Solving a Level-Annuity Value Equation

Set Up the Value Equation Before Solving

A level annuity makes equal payments at regular intervals. When its value is known, the unknown may be the payment amount or the number of payments. The algebra is usually manageable; the important work is matching the annuity factor to the actual payment dates and to the date at which the target value is stated.

Assume an effective interest rate $i$ per payment period, with $i>0$, and a level payment $R$. An annuity-immediate has payments at the ends of periods; an annuity-due has payments at the beginnings. If the payments are annual, the rate must be an effective annual rate. If payments occur on another regular schedule, use the effective rate per that payment period.

Definition: For $n$ level payments of $R$, an annuity-immediate has present value $R a_{\overline{n}|}$ at the date one period before its first payment, and accumulated value $R s_{\overline{n}|}$ at the date of its last payment. An annuity-due has present value $R\ddot{a}_{\overline{n}|}$ at the date of its first payment, and accumulated value $R(1+i)s_{\overline{n}|}$ one period after its last payment. These values assume that the stated payments are made in full and that interest is earned at the same effective rate each period.

The factor formulas are $a_{\overline{n}|}=(1-v^n)/i$, $s_{\overline{n}|}=((1+i)^n-1)/i$, and $\ddot{a}_{\overline{n}|}=(1+i)a_{\overline{n}|}$, where $v=1/(1+i)$. The payment amount enters the value equation linearly. The number of payments enters through a power, so solving for it generally requires logarithms.

Why the Value Equations Work

For an annuity-immediate, the payments occur at times $1,2,\ldots,n$. Discounting each payment to time 0 gives $Rv,Rv^2,\ldots,Rv^n$. This is a finite geometric series:

PV=R\sum_{k=1}^{n}v^k =R\frac{v(1-v^n)}{1-v} =R\frac{1-v^n}{i} =Ra_{\overline{n}|}.

The third expression follows because $1-v=i/(1+i)=iv$, so $v/(1-v)=1/i$. For accumulated value at time $n$, carry each payment forward to time $n$: the first payment earns interest for $n-1$ periods and the last earns none. Summing those terms gives:

AV=R\sum_{j=0}^{n-1}(1+i)^j =R\frac{(1+i)^n-1}{i} =Rs_{\overline{n}|}.

An annuity-due payment occurs one period earlier than the corresponding annuity-immediate payment. Therefore, its present value is multiplied by $1+i$ when compared at the due annuity’s first-payment date, and its accumulated value one period after the last payment is also multiplied by $1+i$. The factor changes because the payment dates change, not because the payment amount changes.

A Reliable Solving Procedure

Identify the payment dates.
Decide whether the payments form an annuity-immediate or an annuity-due. Count the actual payments, not just the number of intervals described in words.

Choose the focal date.
Use present value if the target is before the payments, accumulated value if it is at or after the final payment, or move values to the stated date using the interest rate.

Write one value equation.
Set the correctly timed annuity value equal to the target. Do not switch from an immediate factor to a due factor midway through the equation.

Solve and check.
Isolate $R$ directly, or isolate $(1+i)^n$ and take logarithms to find $n$. Substitute the result into the original equation and check whether the payment schedule permits it.

Solving for the Payment

Worked Example: Payment for a Present-Value Target

A fund is to provide eight annual withdrawals of equal size, each at the end of a year. The effective annual interest rate is 6%, and the fund must have a present value of $50,000 immediately before the first withdrawal. Find the annual withdrawal. State the method and check the result.

Method. The withdrawals form an eight-payment annuity-immediate. Set $Ra_{\overline{8}|}=50{,}000$, then divide by the factor.

First calculate the discount factor power and annuity factor:

v^8=\frac{1}{1.06^8} =\frac{1}{1.593848074} \approx 0.627412372, \qquad a_{\overline{8}|} =\frac{1-0.627412372}{0.06} \approx 6.2097938.

Therefore,

R=\frac{50{,}000}{6.2097938} \approx 8{,}051.80.

The annual withdrawal is $8,051.80. To check, $8{,}051.80(6.2097938)\approx 50{,}000.02$; the two-cent difference is due to rounding the payment to cents. The unrounded quotient gives the target exactly. This calculation avoids treating the first withdrawal as if it occurred immediately: the stated end-of-year timing requires $a_{\overline{8}|}$, not $\ddot{a}_{\overline{8}|}$.

Worked Example: Deposit for a Due-Annuity Accumulation Target

A savings plan makes ten equal deposits at the beginnings of years 1 through 10. The effective annual interest rate is 5%, and the balance is valued at the end of year 10. Find the deposit needed to accumulate $30,000. Name the method and check the value at the target date.

Method. There are ten deposits at times $0,1,\ldots,9$, so this is an annuity-due accumulated for ten periods to time 10. Set the due accumulated value equal to $30,000 and solve for the deposit $R$.

s_{\overline{10}|} =\frac{1.05^{10}-1}{0.05} =\frac{1.628894627-1}{0.05} \approx 12.57789254,

The accumulated-value factor for the due annuity is $1.05s_{\overline{10}|}$, since every deposit earns one more period of interest than the corresponding end-of-period deposit.

1.05s_{\overline{10}|} \approx 1.05(12.57789254) \approx 13.20678717, \qquad R=\frac{30{,}000}{13.20678717} \approx 2{,}271.56.

The required deposit is $2,271.56. Checking, $2{,}271.56(13.20678717)\approx $30,000.01$; the one-cent difference is rounding. The unrounded payment produces exactly $30,000. The common error avoided here is using $s_{\overline{10}|}$ alone, which would value ten end-of-year deposits rather than ten beginning-of-year deposits.

Solving for the Number of Payments

When $n$ is unknown, first isolate the power $(1+i)^n$ or $v^n$. For example, an accumulated-value equation for an annuity-immediate is $AV=R((1+i)^n-1)/i$. Rearranging gives $(1+i)^n=1+iAV/R$, and taking natural logarithms gives $n=\ln(1+iAV/R)/\ln(1+i)$. This algebra gives a real-valued solution; whether that solution is a possible payment count is a separate question.

Worked Example: Term Implied by an Accumulated-Value Target

Equal deposits of $1,000 are made at the end of each year into an account earning an effective annual rate of 10%. The target accumulated value is $3,641, measured at the date of the last deposit. Solve the value equation for the implied term, then determine what happens if only whole annual deposits are permitted.

Method. Use the annuity-immediate accumulated-value equation, $1{,}000s_{\overline{n}|}=3{,}641$. First isolate the annuity factor and then solve its geometric-series formula for $n$.

s_{\overline{n}|}=3.641, \qquad \frac{1.10^n-1}{0.10}=3.641, \qquad 1.10^n=1+0.10(3.641)=1.3641.

Take logarithms of both sides:

n=\frac{\ln(1.3641)}{\ln(1.10)} \approx 3.2577.

Thus, $3.2577$ is the real-valued solution to the equation. It is not a possible number of annual deposits if the contract permits deposits only at year-ends. With three deposits, the accumulated value is $1{,}000s_{\overline{3}|}=1{,}000(3.31)=$3,310$. With four deposits, it is $1{,}000s_{\overline{4}|}=1{,}000(4.641)=$4,641$. There is no whole number of these $1,000 annual deposits that gives exactly $3,641; four is the least whole number that reaches or exceeds the target. The check also guards against reporting a fractional algebraic term as a feasible annual payment count.

Common Mistakes and a Direct Check

A frequent mistake is adding an extra factor of $1+i$ to an immediate-annuity accumulated value. Consider four end-of-year payments of $500 at an effective annual rate of 5%, valued at time 4. The correct factor is $s_{\overline{4}|}$:

s_{\overline{4}|} =\frac{1.05^4-1}{0.05} =4.310125, \qquad AV=500(4.310125)=$2{,}155.06.

A wrong calculation would use $500(1.05)s_{\overline{4}|}=$2{,}262.82$, applying the annuity-due factor even though the payments are at year-ends. The correction is to return to the timeline: the payment at time 4 has no interest period before the value date, so $s_{\overline{4}|}$, without the extra $1.05$, is the appropriate factor.

Also check that the rate is per payment period and that the target is measured at the date used by the factor. If a problem specifies a target at a different date, move the annuity value to that date before comparing it with the target. Rounding should come only after solving; rounded factors can slightly alter the computed payment or term.

Key takeaway: A level-annuity value equation has a simple structure: choose the factor from the payment timing and focal date, set payment times factor equal to the target, and solve. For an unknown term, take logarithms only after isolating the exponential expression, then check whether the resulting term is allowed by the payment schedule. These methods are aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Check Your Understanding

For each problem, identify the payment timing and focal date before solving. Answers are rounded to cents where applicable.

A fund must support six equal withdrawals at the end of each year. At an effective annual rate of 4%, its present value immediately before the first withdrawal is $12,000. Find each withdrawal. (Answer: $2,289.14)
Eight deposits of equal size are made at the beginnings of years 1 through 8 into an account earning 3% effective annually. Find the accumulated value at the end of year 8 if each deposit is $750. (Answer: $6,869.33)
Equal deposits of $500 are made at each year-end into an account earning 8% effective annually. What accumulated value is reached at the date of the fifth deposit? (Answer: $2,933.30)
An account earns 8% effective annually. Equal deposits are made at the end of each year for six years, and the balance immediately after the sixth deposit must be $10,000. Find the deposit. (Answer: $1,363.15)
A hypothetical grant makes five equal payments at the beginnings of five successive years, with the first payment now. At 4% effective annually, the present value at the date of the first payment is $20,000. Find each payment. (Answer: $4,319.75)

What You'll Learn