Multi-Period Dollar-Weighted Return Equations

Why Multi-Period Cash Flows Need One Common Date

When an investment has contributions or withdrawals on several dates, each cash flow is exposed to investment growth for a different length of time. A contribution made early has more time to earn interest than one made later; a withdrawal reduces the amount left invested from its date onward. A dollar-weighted return equation accounts for these timing differences by moving every dated cash flow to the same date before balancing them.

The common date can be the beginning of the measurement period or its end. The equation is the same in either form when the same return is used throughout. The essential discipline is to use the correct number of periods for each cash flow and to keep the sign convention consistent.

Definition: Let $C_t$ be the investor-perspective cash flow at time $t$, with equally spaced dates $t=0,1,\ldots,T$. A contribution is negative; a withdrawal or final account value received by the investor is positive. The multi-period dollar-weighted return $i$, measured per interval, is a rate satisfying the equation that sets the present values or accumulated values of these signed cash flows to zero.

This setup assumes that cash flows occur at the stated period dates and that one effective rate $i$ applies per interval over the measurement period. The return must satisfy $i>-1$, so the accumulation factor $1+i$ is positive. If the dates are not equally spaced, the integer exponents below do not directly represent the elapsed times; a suitable time-based exponent or other dating convention is needed.

Accumulating Cash Flows to the End

Suppose the return is measured over $T$ periods. A cash flow $C_t$ at time $t$ accumulates for $T-t$ periods to time $T$. The accumulated-value equation is therefore:

\sum_{t=0}^{T} C_t(1+i)^{T-t}=0

A contribution enters with a negative sign, while a withdrawal and the final account value enter with positive signs. Thus the equation balances the accumulated value of investor receipts against the accumulated value of contributions. A cash flow at time $T$ has exponent zero because it is already at the common date.

Worked Example: Two Contributions and an Ending Value

An investor contributes $1,000 at time 0 and another $200 at time 1. The account is worth $1,430 at time 2, when it is liquidated. Find the dollar-weighted return per period. Assume the dates are one period apart and use the same effective rate in both periods.

Method. Use investor-perspective signs and accumulate every flow to time 2. The contributions are negative and the liquidation value is positive:

$$ -1000(1+i)^2-200(1+i)+1430=0 $$

Test $i=0.10$. The time-0 contribution accumulates for two periods: $1000(1.10)^2=1000(1.21)=1210$. The time-1 contribution accumulates for one period: $200(1.10)=220$. The ending value is already at time 2, so it remains $1,430. The signed total is $-1210-220+1430=0$. Therefore, $i=10\%$ per period satisfies the equation.

Check by discounting. At 10%, the time-0 value of the flows is $-1000-200/1.10+1430/(1.10)^2$. The terms are $-1000$, approximately $-181.8182$, and approximately $1181.8182$, respectively. Their sum is zero, apart from displayed rounding.

Common mistake avoided: The $200 contribution is at time 1, not time 0, so it earns only one period of interest before the ending date. Giving it a two-period exponent would overstate its accumulated value.

Discounting Cash Flows to the Beginning

Instead of accumulating to time $T$, discount each flow to time 0. With $v=1/(1+i)$, the present-value equation is:

\sum_{t=0}^{T} C_t v^t=0 \qquad\text{or, equivalently,}\qquad \sum_{t=0}^{T}\frac{C_t}{(1+i)^t}=0

A time-0 cash flow is not discounted, and a cash flow at time $t$ is discounted for exactly $t$ periods. This form is often convenient when the cash flows are listed from the beginning of the investment. It is not a different return definition: it is the same cash-flow balance expressed at a different date.

Result: For equally spaced cash-flow dates $0,\ldots,T$ and $i>-1$, the time-0 and time-$T$ equations are equivalent. Multiplying the discounted equation by $(1+i)^T$ gives the accumulated equation:

(1+i)^T\sum_{t=0}^{T}\frac{C_t}{(1+i)^t} = \sum_{t=0}^{T}C_t(1+i)^{T-t}

The multiplication is valid because $1+i>0$. Each discounted term becomes the value of that same cash flow at time $T$: its exponent changes from $-t$ to $T-t$. Thus either equation can be used to find or verify a return, provided all flows are moved to the same date.

Choose a common date.
Use time 0 for present values or the final date $T$ for accumulated values.

List every cash flow with its date and sign.
Write contributions as negative and investor receipts, including a terminal account value, as positive.

Apply the date-specific exponent.
For accumulation to $T$, use $T-t$; for discounting to 0, use $t$.

Solve or verify the rate.
Substitute the proposed per-period rate into the full equation and check that the signed total is zero.

Mixed Contributions and Withdrawals

Withdrawals are positive under the investor-perspective convention because they are amounts received by the investor. But they reduce the amount left in the account. In an accumulated cash-flow equation, this effect is represented by carrying the withdrawal forward as a positive receipt, alongside the negative contributions and positive terminal value. The signs do not change when the common date changes.

Worked Example: Contributions, a Withdrawal, and Liquidation

An investor contributes $1,000 at time 0, withdraws $100 at time 1, and contributes another $400 at time 2. The account is liquidated for $1,650 at time 3. Verify that a 10% return per period satisfies the investor cash-flow equation. Assume equally spaced dates and a constant effective rate per period.

Method. Accumulate each flow to time 3. The exponents are 3 for the time-0 contribution, 2 for the time-1 withdrawal, 1 for the time-2 contribution, and 0 for the liquidation value:

$$ -1000(1+i)^3+100(1+i)^2-400(1+i)+1650=0 $$

Substitute $i=0.10$. The accumulated time-0 contribution is $1000(1.10)^3=1000(1.331)=1331$. The withdrawal accumulated from time 1 to time 3 is $100(1.10)^2=100(1.21)=121$. The time-2 contribution accumulates one period: $400(1.10)=440$. The liquidation value remains $1,650.

The signed total at time 3 is $-1331+121-440+1650=0$. Equivalently, accumulated contributions total $1331+440=1771$, while the withdrawal and liquidation value total $121+1650=1771$. The 10% rate therefore satisfies the equation.

Check by discounting. The time-0 equation is $-1000+100/1.10-400/1.21+1650/1.331=0$. The discounted terms are approximately $-1000$, $90.909091$, $-330.578512$, and $1239.669421$. Adding them gives zero, subject only to rounding in the displayed decimals.

Common mistake avoided: A withdrawal is not entered as a negative investor cash flow under this convention. It is a positive receipt; the contribution at time 2 is the negative flow.

Finding a Return and Checking the Cash-Flow Equation

If the return is unknown, the multi-period equation may be a polynomial in $1+i$. Its terms must retain their dates and signs. A proposed rate is a solution only if it balances the full equation; checking one contribution or comparing the terminal value with the total deposits is not enough.

The cash-flow sign pattern also matters. If all contributions occur before a final receipt, the equation typically has one transition from negative to positive cash flows. If contributions and withdrawals alternate, there can be more than one sign change, and the equation may have multiple return roots. The earlier material on multiple IRRs explains why a cash-flow equation should not be assumed to have a unique solution just because it represents one investment.

Worked Example: Solving with Two Contribution Dates

An investor contributes $800 at time 0 and $200 at time 1. The account is liquidated for $1,306.80 at time 3. Find the dollar-weighted return per period, assuming equally spaced dates and the same effective rate for each period.

Method. Accumulate all flows to time 3. Let $x=1+i$, where $x>0$:

$$ -800x^3-200x^2+1306.80=0 $$

At $x=1.10$, the first contribution accumulates to $800(1.10)^3=800(1.331)=1064.80$. The second accumulates to $200(1.10)^2=200(1.21)=242.00$. Their total is $1064.80+242.00=1306.80$, exactly the liquidation value. Therefore $x=1.10$, so $i=x-1=0.10$, or 10% per period.

This is the only positive value of $x$ that solves the equation: the left-hand contribution total $800x^3+200x^2$ is strictly increasing for $x>0$, since its derivative is $2400x^2+400x>0$. It equals $1,306.80 at $x=1.10$, so no other positive $x$, and hence no other rate $i>-1$, balances these cash flows.

Check: The accumulated contributions are $1,306.80, equal to the terminal receipt. The signs in the original equation therefore give $-1306.80+1306.80=0$.

Common mistake avoided: The $200 is contributed at time 1, so it accumulates for two periods to time 3, not three. Its accumulated value at 10% is $242.00, not $266.20.

Common Exam Traps

Using the wrong exponent. To accumulate to time $T$, a flow at time $t$ receives exponent $T-t$, not $t$. To discount to time 0, its exponent is $t$.
Changing signs midway. Under the investor perspective, contributions stay negative and withdrawals stay positive in both forms of the equation.
Omitting the ending account value. When the account value is supplied as the final receipt, include it at its date with exponent zero in the accumulated equation.
Mixing rate periods. If dates are one year apart, $i$ is an effective annual rate; if dates are one month apart, $i$ is an effective monthly rate. The exponent counts those intervals.
Assuming a candidate rate is correct without substitution. Evaluate every dated term and check the signed total at the chosen common date. Keep full precision through the check and round only the final reported rate when needed.

Key takeaway: A multi-period dollar-weighted return equation is a signed cash-flow balance at one common date. Put every flow on the timeline, use its correct exponent, and verify that the complete equation balances at the proposed rate. This setup is aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Check Your Understanding

For each problem, use the investor-perspective sign convention and equally spaced dates unless stated otherwise.

An investor contributes $1,000 at time 0 and $200 at time 1. The account is liquidated for $1,430 at time 2. Write the accumulated-value equation and find the return per period. (Answer: $-1000(1+i)^2-200(1+i)+1430=0$; $i=10\%$ per period.)
An investor contributes $1,200 at time 0 and $300 at time 2. The account is liquidated for $1,704.15 at time 3. Write the time-3 equation and find the return per period. (Answer: $-1200(1+i)^3-300(1+i)+1704.15=0$; $i=5\%$ per period.)
An investor contributes $700 at time 0, withdraws $50 at time 1, and contributes $200 at time 2. The account is liquidated for $1,091.20 at time 3. Verify a 10% return by accumulating all flows to time 3. (Answer: $-700(1.10)^3+50(1.10)^2-200(1.10)+1091.20=-931.70+60.50-220+1091.20=0$; 10% per period satisfies the equation.)
An investor contributes $900 at time 0, withdraws $100 at time 1, and liquidates the account for $979 at time 2. Write the time-0 equation and verify the return. (Answer: $-900+100/(1+i)+979/(1+i)^2=0$; at $i=10\%$, $-900+100/1.10+979/1.21=-900+90.909091+809.090909=0$, so 10% per period.)
An investor contributes $500 at time 0, contributes $250 at time 1, withdraws $40 at time 2, and receives $900 at time 3. Write an accumulated-value equation at time 3 and evaluate its signed total at 10%. (Answer: $-500(1+i)^3-250(1+i)^2+40(1+i)+900=0$; at 10%, $-665.50-302.50+44+900=-24$, so 10% does not satisfy the equation.)

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