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Exam FM · Varying Annuity Payments · Tutorial 219 of 998

Worked Problems on Arithmetic-Gradient Annuities

Learn to translate arithmetic payment changes into focal-date equations and value immediate, due, and deferred streams without shifting payments or gradients incorrectly.

Intermediate 10 min read

What You'll Learn

  • Represent an arithmetic payment stream as a level annuity plus fixed increments or decrements.
  • Value increasing and decreasing annuity-immediate payments at the present-value date.
  • Adjust the gradient weights when payments occur at the beginning of each period.
  • Use a timeline and focal-date equation for deferred arithmetic annuities.
  • Check payment timing, gradient indexing, and nonnegative payment amounts.

How to Value an Arithmetic Payment Stream

An arithmetic annuity changes by a fixed dollar amount from one payment to the next. The key is to distinguish the first payment from the increment. If the first payment is \(C\) and each later payment increases by \(G\), the payments are \(C,\ C+G,\ C+2G,\ldots\). For a decreasing stream, the same pattern applies with a negative increment, or equivalently with a positive decrement subtracted each period.

A timeline fixes the payment dates; a focal-date equation then discounts or accumulates each payment to one common date. This is particularly important when the annuity is due or deferred: the same payment amounts at different dates do not have the same value. Unless stated otherwise, the examples use annual payments and a constant effective annual interest rate \(i\), with interest effective at each year-end.

Definition: An arithmetic-gradient annuity-immediate with first payment \(C\), increment \(G\), and \(n\) payments has payment \(C+(k-1)G\) at time \(k\), for \(k=1,\ldots,n\). An arithmetic-gradient annuity-due has those payments at times \(k-1\), for \(k=1,\ldots,n\). A deferred annuity has its payment dates shifted later; the shift must be included when valuing it.

The standard factors \(a_{\overline{n}|}\) and \(\ddot{a}_{\overline{n}|}\) are the present values of \(n\) unit payments at the ends and beginnings of successive periods, respectively. The increasing-gradient factor \((Ia)_{\overline{n}|}\) is the present value of payments \(1,2,\ldots,n\) at times \(1,2,\ldots,n\). Thus

$$ a_{\overline{n}|}=\sum_{k=1}^{n}v^k,\qquad \ddot{a}_{\overline{n}|}=\sum_{k=0}^{n-1}v^k,\qquad (Ia)_{\overline{n}|}=\sum_{k=1}^{n}kv^k,\qquad v=\frac{1}{1+i}. $$

These factors describe payment timing as well as payment amounts. The increasing-gradient factor does not by itself represent the increments \(0,G,2G,\ldots\) added to a base payment: its weights are \(1,2,\ldots,n\). That indexing difference must be handled explicitly.

Immediate Payments: Increasing and Decreasing Gradients

For an annuity-immediate whose payments are \(C,\ C+G,\ldots,C+(n-1)G\), split each payment into a level part \(C\) and an increment part. At time \(k\), the level part is \(C\) and the increment part is \((k-1)G\). Therefore the present value is

$$ C\,a_{\overline{n}|} +G\bigl((Ia)_{\overline{n}|}-a_{\overline{n}|}\bigr). $$

For a decreasing stream with decrement \(D>0\), payments \(C,\ C-D,\ldots,C-(n-1)D\) have present value

$$ C\,a_{\overline{n}|} -D\bigl((Ia)_{\overline{n}|}-a_{\overline{n}|}\bigr). $$

The formulas require a constant effective rate per payment period and payment dates at the ends of successive periods. A decreasing schedule may be valued algebraically even if it eventually has negative payments, but a stream described as nonnegative payments must satisfy \(C-(n-1)D\geq 0\).

Derivation: At time \(k\), the increasing payment is \(C+(k-1)G\). Its present value is \([C+(k-1)G]v^k\). Summing over all payment dates gives \(C\sum_{k=1}^n v^k+G\sum_{k=1}^n(k-1)v^k\). Since \(\sum_{k=1}^n(k-1)v^k=\sum_{k=1}^nkv^k-\sum_{k=1}^nv^k=(Ia)_{\overline{n}|}-a_{\overline{n}|}\), the stated formula follows.

Worked Example: Increasing Annuity-Immediate

A hypothetical equipment-maintenance contract pays \$500 at the end of year 1, with each later annual payment \$100 greater than the preceding payment, for four years. Find the present value at an effective annual interest rate of 10%. Use a timeline and round the final value to the nearest cent.

Method. The payments are \$500, \$600, \$700, and \$800 at times 1, 2, 3, and 4. The timeline is \(t=0:\) valuation date; \(t=1:\$500;\ t=2:\$600;\ t=3:\$700;\ t=4:\$800\). The stream is immediate, its first payment is \(C=500\), and its increment is \(G=100\).

Here \(v=1/1.1\). The level-annuity factor and increasing-gradient factor are

$$ a_{\overline{4}|}=v+v^2+v^3+v^4=3.169865446, \qquad (Ia)_{\overline{4}|}=v+2v^2+3v^3+4v^4=7.547981695. $$

In particular, the increment factor is \(7.547981695-3.169865446=4.378116249\). The focal-date equation at time 0 is

$$ PV=500(3.169865446)+100(4.378116249) =1,584.932723+437.811625 =2,022.744348. $$

The direct discounted-payment check is \(500v+600v^2+700v^3+800v^4=454.545455+495.867769+525.920361+546.410764=2,022.744349\), with the one-millionth difference due to displayed factor rounding. The present value is \$2,022.74. The factor method avoids discounting the entire schedule as if it were level.

Annuity-Due Payments

For an annuity-due, the first payment occurs at time 0, so the payment pattern \(C,\ C+G,\ldots,C+(n-1)G\) has present value \(C\) at time 0, \(C+G\) at time 1, and so on. Its gradient part has weights \(0,1,\ldots,n-1\), not \(1,2,\ldots,n\) at those dates. The due-stream present value is

$$ C\,\ddot{a}_{\overline{n}|} +G\sum_{j=0}^{n-1}jv^j. $$

An equivalent expression for the gradient factor is \((1+i)\bigl((Ia)_{\overline{n}|}-a_{\overline{n}|}\bigr)\). To see why, replace \(j\) by \(k-1\) in the sum: \(\sum_{j=0}^{n-1}jv^j=\sum_{k=1}^n(k-1)v^{k-1}=(1/v)\sum_{k=1}^n(k-1)v^k\), and \(1/v=1+i\). This relationship is useful when factors for annuity-immediate gradients are available, but a timeline is still the safest way to check the indexing.

Worked Example: Increasing Annuity-Due

A hypothetical service plan pays \$300 immediately and makes three further annual payments, each \$50 greater than the previous one. Find the present value at an effective annual interest rate of 5%. Round to the nearest cent.

Method. There are four payments: \$300 at time 0, \$350 at time 1, \$400 at time 2, and \$450 at time 3. The timeline is \(t=0:\$300;\ t=1:\$350;\ t=2:\$400;\ t=3:\$450\). Discount each payment to time 0:

$$ PV=300+\frac{350}{1.05}+\frac{400}{1.05^2}+\frac{450}{1.05^3}. $$

The individual discounted amounts are \(300.000000\), \(333.333333\), \(362.811791\), and \(388.726919\). Thus \(PV=1,384.872043\), or \$1,384.87 to the nearest cent. As a check on timing, there is no discount applied to the immediate \$300 payment. Treating this as an annuity-immediate would incorrectly discount that first payment by one year.

Deferred Streams and Focal Dates

A deferred annuity is valued most simply by first finding its value immediately before its first payment period, then discounting that value to the requested date. For example, if payments occur at times \(m+1,m+2,\ldots,m+n\), the payment pattern is an annuity-immediate shifted \(m\) years later. At effective annual rate \(i\), the present value at time 0 is

$$ v^m\left[C\,a_{\overline{n}|} +G\bigl((Ia)_{\overline{n}|}-a_{\overline{n}|}\bigr)\right]. $$

The bracketed amount is the value at time \(m\); multiplying by \(v^m\) moves that value back to time 0. This expression applies to annual payments at the ends of periods and a constant effective annual rate. If the first payment is at a different date, identify its actual time on the timeline rather than assuming the same deferral.

1
Mark the payment dates.
Write each date and its payment amount, including any payment at time 0.
2
Choose a focal date.
Use time 0 for present value, or a date just before the first deferred payment if that simplifies the arithmetic.
3
Value and shift.
Apply the gradient formula at the chosen date, then discount or accumulate to the required date.
4
Check the equation.
Confirm that every payment is moved to the focal date using the correct number of periods.

Worked Example: Deferred Decreasing Annuity-Immediate

A hypothetical grant program makes three annual payments of \$140, \$120, and \$100 at times 3, 4, and 5. Find the present value at time 0 if the effective annual interest rate is 10%. Round to the nearest cent.

Method. The first payment is at time 3, so this is a two-year deferral of a three-payment annuity-immediate. The timeline is \(t=0:\) valuation date; \(t=1,2:\) no payments; \(t=3:\$140;\ t=4:\$120;\ t=5:\$100\). Value the stream first at time 2, then discount two years to time 0. At time 2, the payments one, two, and three years later are \$140, \$120, and \$100:

$$ V_2=\frac{140}{1.1}+\frac{120}{1.1^2}+\frac{100}{1.1^3} =127.272727+99.173554+75.131480 =301.577761. $$

Then \(PV=V_2v^2=301.577761(0.826446281)=249.237819\), giving \$249.24. The direct check discounts from the actual payment dates: \(140v^3+120v^4+100v^5=105.184072+81.961615+62.092132=249.237819\). Both calculations agree. Discounting the value at time 2 by only one year would omit one year of deferral.

Common Exam Traps and Checks

A frequent error is to use \((Ia)_{\overline{n}|}-a_{\overline{n}|}\) as the increment factor but also treat \((Ia)_{\overline{n}|}\) as if its first gradient payment were zero. In fact, \((Ia)_{\overline{n}|}\) has weights \(1,2,\ldots,n\) at times \(1,2,\ldots,n\). The factor for increments \(0,1,\ldots,n-1\) at those same immediate dates is the difference \((Ia)_{\overline{n}|}-a_{\overline{n}|}\).

For example, in the first worked example the incorrect calculation \(500a_{\overline{4}|}+100(Ia)_{\overline{4}|}\) gives \(1,584.932723+754.798170=2,339.730893\). That calculation adds a \$100 gradient payment at time 1, so it represents payments \$600, \$700, \$800, and \$900—not the stated schedule beginning at \$500. The correction is to subtract the level-annuity factor from the increasing-gradient factor for the increment component, giving \$2,022.74.

  • Check whether the first payment is at time 0 or time 1. That determines whether the stream is due or immediate.
  • Check the increment index. An immediate stream starting with \(C\) has additions \(0,G,\ldots,(n-1)G\).
  • Check the deferral from the actual dates. A stream first paid at time \(m+1\) has a value at time \(m\) that must be discounted \(m\) years to time 0.
  • Check decreasing amounts. If every payment must be nonnegative, verify that the last payment is not below zero.

A sound focal-date equation is more reliable than memorizing a factor in isolation. These valuation methods are aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Check Your Understanding

For each problem, identify the payment dates before calculating. Round money to the nearest cent.

  1. An annuity-immediate pays \$80 at the end of year 1, with each payment increasing by \$20 for a total of three payments. Find its present value at an effective annual rate of 5%. (Answer: \$270.55)
  2. An annuity-immediate pays \$500, \$450, \$400, and \$350 at the ends of years 1 through 4. Find its present value at an effective annual rate of 8%. (Answer: \$1,423.56)
  3. An annuity-due has four payments, beginning with \$200 at time 0 and increasing by \$25 each year. Find its present value at an effective annual rate of 10%. (Answer: \$817.77)
  4. A deferred annuity-immediate pays \$100, \$120, and \$140 at times 4, 5, and 6. Find its present value at time 0 at an effective annual rate of 5%. (Answer: \$280.76)
  5. An annuity-immediate pays \$100, \$100+G, and \$100+2G at the ends of years 1, 2, and 3. Its present value is \$270 at an effective annual rate of 10%. Find the fixed increment \(G\). (Answer: \$9.15)
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