Deriving Return Equations From Cash-Flow Timelines

From a Cash-Flow Timeline to a Return Equation

A return calculation begins with a dated timeline, not with a formula chosen in isolation. The dates determine how long each cash flow earns interest or is discounted, while the signs determine whether each amount is invested by or paid to the investor. A misplaced date or an unstated assumption about an interim receipt changes the equation and can change the return.

In this tutorial, $i$ denotes the unknown effective annual rate. Unless a problem specifies another reinvestment rule, the return equation uses one constant annual effective rate over the measurement period. The rate is the yield that makes the signed cash flows balance at a common valuation date; its equation treats an interim positive cash flow as accumulating at that same rate when expressed at a later date. This is a mathematical convention of the yield calculation, not a guarantee that the investor can actually reinvest at $i$.

Definition: Let $C_k$ be the investor's net cash flow at time $t_k$, measured in years from the valuation date. An amount paid by the investor is negative; an amount received by the investor is positive. A yield equation sets the present values of these signed cash flows to zero: $$ \sum_{k=0}^{n} C_k(1+i)^{-t_k}=0. $$ Equivalently, at a chosen horizon $H$ at or after all the cash-flow dates, the accumulated-value equation is $$ \sum_{k=0}^{n} C_k(1+i)^{H-t_k}=0. $$

The two equations describe the same rate under the same constant-rate convention. They can be used when all cash-flow dates are measured from one common origin and $i>-1$, so the accumulation factors are positive. If the cash flows are not annual, use their actual times in years; for example, a cash flow at six months has $t_k=0.5$, not $t_k=1$. Cash flows on the same date can be combined into one net amount.

Why the Two Forms Agree

A present-value equation moves every cash flow to time 0. An accumulated-value equation moves every cash flow to time $H$. Under a constant annual effective rate, moving a value from time 0 to time $H$ multiplies it by $(1+i)^H$. Multiplying the entire present-value equation by that positive factor produces the accumulated-value equation:

(1+i)^H\sum_{k=0}^{n}C_k(1+i)^{-t_k} = \sum_{k=0}^{n}C_k(1+i)^{H-t_k}.

Thus, for $i>-1$, a rate that makes one equation zero also makes the other zero. Choosing the form that matches the information in the problem can make the timeline easier to read, but it cannot change the underlying rate. The argument depends on the same constant rate being used throughout. If an interim receipt is explicitly reinvested at another rate, the accumulated value of that receipt must use the specified rate instead.

Set the viewpoint and dates.
Write each cash flow from the investor's perspective and record its time in years.

Choose a common valuation date.
Use present value at time 0, or accumulation to a stated horizon $H$.

State the treatment of interim flows.
In a single-rate yield equation, the same unknown $i$ discounts or accumulates every flow. If the problem specifies a different reinvestment rate, use it for the relevant accumulated receipt.

Solve and verify.
Find a rate in the permitted domain $i>-1$, substitute it into the original cash-flow equation, and check that the two sides agree.

Worked Examples

Worked Example: Solving a Present-Value Return Equation

An investor pays \$1,000 at time 0. The investment pays \$600 at the end of year 1 and \$550 at the end of year 2. Find the annual effective yield $i$. Use the yield convention that interim receipts are valued using this same unknown rate.

Method. Set the signed present values to zero. The cash flows from the investor's perspective are $-1000$, $+600$, and $+550$, at times 0, 1, and 2. Therefore:

-1000+\frac{600}{1+i}+\frac{550}{(1+i)^2}=0.

Let $y=1+i$, with $y>0$. Multiplying through by $y^2$ gives:

$$ 1000y^2-600y-550=0. $$

The discriminant is $ (-600)^2-4(1000)(-550)=360{,}000+2{,}200{,}000=2{,}560{,}000$, whose square root is $1{,}600$. The roots are:

y=\frac{600\pm1{,}600}{2{,}000}, \qquad y=1.1\ \text{or}\ y=-0.5.

Since $y=1+i$ must be positive, reject $y=-0.5$. Thus $i=1.1-1=0.10$, or 10% per year.

Check. At 10%, the present value of the receipts is $600/1.1=545.4545\ldots$ and $550/1.21=454.5454\ldots$. Their sum is \$1,000. Equivalently, at time 2 the initial \$1,000 accumulates to $1000(1.1)^2=\$1,210$, while the receipts accumulate to $600(1.1)+550=\$1,210$. The equation balances in both forms.

Common mistake avoided: leaving the year-1 receipt undiscounted in the present-value equation. Its time-0 value is $600/(1+i)$, not \$600.

Worked Example: Building an Accumulated-Value Equation

An investor pays \$1,500 at time 0, receives a distribution of \$500 at the end of year 1, and receives a final payment of \$1,265 at the end of year 2. Find the annual effective yield using an accumulated-value equation at time 2. Assume the interim distribution is accumulated at the unknown yield.

Method. Accumulate the initial outflow for two years and the year-1 distribution for one year. The year-2 payment is already at the valuation date. The signed equation is:

$$ -1500(1+i)^2+500(1+i)+1265=0. $$

Set $y=1+i$, where $y>0$. Rearranging gives:

$$ 1500y^2-500y-1265=0. $$

The discriminant is $ (-500)^2-4(1500)(-1265)=250{,}000+7{,}590{,}000=7{,}840{,}000$, and its square root is $2{,}800$. Hence:

y=\frac{500\pm2{,}800}{3{,}000}, \qquad y=1.1\ \text{or}\ y=-\frac{23}{30}.

The negative value is inadmissible because $y>0$. Therefore $i=0.10$, or 10% per year.

Check. At 10%, the initial amount accumulates to $1500(1.1)^2=\$1,815$. The year-1 distribution accumulates to $500(1.1)=\$550$, and the year-2 payment is \$1,265. The total is $550+1265=\$1,815$.

If the \$500 distribution were instead explicitly reinvested at a specified 5% rate, its accumulated value at time 2 would be $500(1.05)=\$525$, not \$550. The accumulated amount from the cash flows would then be $525+1265=\$1,790$. That is a different reinvestment assumption; it should not be silently substituted into the single-rate yield equation.

Common mistake avoided: accumulating the year-1 receipt for two years. It is received at time 1, so it earns for one year—not two—before the time-2 valuation date.

Worked Example: Including a Later Contribution

An investor puts \$1,000 into an account at time 0, contributes a further \$200 at the end of year 1, and withdraws \$1,430 at the end of year 2. Find the annual effective yield, assuming both invested amounts accumulate at the same unknown rate.

Method. The contribution is an additional investor outflow, so it is negative in the signed equation. At time 2, the initial \$1,000 has accumulated for two years, the \$200 contribution for one year, and the \$1,430 withdrawal is already at time 2. The accumulated-value equation is:

$$ -1000(1+i)^2-200(1+i)+1430=0. $$

Let $y=1+i>0$. Rearranging:

$$ 1000y^2+200y-1430=0. $$

The discriminant is $200^2-4(1000)(-1430)=40{,}000+5{,}720{,}000=5{,}760{,}000$, with square root $2{,}400$. Thus:

y=\frac{-200\pm2{,}400}{2{,}000}, \qquad y=1.1\ \text{or}\ y=-1.3.

Only $y=1.1$ is positive, so the yield is $i=0.10$, or 10% per year.

Check. At 10%, the initial contribution grows to $1000(1.21)=\$1,210$, and the later contribution grows to $200(1.1)=\$220$. Together they equal $1210+220=\$1,430$, the withdrawal at time 2. The equivalent present-value equation is $-1000-200/1.1+1430/1.21=0$.

Common mistake avoided: treating the year-1 contribution as an investor receipt. It is money added by the investor, so its sign is negative; reversing that sign describes a different set of cash flows.

Reading the Assumption From the Equation

The exponent in a return equation records the length of time between a cash flow and the chosen valuation date. In a present-value equation, a flow at $t_k$ is multiplied by $(1+i)^{-t_k}$. In an accumulated-value equation at $H$, it is multiplied by $(1+i)^{H-t_k}$. These factors encode the single-rate assumption directly.

A problem may instead specify that a receipt is held without interest, reinvested at a separate rate, or paid out of the investment. Those instructions alter how the receipt is handled in an accumulation calculation. For example, if a receipt $C$ at time $t$ is reinvested to time $H$ at a specified effective annual rate $j$, its accumulated amount is $C(1+j)^{H-t}$. It is not $C(1+i)^{H-t}$ unless the specified reinvestment rate equals the yield $i$. Keep the return being solved for distinct from any separately stated reinvestment assumption.

Cash-flow feature	How it enters the equation	Question to ask
Initial investment by the investor	Negative signed cash flow	At what date was the amount paid?
Interim receipt under a single-rate yield convention	Positive flow, discounted or accumulated at $i$	How long does it earn or discount?
Additional investor contribution	Negative signed cash flow	Is it an outflow from the investor's perspective?
Receipt with a separately stated reinvestment rate	Accumulate at the specified rate for the stated interval	Does the problem override the single-rate convention?

Solving a return equation does not always guarantee a unique economically admissible rate. Substituting $y=1+i$ often turns the equation into a polynomial, and a polynomial can have multiple roots. Check every root against $y>0$, then substitute any admissible candidate into the original equation. A numerical answer that does not balance the original cash flows is not a valid solution.

This timeline-first approach is aligned with the SOA Exam FM (Financial Mathematics) syllabus. It is especially useful when a problem mixes receipts, withdrawals, or contributions: first identify the investor's cash flows and their dates, then state the treatment of interim amounts, and only then solve for the rate.

Key takeaway: A return equation is a dated, signed cash-flow balance. The dates determine the exponents, the signs reflect the investor's viewpoint, and the treatment of each interim receipt must be explicit. Present-value and accumulated-value equations give the same yield when they use the same constant effective rate.

Check Your Understanding

For each problem, identify the cash-flow dates and use the stated return or reinvestment convention.

An investor pays \$1,000 at time 0 and receives \$550 at the end of year 1 and \$605 at the end of year 2. Write the present-value yield equation and find $i$. (Answer: $-1000+550/(1+i)+605/(1+i)^2=0$; $i=10\%$.)
An investor pays \$800 at time 0, receives \$200 at the end of year 1, and receives \$748 at the end of year 2. Write the accumulated-value equation at time 2 and solve for the admissible yield. (Answer: $-800(1+i)^2+200(1+i)+748=0$; $i=10\%$.)
An investor pays \$500 at time 0, adds \$100 at the end of year 1, and receives \$715 at the end of year 2. Find the annual effective yield using an accumulated-value equation. (Answer: $-500(1+i)^2-100(1+i)+715=0$; $i=10\%$.)
An investor pays \$900 at time 0, receives \$200 at the end of six months, and receives \$800 at the end of year 1. Write the present-value yield equation using annual effective rate $i$. (Answer: $-900+200/(1+i)^{0.5}+800/(1+i)=0$.)
An investor pays \$1,200 at time 0, receives \$300 at the end of year 1, and receives \$1,000 at the end of year 2. The year-1 receipt is reinvested at an explicitly specified 4% effective annual rate. State its accumulated value at time 2, and write the single-rate yield equation for the cash flows. (Answer: accumulated receipt $=\$312$; yield equation $-1200+300/(1+i)+1000/(1+i)^2=0$.)

What You'll Learn