Deriving the Arithmetic-Gradient Annuity Factor

Why Derive the Gradient Factor?

A level annuity-immediate discounts the same amount at each year-end. In many payment schedules, however, the amount changes by a fixed dollar increment each year. The present value of the schedule depends not only on the number of payments and the interest rate, but also on how the payment amounts are indexed. A factor derived directly from discounted payments makes that indexing clear.

This tutorial derives the present-value factor for payments of 1 at time 1, 2 at time 2, and so on through $n$ at time $n$. The result can then be scaled by a dollar increment or combined with a level annuity. The focus is the derivation and interpretation of this factor, not a general treatment of other changing-payment patterns. The methods are aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Definition: For an annuity-immediate with $n$ annual payments, the arithmetic-gradient annuity factor $(Ia)_{\overline{n}|}$ is the present value at time 0 of payments $1,2,\ldots,n$ made at times $1,2,\ldots,n$, respectively. With effective annual interest rate $i$ and $v=1/(1+i)$, $$ (Ia)_{\overline{n}|}=\sum_{k=1}^{n} k v^k. $$

The first payment is 1 at the end of year 1; the last is $n$ at the end of year $n$. This timing is part of the definition. The factor is not for payments $1,2,\ldots,n$ at times $0,1,\ldots,n-1$. The ordinary annuity-immediate factor is $a_{\overline{n}|}=\sum_{k=1}^{n}v^k$, and the effective annual discount rate is $d=1-v=iv$.

Deriving the Factor From the Discounted Payments

The derivation uses the finite sum in the definition and a shifted copy of that sum. Subtracting the shifted sum leaves an ordinary annuity sum and one boundary term. This is why the result can be written using $a_{\overline{n}|}$.

Write the present value.
Let $G=\sum_{k=1}^{n}kv^k$. The $k$th payment is $k$, discounted $k$ years.

Shift the sum.
Multiplying by $v$ gives $vG=\sum_{k=1}^{n}kv^{k+1}$. This shifts every term one power of $v$ higher.

Subtract term by term.
In $G-vG$, the coefficients of $v^2,\ldots,v^n$ reduce to 1, leaving $v+v^2+\cdots+v^n-nv^{n+1}$.

Recognize the annuity factor.
The first $n$ terms are $a_{\overline{n}|}$. Since $1-v=d$, division gives the required relation.

\begin{aligned} G-vG &=(v+2v^2+\cdots+nv^n)-(v^2+2v^3+\cdots+nv^{n+1})\\ &=v+v^2+\cdots+v^n-nv^{n+1}\\ &=a_{\overline{n}|}-nv^{n+1}. \end{aligned}

Because $G-vG=(1-v)G=dG$, when $i\ne 0$ the derived formula is

(Ia)_{\overline{n}|} =\frac{a_{\overline{n}|}-nv^{n+1}}{d} =\frac{a_{\overline{n}|}-nv^{n+1}}{iv}.

The model is a finite annuity-immediate with equal one-year periods and a constant effective annual rate. The formula assumes $v=1/(1+i)$, with $i>-1$, and applies as written when $i\ne0$. If $i=0$, the division by $d$ is not defined; use the original sum instead:

(Ia)_{\overline{n}|}\big|_{i=0} =1+2+\cdots+n =\frac{n(n+1)}{2}.

This zero-rate result is also the limit of the factor as $i$ approaches zero, but the direct sum is simplest to use. If the payments are made at the beginnings of years, or if the intervals are not one year, the discount powers change; this annuity-immediate formula does not apply without adjusting the valuation model.

Applying the Factor to Payment Schedules

A factor is most useful when its payment pattern matches the schedule exactly. Suppose payments at times $1,\ldots,n$ are $1,2,\ldots,n$ dollars multiplied by a constant amount $g$. Their present value is $g(Ia)_{\overline{n}|}$. For a schedule that starts at $P$ and increases by $g$ each year, the payments are $P,P+g,\ldots,P+(n-1)g$. At time $k$, this amount can be written as $(P-g)+gk$, so its present value is a level component plus a gradient component:

\text{PV}=(P-g)a_{\overline{n}|}+g(Ia)_{\overline{n}|}.

This decomposition works even when $P-g$ is negative: it is an algebraic representation of the cash flows, not necessarily a separate positive payment stream. Always verify the first and last payment after decomposing a schedule.

Worked Example: Derive and Evaluate a Three-Payment Factor

A hypothetical fund pays 1 at the end of year 1, 2 at the end of year 2, and 3 at the end of year 3. The effective annual interest rate is 10%. Find the present value factor using both the discounted payments and the derived relation. Report the final factor to four decimal places.

Method. First use the definition to confirm the payment dates and amounts; then verify the factor with the ordinary annuity relation. This avoids applying a factor whose gradient starts at a different date.

Here $i=0.10$, $v=1/1.10=10/11$, $n=3$, and $d=1-v=1/11$. Discounting each payment gives

\begin{aligned} (Ia)_{\overline{3}|} &=v+2v^2+3v^3\\ &=\frac{10}{11}+2\left(\frac{100}{121}\right)+3\left(\frac{1000}{1331}\right)\\ &=\frac{1210+2200+3000}{1331}\\ &=\frac{6410}{1331}\\ &\approx 4.8159. \end{aligned}

For the check, $a_{\overline{3}|}=v+v^2+v^3=3310/1331$, and $3v^4=30000/14641$. Using the relation, or equivalently keeping the common denominator, gives

\frac{a_{\overline{3}|}-3v^4}{d} =\frac{6410}{1331} \approx 4.8159.

The two methods agree. Thus, a unit gradient with these three year-end payments has present value approximately 4.8159 at 10% interest.

Worked Example: Combine a Level Payment and a Gradient

A hypothetical equipment-replacement account receives four year-end deposits: $500, $650, $800, and $950. The effective annual interest rate is 10%. Find the present value of the deposits, to the nearest cent.

Method. Write the amount at time $k$ as a level amount plus a multiple of $k$. This method avoids treating the first deposit as the first gradient increment.

The deposits are $350+150k$ for $k=1,2,3,4$: at $k=1$, this is $350+150=500$, and at $k=4$, it is $350+600=950$. Therefore,

\text{PV}=350a_{\overline{4}|}+150(Ia)_{\overline{4}|}.

At $i=0.10$, $v=10/11$. The ordinary annuity factor and gradient factor are

\begin{aligned} a_{\overline{4}|}&=v+v^2+v^3+v^4\approx3.169865446,\\ (Ia)_{\overline{4}|}&=v+2v^2+3v^3+4v^4\approx7.547981683. \end{aligned}

Substituting these unrounded factors gives

\begin{aligned} \text{PV} &=350(3.169865446)+150(7.547981683)\\ &\approx1109.452906+1132.197252\\ &\approx \$2{,}241.65. \end{aligned}

As a check, the decomposition reproduces all four deposits in order. The final present value is rounded to the nearest cent.

Worked Example: Scale a Gradient and Check the Timing

A hypothetical research account makes year-end contributions of $200, $400, $600, and $800. The effective annual interest rate is 25%. Find the present value, to the nearest cent, and verify the result using the ordinary annuity relation.

Method. Recognize the payments as 200 times the unit gradient. Then check the answer by evaluating the relation. This prevents an accidental one-period shift, which would value the same amounts at times 0 through 3 instead of times 1 through 4.

Here $i=0.25$, $v=1/1.25=0.8$, and $d=1-v=0.2$. The factor from its defining discounted payments is

\begin{aligned} (Ia)_{\overline{4}|} &=0.8+2(0.8^2)+3(0.8^3)+4(0.8^4)\\ &=0.8+1.28+1.536+1.6384\\ &=5.2544. \end{aligned}

The present value is $200(5.2544)=\$1{,}050.88$. For the relation check, $a_{\overline{4}|}=0.8+0.64+0.512+0.4096=2.3616$, and $4v^5=4(0.32768)=1.31072$. Hence

\frac{a_{\overline{4}|}-4v^5}{d} =\frac{2.3616-1.31072}{0.2} =\frac{1.05088}{0.2} =5.2544.

Multiplying this factor by $200 gives the same present value, $1,050.88. The factor has not been shifted: its first term is discounted once, because the first contribution is made at the end of year 1.

Common Mistake: Shifting the Gradient

A frequent error is to use $v^{k-1}$ for the payment $k$. That values payment 1 at time 0, payment 2 at time 1, and so on. It is an annuity-due timing pattern, not the annuity-immediate pattern in the definition. For payments $1,2,\ldots,n$ at year-ends, the correct term for payment $k$ is $kv^k$.

Another indexing error occurs when a schedule begins at $P$ and increases by $g$: it is not generally $P+gk$ at time $k$, since that would make the first payment $P+g$. The correct expression is $P+g(k-1)=(P-g)+gk$. Test the expression at $k=1$ and $k=n$ before calculating.

Key takeaway: The factor $(Ia)_{\overline{n}|}$ discounts the specific year-end payments $1,2,\ldots,n$. For $i\ne0$, it equals $(a_{\overline{n}|}-nv^{n+1})/d$; at zero interest, it equals $n(n+1)/2$. Match both the payment indexing and the timing before using it.

Check Your Understanding

For each problem, identify the payment timing before selecting a factor. Answers are rounded only as stated.

A unit gradient pays 1 and then 2 at the ends of years 1 and 2. At an effective annual rate of 5%, find its present-value factor to four decimal places. (Answer: 2.7664)
A unit gradient pays 1, 2, 3, 4, and 5 at successive year-ends. Find its present-value factor if the effective annual interest rate is 0%. (Answer: 15)
A hypothetical laboratory account makes year-end contributions of $300, $600, and $900. If the effective annual interest rate is 20%, find the present value to the nearest cent. (Answer: $1,187.50)
A hypothetical service reserve receives four year-end payments of $250, $350, $450, and $550. At an effective annual interest rate of 10%, express the payments as a level component and a gradient component, then find the present value to the nearest cent. (Answer: $1,230.28)

What You'll Learn