Workshop On Missing Payments And Terms

Putting a Missing Payment or Term on a Timeline

A value equation becomes manageable when every cash flow is placed on a timeline and valued at the same date. In a problem with one missing payment, the equation is usually linear in that payment. With one missing term under compound interest, the unknown appears in an exponent, so the equation is solved by recognizing a power or using logarithms. In both cases, the essential first step is the same: identify the cash-flow dates, the interest model, and a focal date.

The examples below assume the stated interest rate is known and applies over the intervals shown. Time is measured in years, and the rate and time units agree. A payment made at the focal date is already valued there; a payment made earlier is accumulated forward to that date. These equations concern value equivalence, not the solution of an unknown interest rate.

Definition: A focal date is the date at which all cash flows in a value equation are expressed. Under compound interest at effective annual rate $i$, a cash flow $C$ at time $t$, accumulated to a later focal time $T$, has value $C(1+i)^{T-t}$. Under simple interest at annual rate $i$, its value at $T$ is $C[1+i(T-t)]$. These formulas assume $T\ge t$; if a cash flow must be valued backward, use the corresponding present-value relation for the stated model.

A Reliable Setup Process

Choose a focal date that makes the equation easy to write. A date at which a known payment occurs is often convenient, but it is not required. Draw and label a timeline before calculating: this makes it easier to see how long each cash flow earns interest. Then value each cash flow separately at the focal date and equate the values on the two sides of the transaction or schedule.

Mark dates and cash-flow directions.
Record each payment time and amount. Keep an unknown payment as a variable, such as $X$, rather than guessing its value.

State the interest model.
Use the stipulated simple or compound interest rule. Do not replace one model with the other, even if the numerical rate is the same.

Choose one focal date.
Move every payment to that date using the correct elapsed time. A payment at the focal date receives a factor of 1.

Solve and check.
Isolate the single unknown, then substitute the result into the original value equation and verify that both sides agree.

Why Each Cash Flow Is Valued Separately

Under compound interest, a principal amount $C$ earns interest for $T-t$ years, so its accumulated value is $C(1+i)^{T-t}$. If several payments are made at different times, each payment has its own starting date and its own accumulation period. Since values at a common date can be added, the focal-date value of payments $C_1,\ldots,C_n$ made at times $t_1,\ldots,t_n$ is the sum of their individual accumulated values.

\text{Value at }T=\sum_{k=1}^{n} C_k(1+i)^{T-t_k}, \qquad T\ge t_k.

The same additivity applies under simple interest, but the accumulation factor is $1+i(T-t_k)$ for each payment. Simple-interest accumulation is not obtained by repeatedly applying a one-year factor: the model uses the total elapsed time from the payment date to the focal date. In either model, the equation expresses that the total value of one set of cash flows equals the total value of the other set at the chosen focal date.

Worked Example: Finding a Missing Payment Under Compound Interest

At an effective annual interest rate of 5% compounded annually, a schedule contains a payment of $1,200 at time 1, an unknown payment $X$ at time 2, and a payment of $1,609.65 at time 4. Find $X$, given that the first two payments accumulate to the value of the payment at time 4.

Method. Accumulate the two earlier payments to time 4. This focal date is convenient because the final payment is already stated at that date.

The $1,200 payment earns interest for $4-1=3$ years. The unknown payment earns interest for $4-2=2$ years. Therefore the value equation is

$$ 1200(1.05)^3+X(1.05)^2=1609.65. $$

Evaluate the known accumulation factors:

(1.05)^3=1.157625,\qquad (1.05)^2=1.1025.

The first payment accumulates to $1200(1.157625)=1389.15$. Subtract this value from both sides:

$$ 1.1025X=1609.65-1389.15=220.50. $$

Thus $X=220.50/1.1025=200$, so the missing payment is $200.00.

Substitution check. At time 4, the $1,200 payment is worth $1,389.15 and the $200 payment is worth $200(1.1025)=\$220.50$. Their total is $1389.15+220.50=\$1,609.65$, exactly the stated value. This check also confirms that each payment was accumulated for the correct number of years.

Worked Example: Finding a Missing Payment Under Simple Interest

A fund is valued at time 3. It received $500 at time 0 and an unknown amount $X$ at time 1. Using simple interest at 6% per year, the accumulated value of those deposits at time 3 is $814. Find $X$.

Method. Use time 3 as the focal date and apply simple interest separately to each deposit. The $500 deposit earns simple interest for 3 years, while $X$ earns it for $3-1=2$ years.

$$ 500[1+0.06(3)]+X[1+0.06(2)]=814. $$

The first factor is $1+0.18=1.18$, so the first deposit accumulates to $500(1.18)=590$. The second factor is $1+0.12=1.12$. Subtract $590 from both sides:

1.12X=814-590=224, \qquad X=\frac{224}{1.12}=200.

The missing deposit is $200.00.

Substitution check. The first deposit accumulates to $500(1.18)=\$590$. The second accumulates to $200(1.12)=\$224$. Their sum is $814, as required.

Finding an Unknown Term Under Compound Interest

If the interest rate and payment amounts are known but a date is unknown, the unknown elapsed time appears in the exponent of a compound-interest factor. First use the timeline to write the correct exponent for each cash flow. If the terms can be factored into one power, the problem may reduce to an equation of the form $(1+i)^n=K$. When no convenient power is apparent and $i>0$, logarithms can isolate the exponent:

(1+i)^n=K \quad\Longrightarrow\quad n=\frac{\ln K}{\ln(1+i)}, \qquad K>0.

This formula applies when the equation has been correctly reduced to a single exponential term and $K$ is positive. If the original equation contains a sum of powers with the unknown in different exponents, it may require algebraic factoring or a numerical solution; do not apply the logarithm formula to only part of that equation.

Worked Example: Finding a Term From Two Deposits

A deposit of $500 is made at time 0, and a second deposit of $600 is made at time 1. At an effective annual rate of 8% compounded annually, their combined accumulated value is $1,329.696 at an unknown time $T$, where $T>1$. Find $T$.

Method. Accumulate each deposit to time $T$. The first deposit earns interest for $T$ years; the second earns interest for $T-1$ years.

500(1.08)^T+600(1.08)^{T-1}=1329.696.

Factor out the common power $(1.08)^{T-1}$. The first term contributes $500(1.08)(1.08)^{T-1}=540(1.08)^{T-1}$, so

(540+600)(1.08)^{T-1}=1329.696, $$ $$ 1140(1.08)^{T-1}=1329.696.

Divide by 1,140:

(1.08)^{T-1}=\frac{1329.696}{1140}=1.1664=(1.08)^2.

Because the base 1.08 is greater than 1, its powers are strictly increasing. Therefore $T-1=2$, giving $T=3$ years.

Substitution check. At time 3, the first deposit is worth $500(1.08)^3=500(1.259712)=\$629.856$. The second is worth $600(1.08)^2=600(1.1664)=\$699.84$. Their sum is $629.856+699.84=\$1,329.696$, which matches the stated accumulated value.

Common Mistake: Switching the Interest Model

The interest model is part of the problem data, not a choice made for convenience. In the simple-interest example, using compound interest by mistake would produce

500(1.06)^3+X(1.06)^2=814, \qquad X=\frac{814-595.508}{1.1236}\approx194.46.

That answer does not satisfy the stated simple-interest model. The correct calculation uses $500[1+0.06(3)]$ and $X[1+0.06(2)]$, giving $X=\$200$. The incorrect result is not fixed by rounding: it comes from valuing the cash flows under a different accumulation rule.

Other errors often arise before the arithmetic begins. A payment at time 1 accumulated to time 4 earns interest for 3 years, not 4. A payment already at the focal date has no accumulation period. And under simple interest, each payment must be valued from its own payment date to the focal date; do not repeatedly apply a one-year simple-interest factor as if the model were compound interest.

Key takeaway: Draw the timeline, state whether the model is simple or compound interest, and write one focal-date value for each payment. After solving for the single unknown, substitute it into that same equation to confirm both the timing and the arithmetic.

Check Your Understanding

For each problem, write the focal-date equation before solving. Check that the stated interest model and elapsed time for every payment are used consistently.

At 5% simple interest, $1,200 is deposited at time 0 and $X$ at time 2. Their accumulated value at time 5 is $1,730. Find $X$. (Answer: $200.00)
At 4% compound interest, $700 is deposited at time 0 and $X$ at time 1. Their combined value at time 3 is $895.5648. Find $X$. (Answer: $100.00)
At 5% compound interest, $400 is deposited at time 0 and $500 at time 1. Their combined accumulated value is $1,014.30 at time $T>1$. Find $T$. (Answer: 3 years)
A single deposit of $900 accumulates to $1,089 under compound interest at 10% per year. Find the number of years. (Answer: 2 years)
At 4% compound interest, $600 is deposited at time 0 and $X$ at time 2. Their combined value at time 5 is $842.47814144. Find $X$. (Answer: $100.00)

What You'll Learn