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Exam FM · Interest and Annuity Applications · Tutorial 907 of 998

Payment Escalation That Resets at a Stage Boundary

Model and value staged payment streams whose arithmetic or geometric escalation restarts from a new base amount at a specified payment date.

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What You'll Learn

  • Define payment stages with separate escalation rules and starting amounts.
  • Distinguish a reset at the boundary from continuing the prior stage’s pattern.
  • Write a present-value equation by splitting payments into their correct stages.
  • Value arithmetic and geometric stages by discounting each payment date.
  • Solve for an unknown reset amount and check a staged cash-flow equation.
  • Identify timing and escalation errors that can materially change an answer.

Why a Reset Needs Its Own Payment Rule

A varying annuity may follow one escalation pattern for several years and then restart from a new base amount. The reset can change the payment level, the size of each arithmetic increase, or the geometric growth rate. The key modeling task is to assign each payment to the correct stage before calculating its value.

A stage boundary does not, by itself, mean the old pattern continues. If the second stage begins with a stated new payment, use that amount as the new base. For example, if payments of \(800, 1{,}000, 1{,}200\) are followed by a stage that begins at \(900\), the next stage begins at \(900\)—not at \(1{,}400\).

Definition: A resetting staged annuity is a series of payments divided into consecutive stages, with each stage having its own stated base payment and escalation rule. In an arithmetic stage, each successive payment changes by a fixed dollar amount. In a geometric stage, each successive payment is multiplied by a fixed ratio. A reset replaces the prior stage’s pattern with the new stage’s base and rule.

Set Up the Payment Timeline First

Assume payments are made annually at the ends of years \(1,2,\ldots,n\), and the effective annual interest rate \(i\) is constant. Let \(v=1/(1+i)\). Suppose the first stage contains payments at times \(1\) through \(m\), and the second stage contains payments at times \(m+1\) through \(n\). The stages have no overlapping payment date.

For an arithmetic first stage with base payment \(A\) and increase \(g\), its payment at time \(t\) is \(A+(t-1)g\). If the second stage resets to base \(B\) and increase \(h\), its payment at time \(t\) is \(B+(t-m-1)h\). The first payment in that second stage, at time \(m+1\), is therefore \(B\).

For a geometric first stage with base payment \(A\) and ratio \(r\), its payment at time \(t\) is \(Ar^{t-1}\). If the second stage resets to base \(B\) and ratio \(q\), its payment at time \(t\) is \(Bq^{t-m-1}\). In particular, the first payment after the boundary is \(B\), not the amount obtained by extending the first stage’s ratio.

The present value at time 0 is the sum of each payment multiplied by its own discount factor. Splitting the sum at the stage boundary makes the reset explicit:

$$ PV_0=\sum_{t=1}^{m} C_t v^t+\sum_{t=m+1}^{n} C_t v^t. $$

Here \(C_t\) is the payment actually specified for time \(t\). For a nonconstant interest environment, replace \(v^t\) with the discount factor from time \(t\) to time 0 implied by the applicable rates. The payment-stage definitions remain separate from the interest-rate assumptions.

Why the Split-Sum Method Works

The present value principle says that the value at time 0 of a payment \(C_t\) due at time \(t\) is \(C_t v^t\). The stages partition the payment dates into two disjoint sets: \(1,\ldots,m\) and \(m+1,\ldots,n\). Adding the values for those dates gives the value of the whole stream.

$$ PV_0 =\sum_{t=1}^{m} C_t v^t+\sum_{t=m+1}^{n} C_t v^t =\sum_{t=1}^{n} C_t v^t. $$

This identity also explains why a reset changes the payment equation, not the discounting rule. Each payment is still discounted from its own date; only the expression used to determine \(C_t\) changes after the boundary. Directly summing the payments is especially reliable when stages have different bases or escalation rules.

Worked Example: Two Arithmetic Stages With a New Base

An account receives end-of-year payments for six years. Payments in years 1–3 begin at $800 and increase by $200 each year. In years 4–6, the payment resets to $900 and then increases by $150 each year. At an effective annual interest rate of 10%, find the present value at time 0.

Method. List the payments by stage, then discount each payment by \(v^t\), where \(v=1/1.10\). This avoids carrying the first stage’s $200 increase into the second stage.

The first stage payments are $800, $1,000, and $1,200. The second stage payments are $900, $1,050, and $1,200. The discount factors, rounded here to nine decimal places for the intermediate calculations, are:

$$ v=0.909090909,\quad v^2=0.826446281,\quad v^3=0.751314801,\quad v^4=0.683013455,\quad v^5=0.620921323,\quad v^6=0.564473930. $$

The stage values are:

$$ \begin{aligned} PV_{\text{stage 1}} &=800v+1{,}000v^2+1{,}200v^3\\ &=727.272727+826.446281+901.577761\\ &=2{,}455.296769, \end{aligned} $$
$$ \begin{aligned} PV_{\text{stage 2}} &=900v^4+1{,}050v^5+1{,}200v^6\\ &=614.712110+651.967389+677.368716\\ &=1{,}944.048215. \end{aligned} $$

Thus \(PV_0=2{,}455.296769+1{,}944.048215=4{,}399.344984\). Rounded to the nearest cent, the present value is $4,399.34. As a check, the six discounted payments sum to the same amount, and the year-4 payment is the stated reset amount of $900.

Worked Example: A Geometric Stage Resets to a Different Base and Rate

A hypothetical grant pays $1,000 at the end of year 1 and increases by 20% for the year-2 payment. A new stage begins in year 3 with a reset payment of $2,000; payments in years 4 and 5 increase by 5% per year from that new base. If the effective annual interest rate is 10%, find the present value.

Method. Apply each geometric rule only within its own stage. The first two payments are \(1{,}000\) and \(1{,}000(1.20)=1{,}200\). The reset stage payments are \(2{,}000\), \(2{,}000(1.05)=2{,}100\), and \(2{,}100(1.05)=2{,}205\).

Using the 10% discount factors \(v=0.909090909\), \(v^2=0.826446281\), \(v^3=0.751314801\), \(v^4=0.683013455\), and \(v^5=0.620921323\):

$$ \begin{aligned} PV_0 &=1{,}000v+1{,}200v^2+2{,}000v^3+2{,}100v^4+2{,}205v^5\\ &=909.090909+991.735537+1{,}502.629602\\ &\quad+1{,}434.328256+1{,}369.131517\\ &=6{,}206.915821. \end{aligned} $$

The present value is $6,206.92, rounded to the nearest cent. The reset is evident in year 3: its payment is $2,000, not the $1,440 that would result from continuing the first stage’s 20% growth from $1,200.

Worked Example: Solve for a Reset Base Amount

A contract pays $700 at the end of year 1 and $900 at the end of year 2. A second stage begins at the end of year 3 with payment \(B\), then increases geometrically by 10% per year for two more payments. The effective annual interest rate is 10%, and the present value of all five payments is $5,000. Find \(B\).

Method. Value the known first stage, then write the second-stage payments from its own base \(B\). The cash flows are $700, $900, \(B\), \(1.10B\), and \(1.21B\), at times 1 through 5.

The present value of the known payments is:

$$ 700v+900v^2 =636.363636+743.801653 =1{,}380.165289. $$

The remaining present value must be \(5{,}000-1{,}380.165289=3{,}619.834711\). At a 10% interest rate, the second-stage payment growth rate equals the interest rate, so each of the three discounted second-stage payments has the same value per unit of \(B\):

$$ Bv^3+1.10Bv^4+1.21Bv^5 =Bv^3+Bv^3+Bv^3 =3Bv^3. $$

Since \(v^3=0.751314801\), the equation is \(3B(0.751314801)=3{,}619.834711\), so \(B=1{,}606\). As a check, the second-stage payments are $1,606, $1,766.60, and $1,943.26. Their present values sum to $3,619.834711; adding the first-stage value gives $5,000. The required reset payment is $1,606.

Worked Example: Check the Boundary Before Discounting

A five-payment stream has an effective annual interest rate of 5%. Payments in years 1–3 are a geometric sequence beginning at $1,000 and growing by 10% per year. In year 4 the stream resets to $1,500, and the year-5 payment is 10% above the reset payment. Find the present value.

Method. Identify the year-4 payment from the reset condition, rather than extending the first stage. The payments are $1,000, $1,100, $1,210, $1,500, and $1,650.

At 5%, the discount factors for years 1 through 5 are \(0.952380952\), \(0.907029478\), \(0.863837599\), \(0.822702475\), and \(0.783526166\). Therefore:

$$ \begin{aligned} PV_0 &=1{,}000(0.952380952)+1{,}100(0.907029478)\\ &\quad+1{,}210(0.863837599)+1{,}500(0.822702475)\\ &\quad+1{,}650(0.783526166)\\ &=952.380952+997.732426+1{,}045.243495\\ &\quad+1{,}234.053713+1{,}292.818174\\ &=5{,}522.228760. \end{aligned} $$

The present value is $5,522.23, rounded to the nearest cent. As a check, the year-5 payment is 10% above the reset amount: \(1{,}500(1.10)=1{,}650\).

Common Exam Trap: Continuing the Old Escalation

A frequent error is to identify the first stage’s pattern correctly, then carry it across the boundary even though a new base is stated. In the first worked example, continuing the $200 annual increase would incorrectly assign payments of $1,400, $1,600, and $1,800 to years 4–6. Discounting those incorrect payments gives:

$$ 1{,}400v^4+1{,}600v^5+1{,}800v^6 =956.218837+993.474117+1{,}016.053074 =2{,}965.746028. $$

Adding the correct first-stage value would produce the wrong total, $5,421.04, instead of $4,399.34. The correction is to use the second stage’s stated base and increase: $900, $1,050, and $1,200. Before discounting, write every payment date and payment amount in a short timeline. This makes the reset visible and helps prevent a timing error as well as an escalation error.

A Reliable Calculation Sequence

1
Mark the payment dates.
State which years belong to each stage, and determine whether a payment occurs at the boundary date or after it.
2
Write each stage’s rule.
Record the base payment and the arithmetic increase or geometric ratio separately for each stage.
3
Generate the payment list.
Check the first payment after each reset against the stated new base before calculating any present value.
4
Discount by actual payment time.
Apply the discount factor for each payment date; do not restart the discount clock when the payment pattern resets.
5
Check the result.
Verify the payment count, stage boundary, and total value. If a base amount is unknown, substitute the solution back into the original cash-flow equation.

Resetting staged annuities arise whenever a payment schedule changes its escalation basis after a specified period. The central discipline is to keep the payment rule and the discount timeline distinct: a reset changes the payment amounts, but each cash flow remains attached to its original payment date. This approach is aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Check Your Understanding

For each problem, identify the stage payments before valuing the stream. Unless stated otherwise, payments are made at the ends of the indicated years.

  1. An annuity pays $400 and $600 in years 1 and 2. It resets to $1,000 in year 3, then increases by $100 for the year-4 payment. At an effective annual interest rate of 10%, find the present value. (Answer: $2,362.13)
  2. At an effective annual interest rate of 8%, a three-year stage pays $500 in year 1 and increases by $200 each year. A two-year stage then resets to $1,200 in year 4 and increases by $300 for the year-5 payment. Find the present value. (Answer: $3,680.46)
  3. At an effective annual interest rate of 5%, a two-year geometric stage begins at $800 and grows by 10% per year. A three-year arithmetic stage then resets to $1,000 and increases by $100 each year. Find the present value. (Answer: $4,269.13)
  4. An annuity earns an effective annual interest rate of 10%. Payments in years 1 and 2 are $300 and $400. In year 3, a new stage resets to \(B\) and grows by 10% for the payments in years 4 and 5. If the present value is $3,000, find \(B\). (Answer: $1,063.33)

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