Extra Deposits Into a Sinking Fund

Why Extra Deposits Need Separate Accumulation

A sinking fund may receive regular scheduled deposits and also one-time contributions. To decide whether the fund will meet a redemption target, each deposit must be valued at the same date: the date the fund is needed. A deposit made earlier earns interest for longer than one made later, so it is not enough to add the dollar amounts deposited.

This tutorial assumes the fund earns a constant effective rate per period and that each deposit is made at a specified time. The scheduled deposits and extra deposits are all fund cash flows. They are not repayments of loan principal: the borrower may still owe the loan principal until redemption, even while the separate fund grows.

Definition: The accumulated fund value at a chosen date is the sum of each deposit multiplied by the factor that accumulates it from its deposit date to that chosen date. A surplus is the accumulated value minus the required redemption target when this difference is positive; a shortfall is the target minus the accumulated value when the difference is positive.

For the formulas below, let $j$ be the fund's effective rate per period, let the redemption date be time $n$, and suppose a level scheduled deposit $R$ is made at the end of each period $1,2,\ldots,n$. Let extra deposits $E_k$ be made at the end of period $k$, where $1\leq k\leq n$. If an extra deposit occurs on the same date as a scheduled deposit, it is additional to that scheduled deposit.

F_n=R s_{\overline{n}|j}+\sum_k E_k(1+j)^{n-k}, \qquad s_{\overline{n}|j}=\frac{(1+j)^n-1}{j}\quad (j\ne 0).

Here $F_n$ is the fund value at time $n$. The summation includes each extra deposit at its own time $k$; if there are no extra deposits, the sum is zero. If $j=0$, the accumulated value is simply the sum of the deposits, so the scheduled-deposit term is $nR$. The formula assumes the stated deposits are made in full and on time, and that the fund earns the same effective rate in every period. If yields change, the accumulation factor for each deposit must instead reflect the rates earned over its actual remaining periods.

Why the Formula Works

The formula follows by valuing every cash flow at the redemption date. A deposit made at time $k$ has $n-k$ full periods to earn interest, so its value at time $n$ is the deposit multiplied by $(1+j)^{n-k}$. The scheduled deposits therefore accumulate as $R(1+j)^{n-1}+R(1+j)^{n-2}+\cdots+R$.

R\sum_{r=0}^{n-1}(1+j)^r = R\frac{(1+j)^n-1}{j} = R s_{\overline{n}|j}.

The first equality is the geometric sum of the scheduled deposits' accumulation factors. Each extra deposit is handled by the same reasoning, but with its own date. Adding the scheduled and extra-deposit values is valid because accumulation is linear: the value of the combined deposits is the sum of their separately accumulated values.

Fix the valuation date.
Use the redemption date as the common date for all values, and identify the number of periods between each deposit date and that date.

Accumulate the scheduled deposits.
For level end-of-period deposits, use $R s_{\overline{n}|j}$. Check that the timing is end-of-period, not beginning-of-period.

Accumulate each extra deposit.
Multiply each extra amount by $(1+j)^{\text{remaining periods}}$. A deposit made at the redemption date has zero remaining periods and receives no further accumulation.

Compare with the target.
Add the accumulated values, then subtract the target. A positive difference is a surplus; a negative difference is a shortfall.

Calculating the Fund Value With Extra Deposits

Worked Example: Testing a Target With Two Extra Deposits

A fund earns an effective annual rate of $4\%$. A borrower deposits $\$800$ at the end of each of five years and makes additional deposits of $\$500$ at the end of year 2 and $\$300$ at the end of year 4. Will the fund meet a $\$5{,}200$ redemption target at the end of year 5? Find the surplus or shortfall.

Method. Accumulate the five scheduled deposits to year 5, then accumulate each extra deposit from its own date. The common mistake avoided here is treating every deposit as if it had the same time to earn interest.

The scheduled deposits accumulate by the factor $s_{\overline{5}|0.04}=\big((1.04)^5-1\big)/0.04$. First, $(1.04)^5=1.2166529024$, so $s_{\overline{5}|0.04}=0.2166529024/0.04=5.41632256$. Thus the scheduled-deposit value is $\$800(5.41632256)=\$4{,}333.058048$.

The $\$500$ deposit at year 2 earns interest for three years: $\$500(1.04)^3=\$500(1.124864)=\$562.432$. The $\$300$ deposit at year 4 earns interest for one year: $\$300(1.04)=\$312$.

The accumulated fund value is $\$4{,}333.058048+\$562.432+\$312=\$5{,}207.490048$. The difference from the target is $\$5{,}207.490048-\$5{,}200=\$7.490048$.

Answer: The fund meets the target, with a surplus of $\$7.49$, rounded to the nearest cent. As a check, all deposits have been valued at year 5; the year-4 extra deposit receives one year of interest, not three or five.

Solving for a Scheduled Deposit

An extra contribution can reduce the regular deposit needed to reach a target. If the target is $T$, the same accumulated-value equation can be rearranged to solve for $R$:

R=\frac{T-\sum_k E_k(1+j)^{n-k}}{s_{\overline{n}|j}}.

This expression assumes the extra deposits and target are fixed and that the remaining target is nonnegative. If the extra deposits alone accumulate to more than $T$, the equation gives a negative value for $R$; that means no positive scheduled deposit is needed to meet the target, not that a negative deposit should be made. The fund would already exceed the target under the stated cash flows.

Worked Example: Finding the Level Deposit After an Extra Contribution

A fund earns $5\%$ effective annually. It will receive a level deposit $R$ at the end of each of six years and an extra $\$1{,}000$ at the end of year 2. Find $R$ so the fund has $\$12{,}000$ at the end of year 6.

Method. First remove the extra deposit's accumulated value from the target, then divide the remaining amount by the accumulation factor for six level end-of-year deposits. This avoids counting the extra deposit as though it were part of each scheduled payment.

The extra deposit has four years to accumulate: $(1.05)^4=1.21550625$, so its value at year 6 is $\$1{,}000(1.21550625)=\$1{,}215.50625$. The target still to be accumulated by the level deposits is $\$12{,}000-\$1{,}215.50625=\$10{,}784.49375$.

Next, $(1.05)^6=1.340095640625$, and therefore $s_{\overline{6}|0.05}=(1.340095640625-1)/0.05=6.8019128125$.

Thus $R=\$10{,}784.49375/6.8019128125=\$1{,}585.50896\ldots$. Rounded to the nearest cent, $R=\$1{,}585.51$.

Check: Using the unrounded deposit, the scheduled deposits accumulate to $\$10{,}784.49375$; adding the extra deposit's $\$1{,}215.50625$ gives exactly $\$12{,}000$. Rounding the deposit to cents can produce a difference of less than one cent in the calculated fund value.

Finding the Extra Deposit Required

Sometimes the scheduled contribution is already fixed, and the question is how much must be added on a specified date. If there is one unknown extra deposit $E$ at the end of period $k$, then its accumulation factor is $(1+j)^{n-k}$. The amount needed is the target shortfall after valuing the other deposits, divided by this factor.

E= \frac{T-Rs_{\overline{n}|j}-\displaystyle\sum_{\text{known }r} E_r(1+j)^{n-r}} {(1+j)^{n-k}}.

The equation applies when the unknown deposit is made at the specified time $k$, and the other amounts and dates are known. If the numerator is zero, no extra deposit is needed. If it is negative, the known deposits already exceed the target; a positive extra deposit would increase the surplus.

Worked Example: Solving for a One-Time Extra Deposit

A fund earns an effective annual rate of $4\%$. The scheduled deposits are $\$1{,}500$ at the end of each of five years. What additional deposit at the end of year 2 is needed for a $\$9{,}000$ fund at the end of year 5?

Method. Find the scheduled deposits' value at year 5, calculate the remaining target, and discount that shortfall back to the date of the extra deposit. The common mistake avoided is adding the shortfall as a year-5 amount directly to a year-2 deposit.

The scheduled deposits accumulate to $\$1{,}500s_{\overline{5}|0.04}=\$1{,}500(5.41632256)=\$8{,}124.48384$. The required extra deposit's value at year 5 is therefore $\$9{,}000-\$8{,}124.48384=\$875.51616$.

A year-2 deposit accumulates for three years, with factor $(1.04)^3=1.124864$. If $E$ is the deposit, $E(1.124864)=\$875.51616$, so $E=\$875.51616/1.124864=\$778.33068\ldots$.

Answer: The required extra deposit is $\$778.33$, rounded to the nearest cent. The check is $\$778.33068\ldots(1.124864)=\$875.51616$; adding this to the scheduled-deposit value gives the $\$9{,}000$ target.

Timing Conventions and Common Traps

The exponent in an extra deposit's accumulation factor is determined by the number of complete periods from its deposit date to the valuation date. For example, a deposit at the end of year 2 has three periods to grow to the end of year 5. A deposit at the end of year 5 has no time to grow. If deposits are made at the beginning of each period rather than at the end, each scheduled deposit is made one period earlier; the end-of-period formula $R s_{\overline{n}|j}$ no longer describes those dates without adjustment.

For instance, if a level deposit $R$ is made at times $0,1,\ldots,n-1$ and valued at time $n$, the scheduled deposits accumulate to $R(1+j)s_{\overline{n}|j}$. Each deposit has one additional period to earn interest compared with the corresponding end-of-period schedule. Always draw or list the dates before applying an annuity factor.

Common exam trap: Do not add an extra deposit to the scheduled deposit amount and then treat the combined amount as if it were deposited every period. An extra deposit is made once, on its stated date, so it has its own accumulation factor. For example, an extra deposit at the end of year 2 in a five-year fund accumulates by $(1+j)^3$, not by $s_{\overline{5}|j}$.

A second trap is to compare the sum of nominal deposits with the redemption target. The target is a fund value at a future date, so each deposit must first be accumulated to that date. A third is to treat the sinking-fund contribution as a reduction of the loan principal. The fund contribution increases a separate asset; it does not, by itself, reduce the amount owed to the lender.

This method is useful whenever a fund has irregular contributions in addition to a regular schedule. It also provides a direct audit: list each deposit date, its remaining accumulation periods, and its value at the target date. If the rate or deposit timing changes, revise the relevant accumulation factors rather than using a formula whose assumptions no longer hold.

Check Your Understanding

For each problem, value all deposits at the stated target date. Money answers are rounded to the nearest cent.

A fund earns $3\%$ effective annually. It receives $\$600$ at the end of each of four years and an extra $\$200$ at the end of year 2. Find the fund value at the end of year 4 and its surplus over a $\$2{,}700$ target. (Answer: $\$2{,}722.36$, a $\$22.36$ surplus.)
A fund earns $4\%$ effective annually. It receives $\$1{,}200$ at the end of each of five years and an extra $\$500$ at the end of year 1. Does it meet a $\$7{,}100$ target at the end of year 5? (Answer: No; the fund is $\$7{,}084.52$, a $\$15.48$ shortfall.)
A fund earns $5\%$ effective annually. Scheduled deposits of $\$1{,}000$ are made at the end of each of four years. Find the extra deposit required at the end of year 2 to reach $\$5{,}000$ at the end of year 4. (Answer: $\$625.74$.)
A fund earns $5\%$ effective annually. It receives $\$700$ at each year-end for four years, plus $\$300$ at the end of year 1 and $\$200$ at the end of year 3. Find its value at the end of year 4 and the shortfall from a $\$3{,}600$ target. (Answer: $\$3{,}574.38$, a $\$25.63$ shortfall.)
A fund earns $4\%$ effective annually. It receives $\$1{,}000$ at times 0, 1, 2, and 3, plus an extra $\$250$ at time 2. Find the fund value at time 4. (Answer: $\$4{,}686.72$.)

What You'll Learn