Amortization at Zero Interest

What Changes When the Interest Rate Is Zero?

In a standard level-payment amortizing loan, each period’s payment covers the interest due on the outstanding balance and repays part of the principal. If the interest rate per payment period is zero, there is no interest charge to allocate. Each payment therefore goes entirely toward principal. With a fixed term and payments at the end of each period, the borrower repays the original principal in equal amounts.

The timing assumption matters: throughout this tutorial, payments are made at the end of each payment period. The loan has no fees or other charges, and no payments are missed. The interest rate is zero for every period in the schedule. A loan that has a zero rate for only part of its term, or that includes fees, does not satisfy this model.

Definition: A zero-interest amortizing loan has an initial principal $L$, a term of $n$ equal payment periods, and an interest rate of zero per payment period. If payments are made at each period’s end and fully repay the loan, the principal repaid in each period is $L/n$. Because there is no interest, the total payment in each period is also $L/n$.

Let $B_k$ denote the outstanding balance immediately after the $k$th payment, where $B_0=L$. If the level payment is $R$, the zero-rate balance recurrence is $B_k=B_{k-1}-R$. The final balance must be zero after $n$ payments. These conditions determine the payment and every intermediate balance.

B_k=L-kR,\qquad B_n=0 \quad\Longrightarrow\quad R=\frac{L}{n} \quad\text{and}\quad B_k=L\left(1-\frac{k}{n}\right).

The formula gives the balance after payment $k$. Immediately before that payment, the balance is the same as immediately after payment $k-1$, since the rate is zero and no interest accrues between those two points. The principal portion of every payment is $R$, the interest portion is zero, and the balance falls by the same amount each period.

Why the Zero-Rate Formula Holds

The result follows directly from the balance recurrence. Since there is no interest, the balance before a payment is not increased by an interest factor; subtracting the payment is the only change. Applying this step repeatedly gives a balance after $k$ payments of $L-kR$. Full repayment requires that the balance after payment $n$ be zero.

B_1=L-R,\quad B_2=L-2R,\quad \ldots,\quad B_k=L-kR, \qquad 0=B_n=L-nR.

Solving the final equation yields $R=L/n$. Substituting that amount into the balance after $k$ payments gives $B_k=L-kL/n$, or $L(1-k/n)$. This also explains the schedule’s shape: each equal payment removes one $n$th of the original principal, so the balance declines linearly rather than geometrically.

The same payment can be obtained by valuing the payments at the start of the loan. At a zero rate, a dollar paid at any future date has the same value as a dollar now under this model. Thus the value of $n$ equal payments is $R+R+\cdots+R=nR$, which must equal $L$. This direct argument applies at zero without dividing by an interest rate.

Important: For a positive rate $i$ per payment period, the level-payment formula is $R=L/a_{\overline{n}|}$, with $a_{\overline{n}|}=(1-v^n)/i$ and $v=1/(1+i)$. Substituting $i=0$ into $(1-v^n)/i$ gives $0/0$, not a usable value. At zero interest, derive the payment from the cash flows directly: $L=nR$, so $R=L/n$.

Constructing a Zero-Interest Schedule

A schedule records, period by period, the opening balance, interest, payment, principal repaid, and closing balance. Under the assumptions above, interest is zero in every row, and the payment equals principal repaid. The table is therefore particularly easy to check: each closing balance must equal the previous balance minus $L/n$, and the final balance must be zero.

Worked Example: Four Semiannual Payments With No Interest

A hypothetical lender advances \$9,600 on a loan with a zero effective rate per semiannual payment period. The borrower makes four equal payments at the end of successive half-years. Find the payment and construct the full schedule. Also determine the balance immediately after the third payment.

Method. There are $n=4$ payment periods, and the interest rate per period is zero. The equal principal installment is the original principal divided by the number of payments:

R=\frac{\$9{,}600}{4}=\$2{,}400.

In every period, interest is $\$0$, so the entire $\$2,400$ payment repays principal. Subtracting $\$2,400$ successively gives the schedule:

Payment	Opening balance	Interest	Total payment	Principal repaid	Closing balance
1	\$9,600	\$0	\$2,400	\$2,400	\$7,200
2	\$7,200	\$0	\$2,400	\$2,400	\$4,800
3	\$4,800	\$0	\$2,400	\$2,400	\$2,400
4	\$2,400	\$0	\$2,400	\$2,400	\$0

After the third payment, the balance is $\$2,400$. As a check, three payments have repaid $3(\$2,400)=\$7,200$, leaving $\$9,600-\$7,200=\$2,400$. The four payments total $\$9,600$, exactly the amount advanced, so total interest is zero.

Common mistake avoided: Do not count the initial advance as one of the four payments. There are four payment periods after the loan is issued, so the principal is divided by four, not by five.

Balances and Repayment at an Intermediate Date

For any completed payment number $k$, the cumulative principal repaid is $kL/n$, and the remaining balance is $L-kL/n$. The number of payments still due is $n-k$, each of size $L/n$. These provide independent checks on a calculation: the remaining balance must equal both the balance formula and the total of the remaining zero-interest payments.

Worked Example: Finding the Balance and Remaining Payments

A hypothetical interest-free equipment loan has principal \$32,400 and is repaid by 18 equal monthly payments at month-end. Find the monthly payment, the balance immediately after payment 11, the total principal repaid by then, and the total of the remaining payments. Assume no fees or other charges.

Method. The rate per monthly period is zero, so the payment is the principal divided by 18:

R=\frac{\$32{,}400}{18}=\$1{,}800.

Eleven payments have repaid $11(\$1,800)=\$19,800$ of principal. Therefore the balance immediately after payment 11 is:

B_{11}=\$32{,}400-\$19{,}800=\$12{,}600.

There are $18-11=7$ payments left. Their total is $7(\$1,800)=\$12,600$, which agrees with the outstanding balance. The full repayment check is $18(\$1,800)=\$32,400$, and total interest is $\$0$.

Common mistake avoided: The balance after payment 11 is based on 11 payments already made, not 11 payments remaining. Using the remaining-payment count, seven, gives the same balance only when it is multiplied by the payment amount.

Comparison With Positive-Rate Amortization

A zero rate makes the total payment and principal portion identical. At a positive rate, each payment first covers the interest accrued on the opening balance; only the remainder repays principal. For a level-payment loan, the total payment can stay constant even though the interest and principal portions change from period to period. As the balance declines, interest charges decline, and the principal portion of the level payment increases.

The next example uses a positive rate only to make this contrast explicit. Assume an annual effective interest rate of 10%, two annual payments at year-end, no fees, and full repayment with level payments.

Worked Example: Comparing Two Annual Payments at 10%

A hypothetical borrower receives \$10,000 and repays it with two equal annual payments. Find the payment and the interest and principal portions in each year. Compare the result with the same principal and term at zero interest.

Method. At 10% per year, the value at loan issue of the two payments is $R/1.10+R/1.10^2$. Equating this value to the \$10,000 advance gives:

\$10{,}000 =R\left(\frac{1}{1.10}+\frac{1}{1.21}\right) =R\left(\frac{110}{121}+\frac{100}{121}\right) =R\frac{210}{121}.

Thus $R=\$10,000(121/210)=\$121,000/21=\$5,761.904761\ldots$. Keep this exact payment for the schedule calculations; the payment reported to the nearest cent is \$5,761.90.

During year 1, interest is 10% of the opening balance: $0.10(\$10,000)=\$1,000$. The first payment’s principal portion is therefore:

\frac{\$121{,}000}{21}-\$1{,}000 =\frac{\$100{,}000}{21} =\$4{,}761.904761\ldots.

The balance after payment 1 is $\$10,000-\$100,000/21=\$110,000/21=\$5,238.095238\ldots$. Year 2 interest is 10% of that balance, or $\$11,000/21=\$523.809523\ldots$. The second payment’s principal portion is the balance then due, $\$110,000/21$, leaving a zero balance after payment.

Total interest over the two years is $\$1,000+\$11,000/21=\$32,000/21=\$1,523.809523\ldots$, or \$1,523.81 to the nearest cent. The two unrounded payments total $\$242,000/21=\$11,523.809523\ldots$, equal to the original \$10,000 plus total interest.

At zero interest, the same \$10,000 loan over two annual periods would instead require payments of $\$10,000/2=\$5,000$ each, with zero interest in both years. The positive-rate payment is larger because it covers interest as well as principal.

Common mistake avoided: Using the zero-rate payment of \$5,000 for each year at 10% would not repay the loan in two equal payments. After year 1 the balance would be $\$10,000(1.10)-\$5,000=\$6,000$; after another year it would be $\$6,600$ before the second payment. The positive-rate payment must account for interest over the full term.

Exam Traps and Practical Checks

Zero-interest problems look simple, but they still test careful interpretation of the payment period, the number of payments, and the point in time at which a balance is measured. In particular, the rate must be zero for the relevant payment period. If a problem gives a rate for another time unit, establish the rate per payment period before using an amortization formula. A zero annual rate is also zero for shorter periods; a positive annual rate is not zero merely because payments are more frequent.

Do not evaluate a positive-rate formula by direct substitution. The expression $(1-v^n)/i$ is undefined at $i=0$. Use the direct sum of payment values at zero interest, or the balance recurrence.
Do not confuse payment count and elapsed time. With 12 monthly payments, there are 12 installments even though the first payment is not made at the loan’s issue date.
Check what “after payment” means. After $k$ payments, the balance is $L-kR$. Before the next payment, the balance is unchanged at a zero rate.
Separate principal and interest explicitly. At zero interest, principal repaid equals the whole payment. At a positive rate, the payment includes interest and its principal portion is smaller than the total payment.
Verify the terminal balance. A correct full-term schedule ends at zero, and the total principal repaid equals the original advance.

Key takeaway: For a fully amortized loan with zero interest per payment period and equal end-of-period payments, each payment is $L/n$, all of it repays principal, and the balance after $k$ payments is $L(1-k/n)$. Positive-rate schedules require interest to be included and do not follow this linear balance pattern.

Check Your Understanding

For each problem, assume payments are made at period-end and there are no fees or other charges unless stated otherwise.

A hypothetical loan of \$15,750 bears zero interest per annual period and is repaid in nine equal annual payments. Find the annual payment and the balance immediately after payment 4. (Answer: \$1,750 per payment; balance \$8,750.)
A \$46,800 interest-free loan is repaid in 24 equal monthly payments. How many payments remain immediately after payment 17, and what is the balance then? (Answer: 7 payments remain; balance \$13,650.)
A hypothetical \$8,400 loan bears zero interest per quarterly period and is repaid in seven equal quarterly payments. Find the payment, principal repaid in the first three quarters, and total interest. (Answer: \$1,200 per payment; \$3,600 principal repaid; \$0 interest.)
A \$18,000 loan is repaid with three equal annual payments. First find the payment at zero interest. Then find the level annual payment if the effective annual rate is 5%. (Answer: \$6,000 at zero interest; \$6,609.75 at 5%.)
A \$21,600 zero-interest loan is repaid in 12 equal monthly payments. After which payment is the balance exactly \$9,000, and how much principal has been repaid by then? (Answer: After payment 7; \$12,600 principal repaid.)

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