Accumulated Values for Regular Savings Deposits

Why Deposit Timing Determines the Fund Value

A savings plan can involve the same deposit amount and interest rate but produce different final values solely because deposits occur at different times. A deposit made at the beginning of a period earns interest during that period; one made at the end does not. To calculate a fund value correctly, identify the deposit dates, the final valuation date, and the effective interest rate for one deposit period.

This tutorial treats equal deposits made at regular intervals, with no withdrawals, fees, or changes in the interest rate. The accumulation methods are aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Definition: An annuity-immediate with $n$ payments has equal payments at the ends of $n$ consecutive periods. An annuity-due with $n$ payments has equal payments at the beginnings of $n$ consecutive periods. The accumulated value is the value of all payments, with interest, at a specified valuation date.

Matching the Factor to the Payment Schedule

Let $i$ be the effective interest rate per deposit period, and let $v=1/(1+i)$. For $n$ equal deposits of 1 at the ends of periods, the accumulated value immediately after the $n$th deposit is $s_{\overline{n}|}$. For deposits at the beginnings of periods, the accumulated value at the end of the $n$th period is $(1+i)s_{\overline{n}|}$.

s_{\overline{n}|}=\frac{(1+i)^n-1}{i}

\text{Period-end deposits: } R s_{\overline{n}|} \qquad \text{Period-beginning deposits: } R(1+i)s_{\overline{n}|}

Here $R$ is the amount of each deposit. The first expression applies when deposits occur at times $1,2,\ldots,n$ and the fund is valued at time $n$. The second applies when deposits occur at times $0,1,\ldots,n-1$ and the fund is valued at time $n$. In both cases there are $n$ deposits, but the due deposits each earn interest for one more period than the corresponding immediate deposits.

Assumptions: The deposit amount is constant; deposits are made exactly once per period at the stated beginning or end; the effective rate $i$ is constant for each period; and the valuation date is the one specified. If the rate changes, deposits are irregular, or the rate period does not match the deposit period, the single-factor formulas cannot be used without adjustment.

Why the Accumulated-Value Formulas Hold

The annuity-immediate formula follows by accumulating each deposit separately to time $n$. The deposit at time 1 earns interest for $n-1$ periods, the deposit at time 2 earns interest for $n-2$ periods, and the final deposit earns none before the valuation date. Thus the accumulated value per unit deposit is a geometric sum:

1+(1+i)+(1+i)^2+\cdots+(1+i)^{n-1} =\frac{(1+i)^n-1}{(1+i)-1} =\frac{(1+i)^n-1}{i} =s_{\overline{n}|}.

For an annuity-due, the deposits occur one period earlier: they earn for $n,n-1,\ldots,1$ periods. Its accumulated value per unit deposit is therefore

(1+i)+(1+i)^2+\cdots+(1+i)^n =(1+i)s_{\overline{n}|}.

This also explains the one-period difference between the two schedules. Each period-beginning deposit is the corresponding period-end deposit shifted one period earlier, so each earns one additional period of interest. Multiplying the immediate accumulated value by $1+i$ makes that shift.

Worked Examples

Worked Example: Equal Deposits at Each Year-End

A saver deposits $1,000 at the end of each year for 5 years into a fund earning an effective annual interest rate of 5%. Find the fund value immediately after the fifth deposit. Assume the rate is constant and there are no withdrawals or fees.

Method: use an annuity-immediate. The deposits are made at times 1, 2, 3, 4, and 5, and the valuation date is time 5. Thus there are $n=5$ annual periods and $i=0.05$ per period.

First calculate the factor:

s_{\overline{5}|} =\frac{(1.05)^5-1}{0.05} =\frac{1.2762815625-1}{0.05} =\frac{0.2762815625}{0.05} =5.52563125.

The accumulated value is

1{,}000(5.52563125)=\$5{,}525.63125.

Rounded to the nearest cent, the fund is $5,525.63. As a check, the deposits accumulate for 4, 3, 2, 1, and 0 years, respectively:

1{,}000\left((1.05)^4+(1.05)^3+(1.05)^2+1.05+1\right) =1{,}000(5.52563125) =\$5{,}525.63.

The check confirms that the first deposit earns the most interest and the last earns none by the valuation instant. A common mistake is to use the annuity-due factor, which would give every deposit an extra year of interest.

Worked Example: Deposits at Each Year-Beginning

A hypothetical education fund receives four deposits of $800, the first immediately and the remaining deposits at the beginning of each following year. The fund earns an effective annual rate of 4%. Find its value at the end of year 4. Assume no other cash flows.

Method: use an annuity-due. The four deposits occur at times 0, 1, 2, and 3. The valuation date is time 4, so their accumulation periods are 4, 3, 2, and 1 years. The rate per deposit period is $i=0.04$, and $n=4$.

Calculate the annuity-immediate factor first:

s_{\overline{4}|} =\frac{(1.04)^4-1}{0.04} =\frac{1.16985856-1}{0.04} =\frac{0.16985856}{0.04} =4.246464.

Shift the deposits one period earlier by multiplying by $1.04$:

(1.04)s_{\overline{4}|} =1.04(4.246464) =4.41632256.

The fund value is

800(4.41632256)=\$3{,}533.058048.

Rounded to the nearest cent, the value is $3,533.06. A direct check uses each deposit’s actual accumulation period:

800\left((1.04)^4+(1.04)^3+(1.04)^2+1.04\right) =800(4.41632256) =\$3{,}533.06.

This check also avoids the common mistake of treating the first deposit, made immediately, as if it were made one year later.

Worked Example: Monthly Deposits with a Nominal Rate

A saver deposits $300 at the end of every month for 2 years. The account earns a nominal annual interest rate of 6% compounded monthly. Find the balance immediately after the 24th deposit. Assume the stated rate applies throughout and there are no fees or withdrawals.

Method: convert to the effective monthly rate, then use an annuity-immediate. Because the nominal rate is compounded monthly, the effective rate per monthly deposit period is

i=\frac{0.06}{12}=0.005.

There are $n=2(12)=24$ deposits. The final deposit is made at the valuation date, so it earns no interest before that date. Compute the factor:

s_{\overline{24}|} =\frac{(1.005)^{24}-1}{0.005} =\frac{1.127159776-1}{0.005} =\frac{0.127159776}{0.005} =25.4319552.

The balance is

300(25.4319552)=\$7{,}629.58656.

Rounded to the nearest cent, the balance is $7,629.59. The 24 deposits earn interest for 23, 22, and successively fewer months down to zero months for the last deposit; their month-by-month accumulation is exactly the geometric sum represented by $s_{\overline{24}|}$.

The common rate-conversion error is to use $0.06$ as the monthly rate. That treats the account as earning 6% each month, not 6% nominal per year compounded monthly. The correct monthly rate is $0.06/12=0.005$.

Choosing the Valuation Date and Checking the Result

The accumulated-value factor is tied to a particular date, not merely to the number of deposits. For an immediate annuity, $R s_{\overline{n}|}$ gives the value at the time of the $n$th deposit. For a due annuity, $R(1+i)s_{\overline{n}|}$ gives the value one full period after the first deposit and at the end of the $n$th period. If the requested date is later than the factor’s valuation date, accumulate the resulting balance for the additional full periods.

For example, if an annuity-immediate has finished making its deposits and the balance remains invested for $k$ more periods at the same effective rate $i$, the value at the later date is

R s_{\overline{n}|}(1+i)^k.

This adjustment assumes the rate remains unchanged during those additional periods and that no further payments occur. If the stated final date does not fall on a deposit-period boundary, use the interest assumptions given for the fractional period rather than silently treating it as a full period.

Common exam trap: Counting deposits is not the same as counting interest periods. A final period-end deposit is included in the number of deposits, but it earns zero periods of interest at the instant it is made. A first period-beginning deposit earns a full period if the fund is valued at the end of the $n$th period.

Before applying a factor, check three items: the number of deposits, whether the deposits occur at period-end or period-beginning, and whether the rate used is effective for one deposit period. Then compare the factor’s valuation date with the date requested. These checks often reveal an incorrect factor, an off-by-one-period error, or a rate expressed in the wrong time unit.

Check Your Understanding

For each problem, identify the payment timing, use the rate per deposit period, and round the final fund value to the nearest cent.

A saver deposits $400 at each year-end for 5 years. The effective annual rate is 5%. Find the value immediately after the fifth deposit. (Answer: $2,210.25)
A fund receives $1,000 at the beginning of each year for 3 years and earns 6% effective annually. Find its value at the end of year 3. (Answer: $3,374.62)
A saver deposits $200 at each quarter-end for 8 quarters. The account earns a nominal annual rate of 8% compounded quarterly. Find the value immediately after the eighth deposit. (Answer: $1,716.59)
A hypothetical reserve receives twelve deposits of $150 at the beginning of each month. Its nominal annual rate is 12% compounded monthly. Find the value at the end of month 12. (Answer: $1,921.40)
A saver deposits $1,000 at each year-end for 5 years into a fund earning 4% effective annually. No deposits are made in years 6 or 7, and the balance remains invested at the same rate. Find the value at the end of year 7. (Answer: $5,858.29)

What You'll Learn