The Single-Period Dollar-Weighted Return Equation

Why the Date of a Cash Flow Matters

An account's dollar-weighted return depends on both its investment performance and the timing of the investor's cash flows. A contribution made early has more time to earn a return than the same contribution made shortly before the account is valued. A withdrawal has the opposite effect: money removed early no longer participates in the account's later investment experience.

This tutorial considers one measurement period of one year, with an initial account value, one contribution or withdrawal at a specified time during the year, and an ending account value. The dollar-weighted return is the effective annual rate that makes the investor's dated cash flows and ending value consistent.

Definition: Let $V_0$ be the account value at time 0 and $V_1$ its value at time 1. Let $t$ be the time, in years, of a single interim cash flow, where $0<t<1$. Let $X$ be the net amount entering the account at time $t$: a contribution is positive and a withdrawal is negative. The single-period dollar-weighted return $r$ is a rate greater than $-1$ satisfying $$ V_1=V_0(1+r)+X(1+r)^{1-t}. $$ This equation treats $r$ as an effective annual rate and uses the fractional-period accumulation factor $(1+r)^s$ for $s$ years.

The sign convention above describes flows from the account's perspective. In the investor's cash-flow record, the signs are reversed: the initial investment is a payment by the investor, a contribution is another payment, a withdrawal is a receipt, and the ending account value is a receipt. The dollar-weighted return is the internal rate of return of those investor cash flows. Keeping the two perspectives distinct prevents sign errors.

Building the One-Period Equation

There are two equivalent ways to construct the equation: accumulate each account flow to time 1, or discount the investor's cash flows to time 0. The accumulation view is often quickest when the account values and interim flow are given.

Accumulate the initial balance.
The initial value $V_0$ remains invested for the full year, so its value at time 1 under rate $r$ is $V_0(1+r)$.

Accumulate the interim flow for the time it is invested.
A flow at time $t$ has $1-t$ years remaining. Its accumulated amount is $X(1+r)^{1-t}$, positive for a contribution and negative for a withdrawal.

Match the account's ending value.
Add the two accumulated amounts and set their total equal to the observed ending value $V_1$.

The same result follows from the investor's internal-rate-of-return equation. Write a contribution as $C>0$ and a withdrawal as $W>0$. If both occur on the same date, their net account flow is $X=C-W$. At rate $r$, the present value of the investor's cash flows must be zero:

-V_0-C(1+r)^{-t}+W(1+r)^{-t}+V_1(1+r)^{-1}=0.

Since $X=C-W$, rearranging and multiplying by $1+r$ gives the account equation:

V_1(1+r)^{-1}=V_0+X(1+r)^{-t} \quad\Longleftrightarrow\quad V_1=V_0(1+r)+X(1+r)^{1-t}.

Important: The exponent on the interim flow is the time remaining after the flow, $1-t$, not the elapsed time $t$. A contribution at time $t$ earns a return only from $t$ to time 1. For a withdrawal, the same timing applies, but its account-flow amount $X$ is negative.

Worked Examples

Worked Example: A Midyear Contribution

An account is worth $\$10{,}000$ at the start of a one-year period. The investor contributes $\$2{,}000$ exactly halfway through the year, and the account is worth $\$14{,}300$ at year-end. Find the dollar-weighted return. Assume a constant effective annual rate with fractional-year accumulation $(1+r)^s$.

Method. Use the account-flow equation. The interim amount is a contribution, so $X=2{,}000$, and $t=0.5$:

14{,}300=10{,}000(1+r)+2{,}000(1+r)^{0.5}.

Set $y=(1+r)^{0.5}$. Because $r>-1$, $y>0$, and $1+r=y^2$. The equation becomes

14{,}300=10{,}000y^2+2{,}000y, \qquad 100y^2+20y-143=0.

Factor the quadratic:

$$ 100y^2+20y-143=100(y-1.1)(y+1.3)=0. $$

The roots are $y=1.1$ and $y=-1.3$. Since $y$ must be positive, use $y=1.1$. Therefore

1+r=y^2=(1.1)^2=1.21, \qquad r=0.21.

The dollar-weighted return is 21%. Check: the initial balance accumulates to $\$10{,}000(1.21)=\$12{,}100$, and the midyear contribution accumulates to $\$2{,}000(1.1)=\$2{,}200$. Their sum is $\$14{,}300$, as required. The common mistake this method avoids is allowing the midyear contribution to earn a full year's return.

Worked Example: A Midyear Withdrawal

An account begins the year at $\$20{,}000$. The investor withdraws $\$3{,}000$ at midyear, and the ending account value is $\$20{,}900$. Find the dollar-weighted return, using the same effective annual and fractional-year assumptions.

Method. From the account's perspective, the withdrawal is negative: $X=-3{,}000$. With $t=0.5$,

20{,}900=20{,}000(1+r)-3{,}000(1+r)^{0.5}.

Let $y=(1+r)^{0.5}>0$. Then

20y^2-3y-20.9=0, \qquad 20(y-1.1)(y+0.95)=0.

The roots are $y=1.1$ and $y=-0.95$. The negative root is inadmissible because $y$ is a positive square root. Thus $y=1.1$, so

$$ r=y^2-1=(1.1)^2-1=0.21. $$

The dollar-weighted return is 21%. Check: the initial account value accumulates to $\$20{,}000(1.21)=\$24{,}200$. The withdrawal reduces the year-end balance by $\$3{,}000(1.1)=\$3{,}300$, leaving $\$24{,}200-\$3{,}300=\$20{,}900$. The withdrawal must be subtracted, but it is accumulated only for the half-year after it leaves the account.

Worked Example: A Contribution One Quarter of the Way Through the Year

An account starts at $\$5{,}000$. A contribution of $\$1{,}000$ is made three months into the year, and the ending value is $\$8{,}651.50$. Find the dollar-weighted return. Assume the fractional-year accumulation factor is $(1+r)^s$.

Method. Three months is $t=0.25$ years, so the contribution remains invested for $1-t=0.75$ years. The equation is

8{,}651.50=5{,}000(1+r)+1{,}000(1+r)^{0.75}.

Set $y=(1+r)^{0.25}$, so $1+r=y^4$ and $(1+r)^{0.75}=y^3$. This gives

8{,}651.50=5{,}000y^4+1{,}000y^3.

The right-hand side increases strictly for $y>0$, because its derivative is $20{,}000y^3+3{,}000y^2>0$. Thus there can be at most one positive solution. Test $y=1.1$:

5{,}000(1.1)^4+1{,}000(1.1)^3 =5{,}000(1.4641)+1{,}000(1.331) =7{,}320.50+1{,}331.00 =8{,}651.50.

The positive solution is $y=1.1$. Therefore $1+r=y^4=1.4641$, and $r=0.4641$. The dollar-weighted return is 46.41%. The check reproduces the given ending value exactly. The key timing detail is that the contribution earns three quarters of a year, not one quarter.

Common Timing and Sign Traps

A reliable setup starts by drawing a one-year timeline and labelling the account value at time 0, the dated flow at $t$, and the ending value at time 1. Then write the flow from the account's perspective and identify how long it remains invested.

For the first example, the incorrect full-year calculation would be

14{,}300=(10{,}000+2{,}000)(1+r), \qquad r=\frac{14{,}300}{12{,}000}-1=0.191666\ldots.

This gives about $19.17\%$, but it wrongly assumes the $\$2{,}000$ contribution was invested at the start of the year. The correct equation gives $21\%$, because the contribution receives only a half-year of accumulation. Other frequent errors are using $t$ instead of $1-t$ in the exponent, adding a withdrawal to the account equation, or treating the effective annual rate as a simple interest rate for a fraction of a year.

A quick reasonableness check is to recompute the ending account value using the proposed rate and the actual dates of the flows. If the accumulated amounts do not sum to $V_1$, the equation, timing, or arithmetic needs review. For an investor cash-flow equation, check the signs separately: payments by the investor are negative and receipts are positive.

Key takeaway: For a one-year period with one interim flow, accumulate the initial value for one year and the net account flow for the remaining $1-t$ years. Set their signed total equal to the ending value; the rate that satisfies this equation is the dollar-weighted return.

Check Your Understanding

For each problem, use an effective annual return and the fractional-period accumulation convention stated in the definition.

An account begins at $\$6{,}000$, receives a $\$1{,}000$ contribution at midyear, and ends at $\$8{,}360$. Find the dollar-weighted return. (Answer: 21%)
An account begins at $\$12{,}000$, pays a $\$2{,}000$ withdrawal at midyear, and ends at $\$12{,}320$. Find the dollar-weighted return. (Answer: 21%)
An account begins at $\$9{,}000$, receives a $\$1{,}000$ contribution at midyear, and ends at $\$8{,}190$. Find the dollar-weighted return. (Answer: $-19\%$)
An account begins at $\$4{,}000$, receives a $\$500$ contribution three months into the year, and ends at $\$6{,}521.90$. Find the dollar-weighted return. (Answer: 46.41%)
An account begins at $\$10{,}000$, has a $\$1{,}000$ withdrawal at midyear, and ends at $\$11{,}000$. Write the account-flow equation and find the dollar-weighted return. (Answer: $11{,}000=10{,}000(1+r)-1{,}000(1+r)^{0.5}$; 21%)

What You'll Learn