Midperiod Contributions And Withdrawals

Why Midperiod Timing Matters

A contribution made halfway through a year does not have the same opportunity to earn a return as money already in the account at the start of the year. A withdrawal made halfway through the year likewise removes money that would otherwise have remained invested for only the rest of that year. A dollar-weighted return must reflect these timing differences.

This tutorial considers a one-year measurement period and cash flows occurring within that year. The key calculation is the fraction of a year for which each cash flow is invested. The method is aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Definition: The dollar-weighted return for a period is the internal rate of return (IRR) on the investor’s cash flows, including the account’s beginning value and ending value. A contribution is an outflow from the investor and an inflow to the account; a withdrawal is an inflow to the investor and an outflow from the account.

Let $B$ be the account value at the start of a one-year period and $E$ its value at the end. Suppose the annual effective dollar-weighted return is $i$. A cash flow at time $t$, measured in years after the start, has $1-t$ years left to accumulate to the end of the period. Under the annual-effective-rate model, its accumulation factor over that remaining fraction is $(1+i)^{1-t}$.

Model assumption: The return is represented by a constant annual effective rate $i$, with fractional-year accumulation given by $(1+i)^s$ for a period of $s$ years. The one-year period runs from time $0$ to time $1$. Use this model when the problem asks for an annual effective dollar-weighted return and does not specify a different fractional-period convention.

Equations for Contributions and Withdrawals

Write the account accumulation equation by treating a contribution as positive money entering the account. If a contribution $C$ is made at time $t$, it accumulates for $1-t$ years. If instead a withdrawal of $W$ is made at that time, it reduces the amount left to accumulate.

E=B(1+i)+C(1+i)^{1-t}

This is the account-balance form for a contribution. For a withdrawal, replace the contribution term with a subtraction:

E=B(1+i)-W(1+i)^{1-t}

The same equations follow from the investor’s IRR perspective. At time $0$, the investor’s cash flow is $-B$. A contribution $C$ at time $t$ is an investor outflow $-C$, while a withdrawal $W$ is an investor inflow $+W$. The ending account value $E$ is an investor inflow at time $1$. Discounting those cash flows to time $0$ gives the corresponding IRR equation.

-B-\frac{C}{(1+i)^t}+\frac{E}{1+i}=0 \qquad\text{or}\qquad -B+\frac{W}{(1+i)^t}+\frac{E}{1+i}=0

The account-balance equations are often convenient when the problem gives a return and asks for an ending value. The investor-cash-flow equations emphasize the sign convention and are useful when solving for the return. Both describe the same dollar-weighted return.

Set the timeline.
Mark the beginning at time $0$, the ending at time $1$, and the cash-flow time $t$ as a fraction of a year.

Choose a consistent sign convention.
For the account equation, contributions are positive and withdrawals are negative. For investor cash flows, contributions are negative and withdrawals are positive.

Use the time remaining.
The beginning value accumulates for one full year; a cash flow at time $t$ accumulates for $1-t$ years to the end.

Solve and check.
Substitute the candidate return into the equation and verify that the calculated ending value agrees with the stated ending value.

Why the Fractional-Year Factor Is Correct

The formula follows by separating the account into amounts that are invested for different lengths of time. At the start, $B$ is invested for a full year, so its end value is $B(1+i)$. A contribution $C$ made at time $t$ is invested only from $t$ to $1$, a duration of $1-t$ years, so its end value is $C(1+i)^{1-t}$. Adding the two accumulated amounts gives the account’s ending value.

\begin{aligned} \text{Ending value} &=\text{accumulated beginning value} +\text{accumulated contribution}\\ &=B(1+i)+C(1+i)^{1-t}. \end{aligned}

For a withdrawal, the withdrawn amount is no longer in the account after time $t$. Its corresponding end-of-year value would have been $W(1+i)^{1-t}$, so that amount is subtracted from the value the account would otherwise have reached. This reasoning depends on the stated constant-rate accumulation model; if a problem specifies a different return convention or actual subperiod rates, use that information rather than imposing this formula.

Worked Examples

Worked Example: Finding the Ending Value After a Midyear Contribution

An account begins the year with $B=\$8{,}000$. The investor contributes $C=\$2{,}000$ exactly halfway through the year. Assume an annual effective dollar-weighted return of $i=10\%$, with fractional-year accumulation $(1+i)^s$. Find the ending value.

Method. Use the account-balance equation for a contribution. The contribution occurs at $t=0.5$, so it is invested for $1-0.5=0.5$ years.

The beginning balance accumulates for the full year:

\$8{,}000(1.10)=\$8{,}800.

The contribution accumulates for half a year:

\$2{,}000(1.10)^{0.5} =\$2{,}000(1.048808848\ldots) =\$2{,}097.617696\ldots.

Add the accumulated amounts:

E=\$8{,}800+\$2{,}097.617696\ldots =\$10{,}897.617696\ldots.

Rounded to the nearest cent, the ending value is $\$10{,}897.62$. Check: the contribution earns a half-year factor, not the full-year factor of $1.10$; substituting the unrounded ending value into the account equation reproduces $E$.

Common mistake avoided: Giving the new contribution a full year of return would use $\$2{,}000(1.10)=\$2{,}200$, overstating the ending value by $\$102.38$.

Worked Example: Solving for the Return With a Midyear Contribution

An account has a beginning value of $\$5{,}000$. The investor adds $\$1{,}000$ at the midpoint of the year, and the ending value is $\$7{,}150$. Find the annual effective dollar-weighted return, assuming fractional-year accumulation follows $(1+i)^s$.

Method. Use the contribution equation with $t=0.5$:

\$7{,}150=\$5{,}000(1+i)+\$1{,}000(1+i)^{0.5}.

Set $x=\sqrt{1+i}$, so $1+i=x^2$. The equation becomes

7{,}150=5{,}000x^2+1{,}000x,

where dollar amounts are in dollars. Rearranging gives

5{,}000x^2+1{,}000x-7{,}150=0.

The positive root is $x=1.10$, since

5{,}000(1.10)^2+1{,}000(1.10) =5{,}000(1.21)+1{,}100 =6{,}050+1{,}100 =7{,}150.

Therefore $1+i=x^2=1.21$, and $i=0.21$. The annual effective dollar-weighted return is 21%. Check: the beginning value grows to $\$6{,}050$, and the midpoint contribution grows to $\$1{,}100$; together they equal the stated ending value.

Common mistake avoided: The contribution is not worth $\$1{,}000(1+i)$ at year-end. It is worth $\$1{,}000\sqrt{1+i}$, because only half a year remains.

Worked Example: Solving for the Return With a Midyear Withdrawal

An account begins the year with $\$10{,}000$. The investor withdraws $\$2{,}500$ halfway through the year, and the account ends the year at $\$11{,}400$. Find the annual effective dollar-weighted return using the same fractional-year model.

Method. Use the withdrawal equation. The withdrawal reduces the account by its end-of-year accumulated value:

\$11{,}400=\$10{,}000(1+i)-\$2{,}500(1+i)^{0.5}.

Set $x=\sqrt{1+i}$, so $1+i=x^2$. Then

11{,}400=10{,}000x^2-2{,}500x, \qquad 10{,}000x^2-2{,}500x-11{,}400=0.

The positive root is $x=1.20$, verified by substitution:

10{,}000(1.20)^2-2{,}500(1.20) =14{,}400-3{,}000 =11{,}400.

Thus $1+i=(1.20)^2=1.44$, giving $i=0.44$. The annual effective dollar-weighted return is 44%. Check: the original $\$10{,}000$ accumulates to $\$14{,}400$, while the midpoint withdrawal removes $\$3{,}000$ of end-of-year value; the result is $\$11{,}400$.

Common mistake avoided: Subtracting only the $\$2{,}500$ withdrawal from the end value ignores the return that the withdrawn amount would have earned over the remaining half-year.

Sign and Timing Traps

A reliable setup keeps the account perspective separate from the investor perspective. In the account equation, a contribution is added and a withdrawal is subtracted. In the investor’s cash-flow equation, those signs reverse: a contribution is money paid into the account by the investor, while a withdrawal is money received.

Cash flow	Account-balance equation	Investor cash-flow equation
Contribution $C$ at time $t$	$+C(1+i)^{1-t}$ at the end	$-C$ at time $t$
Withdrawal $W$ at time $t$	$-W(1+i)^{1-t}$ at the end	$+W$ at time $t$

A second trap is confusing elapsed time with time remaining. If a contribution occurs three months after the start, then $t=3/12$, but the contribution has $1-3/12=9/12$ years to accumulate to year-end. Its end value is $C(1+i)^{9/12}$, not $C(1+i)^{3/12}$.

If several cash flows occur at different times, each has its own remaining duration. The direct account-balance extension is to add each contribution’s accumulated value and subtract each withdrawal’s accumulated value. Do not assign every flow the same fraction of a year merely because all occur in the same annual period.

Key takeaway: For a one-year period, the beginning balance earns a full year of return. A cash flow at time $t$ earns return for the remaining $1-t$ years. Apply the correct sign for its effect on the account, and verify the resulting ending value.

Check Your Understanding

For each problem, use an annual effective return with fractional-year accumulation $(1+i)^s$. Treat the stated dates as exact fractions of a year.

An account starts with $\$6{,}000$, earns an annual effective return of 8%, and receives a $\$1{,}000$ contribution halfway through the year. Find its ending value to the nearest cent. (Answer: $\$7{,}519.23$)
An account starts with $\$4{,}000$, receives a $\$1{,}000$ withdrawal halfway through the year, and earns an annual effective return of 21%. Find its ending value. (Answer: $\$3{,}740$)
An account starts with $\$5{,}000$, receives a $\$1{,}000$ contribution at midyear, and ends at $\$8{,}400$. Find the annual effective dollar-weighted return. (Answer: 44%)
An account starts with $\$10{,}000$, has a $\$2{,}000$ withdrawal at midyear, and ends at $\$9{,}900$. Find the annual effective dollar-weighted return. (Answer: 21%)
An account starts with $\$8{,}000$ and earns 10%. At midyear, the investor contributes $\$1{,}500$ and withdraws $\$500$. Find the ending value, treating the two same-time flows separately or as one net contribution. (Answer: $\$9{,}848.81$)

Cash flow	Account-balance equation	Investor cash-flow equation
Contribution \(C\) at time \(t\)	\(+C(1+i)^{1-t}\) at the end	\(-C\) at time \(t\)
Withdrawal \(W\) at time \(t\)	\(-W(1+i)^{1-t}\) at the end	\(+W\) at time \(t\)

What You'll Learn