Common Errors in Varying Annuity Calculations

Why Small Setup Errors Matter

A varying annuity can be valued efficiently once its cash flows are described correctly. But a concise formula will not repair a misread timeline: starting the first payment one period too early, using a growth exponent of one instead of zero, or treating a fixed dollar increase as a percentage can all produce plausible-looking but incorrect answers.

The most reliable first step is to write down the payment dates and amounts. Then identify whether the pattern is arithmetic or geometric, locate the valuation date, and check that the formula's assumptions match the stream. This tutorial concentrates on those checks, not on combining unrelated cash-flow structures.

Definition: An arithmetic-gradient annuity has payments that change by a fixed amount $D$ from one payment to the next. A geometric-gradient annuity has payments that change by a fixed proportion $g$ from one payment to the next. These descriptions refer to the change between consecutive payments, not to the payment number itself.

Start With Payment Dates and Indices

For an annuity-immediate with $n$ payments at times $1,2,\ldots,n$, the payment at time $k$ is discounted by $v^k$, where $v=1/(1+i)$. If the first payment is $C_1$ and payments grow geometrically by rate $g$, the payment at time $k$ is $C_1(1+g)^{k-1}$. The exponent is $k-1$: the first payment has not yet undergone a growth step.

If the same first payment is made at time $m+1$, after $m$ periods with no payments, payment $k$ in the payment sequence occurs at time $m+k$. Its amount is still $C_1(1+g)^{k-1}$, while its discount factor is $v^{m+k}$. The payment index and the calendar date are related, but they are not interchangeable.

Key check: For a stream whose first payment is $C_1$, write the first term explicitly as $C_1$ times the discount factor for its actual payment date. Only then write a general term. The first growth exponent is zero; the first discount exponent is the first payment date.

Worked Example: A Geometric Stream Beginning at Time 2

A fund will pay four annual amounts at times 2, 3, 4, and 5. The first payment is $800, and each later payment is 10% greater than the preceding one. At an effective annual interest rate of 5%, find the present value at time 0. State the payment amounts before applying a formula.

Method. Index the four payments by $k=1,2,3,4$. Their dates are $k+1$, and their amounts are $800(1.10)^{k-1}$. Thus the amounts are $800, $880, $968, and $1,064.80, at times 2, 3, 4, and 5 respectively.

The present value is

800v^2+880v^3+968v^4+1064.80v^5,\qquad v=\frac{1}{1.05}.

For a check using a geometric ratio, each successive discounted term is multiplied by $1.10v=1.10/1.05=22/21$. The first discounted term is $800v^2=725.6236$. Therefore the four discounted terms, rounded to four decimal places, are $725.6236$, $760.1771$, $796.3760$, and $834.2987$. Their sum, using unrounded values, is approximately $3,116.4753$. The present value is $3,116.48, rounded to the nearest cent.

Common mistake avoided. Using $800v$ as the first term would place the first payment at time 1, not time 2. That incorrect shift gives a value 5% higher: $3,116.4753(1.05)\approx3,272.2991$, or $3,272.30. The actual first payment date requires $800v^2$.

Fixed Increments Are Not Percentage Growth

An arithmetic gradient and a geometric gradient describe different payment patterns. If a payment increases by $100 each year, the increase is $100 regardless of the current payment. If it increases by 10% each year, the dollar increase depends on the current payment and generally becomes larger over time.

For an annuity-immediate with first payment $A$, fixed increment $D$, and $n$ payments, the amount at time $k$ is $A+(k-1)D$. Its present value is the sum of those amounts discounted at their actual dates:

\sum_{k=1}^{n}\bigl(A+(k-1)D\bigr)v^k.

Equivalently, the level portion is $A a_{\overline{n}|}$, and the incremental portion is $D\sum_{k=1}^{n}(k-1)v^k$. The second sum starts with zero at time 1 because there is no increment in the first payment. A fixed increment should not be substituted for a geometric growth rate.

Worked Example: Fixed Dollar Increases Over Five Years

A hypothetical grant pays $1,000 at the end of year 1 and then increases by a fixed $100 at each year-end for four more years. The effective annual interest rate is 5%. Find the present value.

Method. The wording specifies a fixed dollar increment, so the five payments are $1,000, $1,100, $1,200, $1,300, and $1,400, at times 1 through 5. With $v=1/1.05$, the discounted level portion is $1,000a_{\overline{5}|}$. The increment portion has weights 0, 1, 2, 3, and 4:

\begin{aligned} a_{\overline{5}|}&=v+v^2+v^3+v^4+v^5=4.3294767,\\ \sum_{k=1}^{5}(k-1)v^k &=v^2+2v^3+3v^4+4v^5\\ &=0.9070295+1.7276752+2.4681074+3.1341047\\ &=8.2369168. \end{aligned}

Thus the present value is

1,000(4.3294767)+100(8.2369168) =4,329.4767+823.6917 \approx5,153.1684.

The present value is $5,153.17, rounded to the nearest cent. As a check, the direct cash-flow expression is $1000v+1100v^2+1200v^3+1300v^4+1400v^5$, which gives the same value using unrounded terms.

Common mistake avoided. Treating the $100 increment as 10% growth would produce payments of $1,000, $1,100, $1,210, $1,331, and $1,464.10. That is a geometric pattern, not the stated arithmetic pattern; its present value is about $5,237.54, not $5,153.17.

Deferral and the Geometric-Annuity Formula

For $n$ annual payments, let the first payment $C_1$ occur at time $m+1$, followed by payments growing at rate $g$. Under a constant effective annual interest rate $i$, the present value at time 0 is

\begin{aligned} \operatorname{PV} &=\sum_{k=1}^{n} C_1(1+g)^{k-1}v^{m+k}\\ &=C_1v^{m+1}\sum_{j=0}^{n-1}\bigl((1+g)v\bigr)^j. \end{aligned}

The first equality discounts each payment from its own date. In the second, $j=k-1$, so the growth exponent begins at zero and the discount exponent begins at $m+1$. This change of index is a useful way to expose both common sources of error.

Formula and condition: Let $r=(1+g)v=(1+g)/(1+i)$. If $r\ne1$, then $$ \operatorname{PV}=C_1v^{m+1}\frac{1-r^n}{1-r}. $$ If $r=1$, each term in the sum has the same value and $\operatorname{PV}=nC_1v^{m+1}$. With the usual positive annual effective interest rate, $r=1$ occurs when $g=i$. Do not use the fraction when its denominator is zero.

The formula follows directly from the finite geometric-series identity. For $r\ne1$, multiplying $1+r+\cdots+r^{n-1}$ by $1-r$ cancels the interior terms and leaves $1-r^n$. Hence the sum is $(1-r^n)/(1-r)$. When $r=1$, the sum is $1+1+\cdots+1=n$, which gives the separate case. This also explains why substituting $g=i$ into the fraction is invalid: it creates $0/0$, even though the cash-flow value itself is well-defined.

Worked Example: A Deferred Stream With Growth Equal to Interest

A hypothetical scholarship fund pays four annual awards at times 3, 4, 5, and 6. The first award is $500, and each later award is 6% larger than the one before it. The effective annual interest rate is also 6%. Find the present value at time 0.

Method. The first payment is at time 3, so $m=2$. The payment amounts are $500$, $500(1.06)=530$, $530(1.06)=561.80$, and $561.80(1.06)=595.508$. Here $r=(1.06)/(1.06)=1$, so use the special case rather than the fraction formula:

\operatorname{PV}=4(500)v^3 =\frac{2,000}{1.06^3} =\frac{2,000}{1.191016} \approx1,679.2386.

Each discounted payment is equal: $500v^3=530v^4=561.80v^5=595.508v^6$, since each payment grows at exactly the interest rate. The four values sum to approximately $1,679.2386, so the present value is $1,679.24, rounded to the nearest cent.

Common mistake avoided. The first payment has growth exponent zero and discount exponent 3. Applying the no-deferral formula with $m=0$ would value payments at times 1 through 4. Equivalently, incorrectly discounting the correct value by one extra period gives $1,679.2386v\approx1,584.19$, which is too small.

A Reliable Error-Checking Routine

A formula should be the last step in the setup, not the first. The following checks catch most errors in varying-annuity calculations before they affect the final amount.

Write payment dates.
Mark the actual time of the first payment and the spacing between payments. Do not infer payment timing from whether the stream is increasing or decreasing.

Write the first few amounts.
Use the problem wording to decide between a fixed dollar change and a fixed percentage change. The listed amounts should visibly follow the stated pattern.

Check the indices.
For first payment $C_1$ at time $m+1$, payment $k$ has amount $C_1(1+g)^{k-1}$ and discount factor $v^{m+k}$.

Check the formula conditions.
Confirm constant interest and regular annual spacing when using the geometric-annuity formula, and separate the case $r=1$ from the case $r\ne1$.

Check the result.
Discount the individual cash flows or compare the formula with their sum. For positive payments and a positive interest rate, the present value must be less than the undiscounted total.

A formula for a regular annual stream does not automatically apply to irregular payment dates or changing interest rates. In those settings, value each payment using the interest accumulation applicable to its date, rather than forcing the stream into a constant-rate annuity formula. Likewise, a formula based on a fixed increment should not be used for percentage growth, or vice versa.

These checks matter because several wrong setups can still produce positive, plausible-looking answers. A one-period timing shift may change a result by only a few percent; an incorrect growth convention may change the payment pattern gradually. Writing the dates and first terms makes the error visible before arithmetic begins. This disciplined setup is aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Key takeaway: Match every payment amount to its payment index, every discount factor to its calendar date, and every formula to the pattern and conditions stated in the problem. When in doubt, value the cash flows individually as a check.

Check Your Understanding

For each problem, identify the payment pattern and timing before calculating.

A fund pays $500 at the end of year 1, then increases the payment by $50 at each year-end for two more years. The effective annual interest rate is 10%. Find the present value. (Answer: $1,359.88)
Three payments of $400, growing by 5% annually, are made at the ends of years 1, 2, and 3. The effective annual interest rate is 5%. Find the present value and explain why the usual geometric-series fraction needs a special case. (Answer: $1,142.86; the ratio is 1, so the three discounted payments are each $400v$)
Four payments occur at times 3, 4, 5, and 6. The first is $100, and each later payment is $20 more than the preceding one. At an effective annual rate of 5%, write the correct present-value expression. (Answer: $100v^3+120v^4+140v^5+160v^6$, with $v=1/1.05$)
An annuity-immediate pays $100 at the end of year 1, with a fixed $20 increase each year for four payments total. At an effective annual rate of 5%, calculate its present value. (Answer: $456.65)
A hypothetical benefit pays three annual amounts at times 2, 3, and 4. The first amount is $900, and each later amount is 10% greater than the preceding amount. The effective annual interest rate is 10%. Calculate the present value and state the first payment's growth and discount exponents. (Answer: $2,231.40; growth exponent 0 and discount exponent 2)

What You'll Learn