Piecewise Geometric Payment Growth

Why Piecewise Geometric Growth Needs Segmentation

A payment stream may grow at one proportional rate for a while and then change pace. For example, a hypothetical service contract might have payments that increase by a fixed percentage during an introductory period, followed by level payments. Applying a single geometric-annuity factor to the entire stream would misstate its value because the payments do not follow one growth rate throughout.

The reliable method is to split the stream into segments, one for each run of payments with a common growth rate. Value each segment using its own first payment, number of payments, and growth rate. Then move every segment’s value to the same valuation date before adding the results.

Definition: A piecewise geometric payment stream is a series of payments divided into successive stages, such that payments within each stage change by a constant proportional rate. The growth rate may differ between stages. Unless stated otherwise, this tutorial assumes payments occur at the ends of annual periods and use a constant effective annual interest rate $i$.

A stage’s first payment must be identified separately from its growth rate. The growth rate tells how payments change within that stage; it does not, by itself, determine the first payment in the stage. A problem may specify the transition from the previous stage, or give the new stage’s first payment directly. Be explicit about that transition before valuing the segment.

Present Value of One Segment

Suppose a segment contains $n$ payments. Its first payment is $C$, made at time $k$, and each subsequent payment within the segment increases at annual rate $g$. Thus its payments occur at times $k,k+1,\ldots,k+n-1$, and the payment at time $k+j$ is $C(1+g)^j$, for $j=0,\ldots,n-1$. The interest rate is the same effective annual rate $i$ throughout.

PV_0=Cv^k\sum_{j=0}^{n-1}\bigl((1+g)v\bigr)^j

When $g\ne i$, the finite geometric-series formula gives an equivalent expression:

PV_0=Cv^k\frac{1-\left(\frac{1+g}{1+i}\right)^n}{1-\frac{1+g}{1+i}} =Cv^k\frac{1-\left(\frac{1+g}{1+i}\right)^n}{\frac{i-g}{1+i}}

The factor $v^k$ is essential: the first payment in this segment is at time $k$, not at time 1. If $g=i$, the ratio $(1+g)v$ equals 1, so the geometric-series expression has a zero denominator. In that case each payment has the same present value, and the segment’s present value is $nCv^k$.

Special case: For a segment with first payment $C$ at time $k$ and $n$ payments, its present value is $nCv^k$ when its within-stage growth rate equals the effective interest rate, $g=i$. Otherwise, use the geometric-series formula.

Why the Segment Formula Works

The present value of the payment at time $k+j$ is its amount multiplied by the discount factor for that date. The payment amount is $C(1+g)^j$, and discounting $k+j$ years uses $v^{k+j}$. Multiplying these terms produces the series in the segment formula:

C(1+g)^jv^{k+j} =Cv^k\bigl((1+g)v\bigr)^j

Summing over $j=0$ through $n-1$ gives a finite geometric series with first term $Cv^k$ and common ratio $(1+g)v$. If the ratio differs from 1, the sum is the first term times $\bigl(1-r^n\bigr)/(1-r)$. If the ratio is 1, all $n$ terms are equal, which gives $nCv^k$. This also explains the special case $g=i$: each payment’s growth exactly offsets the extra discounting from one year to the next.

Valuing a Multi-Stage Stream

For a stream with several stages, calculate each segment’s present value using its own first payment and within-stage growth rate. Keep the payment dates on the original timeline. In particular, a segment beginning later than time 1 needs the appropriate deferral factor, either explicitly as $v^k$ or implicitly through the dates of its payments.

Mark the payment dates and stage boundaries.
Record which payments belong to each constant-growth stage, and identify the date of each stage’s first payment.

Determine each first payment.
Use the stated amount or calculate it from the previous payment and the specified transition rate. Do not assume a stage restarts at the original base amount.

Value each segment.
Apply the geometric-series formula using that segment’s own growth rate and number of payments. Use the equal-growth special case if $g=i$.

Combine values at one date.
For present value, discount every segment to time zero. For an accumulated value, move every segment to the requested future date before adding.

Worked Example: Present Value When One Stage Grows at the Interest Rate

A hypothetical fund will pay six annual amounts, beginning at the end of year 1. The first payment is $1,000. Payments at the ends of years 1 through 3 increase by 10% per year. The transition from the year-3 payment to the year-4 payment has a 0% increase, and payments at the ends of years 4 through 6 remain level. Find the present value at an effective annual interest rate of 10%.

Method and stage amounts. Segment the stream into years 1–3 and years 4–6. The first stage’s payments are $1,000, $1,100, and $1,210. The specified 0% transition makes the year-4 payment $1,210, so the second stage’s payments are $1,210, $1,210, and $1,210.

First segment. Its growth rate equals the interest rate, so each payment has the same present value:

PV_{1\text{–}3}=3\left(\frac{1000}{1.10}\right) =3(909.090909\ldots) =2727.272727\ldots

Second segment. Discount its payments at their actual dates:

PV_{4\text{–}6} =\frac{1210}{1.10^4}+\frac{1210}{1.10^5}+\frac{1210}{1.10^6}

The discount denominators are $1.10^4=1.4641$, $1.10^5=1.61051$, and $1.10^6=1.771561$. Therefore,

PV_{4\text{–}6} =826.446281+751.314800+683.013455 =2260.774536

The total present value is $2727.272727+2260.774536=4988.047263$, or $4,988.05 to the nearest cent.

Check and common mistake avoided. The six payments are $1,000, $1,100, $1,210, $1,210, $1,210, and $1,210; the calculation discounts each at its own date. Omitting the deferral of the second stage would incorrectly value its first payment at time 1 rather than time 4.

Worked Example: A Deferred Stage with a Different Growth Rate

A hypothetical benefit pays $500 at the end of year 1 and $500 at the end of year 2. The payment at the end of year 3 is 10% greater than the year-2 payment. Payments at the ends of years 3 through 5 then increase by 10% per year. Find the present value at an effective annual interest rate of 10%.

Method and stage amounts. The first two payments form a level segment. The 10% transition makes the year-3 payment $500(1.10)=$550$; subsequent payments are $605 and $665.50. The second segment begins at time 3.

First segment:

PV_{1\text{–}2} =\frac{500}{1.10}+\frac{500}{1.10^2} =454.545455+413.223140 =867.768595

Second segment: Its growth rate equals the interest rate, so each of its three payments has the same present value:

PV_{3\text{–}5} =3\left(\frac{550}{1.10^3}\right) =3(413.223140) =1239.669421

Thus the total present value is $867.768595+1239.669421=2107.438016$, or $2,107.44 to the nearest cent.

Check and common mistake avoided. The year-3 payment is $550, not $500: the transition increase applies before the second stage begins. Also, the segment starts at time 3, so the equal-present-value calculation uses $550/1.10^3$, not $550/1.10$.

Accumulated Values of Piecewise Streams

The same segmentation applies when the requested value is at a future date. One approach is to accumulate each payment directly to that date. Another is to find the accumulated value of each segment at its own endpoint and then accumulate that amount forward. Whichever method is used, every part of the stream must be moved to the same valuation date before summing.

Worked Example: Accumulated Value at the End of the Stream

A hypothetical account receives four annual deposits. The first deposit is $100 at the end of year 1, and the year-2 deposit is 10% larger. The deposits at the ends of years 3 and 4 are level with the year-2 deposit. Find the accumulated value at the end of year 4 at an effective annual interest rate of 10%.

Method and amounts. The deposits are $100, $110, $110, and $110. The first stage is at years 1–2; the second stage begins with no increase between years 2 and 3.

Accumulate the first stage to the end of year 2: the year-1 deposit earns interest for one year, and the year-2 deposit earns none before that date.

AV_{2,\ 1\text{–}2}=100(1.10)+110=220

Move this amount forward two more years to year 4:

AV_{4,\ 1\text{–}2}=220(1.10)^2=266.20

Accumulate the second stage directly to year 4. The year-3 deposit earns one year of interest, while the year-4 deposit earns none:

AV_{4,\ 3\text{–}4}=110(1.10)+110=231.00

The accumulated value is $266.20+231.00=\mathbf{\$497.20}$.

Check and common mistake avoided. Direct accumulation gives $100(1.10)^3+110(1.10)^2+110(1.10)+110=133.10+133.10+121.00+110.00=$497.20$. The year-4 deposit does not earn a year of interest by the valuation date.

Common Errors and Practical Checks

Piecewise streams are usually difficult not because the geometric-series formula is complicated, but because a payment is assigned to the wrong stage or date. Before calculating, write the first payment of every stage and its time on the timeline. Then check that the stated transition between stages produces that amount.

Resetting the base payment. A new stage does not automatically restart at the initial payment. Its first payment must follow the problem’s transition rule or be taken from the stated amount.
Applying a stage rate to the wrong interval. Keep separate the growth within a stage and the rate used to move from the preceding stage to its first payment.
Forgetting the segment’s date. A segment beginning at time $k$ has a factor $v^k$ for its first payment. Valuing it as though it began at time 1 overstates its present value.
Using one growth rate for the whole stream. A single geometric-annuity factor applies only if the same proportional change holds across every payment interval.
Forgetting the equality case. When $g=i$, use $nCv^k$, or sum the equal discounted payments directly; do not substitute into a formula with a zero denominator.

Key takeaway: Segment first, then value. For each segment, identify its first payment, payment dates, number of payments, and within-stage growth rate. Apply the transition rule explicitly, and combine segment values only after bringing them to a common date.

This segmentation method is aligned with the SOA Exam FM (Financial Mathematics) syllabus. It is useful whenever the payment pattern changes over time but the interest assumption remains constant. If the interest rate itself changes across periods, the constant-rate segment formula no longer applies directly.

Check Your Understanding

For each problem, identify the segment dates and transition amounts before calculating.

A fund pays $200 at the ends of years 1 and 2. The year-3 payment is 5% greater than the year-2 payment, and the year-4 payment is 5% greater than the year-3 payment. Find the present value at an effective annual interest rate of 5%. (Answer: $734.69)
An account receives $250 at the end of year 1, with a 10% increase to the year-2 deposit. The year-3 deposit is level with the year-2 deposit, and the year-4 deposit is also level. Find the present value at an effective annual interest rate of 10%. (Answer: $848.99)
At an effective annual interest rate of 4%, deposits at the ends of years 1, 2, and 3 are $500, $520, and $540.80. Deposits at the ends of years 4 and 5 are each $540.80. Find the accumulated value at the end of year 5. (Answer: $2,858.02)
A stream pays $C$ at the end of year 1 and $1.05C$ at the end of year 2. The year-3 payment is level with the year-2 payment, and the year-4 payment increases by 5%. If the present value at an effective annual interest rate of 5% is $1,000, find $C$. (Answer: $268.90)
A stream pays $100 at the end of year 1 and $90 at the end of year 2. The year-3 payment is level with the year-2 payment, and the year-4 payment increases by 10%. Find the present value at an effective annual interest rate of 10%. (Answer: $300.53)

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