The Balance Recurrence for an Amortizing Loan

Why a Balance Recurrence Is Useful

An amortization schedule records how a loan balance changes from one payment date to the next. The key operation is repeated: accrue interest on the opening balance, then subtract the payment. Writing that operation as a recurrence makes the schedule easier to generate and helps identify errors in signs, timing, and rate periods.

This tutorial concerns the balance update for a level-payment loan. It does not require a closed-form formula for the balance after many payments: each row follows directly from the balance at the start of that period.

Definition: Let $B_{k-1}$ be the loan balance immediately after payment $k-1$, and let $B_k$ be the balance immediately after payment $k$. If the effective interest rate per payment period is $i$, and a level payment $R$ is made at the end of each period, then $$ B_k=(1+i)B_{k-1}-R. $$ The opening balance for period $k$ is $B_{k-1}$; the closing balance is $B_k$. In this convention, $B_0$ is the original loan principal before any payments.

Deriving the Update

Assume that the balance earns interest at the effective rate $i$ for one complete payment period, and that no fees, additional advances, or other balance changes occur during that period. The balance just before the payment is therefore the opening balance plus the interest earned on it. Subtract the payment made at the end of the period to obtain the closing balance.

\begin{aligned} \text{Interest in period }k &= iB_{k-1},\\ \text{Balance before payment }k &= B_{k-1}+iB_{k-1}=(1+i)B_{k-1},\\ B_k &= (1+i)B_{k-1}-R. \end{aligned}

This derivation also explains why the balance recurrence is not simply “opening balance minus payment.” The payment first covers the interest accrued during the period; only the remaining amount reduces principal. In the earlier material on splitting each payment into interest and principal, these amounts are the interest portion $iB_{k-1}$ and the principal portion $R-iB_{k-1}$. Subtracting that principal portion from the opening balance gives the same update:

B_k=B_{k-1}-(R-iB_{k-1})=(1+i)B_{k-1}-R.

Assumptions to check: This form applies when the payment is made at the end of the period, the interest rate $i$ is the effective rate for that same period, and the payment $R$ is made in full. If the payment is made at the beginning of a period, or if the period's rate or payment differs, adjust the update to reflect the actual timing and amounts.

Generating Schedule Rows

For each period, a useful calculation order is: identify the opening balance, calculate interest on that balance, add the interest, subtract the payment, and record the closing balance. The same payment is used in each row for a level-payment loan, but the interest charge can change because it is calculated on the current opening balance.

Set the opening balance.
Use the previous row's closing balance as the current row's opening balance.

Accrue interest.
Multiply the opening balance by the effective rate per payment period.

Apply the payment.
Add the interest to the opening balance, then subtract the level payment.

Check the row.
Confirm that the closing balance equals opening balance plus interest minus payment.

Worked Example: Three Quarterly Rows

A hypothetical loan has an opening balance of $10,000 immediately after origination. Payments of $3,000 are made at the end of each quarter, and the effective interest rate is 2% per quarter. Find the balances after the first three payments. Retain exact values in the calculations and report money to the nearest cent.

Method. Use the end-of-period recurrence with $i=0.02$ and $R=3{,}000$. The rate is explicitly quarterly, so each row represents one quarter.

Quarter 1. The opening balance is $10,000.00. Interest is $0.02(10{,}000)=\$200.00$, so the balance before payment is $\$10{,}000+\$200=\$10{,}200$. Subtract the payment:

B_1=1.02(10{,}000)-3{,}000=10{,}200-3{,}000=\$7{,}200.00.

Quarter 2. The opening balance is $7,200.00. Interest is $0.02(7{,}200)=\$144.00$. Therefore:

B_2=1.02(7{,}200)-3{,}000=7{,}344-3{,}000=\$4{,}344.00.

Quarter 3. The opening balance is $4,344.00. Interest is $0.02(4{,}344)=\$86.88$. Therefore:

B_3=1.02(4{,}344)-3{,}000=4{,}430.88-3{,}000=\$1{,}430.88.

Check. In the third row, the payment covers $86.88 of interest and reduces principal by $\$3{,}000-\$86.88=\$2{,}913.12$. Subtracting that principal reduction from the opening balance gives $\$4{,}344-\$2{,}913.12=\$1{,}430.88$, as required. A common mistake is to subtract interest from the opening balance as though it reduced the debt; that would give $10{,}000-200-3{,}000=\$6{,}800$ in the first quarter. Interest increases the amount owed, so the correct first balance is $7,200.00.

Keeping the Rate and Payment Period Aligned

The $i$ in the recurrence is not automatically an annual rate. It must be the effective rate over the period between the payments in the row. If payments are monthly, use the effective monthly rate; if payments are quarterly, use the effective quarterly rate. Do not use an annual rate in a monthly row unless it has first been converted to the effective rate for one month.

The recurrence also works row by row when the effective rate changes. The general update for period $k$, with that period's rate $i_k$ and payment $R_k$, is:

B_k=(1+i_k)B_{k-1}-R_k.

This is the same period-by-period accounting identity. For a level payment, set $R_k=R$ in every period; if rates vary, use the rate that applies to the period being calculated. The formula does not assume that all periods have the same interest rate, but the simpler form with one fixed $i$ does.

Worked Example: A Level Payment With Changing Periodic Rates

A hypothetical loan has a balance of $25,000 immediately after a payment. A level payment of $1,800 is due at the end of each month. The effective monthly rates for the next three periods are 1.0%, 0.8%, and 1.2%, respectively. Find the balance after each of those payments.

Method. Apply $B_k=(1+i_k)B_{k-1}-R$, using the rate specified for each month and $R=\$1{,}800$.

Month 1. Interest is $0.01(25{,}000)=\$250.00$. The balance before payment is $\$25{,}250.00$, and after payment:

B_1=1.01(25{,}000)-1{,}800=\$23{,}450.00.

Month 2. Use 0.8%, not the first month's rate. Interest is $0.008(23{,}450)=\$187.60$. Thus:

B_2=1.008(23{,}450)-1{,}800=23{,}637.60-1{,}800=\$21{,}837.60.

Month 3. Use 1.2%. Interest is $0.012(21{,}837.60)=\$262.0512$, so the balance before payment is $22,099.6512. Subtracting the payment gives:

B_3=1.012(21{,}837.60)-1{,}800=20{,}299.6512\approx \$20{,}299.65.

Check. The third-period interest is $262.0512, and the payment reduces principal by $\$1{,}800-\$262.0512=\$1{,}537.9488$. Hence $21{,}837.60-1{,}537.9488=20{,}299.6512$. The final reported balance is rounded to cents; the unrounded value was carried through the calculation.

When Payment Timing Changes

The recurrence $B_k=(1+i)B_{k-1}-R$ places each payment at the end of its period. If a payment is instead made at the beginning, it is subtracted before that period's interest accrues. The order of operations changes, so applying the end-of-period formula without adjustment gives the wrong balance.

Worked Example: Detecting a Beginning-of-Period Payment

A hypothetical loan has a balance of $12,000 immediately before a payment made at the beginning of the month. The payment is $1,500, and the effective monthly interest rate is 1%. Find the balance at the end of the month, after that month's interest has accrued. Also find the balance at the end of the next month if a second $1,500 payment is made at the beginning of that month.

Method. Subtract each beginning-of-period payment first, then accrue interest on the amount that remains during the month.

First month. Immediately after the first payment, the balance is $\$12{,}000-\$1{,}500=\$10{,}500$. Interest for the month is $0.01(10{,}500)=\$105$. The month-end balance is therefore $\$10{,}500+\$105=\$10{,}605.00$.

Second month. The second payment is made at the beginning of the month, reducing the balance to $\$10{,}605-\$1{,}500=\$9{,}105$. Interest is $0.01(9{,}105)=\$91.05$, so the second month-end balance is $\$9{,}105+\$91.05=\$9{,}196.05$.

Check and trap avoided. The first month's end balance is $1.01(12{,}000-1{,}500)=\$10{,}605$. Using the ordinary end-of-period recurrence instead would give $1.01(12{,}000)-1{,}500=\$10{,}620$, which is $15 too high because it charges one month's interest on the $1,500 already paid at the beginning.

Common Checks Before Accepting a Row

A schedule row should satisfy both the balance recurrence and the payment split. With a period-end payment, the interest portion is $iB_{k-1}$, and the principal portion is $R-iB_{k-1}$. The row can be checked in either of two equivalent ways:

B_k=(1+i)B_{k-1}-R \qquad\text{and}\qquad B_k=B_{k-1}-(R-iB_{k-1}).

Confirm the time unit. The rate and the row must refer to the same period.
Confirm the timing. The standard recurrence assumes the payment occurs at the end of the period.
Check the direction of interest. Interest is added to the amount owed before the payment is subtracted.
Keep precision consistent. For an exact mathematical schedule, retain unrounded values between rows and round reported money at the end. A lender's posted schedule may round each row and adjust a final payment.
Check the principal portion. If the payment exceeds the period's interest, the balance decreases by the difference. If the payment is less than interest, the balance increases under this model.

These checks are useful in exam problems because a balance can look plausible while still reflecting a rate from the wrong period or a payment applied at the wrong time. Writing the opening balance, interest, payment, and closing balance separately makes the source of any discrepancy visible.

Key takeaway: For a period-end level payment, grow the opening balance by the effective rate for one payment period, then subtract the payment: $B_k=(1+i)B_{k-1}-R$. Each closing balance becomes the next row's opening balance.

Check Your Understanding

For each problem, use the stated payment timing and rate period. Report monetary answers to the nearest cent.

A hypothetical loan has a balance of $9,000 immediately after a payment. The effective quarterly rate is 1.5%, and each $1,400 payment is made at quarter-end. Find the balance immediately after the second payment. (Answer: $6,451.03)
A loan balance is $20,000 after origination. The effective monthly rate is 0.8%, and the month-end payment is $1,800. Find the balance after three payments. (Answer: $15,040.54)
A loan balance is $15,000 immediately after a payment. The rate is 2% per period, and each period-end payment is $2,500. Calculate the balances immediately after payments one, two, and three. (Answer: $12,800.00; $10,556.00; $8,267.12)
A loan has an opening balance of $12,000. The effective rate for the period is 1%, and the end-of-period closing balance is $10,890. Find the payment made at the end of the period. (Answer: $1,230.00)
A loan has a balance of $8,000 immediately after a payment. A level payment of $1,000 is made at each period-end. The effective rates for the next two periods are 1% and 1.5%. Find the balance after the second payment. (Answer: $6,186.20)

What You'll Learn