Decomposing Level Payments and Arithmetic Gradients

Why Decompose a Payment Stream?

A sequence of payments may not be level, but it can still be built from simpler streams. If each payment increases by the same dollar amount, the cash flows consist of a level part and a fixed-step part. Valuing those components separately is often quicker and less error-prone than discounting every payment individually.

The key is to match the payment pattern to the definition of the arithmetic-gradient factor. The standard gradient has payments $1,2,\ldots,n$ at the ends of years $1,2,\ldots,n$. A stream whose first payment is positive and then rises by a fixed amount does not itself start with that pattern; its level component must be adjusted, or the gradient must be described as starting at zero.

Definition: An arithmetic-gradient annuity-immediate of term $n$ has payments $1,2,\ldots,n$ at the ends of years $1,2,\ldots,n$. Its present-value factor is $(Ia)_{\overline{n}|}$. A level annuity-immediate has payments $1$ at those same dates, with factor $a_{\overline{n}|}$. Both factors use the same effective annual interest rate $i$.

Decomposing a Fixed-Step Stream

Assume payments are made at the ends of consecutive years, there are $n$ payments, and interest is earned at a constant effective annual rate $i>-1$. There is no mortality or contingency affecting payment. Let $C_1$ be the first payment, and let $G$ be the fixed dollar change from one payment to the next. Thus, $G$ is positive for an increasing stream and negative for a decreasing stream.

Result: The payment at the end of year $k$ is $C_1+(k-1)G$. Its present value at time 0 is $$ \begin{aligned} PV &=(C_1-G)a_{\overline{n}|}+G(Ia)_{\overline{n}|}\\ &=C_1a_{\overline{n}|}+G\big((Ia)_{\overline{n}|}-a_{\overline{n}|}\big). \end{aligned} $$

The first form uses the standard gradient, whose first payment is $1$; the level component is therefore $C_1-G$. The second form separates out $C_1$ level payments and adds a zero-start gradient with amounts $0,G,2G,\ldots,(n-1)G$. These are two algebraically equivalent descriptions of the same cash flows.

Why the Decomposition Works

For any fixed rate, the present value of a collection of cash flows is the sum of each cash flow multiplied by its discount factor. Therefore, adding two streams payment by payment also adds their present values. To verify the decomposition, write the payment in year $k$ as the sum of a level amount and a gradient amount:

$$ (C_1-G)+Gk=C_1+(k-1)G $$

The left side is the year-$k$ payment from a level annuity of amount $C_1-G$, plus $G$ times the standard gradient payment $k$. The right side is the specified payment in the original stream. Since the amounts agree at every payment date, their present values agree as well. In particular, discounting each year’s payment by $v^k$, where $v=1/(1+i)$, gives:

\begin{aligned} PV &=\sum_{k=1}^{n}\big(C_1+(k-1)G\big)v^k\\ &=C_1\sum_{k=1}^{n}v^k +G\sum_{k=1}^{n}(k-1)v^k\\ &=C_1a_{\overline{n}|} +G\big((Ia)_{\overline{n}|}-a_{\overline{n}|}\big). \end{aligned}

The last equality follows because $(Ia)_{\overline{n}|}=\sum_{k=1}^{n}kv^k$ and $a_{\overline{n}|}=\sum_{k=1}^{n}v^k$. If $i=0$, these summations still define the factors: $a_{\overline{n}|}=n$ and $(Ia)_{\overline{n}|}=n(n+1)/2$. The usual closed forms involving division by $i$ should not be used directly at $i=0$.

Worked Examples

Worked Example: Valuing an Increasing Payment Stream

A fund pays \$100 at the end of year 1, with each later annual payment \$30 greater than the previous one. There are four payments, and the effective annual interest rate is 10%. Find the present value at time 0, to the nearest cent.

Method. The first payment is $C_1=100$, the fixed step is $G=30$, and $n=4$. The payment schedule is \$100, \$130, \$160, and \$190. Using the standard gradient, the level component is $C_1-G=100-30=70$. Thus the stream is a level annuity of \$70 plus \$30 times the standard gradient.

At $i=0.10$, the discount factors for years 1 through 4 are $0.9090909$, $0.8264463$, $0.7513148$, and $0.6830135$. Therefore, computed from the unrounded discount factors, $a_{\overline{4}|}=3.169865446$ and $(Ia)_{\overline{4}|}=7.547981693$.

The present value is $70(3.169865446)+30(7.547981693)=221.890581+226.439451=448.330032$. Rounded to the nearest cent, the present value is \$448.33.

Check. Discounting the original payments gives $100(0.9090909)+130(0.8264463)+160(0.7513148)+190(0.6830135)$, which is approximately \$448.33. The decomposition matches all four payments and the direct valuation. The common mistake to avoid is using a standard gradient with a level component of \$100: that would make the first payment \$130, not \$100.

Worked Example: Decomposing a Decreasing Stream

A hypothetical maintenance contract pays \$240 at the end of year 1, followed by payments that decrease by \$40 each year for three more years. The effective annual interest rate is 5%. Find the present value at time 0, to the nearest cent.

Method. Here $C_1=240$, $G=-40$, and $n=4$. The payments are \$240, \$200, \$160, and \$120. The standard-gradient decomposition has level amount $C_1-G=240-(-40)=280$, plus $-40$ times the standard gradient. The negative multiplier is necessary because the stream decreases.

At $i=0.05$, the discount factors for years 1 through 4 are $0.952380952$, $0.907029478$, $0.863837599$, and $0.822702475$. Their sums give $a_{\overline{4}|}=3.545950504$ and $(Ia)_{\overline{4}|}=8.648762607$, using unrounded factors.

The present value is $280(3.545950504)-40(8.648762607)=992.866141-345.950504=646.915637$. Rounded to the nearest cent, the present value is \$646.92.

Check. Direct discounting gives $240(0.952380952)+200(0.907029478)+160(0.863837599)+120(0.822702475)$, also approximately \$646.92. The common mistake to avoid is treating the decrease as a positive step, which would value an increasing rather than decreasing gradient.

Worked Example: Finding an Unknown Step from Present Value

A three-payment stream pays \$200 at the end of year 1. Each subsequent payment is $G$ dollars greater than the previous one. At an effective annual interest rate of 10%, the present value is reported as \$613.82, rounded to the nearest cent. Find $G$ to the nearest cent.

Method. The payments are $200,\ 200+G,\ 200+2G$. This is \$200 level for three years plus a zero-start gradient of $0,G,2G$ at the payment dates. The zero-start gradient factor is $(Ia)_{\overline{3}|}-a_{\overline{3}|}$, because subtracting the level pattern $1,1,1$ from the standard gradient $1,2,3$ gives $0,1,2$.

At $i=0.10$, computed from unrounded discount factors, $a_{\overline{3}|}=2.486851991$ and $(Ia)_{\overline{3}|}=4.815927871$. The zero-start factor is therefore $4.815927871-2.486851991=2.329075880$. The present-value equation is $613.82=200(2.486851991)+G(2.329075880)$.

The level component is $200(2.486851991)=497.370398$. Hence $G=(613.82-497.370398)/2.329075880=49.9982$ approximately. To the nearest cent, $G=$\$50.00.

Check. With $G=\$50$, the payments are \$200, \$250, and \$300. Their present value is $200(0.909090909)+250(0.826446281)+300(0.751314801)=613.824192$, which rounds to \$613.82. The common mistake to avoid is applying the standard gradient factor without subtracting the level factor when the step pattern begins at zero.

Common Mistakes and Practical Checks

A decomposition is reliable only when both the payment amounts and their dates agree with the original stream. Before calculating a present value, write down the payment sequence and check it against the components. The following distinctions are particularly useful:

Payment pattern	Decomposition	Gradient factor used
$C_1,\ C_1+G,\ldots,C_1+(n-1)G$	$C_1$ level plus a zero-start gradient	$(Ia)_{\overline{n}\|}-a_{\overline{n}\|}$
$C_1,\ C_1+G,\ldots,C_1+(n-1)G$	$C_1-G$ level plus $G$ times the standard gradient	$(Ia)_{\overline{n}\|}$
$C_1,\ C_1-G,\ldots,C_1-(n-1)G$, with $G>0$	$C_1+G$ level minus $G$ times the standard gradient	$(Ia)_{\overline{n}\|}$

The two decompositions in the first two rows are equivalent. In the zero-start form, the gradient has amounts $0,1,\ldots,n-1$ at the ends of years $1,\ldots,n$; its first payment contributes zero, but the first payment date is still year 1. Do not shift the dates when changing the gradient indexing.

This method assumes that the step is fixed throughout the payment stream and that payments occur at the stated regular dates under one constant effective rate. If the step changes partway through, the single-gradient representation no longer describes the entire stream. Likewise, if payments are at the beginnings of years or the interest rate changes, the timing or discount factors must be adjusted rather than applying these immediate-annuity factors unchanged.

Key takeaway: For payments $C_1+(k-1)G$ at the ends of years $1$ through $n$, either use $C_1$ level plus a zero-start gradient, or use $C_1-G$ level plus $G$ times the standard gradient. Check that the component payments reproduce the original schedule before valuing it.

Check Your Understanding

For each problem, identify the level and gradient components before calculating or solving.

An annuity pays \$80, \$100, and \$120 at the ends of years 1, 2, and 3. The effective annual interest rate is 5%. Find its present value at time 0 to the nearest cent. (Answer: \$270.55)
Payments of \$300, \$260, \$220, and \$180 are made at the ends of years 1 through 4. The effective annual interest rate is 10%. Find the present value at time 0 to the nearest cent. (Answer: \$775.83)
A four-payment stream has first payment \$75, with each later payment \$15 greater than the preceding one. The effective annual interest rate is 0%. Find the present value at time 0. (Answer: \$390.00)
A two-payment stream pays \$100 at the end of year 1 and \$100 plus an unknown step $G$ at the end of year 2. At an effective annual interest rate of 10%, its present value is \$190.00. Find $G$ to the nearest cent. (Answer: \$19.90)

Payment pattern	Decomposition	Gradient factor used
\(C_1,\ C_1+G,\ldots,C_1+(n-1)G\)	\(C_1\) level plus a zero-start gradient	\((Ia)_{\overline{n}\|}-a_{\overline{n}\|}\)
\(C_1,\ C_1+G,\ldots,C_1+(n-1)G\)	\(C_1-G\) level plus \(G\) times the standard gradient	\((Ia)_{\overline{n}\|}\)
\(C_1,\ C_1-G,\ldots,C_1-(n-1)G\), with \(G>0\)	\(C_1+G\) level minus \(G\) times the standard gradient	\((Ia)_{\overline{n}\|}\)

What You'll Learn