Solving Unknown Reinvestment Rates From Accumulated Values

From Receipts to an Unknown Reinvestment Rate

When receipts arrive at different dates, each can earn interest for a different length of time before a specified terminal date. If the receipts and terminal value are known but the constant reinvestment rate is not, the task is to find the rate that makes the accumulated receipts equal that target. The central exam skill is setting up the equation with the correct accumulation period for every receipt.

Assume throughout that each receipt is reinvested immediately at the same effective annual rate $i$, from its receipt date to the terminal date. There are no taxes, fees, withdrawals, or changes in the reinvestment rate. Time is measured in years; a receipt at time $t_j$ accumulates to time $T$ for $T-t_j$ years. For a fractional number of years, use the corresponding fractional power, consistent with the effective annual rate model.

Definition: If receipts $C_j$ occur at times $t_j\leq T$, their accumulated value at time $T$, at a constant effective annual reinvestment rate $i$, is $$ AV_T(i)=\sum_j C_j(1+i)^{T-t_j}. $$ To solve for an unknown rate, set $AV_T(i)$ equal to the stated target value and solve the resulting equation. The accumulation factor $1+i$ must be positive, so an admissible effective annual rate satisfies $i>-1$.

This is a terminal-value equation, not a present-value equation: every receipt is moved forward to the same date. A receipt at the terminal date has exponent zero and is included without accumulation. If all receipts are nonnegative, at least one positive receipt occurs before $T$, and the model uses a constant rate, the accumulated value strictly increases as $i$ increases over $i>-1$. Thus, if a solution exists, it is unique.

Why the Terminal-Value Equation Works

The formula follows by accumulating each receipt separately. A receipt $C_j$ earns interest for $T-t_j$ years, so its value at $T$ is $C_j(1+i)^{T-t_j}$. Values at the same date can then be added. For example, two receipts at times $0$ and $1$, accumulated to time $3$, become $C_0(1+i)^3$ and $C_1(1+i)^2$, respectively.

The uniqueness claim for nonnegative receipts can also be seen directly. Write $x=1+i$, so $x>0$. For positive integer accumulation periods $n_j$, the terminal value is a sum of terms $C_jx^{n_j}$. Its derivative with respect to $x$ is $\sum_j n_jC_jx^{n_j-1}$, which is positive when at least one positive receipt has $n_j>0$, and all receipts are nonnegative. Consequently, the sum is strictly increasing for $x>0$. The same monotonicity holds for positive fractional periods because $x^a$ increases with $x>0$ whenever $a>0$. A target outside the values attainable over the permitted rate range has no admissible solution.

Place every receipt on a timeline.
Record its amount and date, and identify the terminal date whose accumulated value is given.

Assign each accumulation period.
For a receipt at $t_j$, use $T-t_j$, not the receipt date itself or the total horizon for every receipt.

Set accumulated receipts equal to the target.
Write one term for each receipt and preserve the stated timing.

Solve and check admissibility.
Use algebra when practical, or a numerical solver when needed. Keep only roots with $1+i>0$, then substitute the rate into the original equation.

Algebraic Solutions

For integer accumulation periods, replacing $1+i$ with $x$ often turns the terminal-value equation into a polynomial. Solve for $x$, then convert back using $i=x-1$. This substitution can make the algebra simpler, but it does not remove the domain condition $x>0$. A polynomial may have more than one real root even when only one root represents a valid accumulation factor.

Worked Example: Two Receipts and a Quadratic Equation

A fund receives $100 at time 0 and $100 at time 1. The receipts are reinvested at a constant effective annual rate until time 2, when their accumulated value is $231. Find the rate. Use the stated no-fee, no-withdrawal reinvestment model.

Method: set up the terminal-value equation and solve using $x=1+i$. The first receipt earns interest for 2 years, while the second earns interest for 1 year:

$$ 100(1+i)^2+100(1+i)=231. $$

Let $x=1+i$. Divide by 100 and rearrange:

x^2+x=2.31,\qquad x^2+x-2.31=0.

The discriminant is $1-4(1)(-2.31)=10.24$, and $\sqrt{10.24}=3.2$. Therefore,

x=\frac{-1\pm3.2}{2}=1.1\quad\text{or}\quad -2.1.

Since $x=1+i$ must be positive, discard $-2.1$. Thus $i=1.1-1=0.10$, or 10% per year.

Check. At 10%, the time-0 receipt accumulates to $100(1.1)^2=$121$, and the time-1 receipt accumulates to $100(1.1)=$110$. Their sum is $231, as required. The common mistake avoided here is treating both receipts as though they earned interest for two years.

Using a Numerical Solver

When several receipts produce a polynomial that is inconvenient to factor, a financial calculator or numerical root solver is appropriate. Define a function as the accumulated receipts minus the target, then solve for the rate that makes the function zero. A bracket of rates with opposite signs for this function helps confirm that a root has been located; after solving, always substitute the rate into the original terminal-value equation.

Worked Example: A Numerical Solution for Several Receipts

A portfolio receives $60 at time 0, $100 at time 1, and $120 at time 2. It reinvests all receipts at one constant effective annual rate until time 4. The target accumulated value is $350. Find the rate to four decimal places as a percentage.

Method: form the rate equation and use Newton's method or a financial-calculator solver. The accumulation periods are 4, 3, and 2 years:

$$ 60(1+i)^4+100(1+i)^3+120(1+i)^2=350. $$

Define

$$ F(i)=60(1+i)^4+100(1+i)^3+120(1+i)^2-350. $$

A useful bracket is 8.0% to 8.3%. At 8.0%, the accumulated value is $60(1.08)^4+100(1.08)^3+120(1.08)^2 =60(1.36048896)+100(1.259712)+120(1.1664) =$347.5685376$, so $F(0.08)=-2.4314624$. At 8.3%, it is $60(1.083)^4+100(1.083)^3+120(1.083)^2 \approx $350.310675$, so $F(0.083)\approx0.310675$. The sign change brackets a solution.

For Newton's method, differentiate:

$$ F'(i)=240(1+i)^3+300(1+i)^2+240(1+i). $$

Starting at $i_0=0.08$, the first iteration is $i_1=i_0-F(i_0)/F'(i_0)$. Here $F'(0.08)=240(1.259712)+300(1.1664)+240(1.08)=911.45088$, giving $i_1=0.08-(-2.4314624)/911.45088\approx0.0826677$. A second iteration gives approximately $i_2=0.0826610$. Thus the rate is about 8.2661%.

Check. At $i=0.082661$, the accumulation factor is $1.082661$, and the three accumulated amounts are approximately $82.44, $126.90, and $140.66. Their total rounds to $350.00. The common mistake avoided is using a single power of $1+i$ for all three receipts instead of matching each exponent to its own time remaining.

Negative Rates and Admissible Roots

A solved rate need not be positive. Under the effective annual model, a rate between $-1$ and $0$ represents an accumulation factor between 0 and 1: a receipt's value decreases as it is carried forward. Negative rates can therefore be valid mathematical solutions when the target is below the total receipts available at a zero rate. The lower boundary is strict: $i=-1$ would give an accumulation factor of zero and is not an admissible effective annual rate.

Worked Example: Finding and Checking a Negative Rate

A hypothetical account receives $80 at time 0 and $100 at time 1. The receipts are accumulated to time 2 at a constant effective annual rate, and the target is $150. Find the admissible rate and verify it.

Method: solve the quadratic in the accumulation factor and apply its domain. The time-0 receipt earns for 2 years and the time-1 receipt for 1 year, so

$$ 80(1+i)^2+100(1+i)=150. $$

Set $x=1+i$. Dividing by 10 gives $8x^2+10x-15=0$. The quadratic formula gives

x=\frac{-10\pm\sqrt{100+480}}{16} =\frac{-10\pm\sqrt{580}}{16} =\frac{-5\pm\sqrt{145}}{8}.

Since $\sqrt{145}\approx12.04159458$, the positive root is $x\approx0.88019932$; the other root is negative and cannot be an accumulation factor. Therefore $i=x-1\approx-0.11980068$, or $-11.9801\%$ to four decimal places as a percentage.

Check. The positive root has $x^2\approx0.77475085$. The accumulated amounts are $80x^2\approx$61.98007$ and $100x\approx$88.01993$, totaling $150. A tempting but incorrect shortcut is to reject the result just because the rate is negative; the relevant condition is $1+i>0$, not $i\geq0$.

Common Setup and Solver Traps

Most errors in this calculation occur before the root-finding step. A calculator can solve the wrong equation accurately, so the timeline and model assumptions deserve as much attention as the numerical result.

Wrong accumulation period: for a receipt at time $t$, use $T-t$. A receipt one year before the terminal date earns for one year, even if the full investment horizon is longer.
Mixing present and terminal values: the equation here moves every receipt forward to $T$. Do not discount some receipts while accumulating others.
Forgetting receipts at the terminal date: include them with exponent zero; their value is unchanged.
Keeping an inadmissible root: solve for $x=1+i$ only with $x>0$. A negative mathematical root does not represent a valid effective accumulation factor.
Confusing the target with total receipts: the target is the sum of the accumulated amounts at $T$, not generally the unaccumulated sum of the receipts.
Rounding too early: retain sufficient calculator precision during root-finding and round only the reported rate or final money amount.

This method is useful whenever a target terminal value is specified and interim receipts are reinvested under a constant-rate assumption. It is aligned with the SOA Exam FM (Financial Mathematics) syllabus. The calculated rate describes the reinvestment assumption that matches the target; it does not by itself establish that such a rate will be realized in practice.

Key takeaway: Write one accumulated-value term for each receipt, using the time from that receipt to the common terminal date. Solve for $1+i>0$, convert to the effective annual rate, and verify the result in the original equation.

Check Your Understanding

For each problem, use the stated constant effective annual reinvestment model and check the receipt dates carefully.

A fund receives $100 at time 0 and $100 at time 1. Their accumulated value at time 2 is $231. Find the constant effective annual rate. (Answer: 10%)
A reserve receives $100 at time 0 and $50 at time 2. The receipts accumulate to time 3, where their value is $179.9712. Find the rate. (Answer: 8%)
An account receives $200 at time 0 and $100 at time 1. Their accumulated value at time 2 is $252. Find the admissible rate, including its sign. (Answer: $-10\%$)
A fund receives $75 at time 1 and $125 at time 3. At time 3, the accumulated value is $210. Find the constant effective annual rate to four decimal places as a percentage. (Answer: 6.4581%)

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