Break-Even Reinvestment Rates

Why a Break-Even Reinvestment Rate Is Useful

An investment may pay cash before the end of the investor’s chosen horizon. Those receipts contribute to terminal wealth only after they are reinvested. A break-even reinvestment rate is the constant rate at which those interim receipts must grow to make terminal wealth equal a specified target. The purchase price and horizon can be held fixed while the reinvestment rate is solved for.

The target might be a required terminal amount, or it might be constructed from a fixed initial investment and a chosen accumulation benchmark. The calculation does not assume that the original investment itself earns the reinvestment rate. It accumulates only the receipts that are actually reinvested, unless the cash-flow description explicitly says otherwise.

Definition: Suppose an investment has nonnegative interim receipts $C_k$ at times $t_k<T$, and those receipts are reinvested at a constant effective annual rate $i$. If time is measured in years, the terminal value at horizon $T$ of receipt $C_k$ is $C_k(1+i)^{T-t_k}$, for $i>-1$. The break-even reinvestment rate is a rate $i$ that makes total terminal wealth equal the specified target.

Let $K_T$ represent all amounts already fixed at the horizon date. This could include a final payment from the investment and any other horizon-date amount that does not depend on the reinvestment rate. If $W^*$ is the target terminal wealth, the equation to solve is

K_T+\sum_{k=1}^{n} C_k(1+i)^{T-t_k}=W^*.

The initial investment $I_0$ is held fixed as part of the scenario, but it does not appear in this terminal-value equation unless some or all of it is itself accumulated at rate $i$. For example, if the target is defined as the initial investment accumulated for $T$ years at a benchmark rate $g$, then $W^*=I_0(1+g)^T$. That defines the target; it does not change how the interim receipts are accumulated.

Building and Solving the Equation

For each receipt, use the time between its receipt date and the horizon. A receipt paid at year 1 and valued at year 4 earns reinvestment interest for 3 years, not 4. A receipt paid at the horizon has no reinvestment period and belongs in $K_T$. When dates are fractional, $T-t_k$ is still measured in years; for a constant effective annual rate, the accumulation factor is $(1+i)^{T-t_k}$.

Fix the horizon and target.
Record $T$, the required terminal wealth $W^*$, and all cash amounts that are already fixed at time $T$.

List interim receipts with their dates.
For each receipt $C_k$, record the elapsed reinvestment time $T-t_k$.

Accumulate each receipt separately.
Use $C_k(1+i)^{T-t_k}$, then add the fixed horizon-date amounts.

Solve and check.
Find an admissible rate $i>-1$, substitute it into the original equation, and confirm that the resulting terminal wealth equals the target.

Existence and Uniqueness

When receipts are nonnegative, the equation has useful structure. For each positive receipt paid before $T$, its terminal value increases as $i$ increases; a zero-valued receipt contributes a constant zero. The rate equation is therefore strictly increasing if at least one interim receipt is positive.

To see why, write the rate-dependent part as $A(i)=\sum_k C_k(1+i)^{T-t_k}$. For $i>-1$, its derivative is

A'(i)=\sum_{k=1}^{n} C_k(T-t_k)(1+i)^{T-t_k-1}.

Every term is nonnegative, and a term is strictly positive when $C_k>0$ and $t_k<T$. Thus $A'(i)>0$ if there is at least one positive interim receipt. As $i$ approaches $-1$ from above, each positive interim receipt’s accumulated value approaches zero. As $i$ increases without bound, the accumulated value of any positive interim receipt increases without bound. Consequently, if $K_T$ includes all fixed horizon-date amounts, there is exactly one admissible solution when $W^*>K_T$. If $W^*\leq K_T$, there is no admissible solution when at least one positive interim receipt must be accumulated: their terminal value is strictly positive for every $i>-1$.

If there are no positive interim receipts, the reinvestment rate cannot be identified from this equation: terminal wealth is simply $K_T$, independent of $i$. If cash flows include negative amounts that are reinvested or financed at the same unknown rate, monotonicity need not hold, so the existence-and-uniqueness conclusion above does not automatically apply.

Worked Example: One Interim Receipt and a Fixed Final Payment

A hypothetical investment requires an initial outlay of $1,000 at time 0. It pays $120 at the end of year 1 and a fixed $1,000 at the end of year 3. The investor’s horizon is 3 years, and the target terminal wealth is $1,270. Find the constant effective annual reinvestment rate for the interim receipt.

Method. The final $1,000 is already paid at the horizon, so it is fixed and does not earn reinvestment interest. The $120 receipt has $3-1=2$ years to accumulate. The initial outlay is fixed and is not accumulated at the reinvestment rate.

Set terminal wealth equal to the target:

$$ 1000+120(1+i)^2=1270. $$

Subtract the fixed final payment and divide by the receipt:

120(1+i)^2=270,\qquad (1+i)^2=\frac{270}{120}=2.25.

Since $i>-1$, the accumulation factor $1+i$ is positive. Therefore $1+i=\sqrt{2.25}=1.5$, giving $i=0.5$, or 50% effective annually.

Check. The receipt accumulates to $120(1.5)^2=120(2.25)=270$. Adding the fixed final payment gives $270+1{,}000=1{,}270$, as required. The common mistake avoided here is accumulating the final payment for an extra period.

Worked Example: Two Receipts and an Algebraic Solution

A hypothetical project has a fixed initial cost of $500. It pays $100 at the end of year 1, $200 at the end of year 2, and a fixed $500 at the end of year 3. The terminal-wealth target at year 3 is $841. Find the constant effective annual reinvestment rate for the interim receipts.

Method. The year-1 receipt is reinvested for 2 years, and the year-2 receipt for 1 year. The final $500 is already at the horizon. Let $x=1+i$, where $x>0$. Then

100x^2+200x+500=841, \qquad 100x^2+200x=341.

Divide by 100 and factor:

x^2+2x-3.41=0, \qquad (x-1.1)(x+3.1)=0.

The roots are $x=1.1$ and $x=-3.1$. Since $x=1+i$ must be positive for an admissible effective annual rate $i>-1$, select $x=1.1$. Thus $i=0.1$, or 10% effective annually.

Check. At 10%, the year-1 receipt accumulates to $100(1.1)^2=121$, and the year-2 receipt accumulates to $200(1.1)=220$. Total terminal wealth is $121+220+500=841$. The common mistake avoided is using the horizon year, rather than the receipt’s remaining time, as the exponent for every cash flow.

Worked Example: Fractional Receipt Dates

A hypothetical asset requires an initial outlay of $600. It pays $100 at time 0.5 years and another $100 at time 1.5 years. A fixed $400 is paid at the horizon of 2.5 years. Find the constant effective annual reinvestment rate that produces terminal wealth of $664.

Method. The first receipt has $2.5-0.5=2$ years to accumulate. The second has $2.5-1.5=1$ year. The rate is effective annually, so the fractional dates are measured in years, but here the remaining terms happen to be whole years.

$$ 100(1+i)^2+100(1+i)+400=664. $$

Let $x=1+i>0$. Subtract 400 and divide by 100:

x^2+x=2.64, \qquad x^2+x-2.64=0.

Factor the quadratic:

$$ (x-1.2)(x+2.2)=0. $$

The positive root is $x=1.2$, so $i=0.2$, or 20% effective annually.

Check. The first receipt accumulates to $100(1.2)^2=144$; the second accumulates to $100(1.2)=120$. Including the fixed $400 gives $144+120+400=664$. The common mistake avoided is assigning the first receipt 2.5 years of accumulation simply because the horizon is at 2.5 years.

When an Equation Needs Numerical Solving

With several distinct receipt dates, the equation may have more terms and may not factor into a convenient expression. When all receipts are nonnegative and at least one positive receipt arrives before the horizon, the strict-increase result gives a reliable way to solve: choose two admissible trial rates whose terminal values lie on opposite sides of the target, then narrow the interval by bisection or use a financial calculator or spreadsheet solver. The rate should always be substituted into the original cash-flow equation afterward.

If the calculated terminal value is below target, the trial rate is too low; if it is above target, the trial rate is too high. This bracketing interpretation follows directly from the increasing terminal value of positive interim receipts. A numerical answer should be reported with appropriate precision, and the final check should use the unrounded rate where possible.

Common Exam Traps

Using the wrong accumulation period. A receipt at $t_k$ earns interest for $T-t_k$ years, not $t_k$ years and not automatically $T$ years.
Accumulating a horizon-date payment. A payment received at $T$ has exponent zero; it is a fixed amount in the terminal-value equation.
Accumulating the initial outlay without justification. The initial investment is not a reinvested receipt. Include its accumulation only if the stated model explicitly says it remains in an account earning the unknown rate.
Choosing an inadmissible algebraic root. With $x=1+i$, an effective annual rate $i>-1$ requires $x>0$. A negative root for $x$ is not an acceptable accumulation factor.
Ignoring feasibility. If the target is at or below the fixed horizon-date amount while positive interim receipts must be accumulated, no rate $i>-1$ can meet it.

Key takeaway: Keep the initial investment and horizon fixed, separate rate-independent horizon-date amounts from interim receipts, and accumulate each interim receipt only for the time remaining until the horizon. Solve the resulting terminal-wealth equation and verify it using the original cash flows.

Check Your Understanding

For each problem, assume a constant effective annual reinvestment rate $i>-1$. Treat stated final payments as fixed horizon-date amounts.

A hypothetical fund pays $200 at the end of year 1 and a fixed $1,000 at the end of year 3. Its target terminal wealth at year 3 is $1,242. Find the break-even reinvestment rate. (Answer: 10% effective annually)
A project pays $100 at the end of year 1, $100 at the end of year 2, and a fixed $500 at the end of year 3. Its target terminal wealth is $731. Find the rate. (Answer: 10% effective annually)
An asset pays $250 at time 0.5 years and $150 at time 1.5 years, with a fixed $300 payment at the 2.5-year horizon. The target terminal wealth is $840. Find the rate. (Answer: 20% effective annually)
A portfolio has a fixed horizon-date payment of $700 and positive interim receipts that must be accumulated to the same horizon. Can any rate $i>-1$ produce target wealth of $695? (Answer: No; the accumulated interim receipts are positive, so terminal wealth must exceed $700.)
A project pays $100 at the end of year 1, $200 at the end of year 2, and a fixed $300 at the end of year 4. Its target terminal wealth is $675.10. Find the rate. (Answer: 10% effective annually)

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