Piecewise Arithmetic Payment Gradients

Why a Piecewise Payment Stream Needs Segments

An arithmetic payment stream has a constant dollar change from one payment to the next. In a piecewise arithmetic payment stream, that change is not the same throughout the term: it changes at a specified payment date. The valuation method is to divide the stream where the increment changes, value each segment on a suitable date, and move those values to one common date.

The key is to get the payment amount at the switch right. If the new increment applies to the change after the last payment of the first segment, the first payment of the next segment equals that last payment plus the new increment. It does not restart at the original base amount.

Definition: Suppose a segment has $m$ payments at the ends of periods $s+1,\ldots,s+m$. If its first payment is $B$ and its fixed increment is $G$, the payment at time $s+j+1$ is $B+jG$, for $j=0,\ldots,m-1$. Here $s$ is the number of periods before the segment’s first payment. The increment $G$ may be positive, zero, or negative.

Throughout the worked examples, payments occur at the ends of annual periods, the stated annual rate is an effective annual interest rate, and the rate is constant over the payment term. These assumptions allow discounting with $v=1/(1+i)$. If rates change by period or payments occur on irregular dates, the factors below cannot be applied unchanged.

Valuing One Segment at a Time

The standard immediate-annuity factors are $a_{\overline{m}|}=\sum_{k=1}^{m}v^k$ and $(Ia)_{\overline{m}|}=\sum_{k=1}^{m}kv^k$. The first is the present value, one period before the first payment, of $m$ level payments of 1. The second is the present value at that same date of payments $1,2,\ldots,m$.

Result: The value at time $s$ of a segment with $m$ end-of-period payments $B,B+G,\ldots,B+(m-1)G$ is $$ B a_{\overline{m}|}+G\big((Ia)_{\overline{m}|}-a_{\overline{m}|}\big). $$ Its value at time 0 is $v^s$ times this amount. To value a complete piecewise stream, add the time-0 values of all its segments.

The formula uses a gradient that starts at zero: the part of each payment due to $G$ is $0,G,2G,\ldots,(m-1)G$. The factor $(Ia)_{\overline{m}|}$, by contrast, corresponds to $G,2G,\ldots,mG$. Subtracting one level payment of $G$ from each period converts that pattern into the required zero-start gradient.

\begin{aligned} \text{Value at time }s &=\sum_{j=0}^{m-1}(B+jG)v^{j+1}\\ &=B\sum_{k=1}^{m}v^k +G\sum_{k=1}^{m}(k-1)v^k\\ &=B a_{\overline{m}|} +G\big((Ia)_{\overline{m}|}-a_{\overline{m}|}\big). \end{aligned}

This also justifies why discounting the segment to time $s$ and then moving it to time 0 works: time $s$ is exactly one period before the segment’s first payment. Discounting its value by $v^s$ accounts for the $s$ periods from time 0 to that date. More generally, if the desired common date is $T$, accumulate or discount each segment value from its own valuation date to $T$, using the same constant rate.

Find the Payment at the Change Date

First determine what the contract or problem statement says about the new increment. If the first $n$ payments begin at $B_1$ and increase by $G_1$, the $n$th payment is $B_1+(n-1)G_1$. If the increment changes to $G_2$ for the following payment changes, then the next payment is

$$ B_1+(n-1)G_1+G_2. $$

The new segment’s first payment is the amount just calculated, not necessarily a new base payment specified at the start of the whole stream. If the problem explicitly gives a separate first payment for the new segment, use that amount instead. A segment boundary divides the cash flows for valuation; it does not itself create an extra payment.

Worked Example: Value a Stream When the Increment Falls

A hypothetical contract pays at the end of each of years 1 through 6. Payments begin at $100 and increase by $20 through year 3. Starting with the next payment, the increment is $10 per year. Find the present value at an effective annual interest rate of 25%.

Method and switch payment. The first three payments are $100, $120, and $140. The new increment applies between years 3 and 4, so the year-4 payment is $140+$10=$150. The remaining payments are $160 and $170.

First segment, valued at time 0. It has $B=100$, $G=20$, and $m=3$. At 25%, $v=0.8$, so $a_{\overline{3}|}=0.8+0.64+0.512=1.952$ and $(Ia)_{\overline{3}|}=0.8+2(0.64)+3(0.512)=3.616$. Thus the first segment is worth $100(1.952)+20(3.616-1.952)=195.2+33.28=\$228.48$.

Second segment, first valued at time 3. Its payments are $150, $160, and $170, so $B=150$, $G=10$, and $m=3$. Its value one period before its first payment is $150(1.952)+10(3.616-1.952)=292.8+16.64=\$309.44$. Discount that amount three years to time 0: $309.44(0.8)^3=309.44(0.512)=\$158.43328$.

Combine and check. The present value is $228.48+158.43328=\$386.91328$, or $386.91 to the nearest cent. Direct discounting gives $100(0.8)+120(0.64)+140(0.512)+150(0.4096)+160(0.32768)+170(0.262144)=\$386.91328$, confirming the segment calculation. The method avoids treating the year-4 payment as a restart at $100 or counting a payment twice at the segment boundary.

Solving for an Unknown Base Payment

When the initial payment is unknown, each segment can be written as a part proportional to that unknown plus a known gradient adjustment. The coefficient of the unknown is the present value of a unit payment at every payment date where that unknown appears. It is often quickest to use the annuity factor for this coefficient, then separately add the known increments.

Worked Example: Solve for a Common Base Payment

A six-payment stream has an unknown base payment $A$. At an effective annual rate of 25%, its payments are $A,A+15,A+30$ in years 1 through 3. From year 4 onward, the increment is $5 per year, so the remaining payments are $A+35,A+40,A+45$. The present value is $654.48448. Find $A$.

Method. Separate the common $A$ in all six payments from the fixed additions. The coefficient of $A$ is the sum of the six discount factors: $0.8+0.64+0.512+0.4096+0.32768+0.262144=2.951424$.

The present value of the additions is $15(0.64)+30(0.512)+35(0.4096)+40(0.32768)+45(0.262144)$. Term by term, this is $9.6+15.36+14.336+13.1072+11.79648=\$64.19968$. Therefore $2.951424A+64.19968=654.48448$, so $2.951424A=590.28480$, and $A=590.28480/2.951424=200$.

Check. The resulting payments are $200, $215, $230, $235, $240, and $245. Their present value is $160+137.6+117.76+96.256+78.6432+64.22528=\$654.48448$, as required. The method avoids omitting the $5 change when moving from the year-3 payment to the year-4 payment.

Accumulated Value at a Common Date

The same segmentation works when the requested value is at a date after one or more payments. A segment may be valued at time 0 and accumulated to the target date, or valued at a date close to its payments and then moved to the target date. The essential requirement is that every segment value be expressed at the same date before adding.

Worked Example: Accumulate a Piecewise Stream to Its Final Payment

A hypothetical fund receives payments at the ends of years 1 through 7. The first four payments are $80, $100, $120, and $140. Beginning with the year-5 payment, the annual increment changes to $5. Find the accumulated value at time 7 at an effective annual interest rate of 25%.

Identify the segments. The year-4 payment is $140. Applying the new $5 increment gives payments of $145, $150, and $155 in years 5, 6, and 7. The first segment consists of years 1–4; the second consists of years 5–7.

Accumulate each payment to time 7. With $1+i=1.25$, the first segment’s value is $80(1.25)^6+100(1.25)^5+120(1.25)^4+140(1.25)^3$. The terms are $305.17578125$, $305.17578125$, $292.96875$, and $273.4375$, giving $\$1,176.7578125$.

The second segment’s value at time 7 is $145(1.25)^2+150(1.25)+155=226.5625+187.5+155=\$569.0625$. Adding the segment values gives $\$1,745.8203125$, or $1,745.82 to the nearest cent.

Check. The seven accumulated payment amounts sum to $305.17578125+305.17578125+292.96875+273.4375+226.5625+187.5+155=\$1,745.8203125$. The check confirms that the switch changes the increment but does not alter the timing of any payment.

Common Mistakes and Practical Checks

Piecewise streams are usually straightforward once the cash flows are written out. Errors most often arise from applying a correct annuity factor to the wrong payment pattern or valuing segments at inconsistent dates.

Restarting the payment amount at the switch. If the increment changes but the payment sequence continues, calculate the next payment from the previous payment and the new increment.
Using the standard gradient factor without adjusting its first payment. $(Ia)_{\overline{m}|}$ represents payments $1,2,\ldots,m$, not $0,1,\ldots,m-1$. For a zero-start gradient, use $(Ia)_{\overline{m}|}-a_{\overline{m}|}$.
Combining values from different dates. A segment value at time 3 cannot be added directly to a present value at time 0. Discount or accumulate first.
Double-counting the boundary payment. Assign every payment date to exactly one segment. The first payment of the new segment follows the last payment of the previous segment; it is not an additional duplicate payment at the switch.
Using an annual factor with a different payment period. The payment period and interest period must match the factor, or the rate must first be converted appropriately.

A useful exam check is to write a short list of actual payment amounts before applying factors. This makes the segment lengths, first payments, increments, and boundary date visible. Then verify that every payment appears exactly once and that every value has been moved to the requested common date. This approach is aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Check Your Understanding

For each problem, first identify the payment amounts and the date at which the value is requested.

At an effective annual rate of 10%, a stream pays $200, $220, and $240 at the ends of years 1–3. The increment then changes to $15, giving payments in years 4 and 5. Find the present value to the nearest cent. (Answer: $885.77)
At an effective annual rate of 5%, payments at the ends of years 1–5 begin at $300 and increase by $30 through year 2. The increment is then $10 per year. Find the present value to the nearest cent. (Answer: $1,448.75)
At an effective annual rate of 25%, payments are $50, $60, $70, and $80 in years 1–4. The increment changes to $-$5$ for the next payment changes. Find the present value of all six payments to the nearest cent. (Answer: $189.93)
At an effective annual rate of 10%, a stream pays $100, $120, and $140 in years 1–3; the increment then changes to $10 per year, giving payments in years 4 and 5. Find the accumulated value at time 5 to the nearest cent. (Answer: $800.53)
At an effective annual rate of 25%, six payments have the pattern $A,A+20,A+40,A+50,A+40,A+30$, where $A$ is unknown. Their present value is $369.87392. Solve for $A$, to the nearest cent. (Answer: $100.00)

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