Accumulated Value of an Annuity-Due by Shifting Payments

Why the Final Payment Date Matters

An annuity-due has payments at the beginning of successive periods. When asked for its accumulated value, first identify the date at which the value is wanted. In this tutorial, that date is the date of the last payment. A payment made on that date is included at face value; it has earned no interest by that instant.

For $n$ equal payments of $R$, made at times $0,1,\ldots,n-1$, the final payment date is time $n-1$. Earlier payments earn interest up to that date. Shifting every payment one period later produces an annuity-immediate with payments at times $1,2,\ldots,n$. Its accumulated value at time $n$ has exactly the same interest exponents as the original stream has at time $n-1$.

Definition: An annuity-due of $n$ equal payments has payments at the beginnings of $n$ consecutive periods. If each payment is $R$ and the effective interest rate per period is $i$, its accumulated value at the date of its last payment is the sum of each payment accumulated from its payment date to that final date. The last payment is included without interest accumulation.

Deriving the Accumulated Value by Shifting

Assume a constant effective rate $i$ per payment period, with $i>-1$, and let $i\ne 0$ for the geometric-series expression below. The $n$ payments occur at times $0$ through $n-1$. By the final payment date, the first payment has accumulated for $n-1$ periods, the next for $n-2$, and so on. The final payment has accumulated for zero periods.

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

Now shift the whole payment stream one period later. The payments occur at times $1$ through $n$, and the accumulated value is measured at time $n$. The first payment accumulates for $n-1$ periods, the second for $n-2$, and the last for zero periods. Thus each payment has exactly the same accumulation period in the shifted stream as in the original stream. The values are equal, even though the dates are one period apart.

The shifted stream is an annuity-immediate. Its accumulated value at time $n$ is $R s_{\overline{n}|}$, where $s_{\overline{n}|}$ denotes the accumulated value at time $n$ of $n$ unit payments at times $1$ through $n$. Summing the geometric series gives the result.

\begin{aligned} \text{Value at time }n-1 &=R\sum_{k=0}^{n-1}(1+i)^k\\ &=R\frac{(1+i)^n-1}{i}\\ &=R s_{\overline{n}|}. \end{aligned}

If $i=0$, no payment earns interest, so the accumulated value is $nR$. The geometric-series quotient is not used in that case; its limiting value as $i$ approaches zero is also $nR$. The shift argument requires the same accumulation rule over each period. If payment periods and interest periods differ, first determine the effective rate for one payment period.

Theorem: For $n$ equal payments $R$ made at times $0,1,\ldots,n-1$, with effective interest rate $i$ per period, the accumulated value at time $n-1$, the date of the last payment, is $R s_{\overline{n}|}=R\big((1+i)^n-1\big)/i$ when $i\ne0$, and $nR$ when $i=0$. This result assumes level payments and a common rate per period.

Mark the payment dates.
For a due annuity, place the payments at times $0$ through $n-1$. Do not place the first payment at time 1.

Choose the focal date.
Here it is time $n-1$, the date of the last payment. Determine how many periods each payment accumulates.

Shift the stream, if useful.
Move each payment one period later and the focal date one period later. The accumulation exponents stay the same, giving the annuity-immediate factor $s_{\overline{n}|}$.

Check the rate period.
Use an effective rate for one payment period. Convert a nominal rate to the payment frequency before applying the factor.

Worked Applications

Worked Example: Accumulating Four Annual Deposits

A saver deposits \$1,000 at the beginning of each year for four years. The effective annual interest rate is 10%. Find the accumulated value on the date of the fourth deposit, immediately after that deposit is made.

Method. The deposits occur at times 0, 1, 2, and 3. The requested date is time 3. The deposits therefore accumulate for 3, 2, 1, and 0 years, respectively. The last deposit is not multiplied by an interest factor.

\begin{aligned} \text{Value at time }3 &=1000(1.10)^3+1000(1.10)^2+1000(1.10)+1000\\ &=1000(1.331)+1000(1.21)+1000(1.10)+1000\\ &=1331+1210+1100+1000\\ &=\$4{,}641. \end{aligned}

As a check using the shift, move the deposits to times 1, 2, 3, and 4, and value them at time 4. Their accumulation periods are still 3, 2, 1, and 0. The result is $1000s_{\overline{4}|}=\$4{,}641$, as required. No rounding is needed.

Common mistake avoided: The requested date is time 3, not time 4. Accumulating the original deposits to time 4 would add one extra year of interest to the entire value.

Worked Example: Finding the Required Beginning-of-Period Payment

A fund must have \$2,400 immediately after its third annual beginning-of-year deposit. The effective annual interest rate is 10%, and the three deposits are equal. Find the deposit amount.

Method. The deposits occur at times 0, 1, and 2, so the target date is time 2. The accumulation exponents are 2, 1, and 0. First calculate the factor per dollar, then divide the target by that factor.

\begin{aligned} \text{Accumulation factor} &=(1.10)^2+1.10+1\\ &=1.21+1.10+1\\ &=3.31. \end{aligned}

R=\frac{2400}{3.31}=725.0755287\ldots

The required payment is \$725.08 to the nearest cent. Before rounding, multiplying the payment by the factor gives \$2,400 exactly. If the payment itself must be made in cents, \$725.08 accumulates to $725.08(3.31)=\$2,400.0148$, or \$2,400.01 to the nearest cent.

Common mistake avoided: Counting three deposits does not mean that the first deposit earns interest for three full years. At the third deposit date, the deposits earn interest for two, one, and zero years.

Worked Example: Matching a Nominal Rate to Semiannual Deposits

A hypothetical savings plan makes five equal deposits of \$400 at the beginning of each half-year. The nominal annual interest rate is $i^{(2)}=8\%$, convertible semiannually. Find the accumulated value immediately after the fifth deposit, and separate the interest earned from the total deposits.

Method. There are two conversion periods per year, so the effective rate per half-year is $0.08/2=0.04$. The deposits occur at half-year times 0, 1, 2, 3, and 4; the fifth deposit date is time 4 in half-year units. The accumulation exponents are 4, 3, 2, 1, and 0.

\begin{aligned} \text{Accumulated value} &=400\left((1.04)^4+(1.04)^3+(1.04)^2+1.04+1\right)\\ &=400(1.16985856+1.124864+1.0816+1.04+1)\\ &=400(5.41632256)\\ &=\$2{,}166.529024. \end{aligned}

The total amount deposited is $5(400)=\$2,000$. Therefore, the interest included in the accumulated value is $\$2,166.529024-\$2,000=\$166.529024$. To the nearest cent, the accumulated value is \$2,166.53 and the interest is \$166.53.

The conversion frequency is essential: using 8% as the rate for each half-year would not match the stated nominal rate. The calculation is also checked by adding the deposits and interest: \$2,000 plus \$166.53 rounds to \$2,166.53.

Timing Traps and Limits

The most important check is whether the focal date is the final payment date or one period later. For four annual payments of \$1,000 at 10%, the value at the fourth payment date, time 3, is \$4,641. A calculation that instead uses time 4 gives

4641(1.10)=\$5{,}105.10.

That is the value one year after the fourth payment, not the value at the date of that payment. The wrong calculation has added an interest period that the question did not request. Similarly, including interest on the final payment when valuing at its payment date is incorrect: that payment has just been made.

The shift is a way to track timing, not a claim that two payment streams occur on the same dates. The original stream is valued at time $n-1$; the shifted stream is valued at time $n$. They have equal values because each corresponding payment has the same number of periods to accumulate. This argument does not justify using the formula for unequal payment amounts, irregular payment dates, or rates that vary by period without adjusting the valuation.

This method is useful because it reduces a beginning-of-period accumulation to a familiar end-of-period annuity calculation while keeping the focal date explicit. On questions aligned with the SOA Exam FM (Financial Mathematics) syllabus, a brief timeline and an exponent check can prevent errors that a correct-looking factor alone may conceal.

Check Your Understanding

For each question, identify the final payment date before calculating. Dollar answers are rounded to the nearest cent where needed.

A plan makes three annual deposits of \$800 at the beginning of each year. The effective annual rate is 5%. Find the value immediately after the third deposit. (Answer: \$2,522.00)
Five beginning-of-year deposits of \$250 earn an effective annual rate of 4%. Find the accumulated value on the date of the fifth deposit. (Answer: \$1,354.08)
A hypothetical fund receives four equal deposits of \$600 at the beginning of each half-year. The nominal annual rate is $i^{(2)}=10\%$, convertible semiannually. Find the accumulated value immediately after the fourth deposit. (Answer: \$2,586.08)
Six equal annual deposits are made at the beginning of successive years. At an effective annual rate of 5%, the value immediately after the sixth deposit must be \$5,000. Find the required deposit to the nearest cent. (Answer: \$735.09)

What You'll Learn