Deriving the Arithmetic-Gradient Annuity Value

Why Derive the Arithmetic-Gradient Value?

A payment stream can increase by a fixed dollar amount each period rather than by a fixed percentage. For example, a hypothetical maintenance contract might pay $100 at the end of the first year, $140 at the end of the second, and $180 at the end of the third. Its present value depends on both the level of the first payment and the repeated dollar increase. Deriving the value from the discounted payments makes the timing convention explicit and provides a useful check on any formula-based calculation.

Throughout this tutorial, payments are made at the ends of periods, the effective interest rate per period is constant, and there are finitely many payments. The rate $i$ satisfies $i>-1$, and $v=1/(1+i)$. Unless stated otherwise, the examples use a positive rate. Mortality or survival contingencies are not involved: each specified payment is assumed to be made in full.

Definition: A finite arithmetic-gradient annuity with initial payment $A$, per-period increase $G$, and $n$ payments pays $A+(k-1)G$ at the end of period $k$, for $k=1,2,\ldots,n$. Its present value at time zero is the sum of those payments discounted to time zero. The unit arithmetic-gradient annuity, denoted $(Ia)_{\overline{n}|}$, has payments $1,2,\ldots,n$ at times $1,2,\ldots,n$, respectively.

The increase $G$ may be positive, zero, or negative. When $G=0$, the stream is level. When $G<0$, it decreases arithmetically; the payment amounts must still be interpreted as specified, and a negative payment represents an outflow in the chosen cash-flow perspective. The standard finite-annuity formula does not by itself impose a nonnegativity condition on payments.

Deriving the Unit Arithmetic-Gradient Value

Discount each payment of the unit gradient separately. The first payment is 1 at time 1, the second is 2 at time 2, and the $n$th is $n$ at time $n$. Thus the present value is a finite weighted geometric sum.

(Ia)_{\overline{n}|}=\sum_{k=1}^{n}k v^k

To obtain a closed form, multiply the sum by $1-v$. Subtracting the same sum shifted by one power of $v$ cancels the intermediate coefficients, leaving the first $n$ powers and the final boundary term:

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

Since $1-v=d$, where $d=i/(1+i)$, division gives the finite unit-gradient formula. The expression in terms of $i$ follows from the equivalent identity $a_{\overline{n}|}=(1-v^n)/i$.

Theorem: For $i\ne 0$, the present value of the unit arithmetic-gradient annuity is

(Ia)_{\overline{n}|} =\frac{a_{\overline{n}|}-n v^{n+1}}{d} =\frac{\ddot{a}_{\overline{n}|}-n v^n}{i}.

The equivalent finite-sum form is often the most reliable way to check a calculation. For a payment stream $A,A+G,\ldots,A+(n-1)G$, separate each payment into a level component $A$ and an increasing component $0,G,2G,\ldots,(n-1)G$. The latter is the unit gradient shifted down by one unit payment, multiplied by $G$. Therefore:

\begin{aligned} PV_0 &=A a_{\overline{n}|} +G\left((Ia)_{\overline{n}|}-a_{\overline{n}|}\right)\\ &=(A-G)a_{\overline{n}|} +G(Ia)_{\overline{n}|}. \end{aligned}

This formula assumes that the first payment $A$ occurs at time 1. The subtraction of $a_{\overline{n}|}$ matters: $G(Ia)_{\overline{n}|}$ alone represents payments $G,2G,\ldots,nG$, not the increases $0,G,\ldots,(n-1)G$ above the initial payment.

Worked Example: Valuing a Level Amount Plus a Gradient

A hypothetical service agreement pays $100 at the end of year 1, with each later annual payment $40 higher than the preceding payment. There are three payments, and the effective annual interest rate is 25%. Find the present value. Use the general arithmetic-gradient formula, then check by direct discounting.

Method and timing. Here $A=100$, $G=40$, $n=3$, and $i=0.25$. The payments are $100, $140, and $180 at times 1, 2, and 3. The discount factor is $v=1/1.25=0.8$.

First calculate the level-annuity factor:

a_{\overline{3}|}=v+v^2+v^3=0.8+0.64+0.512=1.952.

Next calculate the unit-gradient factor from its defining sum:

(Ia)_{\overline{3}|} =v+2v^2+3v^3 =0.8+1.28+1.536 =3.616.

The factor for the increases above the initial level is $3.616-1.952=1.664$. Hence:

PV_0=100(1.952)+40(1.664) =195.20+66.56 =\$261.76.

Check by direct discounting. The present value of the three stated payments is $100(0.8)+140(0.64)+180(0.512)=80+89.60+92.16=\$261.76$. The two methods agree. A common mistake would be to use $40(Ia)_{\overline{3}|}$ as the gradient contribution; that values increases of $40, $80, and $120, incorrectly including a $40 increase in the first payment. The correct gradient contribution is $40((Ia)_{\overline{3}|}-a_{\overline{3}|})$.

Valuing a Deferred Arithmetic-Gradient Stream

A deferred stream is valued by first locating its value at the date just before its first payment, then discounting that value to the required focal date. If the first payment is at time $m+1$, the stream is an $m$-period deferred annuity-immediate: its value at time $m$ is the value of an ordinary $n$-payment annuity-immediate. The whole value is then discounted by $v^m$. This separation keeps the annuity’s payment timing distinct from the deferral period.

Result: If payments $A,A+G,\ldots,A+(n-1)G$ occur at times $m+1,m+2,\ldots,m+n$, with the same effective rate $i$ in every period, then

PV_0 =v^m\left[ A a_{\overline{n}|} +G\left((Ia)_{\overline{n}|}-a_{\overline{n}|}\right) \right].

Worked Example: Discounting a Deferred Gradient

A hypothetical equipment fund will receive three annual payments of $100, $120, and $140 at the ends of years 3, 4, and 5. The effective annual interest rate is 25%. Find the value at time zero, rounding to the nearest cent, and verify it by discounting each payment directly.

Method and timing. The first payment is at time 3, so the stream is deferred for $m=2$ periods. At time 2, the three payments are one, two, and three periods away. Here $A=100$, $G=20$, $n=3$, and $v=0.8$.

Using the factors already obtained from their defining sums, the value at time 2 is:

\begin{aligned} PV_2 &=100a_{\overline{3}|} +20\left((Ia)_{\overline{3}|}-a_{\overline{3}|}\right)\\ &=100(1.952)+20(3.616-1.952)\\ &=195.20+33.28\\ &=\$228.48. \end{aligned}

Discount this two-period value to time zero:

PV_0=v^2PV_2=(0.8)^2(228.48)=0.64(228.48)=\$146.2272.

Rounded only at the final step, the present value is $146.23. For the direct check, discount the payments from their actual dates: $100(0.8)^3+120(0.8)^4+140(0.8)^5=51.2+49.152+45.8752=146.2272$, or $146.23 to the nearest cent. Discounting by $v^3$ after first calculating $PV_2$ would defer the stream one extra period; the focal date $2$ is immediately before the first payment at time $3$.

Limiting Cases and Consistency Checks

The formulas also provide useful checks on boundary cases. These checks are not substitutes for confirming payment dates, but they can expose a misplaced gradient or an incorrect deferral factor.

Zero gradient.
If $G=0$, every payment is $A$. The general formula reduces to $PV_0=Aa_{\overline{n}|}$, as it must.

One payment.
For $n=1$, the gradient has no opportunity to affect a later payment. The formula gives $A a_{\overline{1}|}+G((Ia)_{\overline{1}|}-a_{\overline{1}|})=Av$, because both factors in the difference equal $v$.

Zero interest.
At $i=0$, $v=1$, so the discounted-payment sum is simply the sum of the payments. The quotient formulas involving division by $i$ or $d$ are not used at this rate.

Deferred timing.
If the first payment is at time $m+1$, the value at time zero must include $v^m$ times the value one period before that first payment. Directly discounting each payment is an effective timing check.

At zero interest, a unit gradient has present value $1+2+\cdots+n=n(n+1)/2$. A stream beginning at $A$ and increasing by $G$ therefore has value $nA+G\,n(n-1)/2$. This is also what the general decomposition gives when each payment is counted without discounting. For a deferred stream, multiply that undiscounted sum by $v^m=1$, so deferral has no effect at zero interest.

Worked Example: Recovering the Initial Payment

A three-payment annuity-immediate has payments $A$, $A+40$, and $A+80$. At an effective interest rate of 25% per period, its present value is $261.76. Find $A$, and check that the inferred amount reproduces the stated value.

Method. Here $G=40$, $n=3$, and $v=0.8$. From the factor calculations, $a_{\overline{3}|=1.952$ and $(Ia)_{\overline{3}|}=3.616$. Thus the present value equation is:

$$ 261.76=A(1.952)+40(3.616-1.952). $$

The gradient component is $40(1.664)=66.56$. Subtract it from the given value: $261.76-66.56=195.20$. Divide by the level-annuity factor:

A=\frac{195.20}{1.952}=100.

The inferred payments are $100, $140, and $180. Their present value is $80+89.60+92.16=\$261.76$, as required. The method first isolates the non-level component, then solves the remaining level-annuity equation; treating the entire stream as level would give the wrong initial payment.

Common Exam Traps

Starting the gradient too early. A stream $A,A+G,\ldots$ has no increase added to its first payment. Use $G((Ia)_{\overline{n}|}-a_{\overline{n}|})$, not $G(Ia)_{\overline{n}|}$, for the increase portion.
Using the wrong unit-gradient convention. Here $(Ia)_{\overline{n}|}$ means payments $1,2,\ldots,n$ at times $1,2,\ldots,n$. State the payment sequence before using the symbol.
Miscounting deferral periods. If the first payment is at time $m+1$, discount the value at time $m$ by $v^m$, not by $v^{m+1}$.
Applying a positive-rate quotient at zero interest. The formulas with denominators $i$ or $d$ require a nonzero rate. At $i=0$, sum the payments directly.
Rounding too soon. Retain full values of discount factors and intermediate present values, then round the final monetary result to cents when dollars are requested.

A disciplined solution identifies the first payment date, writes the payment pattern, and selects the focal date before applying a factor. The derivation by discounted summation makes the unit-gradient convention transparent; the decomposition into a level stream and an incremental stream then handles the general arithmetic pattern. Direct discounting remains a compact and dependable check, especially when a deferral or unusual payment date is present. These valuation techniques are aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Check Your Understanding

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

A unit arithmetic-gradient annuity has four payments, and the effective rate per period is 20%. Find its present value, rounded to four decimal places. (Answer: 5.8873)
A six-payment annuity-immediate pays $250 at the end of the first period and increases by $30 each period. The effective rate is zero. Find its present value. (Answer: $1,950.00)
A two-payment stream pays $80 and $100 at the ends of years 5 and 6, respectively. The effective annual rate is 25%. Find its present value at time zero, rounded to cents. (Answer: $52.43)
A three-payment annuity-immediate has payments $120, $150, and $180. The effective rate per period is 25%. Find its present value. (Answer: $284.16)
A two-payment annuity-immediate has payments $100 and $160. The effective rate per period is 10%. Find its present value, rounded to cents. (Answer: $223.14)

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