [1].
On the other hand, [5] assumes you withdraw from your portfolio then immediately put it back in.
Your actual return is whatever is the solution to [5].

Let's recap:

• IRR (by definition) is the solution to [1].
• If you withdrew \$5K and burned that withdrawal money (or bought chocolate bars), your return would be the solution to [1].
That's IRR.
• If you reinvest \$5K at a 0% rate (under the pillow!) and assume the final portfolio is \$12K you get the 6.27% return.
That's the solution to [3], namely 25.8%.
• If you reinvest at some (unknown) rate x (the same as the original investment) and assume the final portfolio is \$7K, you get a negative return.
That's like withdrawing \$5K then immediately putting it back in your portfolio. Your return is the solution to [5].
• Now you reinvest at the IRR rate (putting the withdrawal in some other investment account so as not to confuse it with the original account).
The balance in that other account must be added to the \$7K in the original portfolio, giving you a final balance of 7+5(1+IRR)2.
You get a return which is the solution to [4] with r = IRR, namely 28.4%.

Conclusion?
Be careful what you stick in for your final portfolio value and where you invest the withdrawals.

>So what does IRR give?
I've already told you ... it's the solution to [1].
It'd give the annual return if your final portfolio is just \$7K and you spend the withdrawals on chocolate bars.

>So it should be called CIRR.
Chocolate Internal Rate of Return? I agree.

Let's consider the general case:

• There's an initial investment in a portfolio P, namely A0, and are several others: A1, A2, ...
They occur at times t1, t2, ...
• There are also several withdrawals: B1, B2, ... which occur at times z1, z2, ...
They are invested in some other portfolio Q at investment rates r1, r2, ...
• After n years your portfolio P is worth \$K. Then:
[P]       { A0(1+x)n + A1(1+x)n-t1 + A2(1+x)n-t2 + } - { B1(1+x)n-z1 + B2(1+x)n-z2 + ... } = K
The solution is x = IRR (whatever that turns out to be).
It's just equation equation ** with different notation, right?
• Then ...
>Wait! You mean I get that invest-withdrawals-in-chocolate-bars return?
They become part of your final worth -- to be added to \$K.
• Your other portfolio, Q, is worth:
[Q]       B1(1+r1)n-z1 + B2(1+r2)n-z2 + ...
They're part of your financial gains, too. So how much money do you (and your wife) have?
• [R]       Final worth = K + B1(1+r1)n-z1 + B2(1+r2)n-z2 + ...
>So my actual return isn't IRR?
No, it's the solution to an equation like [4], namely:
 { A0(1+x)n + A1(1+x)n-t1 + A2(1+x)n-t2 + } - { B1(1+x)n-z1 + B2(1+x)n-z2 + ... } = K + B1(1+r1)n-z1 + B2(1+r2)n-z2 + ...
Here's something interesting:
If you start with \$A0 and make no more investments and all withdrawals are invested at IRR, then A(1+IRR)n = K + {sum of final values of investments}

>Huh?
I mean that:
 IRR satisfies (by definition): ! A0(1+IRR)n - { B1(1+IRR)n-z1 + B2(1+IRR)n-z2 + ... } = K which can be written !! A0(1+IRR)n = K + { B1(1+IRR)n-z1 + B2(1+IRR)n-z2 + ... }
>So the withdrawals are invested at IRR!
Okay, look at it this way:

• You invest A0 then fall asleep for n years.
• While sleeping, your mutual fund manager makes a number of withdrawals and invests them at IRR.
That's the solution to equation [P] without any additional investments A1, A2 etc.
• Then the manager sticks these withdrawals in your portfolio (after n years of IRR returns).
• You wake up and say:
I started with A0 and now I have K + { B1(1+IRR)n-z1 + B2(1+IRR)n-z2 + ... }.
That's an n-year return on my original investment of \$A0 which then satisfies:
[6]     A0(1+x)n = K + { B1(1+IRR)n-z1 + B2(1+IRR)n-z2 + ... }
>And the solution is x = IRR, right?
Yes, according to !!.
That's why (some) consider IRR to involve withdrawals to be invested at the IRR rate.
Of course, had you made the withdrawals and just burned the withdrawal monies (no reinvestment), you'd solve ! and get the solution: IRR.
However, suppose you made NO withdrawals, but found some extra money in your portfolio after n years (equivalent to fictitious withdrawals invested at IRR)?
You'd get equation !! with the same solution: IRR.

>You're spending a lot of time rambling on about IRR and ...
That's because I'm trying to understand the various interpretations people might have regarding the "reinvestment" of withdrawals.
For example:

1. You withdraw \$5K and immediately reinvest it at the same rate.
That presumably means you take it out then put it back in.
That'd mean 10(1+x) - 5 + 5 after 1 year (in our example, above).
That'd mean you didn't withdraw it at all. (See [5])
2. You withdraw and burn the money that you withdrew ... or buy chocolate bars. (See [1])
3. You withdraw and reinvest at a different rate. (See [4])
4. You withdraw and keep the \$5K, uninvested -- under your pillow. (See [3])
5. You withdraw and invest at the IRR rate. (See [4] with r = IRR, the solution to [1])
6. You make only one, original investment in a mutual fund. The manager withdraws and invests at IRR. (See [6])
7. etc. etc.

The example on the spreadsheet concerns an initial investment of \$10K followed by 4 annual withdrawals of \$5K ending with a portfolio worth \$5K after 5 years.
 To find IRR, we must solve: f(x) = 10*(1+x)5 - 5*(1+x)4 - 5*(1+x)3 - 5*(1+x)2 - 5*(1+x) - 5 = 0. The solution is x = IRR = 41.0415% ... which can be obtained using Excel's IRR command -- but I like newton! >You like Newton? Yes. There can be more than one solution to such equations and Newton can find 'em ... if you ask nice. In fact, for f(x) = 1000*(1+x)3 - 2500*(1+x)2 + 1700*(1+x) - 200 you get this: That's investing \$1000, withdrawing \$2500, investing an additonal \$1700 ... and ending with a \$200 portfolio after 3 years. You'd have to apply some cerebral gesticulation to identify the "appropriate" solution. >Which is? I'd flip a coin. But, seriously, I'd probably pick 35.2%.
>Why?
I'd like to feel good.

However, I invest \$1000 and, after a year I withdraw \$2500. Can you imagine withdrawing that much if the investment return were lousy?
>Like -85.2%, eh?
 Remember, the Math is quite happy subtracting from a negative portfolio balance. Of course, there may be strange annual returns which may happen. Your portfolio may look like this: You start with \$1000, do your \$2500 withdrawal and \$1700 investment ... and still end up with \$200 after 3 years. So what would you say about your annual return? Would you say 35.2% or -85.2%? You could always say: With annual returns of 200%, -50% and -89% a \$1 investment would end up with: (1 + 2)(1 - 0.5)(1 - 0.89) = 0.165 and that's 16 1/2 cents and that's a return of -83.5%.     >But them annual returns are unlikely, right? Unlikely, perhaps, but possible. My own portfolio lost 80% in 2008, but had a 400% return in 2009.

>Haven't we talked about multiple returns before?
Yes ... here.

I should point out that IRR (or XIRR) assumes you're investing in something that has a constant return over the life of the investment ... like a bank account.
 IRR (or XIRR) is the answer to the question: What constant annual return would give me the same final portfolio, considering all my investments and withdrawals?
Why should we expect that violent changes in annual returns could be summarized in a simple "constant" return like IRR?

Note:
IRR is often used to compare two business ventures. The one with the larger IRR is (presumably) the better choice.
In that scenario, the initial portfolio investment is replaced by a bank loan (to start the business).
Subsequent portfolio withdrawals are replaced by the profits from the business.
In this scenario, you'd be investing the profits at some rate independent of the bank loan.
In such a case, you'd be better served to consider not IRR, but MIRR.
With MIRR you know both the loan rate and the investment rate and would like to calculate some kind of return.
What return? The MIRR, of course -- and there ain't no fancy equation to solve and no multiple solutions.