Zero-Interest and Final-Payment Balance Cases

Why These Balance Cases Need Care

Outstanding-balance formulas often use an annuity factor or divide by an interest rate. Those forms are useful when the rate is positive, but a zero rate makes division by the rate invalid. A final installment that differs from earlier payments creates a different issue: the final payment must be calculated from the balance actually due at its payment date, not assumed to equal the preceding installment.

The reliable approach is to track the balance from one payment date to the next. This works at a zero rate, with unequal payments, and when the final payment is a balloon payment. Throughout, assume a loan is advanced at time 0, interest is credited at a fixed effective rate $i$ per payment period, and payments are made at the ends of periods. Assume no fees, penalties, or other cash flows unless stated. A payment is an amount paid by the borrower to reduce the debt.

Definition: The outstanding balance immediately after the $k$th payment is the amount of debt remaining just after that payment has been applied. If the balance after period $k-1$ is $B_{k-1}$, interest accrues for one period at rate $i$, and the payment at time $k$ is $P_k$, then $$ B_k=B_{k-1}(1+i)-P_k,\qquad B_0=L, $$ where $L$ is the original loan principal. The balance immediately before the payment at time $k$ is $B_{k-1}(1+i)$. These are different amounts.

Zero Interest: Subtract Payments Directly

When $i=0$, the balance earns no interest between payment dates. The recurrence therefore reduces to

B_k=B_{k-1}-P_k.

Repeated application gives a direct sum: after payments $P_1,\ldots,P_k$, the balance is the original principal less the total payments made. Payment timing within the sequence still matters for identifying which payments have occurred by the balance date, but it does not change the amount of interest, since there is none.

B_k=L-\sum_{j=1}^{k}P_j\qquad\text{when }i=0.

This formula does not require equal payments. If the balance immediately before the final payment is $C$, a final payment of $C$ clears the loan exactly. If the borrower instead pays less than $C$, a positive balance remains; paying more than $C$ produces an overpayment under this model.

Worked Example: A Zero-Interest Loan With a Different Final Payment

A lender advances $ \$12{,}000 $ at a zero effective annual rate. The borrower pays $ \$2{,}300 $ at the end of each of years 1 through 4, then makes a final payment at the end of year 5. Find the balance after the third payment and the required final payment.

Method. At a zero rate, subtract only the payments that have already been made. After three payments, $$ B_3=12{,}000-3(2{,}300)=12{,}000-6{,}900=5{,}100. $$

The fourth payment is $ \$2{,}300 $, so the balance immediately after it is $$ B_4=5{,}100-2{,}300=2{,}800. $$ No interest accrues during year 5. Thus the final payment that clears the loan is $ \$2{,}800 $, and the balance immediately after that payment is $$ B_5=2{,}800-2{,}800=0. $$

Check. The five payments total $4(2{,}300)+2{,}800=9{,}200+2{,}800=12{,}000$, equal to the original principal. The common mistake this method avoids is treating the final payment as another $ \$2{,}300 $ installment without checking whether it repays the remaining balance.

Deriving the Balance Recurrence

The recurrence follows directly from the order of events in each period. First the balance after the previous payment accrues interest; then the current payment reduces the amount owed. Starting with $B_0=L$, the balance before payment 1 is $L(1+i)$, so after paying $P_1$ it is $B_1=L(1+i)-P_1$. In the next period, $B_1$ accrues interest and payment $P_2$ is subtracted, giving $B_2=B_1(1+i)-P_2$. The same reasoning applies to each subsequent period, which proves the recurrence.

For a final payment at time $n$, suppose the preceding payments are $P_1,\ldots,P_{n-1}$, and let $Q$ be the payment at time $n$. The balance just after payment $n-1$ is $B_{n-1}$. It accrues one period of interest before the final payment is due. Therefore, to clear the loan,

Q=B_{n-1}(1+i).

This expression is the amount due before the final payment; once $Q$ is paid, the balance is zero. At a zero rate, it simplifies to $Q=B_{n-1}$. The formula assumes the final payment occurs one full payment period after the preceding payment. If payment dates are irregular, the interest factor must match the elapsed time and the rate convention.

Start with the correct balance date.
Identify whether the question asks for the balance just before or just after a payment.

Accrue interest for the period.
Multiply the previous after-payment balance by $1+i$. At a zero rate, this step leaves the balance unchanged.

Subtract the payment made at that date.
The resulting amount is the balance immediately after that payment.

Verify the final payment.
The final payment should equal the balance immediately before it if it is intended to pay off the loan exactly.

Worked Example: Finding a Final Payment at a Positive Rate

A loan of $ \$20{,}000 $ accrues interest at an effective annual rate of $5\%$. The borrower pays $ \$5{,}000 $ at the end of each of years 1, 2, and 3. Find the payoff payment due at the end of year 4.

Method. Apply the recurrence period by period, using $i=0.05$. After year 1, $$ B_1=20{,}000(1.05)-5{,}000=21{,}000-5{,}000=16{,}000. $$ After year 2, $$ B_2=16{,}000(1.05)-5{,}000=16{,}800-5{,}000=11{,}800. $$ After year 3, $$ B_3=11{,}800(1.05)-5{,}000=12{,}390-5{,}000=7{,}390. $$

The balance due immediately before the year-4 payment is $B_3(1.05)$: $$ Q=7{,}390(1.05)=7{,}759.50. $$ So the required final payment is $ \$7{,}759.50 $. The balance immediately after that payment is $ \$7{,}759.50-\$7{,}759.50=0 $.

Check. Interest during year 4 is $0.05(7{,}390)=\$369.50$, so the amount owed rises from $ \$7{,}390 $ to $ \$7{,}759.50 $ before the payoff. The common mistake avoided here is using the after-year-3 balance itself as the year-4 payment and forgetting one more year of interest.

Unequal Payments and the Final Balance

The recurrence is especially useful when installments are unequal, because it makes no level-payment assumption. For example, if the first payments are specified individually, calculate each successive balance before finding the final payment. Do not substitute an annuity formula that represents a different payment pattern.

A useful expanded form follows by repeatedly substituting the recurrence:

B_k=L(1+i)^k-\sum_{j=1}^{k}P_j(1+i)^{k-j}.

To justify the formula, the original loan grows for $k$ periods to $L(1+i)^k$. A payment $P_j$ made at time $j$ would, if left in the account, grow for $k-j$ more periods to $P_j(1+i)^{k-j}$. Since each payment reduces the debt, subtract each accumulated payment from the accumulated original principal. At $i=0$, every factor $(1+i)^{k-j}$ equals 1, recovering the direct subtraction formula.

Worked Example: Unequal Earlier Payments and a Balloon Payment

A borrower receives $ \$12{,}000 $ at time 0. The effective annual interest rate is $6\%$. Payments of $ \$2{,}000 $ and $ \$3{,}000 $ are made at the ends of years 1 and 2, respectively. Determine the final payment at the end of year 3 that pays off the loan.

Method. Calculate the after-payment balance each year. At the end of year 1, $$ B_1=12{,}000(1.06)-2{,}000=12{,}720-2{,}000=10{,}720. $$ At the end of year 2, $$ B_2=10{,}720(1.06)-3{,}000=11{,}363.20-3{,}000=8{,}363.20. $$

The balance accrues one more year of interest before the final payment: $$ Q=8{,}363.20(1.06)=8{,}864.992. $$ The exact amount is $ \$8{,}864.992 $. In dollars, rounded to the nearest cent, the final payment is $ \$8{,}864.99 $. If the contract accepts only cent-denominated payments, the remaining fraction of a cent is handled according to its rounding convention; the unrounded calculation identifies the mathematical payoff amount.

Check. The year-3 interest is $0.06(8{,}363.20)=\$501.792$, and $8{,}363.20+501.792=8{,}864.992$. This check confirms both the interest period and the amount of the balloon payment. The common mistake avoided is subtracting the year-2 payment from the year-1 balance before accruing year-2 interest.

Common Mistakes and Exam Checks

A zero rate is a special case of the balance recurrence, not a reason to force a positive-rate formula into an undefined form. For instance, the annuity factor $a_{\overline{n}|}=(1-v^n)/i$ contains division by $i$, so direct substitution of $i=0$ is invalid. For zero-interest balances, use the payment sum directly. Although a limiting value can be established, it is unnecessary for this calculation.

A second frequent error is labeling a balance before the last payment as the balance after it. Suppose a zero-interest loan starts at $ \$14{,}700 $, and payments of $ \$4{,}200 $ and $ \$5{,}000 $ have been made. The balance before a proposed final payment is $$ 14{,}700-4{,}200-5{,}000=5{,}500. $$ If the borrower then pays $ \$5{,}500 $, the balance after that payment is $ \$5{,}500-\$5{,}500=0 $, not $ \$5{,}500 $. The amount due and the remaining balance are not interchangeable: state the balance date explicitly.

Before finalizing a calculation, check the following:

Is the stated rate per payment period, or must it be converted?
Have you accrued interest before subtracting a payment made at the period end?
Does the question ask for the balance just before a payment or just after it?
Does the proposed final payment reduce the balance to zero, or does it leave a surplus or shortfall?

These checks are useful in any outstanding-balance problem, particularly when a contract has a zero rate or a payment schedule that ends with a different amount. Correctly applying the recurrence keeps the calculation tied to the actual cash flows rather than to an assumed level-payment pattern. The treatment here is aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Check Your Understanding

For each problem, identify the balance date carefully and use the stated rate per period.

A zero-interest loan of $ \$9{,}400 $ has three payments of $ \$1{,}800 $, followed by a final payment at the end of period 4. Find the final payment that clears the loan. (Answer: $ \$4{,}000 $.)
A zero-interest loan of $ \$16{,}500 $ receives payments of $ \$2{,}500 $, $ \$3{,}400 $, and $ \$4{,}100 $ at the ends of periods 1, 2, and 3. Find the payoff payment at the end of period 4. (Answer: $ \$6{,}500 $.)
A $ \$10{,}000 $ loan accrues interest at $7\%$ effective annually. Payments of $ \$2{,}000 $ and $ \$3{,}000 $ are made at the ends of years 1 and 2. Find the payoff payment at the end of year 3. (Answer: $ \$6{,}750.63 $.)
A $ \$25{,}000 $ loan accrues interest at $4\%$ effective annually. The borrower pays $ \$6{,}000 $ at each of the ends of years 1, 2, and 3, then pays the remaining balance at the end of year 4. Find that final payment. (Answer: $ \$9{,}767.68 $.)
A zero-interest loan of $ \$12{,}800 $ has payments of $ \$2{,}500 $, $ \$3{,}000 $, and $ \$2{,}800 $ at the ends of periods 1, 2, and 3, with a payoff at the end of period 4. Find the balance after period 2 and the final payment. (Answer: Balance after period 2 is $ \$7{,}300 $; final payment is $ \$4{,}500 $.)

What You'll Learn