Solving for Base Payments and Growth Parameters

Solving an Equation of Value for a Payment Parameter

A varying annuity is often described by most of its terms: the number and timing of payments, the interest rate, and the rule governing how payments change. The unknown may be the first payment, a fixed dollar increment, or a percentage growth rate. The central task is to translate the payment rule into dated cash flows, value those flows at a common date, and solve the resulting equation.

The equation of value is only as reliable as the payment indexing behind it. For the conventions used here, an annuity-immediate has payments at times $1,2,\ldots,n$. An arithmetic-gradient stream with base payment $B$ and increment $K$ pays $B$ at time 1, $B+K$ at time 2, and so on. A geometric-gradient stream with base payment $B$ and growth rate $g$ pays $B$ at time 1, then $B(1+g)$ at time 2.

Definition: An equation of value equates the values, at one specified valuation date, of two or more cash-flow streams. For an annuity-immediate with payments $C_k$ at times $k=1,\ldots,n$, the present value at time 0, under a constant effective annual interest rate $i>-1$, is $\sum_{k=1}^{n} C_kv^k$, where $v=1/(1+i)$. The usual exam setting has $i\geq 0$.

For a fixed rate and payment schedule, the discounted payment amounts are linear in the unknown dollar amounts. That makes a base payment or fixed increment a linear equation to solve. An unknown geometric growth rate instead appears in powers of $1+g$, so its equation may be nonlinear.

Arithmetic Gradients: Solving for the Base or Increment

Let the payment at time $k$ be $B+(k-1)K$. Here $B$ is the first payment and $K$ is the fixed dollar increase from one payment to the next; $K$ may be negative for a decreasing stream. The present value is

PV=\sum_{k=1}^{n}\left[B+(k-1)K\right]v^k =B\,a_{\overline{n}|}+K\left[(Ia)_{\overline{n}|}-a_{\overline{n}|}\right].

In this expression, $a_{\overline{n}|}=\sum_{k=1}^{n}v^k$, and $(Ia)_{\overline{n}|}=\sum_{k=1}^{n}kv^k$, the present value of payments $1,2,\ldots,n$. The subtraction is important: the stream multiplied by $K$ is $0,1,\ldots,n-1$, not $1,2,\ldots,n$.

Worked Example: Solving for the First Arithmetic Payment

An annuity-immediate has two annual payments. The second payment is \$100 more than the first. At an effective annual interest rate of 25%, the present value is \$1,504. Find the first payment.

Method. Let the first payment be $B$. The payments are $B$ at time 1 and $B+100$ at time 2. With $i=0.25$, $v=1/1.25=0.8$ and $v^2=0.64$. The equation of value at time 0 is

$$ 0.8B+0.64(B+100)=1504. $$

Expand and collect the terms in $B$:

0.8B+0.64B+64=1504,\qquad 1.44B=1440,\qquad B=1000.

The payments are therefore \$1,000 and \$1,100. Checking their present value gives $1000(0.8)+1100(0.64)=800+704=1504$, as required. The answer is \$1,000. The common mistake avoided here is treating the second payment as $B+200$, even though there is only one increase after the first payment.

If $B$ is known and $K$ is unknown, the same equation is rearranged to isolate $K$. More generally, if the payment at time $k$ is $B+(k-1)K$, each dollar of $K$ contributes $(k-1)v^k$ to present value. Summing those weights gives the coefficient of $K$.

Worked Example: Solving for a Fixed Arithmetic Increment

A three-payment annuity-immediate pays \$500 at time 1. Each payment after the first is larger than the preceding payment by the same unknown dollar amount $K$. At an effective annual rate of 25%, the present value is \$1,425.28. Find $K$.

Method. The payments are $500$, $500+K$, and $500+2K$. The discount factors for times 1, 2, and 3 are $0.8$, $0.64$, and $0.512$, respectively. Thus

$$ 1504.28? $$

The equation, using the stated present value, is

$$ 500(0.8)+(500+K)(0.64)+(500+2K)(0.512)=1425.28. $$

The terms independent of $K$ total $400+320+256=976$. The coefficient of $K$ is $0.64+2(0.512)=0.64+1.024=1.664$. Therefore

976+1.664K=1425.28,\qquad 1.664K=449.28,\qquad K=270.

The payments are \$500, \$770, and \$1,040. Their present values are $400$, $770(0.64)=492.80$, and $1040(0.512)=532.48$. The sum is $400+492.80+532.48=1425.28$, which checks. The fixed increment is \$270. A common error is to assign the third payment only one increment; it is two steps after the first payment and is $B+2K$.

Geometric Gradients: Solving for the Growth Rate

For a geometric-gradient annuity-immediate with first payment $B$, the payment at time $k$ is $B(1+g)^{k-1}$, where $g$ is the effective annual percentage growth rate. Its present value is

PV=\sum_{k=1}^{n}B(1+g)^{k-1}v^k.

For $i>-1$ and $g>-1$, this can be written as a geometric series. When $g\neq i$,

PV=Bv\frac{1-\left(\frac{1+g}{1+i}\right)^n} {1-\frac{1+g}{1+i}}.

When $g=i$, the ratio in the series is 1, so the formula with a zero denominator is not used. Every discounted payment is then $Bv$, and $PV=nBv$. This limiting case is also easy to see directly: payment $k$ is $B(1+i)^{k-1}$, and its discount factor is $v^k=(1+i)^{-k}$, giving $Bv$ for each term.

Theorem: Suppose $B>0$, $i>-1$, and $g>-1$. For $n\geq 2$, the present value of the geometric-gradient annuity-immediate is strictly increasing as a function of $g$. Consequently, for a specified present value there is at most one solution for $g$ in this domain. If $n=1$, the present value is $Bv$, independent of $g$, so this uniqueness conclusion does not apply.

To justify the result, write the present value as $\sum_{k=1}^{n}B(1+g)^{k-1}v^k$. The first term does not depend on $g$. For every $k\geq2$, its derivative with respect to $g$ is $B(k-1)(1+g)^{k-2}v^k$, which is positive under the stated assumptions. If $n\geq2$, at least one such term is present, so the derivative of the full sum is positive. This proves strict increase and hence rules out two different solutions in the domain. It does not guarantee a solution for every target value; the target must also be in the range of the present-value function.

Worked Example: Solving for Geometric Growth

A two-payment annuity-immediate has a first payment of \$1,100 and a second payment that is larger by a constant percentage $g$. At an effective annual interest rate of 10%, its present value is \$2,000. Find $g$.

Method. The payments are $1100$ and $1100(1+g)$. The discount factors are $1/1.1$ and $1/1.1^2$, so

\frac{1100}{1.1}+\frac{1100(1+g)}{1.1^2}=2000.

The first term is $1000$. Also, $1100/1.1^2=1100/1.21=10000/11$. Thus

\frac{10000}{11}(1+g)=1000,\qquad 1+g=\frac{1000(11)}{10000}=1.1,\qquad g=0.10.

The growth rate is 10%. The payments are \$1,100 and \$1,210; their present values are $1100/1.1=1000$ and $1210/1.21=1000$, totaling \$2,000. This also illustrates the special case $g=i$. The common mistake avoided is applying the growth rate to the first payment as well: under this indexing, the first payment is $B$, and growth applies only after time 1.

A Reliable Setup and Common Traps

Before solving, identify the valuation date and list the actual payment amounts at their payment times. Then discount each amount to that date, or accumulate each amount to another common date. The equation must use the same valuation date on both sides. For a deferred stream, the payment rule is unchanged, but every payment is discounted from its actual date; do not treat a deferred annuity as though its first payment occurred at time 1.

Define the unknown and indexing.
State whether the unknown is the first payment $B$, the dollar increment $K$, or the percentage rate $g$. Write the first two or three payments to verify the pattern.

Place payments on a timeline.
Record the time of each payment and the valuation date. This prevents a one-period shift in the discount factors.

Write and solve the equation of value.
Discount or accumulate every payment consistently. Collect linear terms for $B$ or $K$; for $g$, retain the powers required by the payment rule.

Check the result.
Reconstruct the payments using the solved parameter and substitute them into the equation of value. Confirm that the stated present or accumulated value is recovered.

Several errors are especially costly. First, confusing $K$, a dollar amount, with $g$, a percentage, changes the payment rule: $B+K$ is not $B(1+g)$ unless the relationship between the parameters has been specified. Second, the geometric base $B$ is the payment at time 1, not the payment at time 0. Third, when solving for $g$, check that the annuity has at least two payments before arguing that present value varies with $g$. With one payment there is no later payment for growth to change.

These methods are useful whenever a contract, savings plan, or hypothetical benefit schedule specifies a present value but leaves one payment parameter unknown. Correctly separating payment timing from payment growth is the main safeguard: the timeline determines the discount powers, while the gradient rule determines the amounts. These techniques are aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Check Your Understanding

For each problem, write the payment stream before solving, then check the result by substitution.

An annuity-immediate pays $B$ at time 1 and $B+50$ at time 2. Interest is 25% effective annually and the present value is \$1,504. Find $B$. (Answer: \$1,000)
A two-payment annuity-immediate pays \$400 at time 1 and \$400+K at time 2. At an effective annual rate of 10%, its present value is \$740. Find $K$. (Answer: \$55)
A three-payment geometric-gradient annuity-immediate has first payment \$100. Interest is 0% effective annually and the sum of the payments is \$364. Find $g$, assuming $g>-1$. (Answer: 20%)
A two-payment arithmetic-gradient annuity-immediate has second payment \$100 greater than its first. Interest is 25% effective annually and the present value is \$1,504. Find the first payment. (Answer: \$1,000)
A two-payment geometric-gradient annuity-immediate has first payment \$1,000. Interest is 10% effective annually and its present value is \$1,900. Find $g$. (Answer: 19.9%)
For a geometric-gradient annuity with $B>0$, $i>-1$, and $g>-1$, explain why present value does not identify $g$ when there is only one payment, but can identify at most one $g$ when there are at least two payments. (Answer: With one payment $PV=Bv$, independent of $g$; with $n\geq2$, each later positive discounted payment increases strictly with $g$, so present value is strictly increasing.)

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