Beginning- and End-of-Period Fund Deposits

Why Deposit Timing Matters

A sinking fund is built by making deposits that earn interest until a specified redemption date. If deposits are made at the end of each period, the first deposit earns interest for fewer periods than it would if deposits were made at the beginning. For the same target and interest rate, beginning-of-period deposits can therefore be smaller.

The key is to match the dates, not merely count the deposits. With $n$ deposits and redemption at time $n$, end-of-period deposits occur at times $1,2,\ldots,n$; beginning-of-period deposits occur at times $0,1,\ldots,n-1$. In both cases there are $n$ deposits, but each beginning deposit is invested for one extra period compared with the deposit made at the corresponding end of a period.

Definition: An annuity-immediate fund deposit is made at the end of each period. An annuity-due fund deposit is made at the beginning of each period. For a level deposit $R$, $n$ deposits, and effective interest rate $i$ per deposit period, the accumulation date is immediately after the $n$th end-of-period date.

Accumulation Formulas for the Two Timings

Assume a fixed effective interest rate $i>0$ per deposit period, equal deposits, no initial fund balance, and no withdrawals or expenses. The target is measured at time $n$, just after the final end-of-period deposit would be made. The same formulas apply to a hypothetical fund: the calculation does not assume that an actual fund earns a guaranteed rate.

For end-of-period deposits, the deposit at time $k$ earns interest for $n-k$ periods. For beginning-of-period deposits, the deposit at time $k$ earns interest for $n-k$ periods when $k=0,\ldots,n-1$. Equivalently, each beginning deposit occurs one period earlier than its corresponding end deposit.

\begin{aligned} \text{End-of-period deposits at }1,2,\ldots,n: \quad F_{\text{imm}} &= R s_{\overline{n}|} = R\frac{(1+i)^n-1}{i},\\[4pt] \text{Beginning-of-period deposits at }0,1,\ldots,n-1: \quad F_{\text{due}} &= R\ddot{s}_{\overline{n}|} = R(1+i)s_{\overline{n}|}. \end{aligned}

Here $F$ denotes the accumulated fund at time $n$. The factor $s_{\overline{n}|}$ accumulates an annuity-immediate, while $\ddot{s}_{\overline{n}|}$ accumulates an annuity-due. These are accumulation factors, not present-value factors.

Why the Due Accumulation Factor Is Larger

The relationship follows directly from the deposit dates. For any one end-of-period deposit, the corresponding beginning-of-period deposit is invested one extra period. At rate $i$, one extra period multiplies that deposit’s accumulated value by $1+i$. This is true for every deposit, so it is true for the sum of their accumulated values.

\begin{aligned} R\ddot{s}_{\overline{n}|} &= R\big((1+i)^n+(1+i)^{n-1}+\cdots+(1+i)\big)\\ &= (1+i)R\big((1+i)^{n-1}+(1+i)^{n-2}+\cdots+1\big)\\ &= (1+i)R s_{\overline{n}|}. \end{aligned}

The same reasoning gives the required deposits for a common target $F$. Solve each accumulation equation for $R$:

R_{\text{imm}}=\frac{F}{s_{\overline{n}|}}, \qquad R_{\text{due}}=\frac{F}{\ddot{s}_{\overline{n}|}} =\frac{F}{(1+i)s_{\overline{n}|}} =\frac{R_{\text{imm}}}{1+i}.

Thus the beginning-of-period deposit is smaller by the factor $1/(1+i)$. The deposit amounts are not equal, even though the target and number of deposits are the same. This relationship requires the same rate per period, the same number of deposits, and the stated one-period shift in timing. It does not apply unchanged if the fund starts with a balance, the deposits vary, or the payment dates do not differ by exactly one period.

Worked Example: Same Target With Three Annual Deposits

A hypothetical fund must contain $3,641 immediately after the end of year 3. The fund earns an effective annual rate of 10%. Find the equal annual deposit if deposits are made at each year-end, and then if they are made at each year-beginning. Use three deposits in either case.

Method. Calculate the annuity-immediate and annuity-due accumulation factors for three annual periods, then divide the target by the appropriate factor. The common mistake to avoid is treating a beginning deposit as if it earns the same number of years as the corresponding end deposit.

For end-of-year deposits, the times are $1,2,3$. The accumulation factor is

s_{\overline{3}|} =\frac{(1.10)^3-1}{0.10} =\frac{1.331-1}{0.10} =3.31.

The required end-of-year deposit is

R_{\text{imm}}=\frac{3641}{3.31}=1100.

For beginning-of-year deposits, the times are $0,1,2$. Their accumulation factor is

\ddot{s}_{\overline{3}|} =1.10(3.31)=3.641, \qquad R_{\text{due}}=\frac{3641}{3.641}=1000.

Check. The end-of-year deposits accumulate to $1100(1.21+1.10+1)=1100(3.31)=3641$. The beginning-of-year deposits accumulate to $1000(1.331+1.21+1.10)=1000(3.641)=3641$. Both schedules meet the target, and $1000=1100/1.10$, as required by the timing relationship.

Keep the Rate Period Consistent With Deposit Timing

The rate $i$ in these formulas is the effective rate for one interval between deposits, not automatically an annual rate. If deposits are semiannual, use the effective rate per half-year. If the stated rate is annual but deposits are more frequent, first determine the effective rate for the deposit interval using the applicable interest-rate convention. Do not insert an annual rate into a monthly or quarterly accumulation formula without conversion.

The number $n$ is also a count of deposit periods. For example, six semiannual deposits span six half-year periods. If they are made at the beginning of those periods, the first is at time 0 and redemption is at the end of the sixth half-year period. If they are made at the end, the first is at the end of the first half-year period.

Worked Example: Semiannual Deposits and a Common Redemption Target

A hypothetical fund needs a balance at the end of three years. Six equal deposits are made at six-month intervals, and the effective rate is 6% per half-year. If each beginning-of-period deposit is $750, find the target balance and the equal end-of-period deposit that would reach the same target. Round money to the nearest cent.

Method. Treat each half-year as one period. First accumulate the beginning deposits using the annuity-due factor, then use the annuity-immediate factor to find the alternative deposit. The common mistake to avoid is treating 6% as an annual rate when it is specified per half-year.

There are $n=6$ periods, and $i=0.06$ per period. The annuity-immediate factor is

s_{\overline{6}|} =\frac{(1.06)^6-1}{0.06} =\frac{1.418519112256-1}{0.06} =6.9753185376.

The annuity-due factor is

\ddot{s}_{\overline{6}|} =1.06(6.9753185376) =7.393837649856.

Therefore, the target is

F=750(7.393837649856) =5545.378237392 \approx \$5{,}545.38.

For end-of-period deposits, divide this target by the annuity-immediate factor:

R_{\text{imm}} =\frac{5545.378237392}{6.9753185376} =795 \approx \$795.00.

Check. The due deposit is $795/1.06=750$, so the two deposits satisfy the required one-period timing relationship. Accumulating either schedule gives $5,545.38 when rounded to cents.

Dates, Deposit Counts, and Common Traps

A timing label should always be translated into dates before choosing a formula. “At the beginning of each year for three years” normally means deposits at times $0,1,2$, followed by the target date at time $3$. “At the end of each year for three years” means deposits at times $1,2,3$, with the target measured just after the last deposit. Both schedules have three deposits; neither has four.

A frequent error is to use the right factor with the wrong target date. If the fund is needed at time $n$, an end-of-period schedule includes the deposit at time $n$, so that last deposit earns no interest before the target is measured. A beginning-of-period schedule has its last deposit at time $n-1$, so that deposit earns one period of interest. If instead a problem measures the balance immediately before a scheduled deposit, the final deposit is not part of that balance. The dates must be reconsidered rather than applying the formulas mechanically.

Worked Example: Verify the One-Period Timing Shift

A hypothetical redemption fund must accumulate to $4,641.53025 at the end of year 5. It earns an effective annual rate of 5%, with five equal deposits. Find the deposits for both timing conventions and verify the result by accumulating the payments.

Method. Use the five-period accumulation factors and then verify by listing the interest earned by each payment. The common mistake to avoid is placing the first beginning deposit at time 1 rather than time 0.

First calculate the annuity-immediate factor:

s_{\overline{5}|} =\frac{(1.05)^5-1}{0.05} =\frac{1.2762815625-1}{0.05} =5.52563125.

The required end-of-year deposit is

R_{\text{imm}} =\frac{4641.53025}{5.52563125} =840.

The annuity-due factor and required beginning-of-year deposit are

\ddot{s}_{\overline{5}|} =1.05(5.52563125)=5.8019128125, \qquad R_{\text{due}} =\frac{4641.53025}{5.8019128125}=800.

Check by deposit dates. The end-of-year deposits accumulate as $840(1.05^4+1.05^3+1.05^2+1.05+1)=840(5.52563125)=4641.53025$. The beginning-of-year deposits accumulate as $800(1.05^5+1.05^4+1.05^3+1.05^2+1.05)=800(5.8019128125)=4641.53025$. The beginning deposits are $840/1.05=800$, confirming the relationship.

A Reliable Calculation Sequence

Mark the target date.
State exactly when the redemption amount is measured, such as immediately after the final period ends.

Count the periods and deposits.
For $n$ level deposits, write the dates as $1,\ldots,n$ for an immediate or $0,\ldots,n-1$ for a due.

Match the rate to one period.
Use the effective rate for the interval between consecutive deposits.

Choose the accumulation factor.
Use $s_{\overline{n}|}$ for end-of-period deposits and $\ddot{s}_{\overline{n}|}$ for beginning-of-period deposits.

Verify the dates and target.
Accumulate the deposits to the stated date and confirm that the result equals the required fund.

Key takeaway: For the same target, rate per period, and number of deposits, an annuity-due deposit is $1/(1+i)$ times the annuity-immediate deposit. The difference comes from one extra period of interest on each beginning-of-period payment.

Check Your Understanding

For each problem, identify the deposit interval and target date before calculating.

A fund needs $649.28 after three annual periods at an effective annual rate of 8%. Find the equal end-of-year deposit and the equal beginning-of-year deposit. (Answer: end-of-year $200.00; beginning-of-year $185.19)
At an effective rate of 10% per half-year, four beginning-of-period deposits of $500 are made. Find the target at the end of the fourth half-year period and the end-of-period deposit that would reach the same target. (Answer: target $2,552.55; end-of-period deposit $550.00)
A hypothetical fund earns 4% per quarter. Eight equal deposits of $400 are made. Find the accumulated value at the end of the eighth quarter if deposits are made at quarter-end, and then if made at quarter-beginning. (Answer: quarter-end deposits accumulate to $3,685.69; quarter-beginning deposits accumulate to $3,833.12)
At an effective annual rate of 5%, four end-of-year deposits of $1,200 are made. Find the target at the end of year 4 and the equal beginning-of-year deposit that would reach the same target. (Answer: target $5,172.15; beginning-of-year deposit $1,142.86)
A fund must contain $843.654 at the end of year 3 and earns 6% effective annually. Find the equal deposit for three beginning-of-year deposits and the equal deposit for three end-of-year deposits. (Answer: beginning-of-year $250.00; end-of-year $265.00)

What You'll Learn