Building a Reserve With Periodic Contributions

Building a Reserve From Regular Contributions

A reserve target is a future amount that must be available at a specified date—for example, funds set aside for a planned equipment replacement or a future operating commitment. If a business makes equal contributions over time, the required contribution depends on more than the target and the interest rate: the date each contribution is deposited matters. A deposit made earlier earns interest for longer.

The calculation is an accumulation problem. First choose the date at which the reserve is measured, then move every contribution to that date using the interest earned over the relevant number of periods. The formulas below assume level deposits and a consistent effective rate per deposit period.

Definition: Let $R$ be a level contribution made once per period, $i$ the effective interest rate per period, and $n$ the number of contributions. An end-of-period contribution plan deposits $R$ at times $1,2,\ldots,n$, with the reserve measured immediately after the deposit at time $n$. A beginning-of-period contribution plan deposits $R$ at times $0,1,\ldots,n-1$, with the reserve measured at time $n$.

Assume there are no withdrawals, fees, or taxes, and that each deposit earns interest at rate $i$ for every full period it remains invested. The rate and deposit frequency must refer to the same period: for annual deposits, $i$ is the effective annual rate. This tutorial keeps the rate and contribution intervals aligned.

Accumulation Factors for the Two Deposit Timings

For end-of-period deposits, the first contribution earns interest for $n-1$ periods, the second for $n-2$ periods, and the last for no time before the valuation date. Thus the accumulated reserve is $R$ times a sum of powers of $1+i$.

R\left((1+i)^{n-1}+(1+i)^{n-2}+\cdots +(1+i)+1\right) =R s_{\overline{n}|}

For $i\ne 0$, the geometric-sum formula gives $s_{\overline{n}|}=((1+i)^n-1)/i$. For beginning-of-period deposits, every contribution is made one period earlier than the corresponding end-of-period contribution. Each therefore earns one additional period of interest, so the accumulated value is multiplied by $1+i$.

Key result: If $F$ is the target reserve at time $n$, the required level contribution is $R=F/s_{\overline{n}|}$ for end-of-period deposits and $R=F/\big((1+i)s_{\overline{n}|}\big)$ for beginning-of-period deposits, where $i\ne 0$. When $i=0$, both plans accumulate to $nR$, so $R=F/n$.

s_{\overline{n}|}=\frac{(1+i)^n-1}{i}, \qquad \ddot{s}_{\overline{n}|}=(1+i)s_{\overline{n}|}

The notation $\ddot{s}_{\overline{n}|}$ denotes the accumulated value of an annuity-due of $n$ equal payments. The formulas assume that the reserve target is measured at the stated date and, for the end-of-period plan, after the final deposit. If the target is measured before that deposit, the deposit is not part of the reserve at that date.

Why the Accumulation Formula Holds

For end-of-period deposits, each deposit grows by the account accumulation factor for the number of periods remaining. To justify the closed form, let $q=1+i$. The finite geometric sum is $1+q+\cdots+q^{n-1}$. Multiplying it by $q-1=i$ cancels all interior terms:

(q-1)(1+q+\cdots+q^{n-1})=q^n-1

Dividing by $i$ gives $s_{\overline{n}|}=((1+i)^n-1)/i$. For beginning-of-period deposits, shift every deposit one period earlier. Its value at time $n$ is consequently multiplied by $1+i$, giving $\ddot{s}_{\overline{n}|}=(1+i)s_{\overline{n}|}$. This timing shift is the reason the two contribution formulas differ.

Fix the target date.
Record when the reserve is measured and whether the final deposit is included.

Identify the deposit timing.
Use $s_{\overline{n}|}$ for deposits at times $1$ through $n$, or $\ddot{s}_{\overline{n}|}$ for deposits at times $0$ through $n-1$.

Allow for any existing reserve.
Accumulate the amount already invested to the target date, then subtract it from the target.

Solve and check.
Divide the remaining target by the appropriate accumulation factor, then accumulate the deposits back to the target date to verify the result.

Worked Contribution Problems

Worked Example: End-of-Year Contributions to a Four-Year Target

A hypothetical firm wants to accumulate $8,620.25 immediately after its fourth year-end deposit. It will deposit the same amount at the end of each year. The account earns an effective annual rate of 5%, with no withdrawals or fees. Find the annual contribution.

Method. There are four end-of-year deposits, so use $s_{\overline{4}|}$. The deposits earn interest for four, three, two, and one? No: because the target is immediately after the fourth deposit, they earn interest for three, two, one, and zero years, respectively. The accumulation factor is

s_{\overline{4}|} =\frac{1.05^4-1}{0.05} =\frac{1.21550625-1}{0.05} =4.310125

Set the accumulated contributions equal to the target:

R=\frac{8{,}620.25}{4.310125}=2{,}000

Check. The four deposits accumulate to $2{,}000(1.05^3+1.05^2+1.05+1)$. The powers are $1.157625$, $1.1025$, $1.05$, and $1$, whose sum is $4.310125$. Thus the reserve is $2{,}000(4.310125)=\$8{,}620.25$. The required contribution is $2,000 per year. The timing trap avoided here is giving the final deposit an extra year of interest.

Worked Example: Beginning-of-Year Contributions to a Five-Year Target

A hypothetical organization makes five equal deposits at the beginning of each year, starting now. The reserve is valued at the end of year 5. The account earns an effective annual rate of 4%, and the target is $11,265.95. Find the beginning-of-year contribution.

Method. The deposits occur at times 0, 1, 2, 3, and 4, so each earns one more year of interest than an end-of-year deposit in the corresponding position. First calculate the ordinary accumulation factor:

s_{\overline{5}|} =\frac{1.04^5-1}{0.04} =\frac{1.2166529024-1}{0.04} =5.41632256

The beginning-of-year factor is $ \ddot{s}_{\overline{5}|}=1.04(5.41632256)=5.6329754624 $. Therefore,

R=\frac{11{,}265.95}{5.6329754624} \approx 1{,}999.99984

Rounded to the nearest cent, the contribution is $2,000.00. To check, the five deposits of $2,000 grow for five, four, three, two, and one years. Their accumulated values are $2,433.3058048, $2,339.71712, $2,249.728, $2,163.20, and $2,080. Their sum is $11,265.9509248, which rounds to the stated target of $11,265.95. The method avoids treating a deposit made at the beginning of a year as if it were made at the end.

Worked Example: Including an Existing Reserve

A hypothetical business already has $800 invested at time 0. It plans three equal contributions at the end of years 1, 2, and 3, and needs $8,807.35 immediately after the third contribution. The account earns an effective annual rate of 5%. Find each contribution.

Method. First accumulate the existing $800 for three years:

800(1.05)^3=800(1.157625)=\$926.10

The new contributions must provide the remaining $8{,}807.35-926.10=\$7{,}881.25$. They are end-of-year contributions, so their factor is

s_{\overline{3}|} =\frac{1.05^3-1}{0.05} =\frac{1.157625-1}{0.05} =3.1525

The required contribution is

R=\frac{7{,}881.25}{3.1525}=\$2{,}500

Check. The three contributions accumulate to $2{,}500(1.05^2+1.05+1)=2{,}500(1.1025+1.05+1)=\$7{,}881.25$. Adding the existing reserve's $926.10 gives $8,807.35. The required year-end contribution is $2,500. The mistake avoided is applying the contribution factor to the full target without first accounting for the reserve already invested.

Timing Errors and Practical Checks

A quick comparison can help detect a timing error. For the same $n$, $i>0$, and target $F$, the beginning-of-period factor is $(1+i)$ times the end-of-period factor. Therefore, the required beginning-of-period contribution is smaller:

\frac{F}{\ddot{s}_{\overline{n}|}} = \frac{F}{(1+i)s_{\overline{n}|}} < \frac{F}{s_{\overline{n}|}}

This comparison assumes a positive interest rate and the same target date and number of deposits. If $i=0$, the two contribution amounts are equal because no deposit earns interest.

A common exam trap is to see five deposits and use the ordinary-annuity factor automatically. For the five beginning-of-year deposits in the previous example, dividing the target by $s_{\overline{5}|}$ gives approximately $11{,}265.95/5.41632256=\$2{,}080.00$. That is the wrong timing factor: those deposits occur at times 0 through 4, not 1 through 5. The correct factor is $\ddot{s}_{\overline{5}|}=5.6329754624$, giving about $2,000 per deposit. Always write the deposit dates before selecting the factor.

When an answer must be stated to cents, round only after solving. If the exact required contribution lies between cent amounts and the reserve must be at least the target, ordinary rounding to nearest cent could leave the reserve slightly short; in that circumstance, round the contribution up to the next cent and verify the accumulated balance.

Key takeaway: Accumulate each contribution to the target date. End-of-period contributions use $s_{\overline{n}|}$; beginning-of-period contributions use $\ddot{s}_{\overline{n}|}=(1+i)s_{\overline{n}|}$. If a reserve already exists, accumulate it separately before solving for the new contributions.

Check Your Understanding

For each problem, use the stated effective rate per year and identify the deposit timing before calculating.

A reserve must equal $3,121.60 immediately after the third year-end deposit. The account earns 4% effective annually. Find the level annual deposit. (Answer: $1,000.00)
Four deposits of equal amount are made at the beginning of each year, and the reserve is measured at the end of year 4. At 5% effective annually, the target is $5,430.76. Find the deposit to the nearest cent. (Answer: $1,200.00)
A target of $15,000 is needed immediately after the fifth year-end deposit. The account earns 6% effective annually. Find the required annual deposit to the nearest cent. (Answer: $2,660.95)
A business has $2,000 invested now and needs $11,051.26 at the end of year 4. It will add equal deposits at the ends of years 1 through 4. The effective annual rate is 5%. Find the required deposit to the nearest cent. (Answer: $2,000.00)
A target of $9,051.26 is needed at the end of year 4, and the account earns 5% effective annually. Compare the equal contribution if four deposits are made at the beginning of years 1 through 4 with the contribution if they are made at the ends of years 1 through 4. Give each to the nearest cent. (Answer: Beginning-of-year deposits: $2,000.00; end-of-year deposits: $2,100.00)

What You'll Learn