Accumulating Equal Sinking-Fund Deposits

Why Equal Deposits Accumulate to More Than Their Total

A sinking fund is set aside to build a specified amount by a future date. When deposits are equal and made at regular intervals, earlier deposits earn interest for longer than later ones. The final balance is therefore more than the sum of the deposits when the fund earns a positive rate. To calculate that balance correctly, first match the interest rate period to the deposit interval, then account for how many periods each deposit earns interest.

This tutorial considers a fund with no initial balance, equal deposits made at the end of each period, and a constant effective interest rate per period. These timing and rate assumptions are essential: deposits made at the beginning of each period earn interest for an additional period, while changing rates require a different accumulation calculation.

Definition: An end-of-period deposit fund with $n$ equal deposits of amount $R$ has deposits at times $1,2,\ldots,n$, and no initial deposit at time 0. If the fund earns effective interest rate $i$ per deposit period, its accumulated value immediately after the $n$th deposit is the sum of each deposit accumulated to time $n$. The rate must be measured over the same period as the deposits.

Accumulated Value Formula

At the valuation date, the deposit made at time 1 has earned interest for $n-1$ periods. The deposit made at time 2 has earned interest for $n-2$ periods, and the final deposit has earned no interest yet. Thus the balance immediately after the last deposit is

R(1+i)^{n-1}+R(1+i)^{n-2}+\cdots+R(1+i)+R.

The terms form a finite geometric series. For $i\ne 0$, its sum gives the standard accumulated-value factor $s_{\overline{n}|}$.

Theorem: If $n$ equal deposits of $R$ are made at the ends of successive periods, the fund earns a constant effective rate $i$ per period, and there is no initial balance, then the balance immediately after the $n$th deposit is

R s_{\overline{n}|} = R\frac{(1+i)^n-1}{i}, \qquad i\ne 0.

Here $s_{\overline{n}|}$ is the accumulated value of an annuity-immediate of 1 per period. It is an accumulation factor, not a payment amount. If $i=0$, each deposit remains unchanged, so the accumulated value is $nR$; the quotient formula is not used because its denominator is zero.

Why the Formula Holds

The formula follows by adding the value of each deposit at the common valuation date. Let $q=1+i$. The geometric series in the balance can be written $R(1+q+\cdots+q^{n-1})$. Multiplying the sum in parentheses by $q-1=i$ makes all interior terms cancel:

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

Since $i\ne 0$, dividing by $i$ gives $1+q+\cdots+q^{n-1}=((1+i)^n-1)/i$. Multiplying by $R$ proves the stated accumulated value. The exponents also show why the last deposit earns zero periods of interest and why the first earns $n-1$, not $n$.

Set the period.
Determine how often deposits are made and identify the effective interest rate for that same interval.

Set the count and timing.
Count the deposits and confirm they occur at period ends. The balance date is immediately after the final deposit.

Calculate the fund balance.
Use $R s_{\overline{n}|}$, or add the individual accumulated deposits as a check.

Compare with the target.
Subtract the target amount from the calculated balance to identify a surplus or shortfall.

Worked Applications

Worked Example: Annual Deposits at a Constant Effective Rate

A hypothetical equipment-replacement fund receives five deposits of $800, each at the end of a year. The fund earns an effective annual rate of 4%. Find the fund balance immediately after the fifth deposit. State whether it reaches a target of $4,300.

Method and assumptions. The deposits are end-of-year payments, the effective rate is 4% per year, and there is no starting balance. Because deposits and interest periods are both annual, use $i=0.04$ and $n=5$. The common mistake to avoid is letting the fifth deposit earn a full year of interest.

The accumulation factor is

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

Therefore the balance is

800(5.41632256)=\$4,333.058048.

Rounded to the nearest cent, the balance is $4,333.06. Compared with the $4,300 target, the fund has a surplus of $\$4,333.06-\$4,300=\$33.06$, so it reaches the target.

As a timing check, the five deposits accumulate for 4, 3, 2, 1, and 0 years. Their accumulated values are $800(1.04)^4=\$935.89$, $800(1.04)^3=\$899.89$, $800(1.04)^2=\$865.28$, $800(1.04)=\$832.00$, and $\$800.00$. Their unrounded sum is $4,333.058048, matching the formula.

Worked Example: Quarterly Deposits and a Quarterly Rate

A hypothetical maintenance reserve receives twelve deposits of $250 at the end of each quarter for three years. The effective interest rate is 2% per quarter. Find its balance at the end of the three years and compare it with a $3,300 target.

Method and assumptions. The fund has no opening balance, the quarterly rate stays constant, and each deposit is made at quarter-end. There are $n=3(4)=12$ deposits, and $i=0.02$ per quarter. Do not substitute an annual rate for a quarterly rate in the formula.

Calculate the accumulation factor and balance:

s_{\overline{12}|} = \frac{(1.02)^{12}-1}{0.02} = \frac{1.2682417946-1}{0.02} = 13.41208973, $$ $$ 250(13.41208973)=\$3,353.02243.

Rounded to the nearest cent, the balance is $3,353.02. The target comparison is $\$3,353.02-\$3,300=\$53.02$, so the fund exceeds its target by $53.02.

The 2% rate is per quarter; it corresponds to an effective annual rate of $(1.02)^4-1=0.08243216$, or 8.243216%. This conversion confirms the rate interpretation, but the quarterly rate is the one used with twelve quarterly deposits.

Worked Example: Semiannual Deposits for a Redemption Target

A hypothetical borrower plans to accumulate $10,500 in a separate fund over four years. The fund earns an effective rate of 3% per half-year, and the borrower deposits $1,200 at the end of each half-year. Determine the accumulated fund balance and whether it meets the target.

Method and assumptions. There are two half-year periods each year, so the schedule has $n=4(2)=8$ end-of-period deposits. The effective rate per deposit period is $i=0.03$. The equivalent effective annual rate is $(1.03)^2-1=0.0609$, or 6.09%. Use the half-year rate with the half-year deposit count. A common mistake is to use 6.09% as though it were the rate for each half-year.

The accumulation factor and the balance are

s_{\overline{8}|} = \frac{(1.03)^8-1}{0.03} = \frac{1.266770081-1}{0.03} = 8.892336033, $$ $$ 1,200(8.892336033)=\$10,670.80324.

Rounded to the nearest cent, the fund balance is $10,670.80. The surplus over the target is $\$10,670.80-\$10,500=\$170.80$. The target is met. This comparison concerns the fund balance only; it does not by itself calculate the borrower’s interest payments or total loan outlay.

Common Timing and Rate Errors

The deposit schedule and the rate must be interpreted together. In particular, do not count an extra period of interest for an end-of-period deposit. The following incorrect calculation illustrates the trap from the first example.

\text{Wrong: } 800\ddot{s}_{\overline{5}|} = 800(1.04)s_{\overline{5}|} = \$4,506.38.

This calculation treats the deposits as though they were made at the beginning of each year. But the problem specifies year-end deposits: at the balance date, the fifth deposit has just been made and has earned no interest. The correction is to use the annuity-immediate accumulation factor:

\text{Correct: }800s_{\overline{5}|}=\$4,333.06.

A separate rate-period error occurs when an annual effective rate is paired directly with monthly or quarterly deposits. If deposits are made $m$ times per year and the effective annual rate is $i$, the equivalent effective rate per deposit period is $(1+i)^{1/m}-1$, provided the annual effective rate applies across the year under a constant equivalent-rate model. Use that per-period rate in $s_{\overline{n}|}$. If the problem instead directly states a nominal rate convertible $m$-thly, identify the contractual periodic rate according to that nominal-rate convention.

Key takeaway: For equal end-of-period deposits, accumulate each deposit to the same date. Under a constant effective rate $i$ per deposit period and no initial fund balance, the result is $R s_{\overline{n}|}$. Check both the deposit timing and the rate unit before calculating.

Check Your Understanding

For each question, assume the fund begins with no balance and deposits are made at period-end unless stated otherwise. Round money to the nearest cent.

A fund receives six annual deposits of $400 and earns an effective annual rate of 5%. Find the balance immediately after the sixth deposit. (Answer: $2,720.77)
A reserve receives ten quarterly deposits of $300. The effective rate is 1% per quarter. Find the balance immediately after the last deposit. (Answer: $3,138.66)
A project fund receives four annual deposits of $2,000 and earns 3% effective annually. Find the balance and state whether it meets a $8,300 target. (Answer: $8,367.25; it meets the target by $67.25)
A hypothetical account receives six semiannual deposits of $900 at an effective rate of 2.5% per half-year. Find its balance immediately after the sixth deposit. (Answer: $5,748.96)
A fund receives five annual deposits of $2,200 and earns an effective annual rate of 4%. Determine whether a $12,000 target is reached and by how much. (Answer: balance $11,915.91; shortfall $84.09)

What You'll Learn