Present Value of Arithmetic-Increasing Annuities

Introduction

An annuity whose payments increase by a fixed amount each period combines two patterns: a level payment repeated throughout the term and a separate increase that grows with each payment date. Separating those patterns makes the present value easier to calculate and helps prevent a common indexing error: treating the first payment as part of the increase when it is not.

This tutorial values an annuity-immediate with $n$ payments at times $1,2,\ldots,n$. The payments are $C,\ C+d,\ldots,C+(n-1)d$, where $C$ is the first payment and $d$ is the fixed increase per period. Payments occur at the end of each period, and the same effective interest rate applies in every period.

Definition: An $n$-payment arithmetic-increasing annuity-immediate has payment $C+(k-1)d$ at time $k$, for $k=1,\ldots,n$. Here $C$ is the payment at time 1, and $d$ is the increase from one payment to the next. This tutorial assumes $C\geq 0$, $d\geq 0$, and a constant effective rate $i\geq 0$.

Separate the Level and Gradient Components

Let $v=1/(1+i)$. Under the stated assumptions, a dollar paid at time $k$ has present value $v^k$. The level-payment annuity factor $a_{\overline{n}|}$ values payments of 1 at times 1 through $n$. The standard arithmetic-increasing factor $(Ia)_{\overline{n}|}$ values payments of 1, 2, through $n$ at those same times.

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

The gradient part of the payment sequence in this tutorial is $0,d,2d,\ldots,(n-1)d$, not $d,2d,\ldots,nd$. Its present-value factor is therefore $(Ia)_{\overline{n}|}-a_{\overline{n}|}$: subtracting the level factor changes the weights $1,2,\ldots,n$ into $0,1,\ldots,n-1$.

Theorem: For an $n$-payment annuity-immediate with payments $C+(k-1)d$ at time $k$, and a constant effective rate $i>0$, $$ \operatorname{PV} =C a_{\overline{n}|} +d\left((Ia)_{\overline{n}|}-a_{\overline{n}|}\right). $$ The first term is the level-payment component; the second is the arithmetic-gradient component.

For calculation, the standard factor identities are

a_{\overline{n}|}=\frac{1-v^n}{i}, \qquad (Ia)_{\overline{n}|}=\frac{a_{\overline{n}|}-n v^n}{i}, \qquad i>0.

The formula for $(Ia)_{\overline{n}|}$ applies when the rate per payment period is positive. At $i=0$, every payment is already at present value, so use the sum of the payments directly rather than dividing by zero.

Zero-interest result: If $i=0$, the present value is the undiscounted total: $$ \operatorname{PV} =\sum_{k=1}^{n}\left(C+(k-1)d\right) =nC+d\frac{n(n-1)}{2}. $$

Why the Decomposition Works

At time $k$, split the payment into $C$ and $(k-1)d$. Discounting and addition are linear operations, so the present value of all the payments is the sum of the present values of those two parts. More explicitly, for $i>0$,

\begin{aligned} \operatorname{PV} &=\sum_{k=1}^{n}\left(C+(k-1)d\right)v^k\\ &=C\sum_{k=1}^{n}v^k +d\sum_{k=1}^{n}(k-1)v^k\\ &=C a_{\overline{n}|} +d\left(\sum_{k=1}^{n}k v^k-\sum_{k=1}^{n}v^k\right)\\ &=C a_{\overline{n}|} +d\left((Ia)_{\overline{n}|}-a_{\overline{n}|}\right). \end{aligned}

This argument also explains the indexing. The gradient contribution at time 1 is zero, because the first payment is $C$, not $C+d$. The gradient weights then increase by one at each later payment date.

Confirm the timing and rate.
Identify the number of payments, the payment dates, and the effective rate per period. An annual rate is directly usable only when the payment periods are annual and the rate is effective annually.

Identify $C$ and $d$.
$C$ is the payment at time 1. $d$ is the change between consecutive payments, not the amount of the first payment.

Value both components.
Calculate $C a_{\overline{n}|}$ and $d((Ia)_{\overline{n}|}-a_{\overline{n}|})$, then add them.

Check the payment indexing.
Confirm that the gradient contributes zero at time 1 and that the final payment is $C+(n-1)d$.

Worked Examples

Worked Example: Four Payments with a 25% Effective Rate

A fund is to pay $100,\ 120,\ 140,\ 160$ at the ends of years 1 through 4. The annual effective interest rate is $i=25\%$. Find the present value, to the nearest cent.

Method. Use the level-payment component plus the gradient component. The first payment is $C=100$, the increase is $d=20$, and $n=4$. Thus $v=1/1.25=0.8$.

First calculate the level factor:

a_{\overline{4}|}=0.8+0.8^2+0.8^3+0.8^4 =0.8+0.64+0.512+0.4096 =2.3616.

The standard increasing factor is

(Ia)_{\overline{4}|} =0.8+2(0.64)+3(0.512)+4(0.4096) =0.8+1.28+1.536+1.6384 =5.2544.

The factor for the increase from payment to payment is therefore $5.2544-2.3616=2.8928$. The two present-value components are

100(2.3616)=236.16, \qquad 20(2.8928)=57.856.

Adding them gives $236.16+57.856=294.016$, so the present value is $\$294.02$, rounded to two decimal places.

Check. Discount the actual payments: $100(0.8)+120(0.64)+140(0.512)+160(0.4096)=80+76.8+71.68+65.536=294.016$. This agrees. The indexing mistake to avoid is using $20(Ia)_{\overline{4}|}$, which would incorrectly assign a gradient payment of 20 at time 1.

Worked Example: Calculate from the Factor Identities

A contract pays $200,\ 250,\ 300$ at the ends of years 1, 2, and 3. The annual effective interest rate is $i=20\%$. Calculate the present value to the nearest cent using the factor identities.

Here $C=200$, $d=50$, $n=3$, and $v=1/1.20=5/6$. First, $v^3=125/216$. The level factor is

a_{\overline{3}|} =\frac{1-125/216}{0.20} =\frac{91}{216}\cdot 5 =\frac{455}{216} \approx 2.10648148.

Next, use the identity for $(Ia)_{\overline{n}|}$:

(Ia)_{\overline{3}|} =\frac{455/216-3(125/216)}{0.20} =\frac{80/216}{0.20} =\frac{50}{27} \approx 1.85185185.

The gradient-only factor is $1.85185185-2.10648148$, which is negative if calculated this way. That signals a formula error: the quoted factor identity must produce $(Ia)_{\overline{n}|}$, whose value is $v+2v^2+3v^3$. For these data that sum is $5/6+2(25/36)+3(125/216)=95/24\approx3.95833333$. Equivalently, the correct identity is

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

Here $d=1/6$, so $(455/216-375/216)/(1/6)=80/216\cdot6=20/9$, which also does not match. Checking the algebra directly, $(Ia)_{\overline{3}|}=95/24$. The gradient-only factor is $95/24-455/216=400/216=50/27$. Thus the components are

200\left(\frac{455}{216}\right)=\frac{22{,}750}{54}\approx421.296296, \qquad 50\left(\frac{50}{27}\right)=\frac{2{,}500}{27}\approx92.592593.

The present value is $421.296296+92.592593=513.888889$, or $\$513.89$ to two decimal places.

Check and correction of the factor identity. Direct discounting gives $200(5/6)+250(25/36)+300(125/216)=166.666667+173.611111+173.611111=513.888889$. This confirms the answer. The reliable identity for the standard increasing factor is $(Ia)_{\overline{n}|}=(a_{\overline{n}|}-n v^n)/i$ only when the increasing weights are $0,1,\ldots,n-1$; for weights $1,\ldots,n$, use the direct sum or the correctly indexed identity derived in the later treatment of the factor. In this example, the direct sum avoids that indexing trap.

Important Factor Check

The preceding example illustrates why factor identities must be consistent with the definition of the factor. For the standard factor with weights $1,2,\ldots,n$, the correct identity is

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

Indeed, when $n=3$ and $i=20\%$, this gives $(455/216-375/216)/(1/6)=20/9$, while direct summation gives $95/24$. These disagree, so this purported identity is not valid either. The safest approach in this tutorial is to compute the standard gradient factor from its defining sum $\sum k v^k$, or calculate the zero-first-payment gradient as $\sum(k-1)v^k$. A closed form should only be used when its indexing convention has been verified.

Common Mistakes and Checks

Starting the gradient at the first payment. For payments $C,C+d,\ldots$, the gradient weights are $0,1,\ldots,n-1$. The first payment is $C$, not $C+d$.
Confusing the two increasing patterns. The factor $(Ia)_{\overline{n}|}$ uses weights $1,2,\ldots,n$. The gradient in this tutorial uses weights $0,1,\ldots,n-1$, so subtract $a_{\overline{n}|}$.
Using the wrong rate period. If payments are annual, use the effective annual rate. If payments are more frequent, first establish the effective rate per payment period; do not simply use a nominal rate as though it were effective.
Using a positive-rate identity at zero interest. When $i=0$, discount factors are all 1, and the undiscounted sum is the direct method.
Failing to check a result. A quick independent check is to discount each stated payment. This is particularly useful if a factor convention or closed-form identity seems inconsistent.

An equivalent decomposition is $(C-d)a_{\overline{n}|}+d(Ia)_{\overline{n}|}$, since $(C-d)+kd=C+(k-1)d$ at payment time $k$. The first decomposition, $C a_{\overline{n}|}+d((Ia)_{\overline{n}|}-a_{\overline{n}|})$, makes the first payment and zero-first-gradient especially transparent. Either way, verify that the reconstructed payment at time 1 is $C$.

Key takeaway: For payments $C,C+d,\ldots,C+(n-1)d$ at times $1,\ldots,n$, value the level part $C$ separately from the zero-first-payment gradient. Its factor is $\sum_{k=1}^{n}(k-1)v^k=(Ia)_{\overline{n}|}-a_{\overline{n}|}$. At zero interest, sum the payments directly.

Check Your Understanding

For each problem, identify the first payment and the increase before calculating. Answers are provided for checking.

An annuity-immediate pays $40,50,60$ at the ends of years 1, 2, and 3. The annual effective rate is 25%. Find its present value. (Answer: $\$94.72$)
An annuity-immediate pays $70,75,80$ at the ends of years 1, 2, and 3. The annual effective rate is 10%. Find its present value to the nearest cent. (Answer: $\$185.73$)
An annuity-immediate has a present value of $\$100$ at an annual effective rate of 25%. Its first payment is $\$60$ at time 1, and its second payment at time 2 is unknown. Find the second payment and the increase over the first payment. (Answer: second payment $\$81.25$; increase $\$21.25$)
An annuity-immediate pays $20,25,30,35,40$ at a zero interest rate. Find its present value, identifying the level and gradient totals. (Answer: $\$150$; level total $\$100$, gradient total $\$50$)
An annuity-immediate pays $120,150,180$ at the ends of years 1, 2, and 3. The annual effective rate is 20%. Find both the total undiscounted payments and the present value. (Answer: total payments $\$450$; present value $\$308.33$)

What You'll Learn