Solving for an Unknown Annuity Term and Deferral

Why Two Value Conditions Can Identify Two Unknowns

A deferred annuity problem may give two values for the same stream: its value immediately before payments begin and its value at an earlier focal date. If the annual payment is known, these conditions can identify both the number of payments and the deferral period. The key is to use the two values in the right order: their ratio isolates the deferral, while the boundary value identifies the payment count.

The method depends on a precise timing model. In this tutorial, payments are level, annual, and made at the end of each year. The deferral is a whole number of annual periods, and the interest rate is a positive effective annual rate. A different payment frequency or a fractional-period boundary requires a separately specified model.

Definition: Suppose an annuity-immediate pays $R$ at times $k+1,k+2,\ldots,k+n$, where $n$ is a positive integer and $k$ is a nonnegative integer. The annuity is deferred $k$ periods: its first payment is one period after time $k$. Let $B$ be its value at time $k$, immediately before the first payment, and let $P_0$ be its value at time zero. At effective interest rate $i>0$, with $v=(1+i)^{-1}$, $$ B=R a_{\overline{n}|} =R(v+v^2+\cdots+v^n), \qquad P_0=v^k B. $$

The payment at time $k+1$ is discounted one period to time $k$, not zero periods. That is why $a_{\overline{n}|}$, the annuity-immediate factor, appears in the boundary value. The deferral then discounts that boundary value by $k$ periods to time zero.

Separate the Unknowns Before Solving

The equations contain different information. Dividing the present value by the boundary value removes the annuity factor and leaves only the deferral. Once that period count is known, the boundary value divided by the payment gives the annuity factor, from which the number of payments can be determined.

\frac{P_0}{B}=v^k \quad\Longrightarrow\quad k=\frac{\ln(P_0/B)}{\ln(v)}

For the payment count, use

\frac{B}{R}=a_{\overline{n}|} =\frac{1-v^n}{i}.

In an exact problem, the computed $k$ must be a nonnegative integer. The value $B/R$ must also equal $a_{\overline{n}|}$ for a positive integer $n$. If the data are rounded, a candidate solution should instead be checked against the stated rounding precision or tolerance.

Uniqueness condition: For a fixed $i>0$, the ratio $P_0/B$ determines at most one real $k$, because $v^k$ is strictly decreasing in $k$. For a fixed $i>0$, $a_{\overline{n}|}$ is strictly increasing in the positive integer $n$, because $$ a_{\overline{n+1}|}-a_{\overline{n}|}=v^{n+1}>0. $$ Therefore, if the two given values are positive, $R$ is known, and both conditions match admissible integers $k$ and $n$, the solution is unique under this model. If $i=0$, then $v=1$ and $P_0/B=1$ regardless of $k$; the value conditions cannot identify the deferral.

Why the Two Equations Hold

At time $k$, the payment at time $k+j$ is $j$ periods away, for each $j=1,\ldots,n$. Its value there is $Rv^j$. Adding the values of all $n$ payments gives the boundary value. Moving that value back $k$ periods multiplies it by $v^k$.

B=\sum_{j=1}^{n}Rv^j =R\frac{v(1-v^n)}{1-v} =R\frac{1-v^n}{i} =R a_{\overline{n}|},

where $1-v=i/(1+i)=iv$, so $v/(1-v)=1/i$. The time-zero value is consequently

P_0=\sum_{j=1}^{n}Rv^{k+j} =v^k\sum_{j=1}^{n}Rv^j =v^kB.

This derivation also explains a useful feature of the method: the ratio of the two values contains no payment amount or payment-count factor. It is the cleanest way to find the deferral before looking for $n$.

A Reliable Solution Procedure

Write the timing model.
State when the first payment occurs relative to the boundary date, and confirm that $n$ and $k$ are whole periods under the problem’s assumptions.

Find the deferral.
Use $P_0/B=v^k$. Check that the resulting $k$ is a nonnegative integer, or that it is consistent with any stated rounding tolerance.

Find the payment count.
Compute $B/R$ and compare it with $a_{\overline{n}|}$ for positive integer values of $n$. Because the factor increases strictly, at most one such $n$ can work.

Substitute back.
Verify both $B=R a_{\overline{n}|}$ and $P_0=v^kB$. A candidate that satisfies only one condition is not a solution.

Worked Example: Recovering Both Unknowns From Exact Values

A hypothetical annuity pays $\$500$ at the end of each year. The effective annual interest rate is $25\%$. Its value immediately before the first payment is $\$976$, and its value at time zero is $\$624.64$. Find the number of payments $n$ and the whole-year deferral $k$.

Method. First take the ratio of the two values to isolate $k$. Then match the boundary value to an annuity-immediate factor.

The annual discount factor is $v=1/1.25=0.8$. The ratio is

\frac{P_0}{B}=\frac{624.64}{976}=0.64.

Since $0.64=0.8^2=v^2$, the deferral is $k=2$ years.

Next, divide the boundary value by the annual payment:

\frac{B}{R}=\frac{976}{500}=1.952.

The first three discounted annual payments give $a_{\overline{3}|}=0.8+0.8^2+0.8^3=0.8+0.64+0.512=1.952$. Thus $n=3$.

Check. The boundary value is $500(1.952)=\$976$. Discounting it two years gives $976(0.8)^2=976(0.64)=\$624.64$, the stated time-zero value. Both conditions hold exactly, so the answer is $n=3$, $k=2$.

Common mistake avoided. The deferral is two years, but the first payment is at time $3$, not time $2$. The boundary date is immediately before the first payment, so the annuity factor at that date begins with $v^1$.

Worked Example: Exact Data With a Nonterminating Decimal

A hypothetical annuity pays $\$100$ at the end of each year at an effective annual interest rate of $10\%$. Its boundary value is exactly $100(10/11+100/121+1000/1331)$, and its time-zero value is exactly that boundary value multiplied by $(10/11)^2$. Find $n$ and $k$. Give decimal values to six places as a check.

Method. Keep the supplied values exact as fractions. Decimal displays are rounded checks, not replacements for the exact data.

Here $v=10/11$. The boundary value simplifies to

B=100\left(\frac{1210+1100+1000}{1331}\right) =\frac{331000}{1331} \approx \$248.685199.

The time-zero value is exactly

P_0=B\left(\frac{10}{11}\right)^2 =\frac{33100000}{161051} \approx \$205.524958.

For the deferral, the exact ratio is $P_0/B=100/121=(10/11)^2=v^2$, so $k=2$.

For the payment count, divide the exact boundary value by $R=100$:

\frac{B}{R}=\frac{3310}{1331}.

The candidate factors are $a_{\overline{1}|}=1210/1331$, $a_{\overline{2}|}=(1210+1100)/1331=2310/1331$, and $a_{\overline{3}|}=(1210+1100+1000)/1331=3310/1331$. Therefore $n=3$.

Check. Three payments of $\$100$ have boundary value $100(10/11+100/121+1000/1331)=331000/1331$. Multiplying by $(10/11)^2$ gives $33100000/161051$, exactly the stated time-zero value. The solution is $n=3$, $k=2$.

Common mistake avoided. Rounding $B$ and $P_0$ first can make their ratio differ slightly from $v^2$. Use the exact expressions when the data are defined exactly, and round only the final decimal displays.

Worked Example: Detecting Conditions That Cannot Both Hold

A hypothetical annuity pays $\$100$ annually at an effective annual interest rate of $25\%$. The stated boundary value is $\$195.20$, and the stated time-zero value is $\$175.68$. Under the whole-year timing model, determine whether an integer payment count and integer deferral exist.

Method. Test the conditions independently. A plausible answer from one condition does not rescue a failure of the other.

Here $v=0.8$, and $P_0/B=175.68/195.20=0.9$. Thus any real-valued deferral inferred from the ratio would be

k=\frac{\ln(0.9)}{\ln(0.8)}\approx 0.4722.

This is not a whole number of annual periods. The deferral condition therefore fails under the stated model.

The boundary condition does identify a payment count: $B/R=195.20/100=1.952$, and $a_{\overline{3}|}=0.8+0.64+0.512=1.952$. So it implies $n=3$, but the value ratio still does not give an admissible $k$.

Check and conclusion. The boundary value is consistent with three payments, but the time-zero-to-boundary ratio is not $0.8^k$ for any whole-number $k$. There is no exact solution under the stated annual-period assumptions. Allowing fractional deferral would require a model that explicitly permits fractional periods; it cannot be silently assumed here.

Common mistake avoided. Stopping after finding $n=3$ would ignore the independent present-value condition. Both equations must be satisfied.

What Makes the Solution Unique—and When It Fails

The two value conditions are independent in a useful algebraic sense. The ratio $P_0/B$ depends on $k$, but not on $n$ or $R$. The quotient $B/R$ depends on $n$, but not on $k$. With $i>0$, increasing $k$ strictly decreases $v^k$; increasing $n$ strictly increases $a_{\overline{n}|}$. As a result, there can be at most one admissible pair $(k,n)$.

Uniqueness does not mean that every pair of supplied values produces a solution. The conditions may fail in several ways:

Invalid deferral: the ratio $P_0/B$ may imply a negative, noninteger, or otherwise inadmissible $k$.
Invalid payment count: $B/R$ may not equal an annuity factor for any positive integer $n$.
Inconsistent data: each value may appear plausible on its own, but the two conditions may not fit the same annuity.
Insufficient information: if one of the two independent value conditions is missing, a single value equation will generally not determine both $n$ and $k$.

If $n$ is known to be an integer but the boundary value is rounded, comparing $B/R$ to a displayed annuity factor is not enough to establish exact equality. Use the precision stated in the problem: calculate the candidate values from the model, round as instructed, and check whether they reproduce the reported data. Do not report an approximate logarithmic value of $k$ as an exact whole-year deferral unless it matches an integer under the required precision.

Common Timing and Solving Traps

Trap	Why it fails	Correction
Using $v^{k+1}B$ for the time-zero value	The boundary value is already at time $k$; discounting it $k+1$ periods counts one extra period.	Use $P_0=v^kB$.
Using $s_{\overline{n}\|}$ at the boundary	The accumulated-value factor applies when payments are accumulated forward, not when future payments are valued immediately before the first one.	Use $a_{\overline{n}\|}=v+\cdots+v^n$.
Solving the boundary equation and stopping	The boundary value identifies $n$, but does not by itself verify the deferral.	Check the ratio $P_0/B=v^k$ as a separate condition.
Rounding exact inputs before taking their ratio	Premature rounding can shift a ratio away from the exact power $v^k$.	Keep exact expressions or full precision through the solution; round at the final step.

Key takeaway: For a level annual annuity-immediate with known payment $R$, first use $P_0/B=v^k$ to find and validate the deferral. Then use $B/R=a_{\overline{n}|}$ to find and validate the payment count. Substitute both integers back into both original value conditions before accepting the result. These techniques are aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Check Your Understanding

For each problem, use the stated annual-period model. Treat supplied expressions as exact unless a rounding instruction is given.

A hypothetical annuity pays $\$200$ at each year-end at an effective annual rate of $25\%$. Its boundary value is $\$390.40$ and its time-zero value is $\$249.856$. Find $n$ and $k$. (Answer: $n=3,\ k=2$.)
A hypothetical annuity pays $\$300$ annually at an effective annual rate of $20\%$. Its boundary value is exactly $300(5/6+25/36)$, and its time-zero value is $5/6$ of its boundary value. Find $n$ and $k$. (Answer: $n=2,\ k=1$.)
A hypothetical annuity pays $\$110$ annually at an effective annual rate of $10\%$. Its boundary value is exactly $110(10/11+100/121)$, and its time-zero value is its boundary value multiplied by $(10/11)^3$. Find $n$ and $k$. (Answer: $n=2,\ k=3$.)
A hypothetical annuity pays $\$100$ annually at an effective annual rate of $25\%$. Its boundary value is $\$180$, and its time-zero value is $\$144$. Is there a solution with a positive integer payment count and a nonnegative integer deferral? (Answer: No. The ratio implies $k=1$, but $B/R=1.8$ is not $a_{\overline{n}|}$ for any positive integer $n$; it lies between $a_{\overline{2}|}=1.44$ and $a_{\overline{3}|}=1.952$.)
A hypothetical annuity pays $\$216$ annually at an effective annual rate of $20\%$. Its boundary value is $\$455$, and its time-zero value is exactly $455(25/36)$. Find $n$ and $k$. (Answer: $n=3,\ k=2$.)

Trap	Why it fails	Correction
Using \(v^{k+1}B\) for the time-zero value	The boundary value is already at time \(k\); discounting it \(k+1\) periods counts one extra period.	Use \(P_0=v^kB\).
Using \(s_{\overline{n}\|}\) at the boundary	The accumulated-value factor applies when payments are accumulated forward, not when future payments are valued immediately before the first one.	Use \(a_{\overline{n}\|}=v+\cdots+v^n\).
Solving the boundary equation and stopping	The boundary value identifies \(n\), but does not by itself verify the deferral.	Check the ratio \(P_0/B=v^k\) as a separate condition.
Rounding exact inputs before taking their ratio	Premature rounding can shift a ratio away from the exact power \(v^k\).	Keep exact expressions or full precision through the solution; round at the final step.

What You'll Learn