Integrated Varying-Annuity Valuation Problems

Why Integrated Varying-Annuity Valuation Requires Care

A varying payment stream may combine a changing payment amount with interest rates that change over time. The valuation method is still an equation of value: move every payment to one chosen date using the rates that apply along its path, then add the resulting values. The main challenge is identifying that path correctly. A rate from a year before a payment is not used to accumulate that payment forward, and a rate after the focal date is not used to discount a payment back to it.

The examples use annual payment dates. Let $i_k$ denote the effective annual rate during the interval from time $k-1$ to time $k$. If a nominal rate is given, first convert it to an effective rate for the period being valued. The effective annual rate equivalent to a nominal annual rate $i^{(m)}$, convertible $m$ times per year, is $(1+i^{(m)}/m)^m-1$.

Definition: An equation of value equates the values of cash flows at a common focal date. To move a payment forward across several periods, multiply by each period's accumulation factor. To move it backward, divide by each period's accumulation factor. The payment amount itself is not changed by the valuation; only its value at the focal date changes.

Choosing the Correct Factors

For annual periods with effective rates $i_1,i_2,\ldots$, a payment $C_j$ at time $j$ has value at time $t$ given by the product of the factors between $j$ and $t$. If $j<t$, it is accumulated forward; if $j>t$, it is discounted backward. In either case, the period immediately after a payment date is the first period used when moving forward from that payment.

V_t=\sum_j C_j \begin{cases} \displaystyle\prod_{k=j+1}^{t}(1+i_k), & j<t,\\[6pt] 1, & j=t,\\[4pt] \displaystyle\frac{1}{\prod_{k=t+1}^{j}(1+i_k)}, & j>t. \end{cases}

An empty product does not arise in the first or third cases because those cases require $j<t$ or $j>t$. A payment at the focal date is included without adjustment. The formula assumes that each $i_k$ is an effective rate for the stated annual interval and that the payment dates align with the period boundaries. For payment dates within a year, use the accumulation factors for the actual subperiods instead.

Why the product is correct: Over one period, a value $V$ grows to $V(1+i_k)$. Over two successive periods, it grows first to $V(1+i_{k})$, then to $V(1+i_{k})(1+i_{k+1})$. Repeating this one-period rule gives the product of every accumulation factor crossed. Reversing the process divides by the same factors, which gives the discounting part of the equation.

Worked Examples

Worked Example: A Geometric Payment Pattern and a Nominal Rate

A fund receives four annual payments of $108.16, $113.568, $119.2464, and $125.20872 at times 1 through 4. The effective annual rate is equivalent to a nominal annual rate of $8\%$ convertible semiannually during year 1. It is $5\%$ during years 2, 3, and 4. Find the value at time 0 and at time 2. The payments grow by $5\%$ each year. State the method and check the two values against each other.

Method. Convert the nominal rate to the effective rate for year 1, then value each payment using only the factors between its date and the focal date.

The year-1 effective rate is

\left(1+\frac{0.08}{2}\right)^2-1 =(1.04)^2-1 =0.0816.

Thus the accumulation factor for year 1 is $1.0816$. Each later year has accumulation factor $1.05$. At time 0, the four discount denominators are $1.0816$, $1.0816(1.05)$, $1.0816(1.05)^2$, and $1.0816(1.05)^3$. The present value is therefore

\begin{aligned} V_0 &=\frac{108.16}{1.0816} +\frac{113.568}{1.0816(1.05)} +\frac{119.2464}{1.0816(1.05)^2} +\frac{125.20872}{1.0816(1.05)^3}\\ &=100+100+100+100\\ &=\$400.00. \end{aligned}

At time 2, the payment at time 1 accumulates through year 2 at $5\%$. The payment at time 2 is unchanged. The payments at times 3 and 4 are discounted through the applicable later years:

\begin{aligned} V_2 &=108.16(1.05)+113.568 +\frac{119.2464}{1.05} +\frac{125.20872}{(1.05)^2}\\ &=113.568+113.568+113.568+113.568\\ &=\$454.272. \end{aligned}

To check, move the time-0 value forward through years 1 and 2: $400(1.0816)(1.05)=454.272$, agreeing with the direct time-2 valuation. The common mistake is to use the year-1 rate to accumulate the time-1 payment to time 2. That payment is already at time 1, so its forward accumulation uses the year-2 rate only.

Worked Example: Solving for an Arithmetic-Gradient Base Payment

Payments of $x$, $x+50$, and $x+100$ are made at times 1, 2, and 3. The effective rate is $4\%$ during year 1 and $25\%$ during each of years 2 and 3. The value of the payments at time 2 is $1,350. Find $x$. Use the focal date directly and verify the result.

Method. Express each payment at time 2. The year-1 rate is provided, but it does not enter this particular equation: none of the cash flows is being moved across year 1 to reach time 2.

The time-1 payment accumulates across year 2 by a factor of $1.25$. The time-2 payment is unchanged. The time-3 payment is discounted by $1.25$:

1.25x+(x+50)+\frac{x+100}{1.25}=1350.

Since $1/1.25=0.8$, collect the terms:

\begin{aligned} 1.25x+x+50+0.8x+80&=1350,\\ 3.05x+130&=1350,\\ 3.05x&=1220,\\ x&=400. \end{aligned}

The payment amounts are therefore $400, $450, and $500. At time 2, they are worth $400(1.25)=500$, $450, and $500/1.25=400$, respectively. Their sum is $1,350, as required. A common mistake is to apply the year-1 $4\%$ rate to the time-1 payment when moving it to time 2; the interval crossed is year 2, so the correct factor is $1.25$.

Worked Example: A Geometric Gradient Under Changing Rates

Four annual payments are made at times 1 through 4. Each payment is $10\%$ greater than the preceding payment. The effective annual rate is $5\%$ during year 1 and $10\%$ during years 2, 3, and 4. The present value of the stream is $4,000. Find the first payment and verify the full stream.

Method. Let the first payment be $x$, so the payments are $x$, $1.1x$, $1.21x$, and $1.331x$. Discount each payment to time 0 using the rates in all the years before that payment.

\frac{x}{1.05} +\frac{1.1x}{1.05(1.10)} +\frac{1.21x}{1.05(1.10)^2} +\frac{1.331x}{1.05(1.10)^3} =4000.

In each term, the powers of $1.10$ in the payment and denominator cancel. For example, $1.21=(1.10)^2$, and $1.331=(1.10)^3$. Thus each discounted payment is $x/1.05$, giving

\frac{4x}{1.05}=4000, \qquad x=\frac{4000(1.05)}{4}=1050.

The payments are $1,050, $1,155, $1,270.50, and $1,397.55. Their time-0 values are

\frac{1050}{1.05}=1000,\quad \frac{1155}{1.05(1.10)}=1000,\quad \frac{1270.50}{1.05(1.10)^2}=1000,\quad \frac{1397.55}{1.05(1.10)^3}=1000.

The sum is $4,000, confirming the result. The cancellation works because the payment growth rate matches the $10\%$ interest rate in years 2 through 4. It does not permit omitting the year-1 factor: every payment is still discounted through year 1.

Common Traps and a Reliable Process

The focal date determines which rates matter. A rate schedule is not a single rate to apply indiscriminately to the entire stream. Each payment has its own route to the focal date, and the rate for each interval on that route must be included once.

For example, suppose a $300 payment occurs at time 1, the effective rate is $4\%$ during year 1 and $6\%$ during year 2, and the desired value is at time 2. The incorrect calculation $300(1.04)=312$ uses the rate in the interval before the payment. The correct calculation is $300(1.06)=318$, because the payment moves from time 1 to time 2 through year 2.

A disciplined setup helps prevent this error and keeps unknown-payment problems manageable:

Mark payment dates and the focal date.
Identify which payments are before, at, and after the focal date.

Convert rates to the needed period basis.
For a nominal rate, use its stated conversion frequency before treating it as an effective rate.

Write a factor for every interval crossed.
Multiply factors when moving forward and divide by them when moving backward.

Substitute the payment pattern.
Write arithmetic or geometric payments in terms of the unknown amount, then solve the equation of value.

Check the result.
Substitute the solved payments into the original equation and, when useful, value the stream at a second focal date.

This process is useful in multi-stage problems because it separates three issues that are easy to confuse: the payment pattern, the rate conversion, and the valuation date. The same approach is aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Check Your Understanding

Use the stated focal date and period rates in each problem. Answers are rounded to cents where needed.

Payments of $200 and $250 occur at times 1 and 2. The effective annual rates are $4\%$ in year 1 and $6\%$ in year 2. Find the value at time 0. (Answer: $419.09)
Payments of $x$, $x+20$, and $150 occur at times 1, 2, and 3. The effective annual rate is $5\%$ in years 2 and 3. Their value at time 2 is $400. Find $x$. (Answer: $115.68)
Payments $x$, $1.1x$, and $1.21x$ occur at times 1, 2, and 3. The effective annual rate is $5\%$ in year 1 and $10\%$ in years 2 and 3. If the present value is $300, find $x$. (Answer: $105.00)
A nominal annual rate of $8\%$, convertible quarterly, applies in each of two successive years. Payments of $500 and $550 occur at times 1 and 2. Find the value at time 2. (Answer: $1,091.22)
Payments of $300, $330, and $363 occur at times 1, 2, and 3. The effective annual rate is $4\%$ in year 2 and $10\%$ in year 3. Find the value at time 2. (Answer: $972.00)

What You'll Learn