Multiple Or Nonexistent Dollar-Weighted Returns

Why a Dollar-Weighted Return May Not Be Unique

For an ordinary investment with an initial contribution followed by a final value, the dollar-weighted return is often a single, intuitive rate. But an investor may contribute more money after the investment begins, withdraw money, or receive distributions and then invest additional funds. The resulting cash-flow pattern can change direction several times. In that setting, the return equation can have more than one solution—or no solution at all.

This is not merely a numerical-solver problem. Before trusting a computed rate, you need to understand how the cash-flow signs constrain possible roots, whether a root actually exists, and whether its accumulation factor makes it a financially admissible effective rate.

Definition: Let $C_t$ denote the investor’s net cash flow at time $t$, with contributions by the investor recorded as negative and receipts to the investor recorded as positive. For cash flows at integer times $t=0,1,\ldots,n$, a dollar-weighted return $i$ is a real solution of $$ \sum_{t=0}^{n} \frac{C_t}{(1+i)^t}=0. $$ The rate is financially admissible when $1+i>0$, or equivalently $i>-1$. A cash-flow stream can have multiple admissible solutions or no admissible solution.

The equation equates the present value of all investor cash flows to zero at the candidate rate. For a solution to represent an effective annual rate with integer-period cash flows, the accumulation factor $1+i$ must be positive. A root that violates this condition is not an admissible dollar-weighted return, even if it solves an algebraic expression produced during the calculation.

For integer times, set $x=1+i$. Admissibility then means $x>0$, and the return equation becomes $$ \sum_{t=0}^{n} \frac{C_t}{x^t}=0. $$ Multiplying by $x^n$, which is positive for every admissible $x$, gives the polynomial equation $$ C_0x^n+C_1x^{n-1}+\cdots+C_n=0. $$ This transformation helps identify candidate roots. It does not make every algebraic root a valid return: the candidate must still have $x>0$.

What Cash-Flow Sign Changes Can Tell You

Read the signs of the cash flows in chronological order, ignoring any zero amounts. Every change from positive to negative or negative to positive is a sign change. For equally spaced cash flows, the transformed polynomial has those cash flows as its coefficients, in reverse chronological order. Reversing a list does not change its number of sign changes.

Descartes’ rule of signs gives a useful upper bound: the number of positive real roots of a polynomial, counted with multiplicity, is at most the number of sign changes in its coefficient list, and differs from that number by an even integer. Here, a positive root $x$ corresponds to an admissible return $i=x-1$. The rule is a bound, not a guarantee that a root exists. A stream with two sign changes might have two positive roots, one repeated positive root, or no positive roots.

This root-counting result applies directly when the dates are equally spaced whole periods and the equation transforms to a polynomial. With irregular dates, the return equation can involve fractional powers instead. Then a polynomial sign-count argument may not apply; numerical methods can locate candidate solutions, but the cash-flow equation and admissibility checks remain essential.

Worked Examples

Worked Example: Two Distinct Admissible Returns

An investor contributes $1,000 at time 0, receives $2,100 at the end of year 1, and contributes a further $1,100 at the end of year 2. Find all financially admissible annual dollar-weighted returns. Assume annual periods and a constant effective annual rate.

Method. Record the investor-perspective cash flows as $C_0=-1000$, $C_1=2100$, and $C_2=-1100$. The signs are negative, positive, negative, so there are two sign changes and at most two positive roots, counted with multiplicity.

The return equation is $$ -1000+\frac{2100}{1+i}-\frac{1100}{(1+i)^2}=0. $$ Set $x=1+i$, with $x>0$, and multiply by $x^2$: $$ -1000x^2+2100x-1100=0. $$ Factoring gives $$ -1000(x-1)(x-1.1)=0. $$ Thus $x=1$ or $x=1.1$, both positive. Converting back to effective annual rates gives $i=1-1=0$, or $i=1.1-1=0.1$. The two returns are therefore $0\%$ and $10\%$.

Check. At $i=0$, the equation is $-1000+2100-1100=0$. At $i=10\%$, it is $-1000+2100/1.1-1100/1.21=-1000+1909.0909\ldots-909.0909\ldots=0$. Both rates satisfy the original equation and are admissible.

Mistake avoided. Reporting only one return because a numerical solver found one misses the second valid root. When signs change more than once, check for additional solutions.

Worked Example: Sign Changes but No Real Return

An account has investor cash flows of $-$1,000$ at time 0, $+$1,000$ at the end of year 1, and $-$1,000$ at the end of year 2. Determine whether an annual dollar-weighted return exists.

Method. The signs are negative, positive, negative, so there are two sign changes. Write the equation and let $x=1+i>0$: $$ -1000+\frac{1000}{x}-\frac{1000}{x^2}=0. $$ Multiplying by $x^2$ gives $$ -1000x^2+1000x-1000=0. $$ The discriminant is $$ 1000^2-4(-1000)(-1000)=1{,}000{,}000-4{,}000{,}000=-3{,}000{,}000. $$ Since the discriminant is negative, the quadratic has no real roots, so it has no positive roots and there is no financially admissible return. Equivalently, its left side can be written as $$ -1000\left((x-0.5)^2+0.75\right), $$ which is strictly negative for every real $x$.

Check. The expression cannot equal zero for any $x>0$, so no annual effective rate solves the original equation.

Mistake avoided. Two sign changes do not mean that there must be two returns. They give an upper bound on the number of positive roots; the equation must still be solved or otherwise shown to have roots.

Worked Example: A Repeated Return

An investor contributes $1,000 at time 0, receives $2,000 at the end of year 1, and contributes $1,000 at the end of year 2. Find the distinct annual dollar-weighted return and explain what the repeated root means.

Method. The investor cash flows are $-1000,+2000,-1000$, again giving two sign changes. With $x=1+i>0$, the return equation becomes $$ -1000+\frac{2000}{x}-\frac{1000}{x^2}=0. $$ Multiplying by $x^2$ and factoring gives $$ -1000x^2+2000x-1000=-1000(x-1)^2=0. $$ The only root is $x=1$, with multiplicity two. It is positive, so the distinct admissible return is $i=1-1=0$, or $0\%$.

Check. Substitution into the original equation gives $-1000+2000-1000=0$. More generally, its left side equals $-1000(1-1/x)^2$, which is nonpositive for $x>0$ and equals zero only at $x=1$.

Interpretation and mistake avoided. A repeated root counts twice in polynomial root-counting, but it is still only one distinct return rate. Do not report two different returns just because a root has multiplicity two.

Worked Example: Rejecting Algebraic Roots Outside the Admissible Range

A dated sequence of receipts is $+$1,000$ at time 0, $+$3,000$ at the end of year 1, and $+$2,000$ at the end of year 2. A calculation produces algebraic roots for the transformed equation. Determine whether either root is a financially admissible annual dollar-weighted return.

Method. The return equation is $$ 1000+\frac{3000}{1+i}+\frac{2000}{(1+i)^2}=0. $$ Set $x=1+i$ and multiply by $x^2$: $$ 1000x^2+3000x+2000=1000(x+1)(x+2)=0. $$ The algebraic roots are $x=-1$ and $x=-2$, corresponding to $i=-2$ and $i=-3$. Neither root has $x>0$, so neither is an admissible effective annual rate.

Check. For any admissible $x>0$, each term $1000$, $3000/x$, and $2000/x^2$ is positive. Their sum cannot equal zero. Thus there is no admissible return, consistent with rejecting both algebraic roots.

Mistake avoided. A root of a transformed polynomial is only a candidate. Check $1+i>0$, and, when possible, substitute the rate into the original return equation.

Why the Return Equation Has These Properties

The polynomial transformation and admissibility condition explain the main diagnostic process. The original equation discounts each cash flow back to time 0. If a candidate rate is valid, the signed present values must sum to zero. Replacing $1+i$ by $x$ makes the condition $x>0$ explicit. Multiplication by $x^n$ preserves equality to zero for such $x$, because $x^n$ is positive and therefore nonzero. It produces a polynomial whose positive roots correspond exactly to admissible returns.

This equivalence is why the restriction matters. Multiplication by $x^n$ can reveal roots at negative values of $x$, but those values were never valid accumulation factors for an annual effective rate. Similarly, a numerical routine may return a candidate without determining whether it is the only root, whether it is accurate, or whether it lies in the admissible range.

The pattern of cash flows is an important first diagnostic, not a substitute for solving the equation. One sign change allows at most one positive root under the polynomial result, while multiple sign changes allow the possibility of multiple roots. However, even with one sign change, the equation must have a root before a return can be reported. For example, a stream consisting only of positive receipts has a strictly positive present value at every admissible rate.

Common Exam Traps

Equating sign changes with the exact number of returns. That is not what the sign rule says. Two sign changes permit two or zero positive roots, among other possibilities, as the no-solution example shows.
Accepting every polynomial root. For an annual effective return, require $x=1+i>0$, or $i>-1$. A root with $x\leq 0$ is not financially admissible.
Counting multiplicity as distinct rates. A repeated root has multiplicity greater than one but gives only one distinct numerical rate.
Ignoring the investor’s perspective. Contributions are negative and receipts are positive in the equation used here. Reversing every sign leaves the roots unchanged, but changing only some signs changes the problem.
Trusting a numerical solver’s first result. With multiple sign changes, use factoring, root analysis, or a search over the admissible range to check whether other roots exist.

Key takeaway: A dollar-weighted return is a root of the investor cash-flow equation, not simply any output from a solver. Use sign changes to gauge how many admissible roots might exist, solve or analyze the equation, and verify every candidate with $1+i>0$ and substitution into the original equation.

Check Your Understanding

For each problem, use investor-perspective cash-flow signs, identify all financially admissible annual returns if they exist, and check the final answer against the equation.

An investor contributes $500 at time 0 and receives $600 at the end of year 1. Find the annual dollar-weighted return. (Answer: $20\%$)
An investor contributes $1,200 at time 0, receives $2,600 at the end of year 1, and contributes $1,400 at the end of year 2. Find all distinct annual dollar-weighted returns. (Answer: $0\%$ and $16.6667\%$)
An account has cash flows $-$800$ at time 0, $+$2,000$ at the end of year 1, and $-$1,250$ at the end of year 2. Find the distinct annual dollar-weighted return and its multiplicity as a polynomial root. (Answer: $25\%$, a repeated root of multiplicity 2)
An investor cash-flow stream is $-$1,500$ at time 0, $+$1,500$ at the end of year 1, and $-$1,500$ at the end of year 2. There are two sign changes. Determine whether a real annual dollar-weighted return exists. (Answer: No real return; the transformed quadratic has discriminant $-6{,}750{,}000$)
A receipt-only stream has $+$1,000$ at time 0, $+$3,000$ at the end of year 1, and $+$2,000$ at the end of year 2. Its transformed polynomial has roots $x=-1$ and $x=-2$. Are either of these admissible dollar-weighted returns? (Answer: No; the corresponding rates $-2$ and $-3$ violate $i>-1$)

What You'll Learn