Why a Project Can Have More Than One IRR
An internal rate of return (IRR) is a discount rate at which a project’s net present value (NPV) equals zero. For a conventional project, the initial cash flow is an outflow followed by inflows. But some projects have later outflows too: a large closure cost, a required restoration payment, or a later investment in replacement equipment. When cash flows switch between positive and negative more than once, the NPV profile can cross zero more than once. The project can therefore have multiple IRRs.
This matters because the usual IRR hurdle comparison assumes a single relevant rate: accept when IRR exceeds the required return. With multiple IRRs, that comparison can give conflicting signals. The NPV at the organization’s required return remains a direct way to assess the project at that rate.
The restriction \(r>-1\) ensures that the discount factor \(1+r\) is positive and the present-value terms are defined. A positive IRR is only one possible kind of IRR: a root between \(-1\) and \(0\) is also a mathematical solution. This tutorial focuses on how repeated sign changes can produce multiple nonnegative roots and conflicting NPV signals.
Repeated sign changes make multiple IRRs possible, not inevitable. A stream with two sign changes might have two, one repeated, or no real roots in the relevant range. The cash-flow signs alone do not give the exact roots.
Turning the IRR Equation Into a Polynomial
Set \(x=1/(1+r)\). For \(r>-1\), \(x\) is positive, and the IRR equation becomes a polynomial:
Once a positive root \(x\) is found, convert it back to a rate using \(r=1/x-1\). A root \(0<x<1\) corresponds to a positive IRR; \(x=1\) corresponds to a zero IRR; and \(x>1\) corresponds to a rate between \(-1\) and \(0\).
The connection between sign changes and the polynomial follows directly: each cash flow \(C_t\) is the coefficient of \(x^t\). The order of coefficients is the chronological order of the cash flows, so the number of sign changes is unchanged. By Descartes’ rule of signs, the number of positive real roots of this polynomial, counted with multiplicity, is at most the number of coefficient sign changes and differs from it by an even number. This is an upper bound on roots \(x>0\), hence on IRRs \(r>-1\); it does not establish how many roots lie in \(0<x\leq1\), which is the range for nonnegative IRRs.
Worked Examples
Worked Example: Two Positive IRRs for a Remediation Project
A hypothetical site project has these cash flows, in thousands of dollars: an immediate investment of \(-100\), a receipt of \(325\) at the end of year 1, and a remediation payment of \(-250\) at the end of year 2. Find its nonnegative IRRs. Then determine whether the project has positive NPV at a required return of 60%.
Method. Write the NPV equation, use \(x=1/(1+r)\), factor the resulting polynomial, and convert each admissible root back to a rate. The cash-flow pattern is negative, positive, negative, with two sign changes, so two IRRs are possible.
In thousands of dollars, the equation is $$ -100+\frac{325}{1+r}-\frac{250}{(1+r)^2}=0. $$ Set \(x=1/(1+r)\). Multiplying the resulting equation by \(-1\) gives $$ 250x^2-325x+100=0 $$ $$ (5x-4)(50x-25)=0. $$ Thus \(x=0.8\) or \(x=0.5\). Convert each root: $$ x=0.8:\quad r=\frac{1}{0.8}-1=0.25=25\%; \qquad x=0.5:\quad r=\frac{1}{0.5}-1=1=100\%. $$
Check each root. At 25%, the NPV is \(-100+325/1.25-250/(1.25)^2=-100+260-160=0\). At 100%, it is \(-100+325/2-250/4=-100+162.5-62.5=0\). Both are valid IRRs.
At the required return of 60%, the NPV is $$ -100+\frac{325}{1.6}-\frac{250}{1.6^2} =-100+203.125-97.65625 =5.46875. $$ The NPV is positive, so the project is acceptable by the NPV rule at 60%. The calculation is in thousands; rounding the final NPV to two decimal places gives \(\$5.47\) thousand.
Common mistake avoided: Treating one IRR as the unique answer and deciding from that rate alone. Here one root is below 60% and one is above it, so a single comparison cannot give an unambiguous IRR signal.
Worked Example: Three IRRs and Alternating NPV Signals
A hypothetical energy-storage project has year 0 through year 3 cash flows, in thousands of dollars, of \(-100,\ 725,\ -1{,}550,\ 1{,}000\). Find its nonnegative IRRs. Then calculate the NPV at required returns of 50% and 200% to show why the roots create different acceptance signals.
Method. Express NPV as a polynomial in \(x=1/(1+r)\), factor it, and convert the roots. The cash flows have three sign changes, so there can be as many as three positive roots, counted with multiplicity.
The polynomial is $$ -100+725x-1550x^2+1000x^3 =1000(x-0.25)(x-0.5)(x-0.8). $$ Therefore the roots are \(x=0.25,\ 0.5,\ 0.8\), all in \(0<x<1\). The corresponding IRRs are $$ x=0.8:\ r=25\%; \qquad x=0.5:\ r=100\%; \qquad x=0.25:\ r=300\%. $$
Check the roots in the original equation. At \(r=25\%\), \(x=0.8\), and the NPV is \(-100+725(0.8)-1550(0.64)+1000(0.512)=-100+580-992+512=0\). At \(r=100\%\), \(x=0.5\), and it is \(-100+362.5-387.5+125=0\). At \(r=300\%\), \(x=0.25\), and it is \(-100+181.25-96.875+15.625=0\).
At 50%, \(x=1/1.5=2/3\), so $$ \operatorname{NPV}(0.5) =-100+725\left(\frac{2}{3}\right)-1550\left(\frac{4}{9}\right)+1000\left(\frac{8}{27}\right) =-\frac{250}{27} \approx -9.26. $$ At 200%, \(x=1/3\), so $$ \operatorname{NPV}(2) =-100+\frac{725}{3}-\frac{1550}{9}+\frac{1000}{27} =\frac{175}{27} \approx 6.48. $$
The NPVs are in thousands of dollars and are rounded to two decimal places. At the 50% hurdle, NPV is negative; at the 200% hurdle, it is positive. The three roots are not three independent acceptance instructions: the appropriate decision is based on NPV at the relevant required return.
Common mistake avoided: Assuming that three sign changes automatically mean three positive IRRs. This example has three, but that conclusion comes from solving and verifying the polynomial, not from counting signs alone.
Worked Example: A Hurdle Rate Between Two IRRs
A hypothetical equipment lease requires an initial outlay of \(\$60{,}000\), produces a year 1 receipt of \(\$156{,}000\), and requires a year 2 return-condition payment of \(\$99{,}000\). Find the IRRs and evaluate the project at a 30% required return.
Method. Work in thousands of dollars and solve the polynomial in \(x=1/(1+r)\). There are two cash-flow sign changes, so multiple IRRs are possible.
The NPV equation and its polynomial form are $$ -60+\frac{156}{1+r}-\frac{99}{(1+r)^2}=0, \qquad -60+156x-99x^2=0. $$ Factoring gives $$ -99\left(x-\frac{10}{11}\right)\left(x-\frac{2}{3}\right)=0. $$ The roots are \(x=10/11\) and \(x=2/3\), giving $$ r=\frac{11}{10}-1=10\%, \qquad r=\frac{3}{2}-1=50\%. $$
Check. At 10%, the NPV is \(-60+156/1.1-99/1.21=-60+141.81818\ldots-81.81818\ldots=0\). At 50%, it is \(-60+156/1.5-99/2.25=-60+104-44=0\).
At the required return of 30%, $$ \operatorname{NPV}(0.3) =-60+\frac{156}{1.3}-\frac{99}{1.3^2} =-60+120-\frac{9900}{169} =\frac{240}{169} \approx 1.42. $$ The NPV is positive, so the project is acceptable at 30%. The amount is in thousands of dollars; rounded to two decimal places, it is \(\$1.42\) thousand.
Common mistake avoided: Rejecting the project because the 50% IRR is above the 30% hurdle, or accepting it because the 10% IRR is below the hurdle. Neither comparison resolves the conflict. The NPV at 30% directly gives the decision.
Why the NPV Can Cross Zero More Than Once
The polynomial representation also explains the changing acceptance signals. As \(r\) increases, \(x=1/(1+r)\) decreases. Each distinct root \(x\) is a zero of the NPV profile, but a root is not necessarily a crossing: if a root has even multiplicity, the profile may touch zero and turn back. With simple roots, the sign changes as the profile passes each root. The order of the roots in \(x\) is the reverse of their order in \(r\), because \(r=1/x-1\) decreases as \(x\) increases.
For example, in the first worked example the two rates are 25% and 100%. At a 60% required return, which lies between them, NPV is positive. At 10%, below both roots, NPV is \(-100+325/1.1-250/1.1^2\), which is negative. Thus comparing each root separately with the hurdle would provide opposite signals, while evaluating NPV at the hurdle gives one answer for that required return.
Include all later outflows as well as inflows, and identify sign changes in chronological order.
Set \(\sum_{t=0}^{n}C_t/(1+r)^t=0\), using consistent periods and cash-flow units.
Use \(x=1/(1+r)\) where helpful. Keep roots corresponding to the rate range being considered.
Substitute each candidate rate into the original NPV equation. For a hurdle-rate decision, calculate NPV at the required return.
Exam Traps and Practical Interpretation
The most reliable habit is to distinguish three tasks: finding roots, verifying roots, and making a decision. A sign-change count is a diagnostic about how many roots might exist, not a solution. Numerical tools can help locate roots, but every candidate should be checked in the original NPV equation. Rounding a root too early can make that check appear not to equal zero.
- Do not omit a later outflow. Leaving out a closure or restoration payment can change the cash-flow sign pattern and the number and location of IRRs.
- Do not treat sign changes as a root count. Three sign changes give an upper bound of three positive polynomial roots, with the difference from the bound even; they do not prove that three IRRs exist.
- Do not confuse polynomial roots with positive IRRs. A positive \(x\) gives \(r>-1\), but a nonnegative IRR requires \(0<x\leq1\).
- Do not apply the conventional single-IRR rule mechanically. When multiple roots exist, compare the project’s NPV with zero at its required return instead.
For capital budgeting, NPV answers whether the project adds value at the specified required return, given the stated cash flows and discount rate. Multiple IRRs describe the zero crossings of the NPV profile; they do not replace that decision calculation. This distinction is aligned with the SOA VEE Accounting and Finance syllabus.
Check Your Understanding
For each problem, identify the relevant roots or sign information and use NPV at the stated hurdle when asked for an acceptance decision.
- A project has cash flows \(-80,\ 200,\ -125\) at years 0, 1, and 2. Use \(x=1/(1+r)\) to find its IRR and state whether it is one distinct root or two. (Answer: \(x=0.8\), so \(r=25\%\); it is one distinct, repeated root.)
- A project has cash flows \(-100,\ 260,\ -160\). Find its nonnegative IRRs and calculate whether NPV is positive or negative at a 30% required return. (Answer: IRRs are \(0\%\) and \(60\%\); NPV at 30% is \(5.3254\), positive, in the cash-flow units.)
- A four-period cash-flow stream has the nonzero sign sequence negative, positive, negative, positive. What is the maximum number of positive polynomial roots counted with multiplicity, and does the sign sequence prove that the maximum occurs? (Answer: At most three positive roots counted with multiplicity; no, the sign sequence does not prove that three roots occur.)
- A project has cash flows \(-100,\ 725,\ -1{,}550,\ 1{,}000\), in thousands. At a 200% required return, determine whether NPV is positive or negative, using the IRRs \(25\%,100\%,300\%\) if helpful. (Answer: NPV is \(175/27\approx 6.48\) thousand, positive; the NPV rule accepts at this required return.)
- A project has cash flows \(-60,\ 156,\ -99\), in thousands, and a required return of 30%. Its IRRs are 10% and 50%. What decision does NPV support, and why is choosing either IRR as the sole hurdle comparison unreliable? (Answer: NPV is \(240/169\approx 1.42\) thousand, so accept at 30%; the two roots lie on opposite sides of the hurdle and give conflicting single-IRR comparisons.)