Finance-Theory · Unit 21 · Video 5 · Interactive Practice

Multiple IRRs, Scale and Timing: Why IRR Can't Rank Projects

IKey Formulas

Formula	Name	What it says
$\text{NPV}(r) = CF_0 + \sum_{t=1}^{T} \dfrac{CF_t}{(1+r)^t}, \quad \text{NPV}(\text{IRR}) = 0$	NPV profile and IRR	Every IRR is a point where the NPV profile crosses zero
$(1+r)^T\,\text{NPV}(r) = CF_0(1+r)^T + CF_1(1+r)^{T-1} + \dots + CF_T$	IRR polynomial	Degree $T$ in $1+r$ : at most $T$ real IRRs, possibly several or none
$(1+r)^3\,\text{NPV}_2 = -500{,}000\,(r-0.04)(r-0.07)(r-0.10)$	The video's Project 2	Three real roots: $4\%$ , $7\%$ and $10\%$
$\Delta CF_t = CF^B_t - CF^A_t, \quad \text{NPV}(\Delta) = \text{NPV}_B - \text{NPV}_A$	Incremental cash flows	Once the smaller project clears the hurdle, take the bigger one only if $\text{IRR}(\Delta)$ exceeds the hurdle

Key Insight: IRR ranks projects by a rate; a firm wants dollars of value. When cash flows change sign more than once, or projects differ in scale or timing, rank by NPV.

IIVisualization 1 — One IRR or Three

Shifting the later cash flows by less than six percent turns one IRR into three.

Project 1 → Project 2 0/25

IIIVisualization 2 — Where a Solver Lands

A root-finder follows tangent lines down from its starting guess, so the guess decides which IRR it reports.

Starting guess 3.0% Steps shown 0

IVVisualization 3 — The Incremental Repair

IRR always ranks A first; the increment B − A asks whether B's extra dollars clear the hurdle.

Example Cost of capital r 10%

VQuiz Questions

Problem 1 · Two Break-Even Rates

Given: A project costs $400,000 today, receives $916,000 in one year and must pay out $524,160 in two years. Find all of its IRRs.

✅ Correct! The cash flows change sign twice (−, +, −), and the quadratic has two real roots:

x = 1.12

and

x = 1.17

, so the IRRs are

12\%

and

17\%

❌ Check the sign. The roots of the quadratic are

x = 1+r = 1.12

and

1.17

; subtracting 1 gives positive rates.

❌ Almost.

1.12

and

1.17

are values of

x = 1 + r

, the gross rate. The IRRs are

x - 1

❌ Not quite.

x = 1.145

is where the quadratic peaks, not where it is zero. A quadratic with a positive discriminant has two real roots, and both are IRRs.

❌ Not quite. Multiply NPV by

(1+r)^2

and solve the quadratic in

x = 1 + r

Show solution

Set NPV to zero and multiply through by $(1+r)^2$ , writing $x = 1 + r$ :

-400{,}000\,x^2 + 916{,}000\,x - 524{,}160 = 0

Divide by $-40{,}000$ :

10x^2 - 22.9x + 13.104 = 0

x = \frac{22.9 \pm \sqrt{22.9^2 - 4(10)(13.104)}}{20} = \frac{22.9 \pm \sqrt{524.41 - 524.16}}{20} = \frac{22.9 \pm 0.5}{20}

So $x = 1.17$ or $x = 1.12$ , and the IRRs are $r = 17\%$ and $r = 12\%$ . Equivalently,

(1+r)^2\,\text{NPV}(r) = -400{,}000\,(r - 0.12)(r - 0.17).

Two sign changes in the cash flows (−, +, −) leave room for two roots, and here both are real.

Problem 2 · Both IRRs Beat the Hurdle

Given: The same project, −$400,000 today, then $916,000 in one year and −$524,160 in two years, with IRRs of $12\%$ and $17\%$ . The opportunity cost of capital is $11\%$ . Decide whether to take it.

✅ Correct! NPV is positive only between the two roots. At

11\%

, below both of them, the project destroys value even though every IRR "clears" the hurdle.

❌ That is the IRR trap. With cash flows −, +, −, NPV is negative below

12\%

and above

17\%

. At

11\%

it is negative, so clearing the hurdle says nothing here.

❌ Check the sign.

(1.11 - 1.12)(1.11 - 1.17) = (-0.01)(-0.06) = +0.0006

, and multiplying by

-400{,}000

makes it negative.

❌ Not quite. No rule picks one of several IRRs, and averaging them is not a rule either. Compute the NPV at

11\%

❌ Not quite. Evaluate the NPV at the cost of capital.

Show solution

Use the factored form from Problem 1 with $r = 0.11$ :

(1.11)^2\,\text{NPV}(11\%) = -400{,}000\,(0.11 - 0.12)(0.11 - 0.17) = -400{,}000\,(0.0006) = -240

\text{NPV}(11\%) = \frac{-240}{1.2321} \approx -194.79

The NPV is negative, about −$194.79, so reject. The leading coefficient is negative, so NPV is positive only for $12\% < r < 17\%$ . An $11\%$ hurdle lies below both IRRs, and the "accept if IRR exceeds the hurdle" rule gives the wrong answer.

Problem 3 · Scale and the Increment

Given: Two mutually exclusive one-year projects at an $11\%$ cost of capital. A: −$200,000 today, $254,000 in a year (IRR $27\%$ ). B: −$500,000 today, $610,000 in a year (IRR $22\%$ ). Find the IRR of the incremental cash flows B − A and choose a project.

IRR of B − A

Which project?

✅ Correct! A clears

11\%

, and the extra $300,000 earns about

18.7\%

, also above

11\%

. B's NPV, about $49,549.55, beats A's, about $28,828.83.

❌ Almost.

356{,}000 / 300{,}000 \approx 1.187

1 + \text{IRR}

; subtract 1.

❌ Not quite. The IRR of the increment is not the difference of the two IRRs. Subtract the cash flows, then find the rate that zeroes them.

❌ Not quite. Averaging IRRs has no meaning. The increment is −$300,000 today and $356,000 in a year.

❌ That is the scale trap. A's higher rate applies to only $200,000. The extra $300,000 in B earns about

18.7\%

, above the

11\%

cost of capital, so B adds more dollars of value.

❌ Check the increment. B − A is −$300,000 today and $356,000 in a year.

❌ Check the decision. Compare the increment's IRR with the

11\%

hurdle.

Show solution

Step 1: Check the smaller project. A's IRR of $27\%$ clears $11\%$ .

Step 2: The increment. B − A: $-500{,}000 - (-200{,}000) = -300{,}000$ today and $610{,}000 - 254{,}000 = 356{,}000$ in a year.

-300{,}000 + \frac{356{,}000}{1 + \text{IRR}} = 0 \;\Rightarrow\; 1 + \text{IRR} = \frac{356{,}000}{300{,}000} = \frac{89}{75} \;\Rightarrow\; \text{IRR} = \frac{14}{75} \approx 18.7\%

Step 3: Decide. $18.7\% > 11\%$ , so make the extra investment: choose B.

Check with NPV:

\text{NPV}_A = -200{,}000 + \frac{254{,}000}{1.11} \approx 28{,}828.83, \qquad \text{NPV}_B = -500{,}000 + \frac{610{,}000}{1.11} \approx 49{,}549.55

\text{NPV}(\Delta) = -300{,}000 + \frac{356{,}000}{1.11} \approx 20{,}720.72 = 49{,}549.55 - 28{,}828.83

Problem 4 · Timing: Early Cash or a Perpetuity

Given: Two mutually exclusive projects at an $11\%$ cost of capital, each costing $300,000 today. A pays $150,000 at the end of each of the next three years (IRR $\approx 23.4\%$ ). B pays $45,000 a year forever, starting next year (IRR $15\%$ ). Find the NPV at $11\%$ of the increment B − A and choose a project.

NPV of B − A at 11%

Which project?

✅ Correct! The increment is worth about $42,533.70 today, exactly

\text{NPV}_B - \text{NPV}_A

, so B wins despite its lower IRR.

❌ Almost. The $45,000 perpetuity in the increment starts in year 4, so

45{,}000/0.11

is its value at year 3. Discount it by

1.11^3

❌ Not quite. That is

\text{NPV}_B

alone. The increment subtracts A's cash flows year by year.

❌ Not quite. That counts only years 1–3 of the increment. From year 4 on, B − A is $45,000 a year forever.

❌ That is the timing trap. A's higher rate comes from paying back early. Discounted at

11\%

, B's long stream is worth more.

❌ Not quite. Paying back sooner is the payback rule's preference, and it ignores everything B pays after year 3. The increment's NPV is positive.

❌ Check the increment. B − A is

0

today, −$105,000 in years 1–3, then $45,000 a year forever.

❌ Check the decision. A positive increment NPV means the switch from A to B adds value.

Show solution

Step 1: Increment B − A. Year 0: $-300{,}000 - (-300{,}000) = 0$ . Years 1–3: $45{,}000 - 150{,}000 = -105{,}000$ . Years 4 onward: $45{,}000$ a year forever.

Step 2: Value it at 11%. The three-year annuity factor is $\frac{1}{1.11} + \frac{1}{1.11^2} + \frac{1}{1.11^3} \approx 2.44371472$ . The perpetuity from year 4 is worth $45{,}000/0.11 \approx 409{,}090.91$ at year 3.

\text{NPV}(\Delta) \approx -105{,}000(2.44371472) + \frac{409{,}090.91}{1.11^3} \approx -256{,}590.05 + 299{,}123.75 = 42{,}533.70

Step 3: Check.

\text{NPV}_A \approx 150{,}000(2.44371472) - 300{,}000 \approx 66{,}557.21, \qquad \text{NPV}_B = \frac{45{,}000}{0.11} - 300{,}000 \approx 109{,}090.91

$109{,}090.91 - 66{,}557.21 = 42{,}533.70$ , the increment's NPV. Its IRR is about $12.6\%$ , above the $11\%$ hurdle. Choose B, even though the IRR rule ranks A first ( $23.4\%$ against $15\%$ ).

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