Why Pooled IRR Requires Cash-Flow Timing

# Computing Pooled IRR from the IRRs of parts


# Computing Pooled IRR from the IRRs of parts












Suppose I have two cash flows:

- CF1: -10001001001100

- CF2: -20020301

I can compute now:

- IRR(CF1) = 10%

- IRR(CF2) =-55%

- IRR(CF1+CF2) = 4.46%

Is there a way to compute (or at least get a fair estimate) of the pooled IRR (i.e. 4.46%) by only knowing the two original IRRs along with, say, the initial investment values (or some additional partial metrics of the CFs)?

## Answer by Nicholas (score 1, accepted)

https://quant.stackexchange.com/a/22094

You could roughly estimate it by approximating the cash flows (which you do not know in full) using some "reasonable" simplified model. One example of such a model would be a cash flow of the form

$$-\mathrm{investment}, 0, 0, \dots{\small (n-1 \text{ zeros})}\dots, 0, \mathrm{returns}$$

Such a simple cash flow has IRR $r$ iff $$ \mathrm{returns} = \mathrm{investment}\cdot(1+r)^n. $$

Now, if you only knew the IRRs $r_1$, $r_2$ and investment amounts $a_1$, $a_2$ for the two cash flows, you could approximate the PIRR by constructing the corresponding "model" cash flows and computing their pooled IRR:

$$ \hat{\mathrm{PIRR}} = \left(\frac{a_1(1+r_1)^n + a_2(1+r_2)^n}{a_1 + a_2}\right)^{\frac{1}{n}} - 1 $$

In your particular example, taking $n=3, a_1 = 1000, a_2 = 200, r_1 = 0.1, r_2 = -0.55$ you would obtain $\hat{\mathrm{PIRR}} = 3.98\%$.

Note that when $n=1$ this approximation corresponds to the weighted average of the two rates (with investment amounts being the weights). As $n\to \infty$ the approximation converges towards $\max(r_1, r_2)$.

The latter observation also illustrates why PIRR is not uniquely defined. In fact, if we used different real-valued lengths $n_1$, $n_2$ in the model representations of the two cash flows, we could have obtained almost any* desired resulting value of PIRR within $(\min(r_1, r_2), \max(r_1, r_2))$.

* It is always the case when both rates are positive. When one or both rates are negative this claim is not as obvious and might require proof.

## Answer by Malick (score 4)

https://quant.stackexchange.com/a/22069

No there is no way since the calculated internal rate of return $r$ is by definition defined as:

$0 = \sum_{i=0}^{I} \frac{C_{i}}{(1+r)^{i}} $

You need to know the entire cash flow distribution and its timing if you want to compute the Pooled IRR.

One advantage of IRR is that it takes the irregular timings of cash flows into account, logically its disadvantage is that you need to know the timing and the associated amount.

## Answer by emcor (score 4)

https://quant.stackexchange.com/a/22070

Exact solution:

Assume we agree that for $y_1:=IRR(CF1)$, $y_2:=IRR(CF2)$, $y:=IRR(CF1+CF2)$, the following equations hold by definition:

$$-1000+\frac{100}{1+y_1}+\frac{100}{(1+y_1)^2}+\frac{1100}{(1+y_1)^3}=0$$ $$-200+\frac{20}{1+y_2}+\frac{30}{(1+y_2)^2}+\frac{1}{(1+y_2)^3}=0$$ $$-1200+\frac{120}{1+y}+\frac{130}{(1+y)^2}+\frac{1001}{(1+y)^3}=0$$

These equations represent the definition of IRR and must hold. It is known that a system of equations can be solved by adding the equations, so it follows: $$-1000-200+\frac{100}{1+y_1}+\frac{20}{1+y_2}+\frac{100}{(1+y_1)^2}+\frac{30}{(1+y_2)^2}+\frac{1100}{(1+y_1)^3}+\frac{1}{(1+y_2)^3}\stackrel{!}{=}-1200+\frac{120}{1+y}+\frac{130}{(1+y)^2}+\frac{1001}{(1+y)^3}$$ So we need to solve $$\frac{100}{1+y_1}+\frac{20}{1+y_2}+\frac{100}{(1+y_1)^2}+\frac{30}{(1+y_2)^2}+\frac{1100}{(1+y_1)^3}+\frac{1}{(1+y_2)^3}=\frac{120}{1+y}+\frac{130}{(1+y)^2}+\frac{1001}{(1+y)^3}$$ Call the left hand side "$X$" and multiply the equation by $(1+y)^3$: $$X(1+y)^3=120(1+y)^2+130(1+y)+1001$$ This is a cubic equation: $$X(1+y)^3-120(1+y)^2-130(1+y)-1001=0$$ The real-valued solution to such equation given by here is: $$1+y= - \frac{1}{3a}\left(b\ +\ C\ +\ \frac{\Delta_0}{C}\right)$$ where $C = \sqrt[3]{\frac{\Delta_1 + \sqrt{\Delta_1^2 - 4 \Delta_0^3}}{2}}$, $\Delta_0 = b^2-3 a c $, $\Delta_1 = 2 b^3-9 a b c+27 a^2 d$.

It follows: $$y= - \frac{1}{3a}\left(b\ +\ C\ +\ \frac{\Delta_0}{C}\right)-1$$ with $a=X$, $b=-120$, $c=-130$, $d=1001$.

This solution incorporates the two IRRs $y_1,y_2$. It happens to have the same expression as when calculating the "normal" IRR based on the cashflows since $X=1200$ by definition. Note that this must be the case, because we are looking for the same $y$ that solves both equations.

As requested by the OP and moderator, I will write out this solution:

$$y= - \frac{1}{3X}\left(-120\ +\ \sqrt[3]{\frac{\Delta_1 + \sqrt{\Delta_1^2 - 4 \Delta_0^3}}{2}}\ +\ \frac{\Delta_0}{\sqrt[3]{\frac{\Delta_1 + \sqrt{\Delta_1^2 - 4 \Delta_0^3}}{2}}}\right)-1$$ $$= - \frac{1}{3\left(\frac{100}{1+y_1}+\frac{20}{1+y_2}+\frac{100}{(1+y_1)^2}+\frac{30}{(1+y_2)^2}+\frac{1100}{(1+y_1)^3}+\frac{1}{(1+y_2)^3}\right)}\left(-120\ +\ \sqrt[3]{\frac{2 b^3-9 a b c+27 a^2 d + \sqrt{\left(2 b^3-9 a b c+27 a^2 d\right)^2 - 4 \Delta_0^3}}{2}}\ +\ \frac{\Delta_0}{\sqrt[3]{\frac{2 b^3-9 a b c+27 a^2 d + \sqrt{\Delta_1^2 - 4 \Delta_0^3}}{2}}}\right)-1$$

I stop writing out the solution here as it would not add more information. This equation infers the PIRR from the IRRs, all other values are constants. It is the exact solution.

However, it shows that the solution must incorporate the cashflows, it is not possible to separate out the solution to depend only on $y_1,y_2$. The third root term in the equation immediately implies that the cashflows cannot be canceled out.

Approximate Solution:

The IRR $y$ of the sum of the cashflows must lie withing the range $[y_1,y_2]$ where $y_1\leq y_2$. If $y$ is higher than $y_1$ and $y_2$, then the above equation has no solution since all cash flows on the right were discounted at a higher rate than the left such that the equation could not hold. By the same argument it follows that $y$ must be larger than the minimum of $y_1,y_2$.

A natural guess would be to take the average of the two: $$y\approx (y_1+y_2)/2=-0.225$$ Since we have a sum of CF1 and CF2, we may further weight the average by the absolute sum of cash flows to get: $$y\approx \frac{2200}{2200+251}0.1+\frac{251}{2200+251}\cdot(-0.55)=0.03345$$ This estimate is very close to the exact value of $0.046$.