Why Interest Charges Fall
For a conventional amortizing loan, each payment covers the interest accrued since the previous payment and repays some principal. When the rate per payment period is fixed and the payment is level, repaying principal reduces the balance on which the next period’s interest is calculated. The interest charge therefore falls as long as the loan is actually paying down.
The key is to compare consecutive periods, rather than to try to infer the change from the payment amount alone. The payment stays level, but the interest and principal portions generally change in opposite directions. This tutorial derives the exact relationship between the interest decrease and the principal repaid in the preceding period.
The rate \(i\) must match the length of a payment period. If payments are monthly, for example, \(i\) in these formulas is the effective monthly rate, not automatically an annual rate. We assume a fixed positive rate, payments at the end of each period, and a level payment that reduces the balance while it is positive. The loan is not in negative amortization: each regular payment exceeds that period’s interest charge.
The Consecutive-Period Relationship
The balance just after payment \(k\) is the opening balance for period \(k+1\). Thus the next interest charge is calculated on \(B_k\), not on \(B_{k-1}\). The balance recurrence makes the change especially clear.
Because \(C_k=R-I_k\), the same recurrence can be written \(B_k=B_{k-1}-C_k\). Multiplying the balance change by the fixed rate gives the next period’s interest charge.
This result says exactly how much the charge falls: multiply the preceding payment’s principal portion by the rate for one payment period. It does not say that the interest charge falls by the principal portion itself, nor by the full payment. The rate scales the change in balance into a change in interest.
Use the effective rate for the interval between payments.
Calculate \(I_k=iB_{k-1}\), then \(C_k=R-I_k\).
Calculate \(B_k=B_{k-1}-C_k\), or equivalently \(B_k=(1+i)B_{k-1}-R\).
Calculate \(I_{k+1}=iB_k\), or check that \(I_k-I_{k+1}=iC_k\).
Worked Examples
Worked Example: Checking a Three-Payment Loan
A loan is issued for the present value of three annual end-of-year payments of $100 at an effective annual rate of 10%. Find the balance and interest charge in each period, and verify the change in interest from one period to the next. Keep full precision in the calculations and report the final schedule amounts to the nearest cent.
Method. First calculate the exact present value of the three payments. Then use the balance recurrence to find each interest charge and principal portion. This avoids treating a rounded starting balance as if it were exact.
The original balance is
For period 1, the interest is \(I_1=0.10B_0\approx 24.8685199098\). Therefore \(C_1=100-I_1\approx 75.1314800902\). The balance after the first payment is
For period 2, \(I_2=0.10B_1\approx 17.3553719008\), so \(C_2=100-I_2\approx 82.6446280992\). The next balance is \(B_2=B_1-C_2\approx 90.9090909091\). For period 3, \(I_3=0.10B_2\approx 9.0909090909\), and \(C_3=100-I_3\approx 90.9090909091\).
The schedule values, rounded only for presentation, are:
| Period | Opening balance | Interest | Principal | Closing balance |
|---|---|---|---|---|
| 1 | $248.69 | $24.87 | $75.13 | $173.55 |
| 2 | $173.55 | $17.36 | $82.64 | $90.91 |
| 3 | $90.91 | $9.09 | $90.91 | $0.00 |
The first decrease in interest is \(I_1-I_2\approx 7.5131480090\). The theorem gives \(0.10C_1\approx 0.10(75.1314800902)=7.5131480090\). The next decrease is \(I_2-I_3\approx 8.2644628099\), which equals \(0.10C_2\). As an exact payoff check, \(B_2=1000/11\), so the third-period interest is \(100/11\), the third principal portion is \(1000/11\), and the payment is exactly \(100\). Thus the balance after payment 3 is zero. The common mistake this method avoids is mixing rounded schedule entries into later calculations and creating a spurious residual balance.
Worked Example: Finding the First Three Interest Decreases
A borrower has a balance of $10,000 immediately before the first of a series of monthly payments. The effective monthly rate is 1%, and each scheduled payment is $1,000 at month-end. Calculate the first three interest charges and the decreases between them. Assume no fees or extra payments.
Method. Apply interest to each period’s opening balance, subtract the principal portion of the $1,000 payment, and carry the resulting balance forward. The payment is greater than interest in each period shown, so the balance falls.
In month 1, \(I_1=0.01(10{,}000)=100\). The principal portion is \(C_1=1{,}000-100=900\), leaving \(B_1=10{,}000-900=9{,}100\). In month 2, \(I_2=0.01(9{,}100)=91\), so \(C_2=1{,}000-91=909\), and \(B_2=9{,}100-909=8{,}191\). In month 3, \(I_3=0.01(8{,}191)=81.91\).
The first interest decrease is \(100-91=\$9\). It also equals \(0.01(900)=\$9\), the rate times the month 1 principal portion. The next decrease is \(91-81.91=\$9.09\), equal to \(0.01(909)=\$9.09\). Thus the interest charges are $100, $91, and $81.91, and the decreases are $9 and $9.09. The common mistake avoided here is multiplying the rate by the payment rather than by the principal portion: \(0.01(1{,}000)=\$10\) is not the first decrease.
Worked Example: Using the Interest Change to Recover Principal Repaid
A loan has a fixed effective quarterly rate of 4%. The interest charge for one quarter is $96, and the charge for the next quarter is $81.84. Find the principal repaid in the first quarter, the balance at the start of that quarter, and the principal portion of the first payment if the level payment is $450.
Method. Use the change in interest to infer the principal reduction, then use the first interest charge to recover the opening balance. This avoids assuming that the interest change equals the principal reduction.
The interest decrease is \(96-81.84=\$14.16\). Since the rate is 4% per quarter, the principal repaid in the first quarter is
The first-quarter interest charge is 4% of the opening balance, so \(B_0=96/0.04=\$2{,}400\). With a $450 payment, the first payment’s principal portion is also \(450-96=\$354\), confirming the inferred amount. The closing balance is \(2{,}400-354=\$2{,}046\), and the next interest charge checks: \(0.04(2{,}046)=\$81.84\). The common mistake avoided is interpreting the $14.16 decrease as $14.16 of principal; the decrease is the rate times principal, so it must be divided by 0.04 to recover principal.
When the Decline Is Not Strict
The condition behind a strict decline is positive principal repayment. Under the standard assumptions, \(C_k=R-I_k\). If \(C_k>0\) and \(i>0\), then \(I_{k+1}-I_k=-iC_k<0\). If a payment exactly equals interest, \(C_k=0\), the balance does not change and neither does the next interest charge. If a payment is less than interest, then \(C_k<0\): the balance increases and, at a positive fixed rate, the next interest charge is higher.
A changing rate also changes the comparison. In that case, the next interest is calculated using both a different balance and a different rate, so \(I_{k+1}-I_k=-iC_k\) need not hold with one fixed \(i\). Similarly, extra payments, fees added to the balance, skipped payments, payment timing changes, and rounding adjustments can alter the schedule. The theorem applies to the stated regular fixed-rate model, not automatically to every loan statement.
This relationship is useful for schedule checks, for interpreting how much of a payment reduces principal, and for recovering a principal portion when consecutive interest charges are known. It also gives a quick sign check: under positive amortization at a positive fixed rate, a computed increase in interest signals either a calculation error or a condition that violates the model assumptions.
Check Your Understanding
For each problem, use the rate per payment period and distinguish interest from principal.
- A loan’s opening balance is $24,000. The effective monthly rate is 0.5%, and the level payment is $800. Find the first two interest charges and their decrease. (Answer: $120 and $116.60; decrease $3.40.)
- A quarterly loan has an opening balance of $5,000, an effective quarterly rate of 2%, and a $600 payment. Find the first principal portion, the balance after the payment, and the next interest charge. (Answer: principal $500; balance $4,500; next interest $90.)
- At a fixed effective rate of 3% per period, consecutive interest charges are $144 and $138.60. How much principal was repaid in the first period? (Answer: $180.)
- A balance is $120,000 immediately before a monthly payment. The effective monthly rate is 0.4%, and the level payment is $900. Calculate the first two interest charges and verify the decrease using the first principal portion. (Answer: $480 and $478.32; first principal $420; decrease $1.68.)
- A borrower owes $10,000 at a fixed effective monthly rate of 1% and makes a level monthly payment of $100. Does the interest charge decline after the first payment? Explain using the balance recurrence. (Answer: No. The first interest is $100, so the principal portion is zero, the balance remains $10,000, and the next interest is also $100.)