How Each Payment Splits Between Interest and Principal

Why a Level Payment Has Changing Parts

A fully amortizing loan can have the same scheduled installment in every period while the amounts applied to interest and principal change from period to period. Interest is calculated on the outstanding balance at the start of the period. As that balance declines, the interest portion generally declines too, leaving more of the fixed installment to repay principal. This tutorial is aligned with the SOA VEE Accounting and Finance syllabus.

The key calculation is not to apply interest repeatedly to the original loan amount. Instead, calculate interest on the balance immediately before each payment, then subtract that interest from the installment. The remainder is the principal portion of that payment.

Definition: Consider a loan with initial balance $L$, $n$ equal installments of $R$, paid at the end of each period, and a fixed effective interest rate $i_p$ per payment period. Let $B_k$ be the balance immediately after the $k$th payment, with $B_0=L$. The interest portion of payment $k$ is $I_k=i_pB_{k-1}$, and its principal portion is $C_k=R-I_k$. The balance then changes by $B_k=B_{k-1}-C_k$.

The rate $i_p$ must be expressed for the same period as the payments. For example, if installments are quarterly, $i_p$ is the effective rate per quarter, not automatically the effective annual rate. Rate conversion is assumed to have been completed before the amortization schedule is calculated.

For a level-payment loan with $i_p>-1$ and $i_p\ne 0$, the installment that exactly repays $L$ in $n$ end-of-period payments is

R=\frac{L}{a_{\overline{n}|}}, \qquad a_{\overline{n}|}=\frac{1-(1+i_p)^{-n}}{i_p}.

Here $a_{\overline{n}|}$ is the present value of $n$ unit payments at the periodic rate $i_p$. If $i_p=0$, the formula’s limiting case is $R=L/n$. The level-payment formula assumes the same rate applies in every period, payments are made at period-end, and there are no fees, skipped payments, or extra principal payments.

Calculating the Two Portions

Use the opening balance for the period to find interest, not the original loan amount or the balance after payment. The principal portion is what remains from the installment once that period’s interest has been covered. Subtracting principal from the opening balance gives the new balance.

Start with the opening balance.
For payment $k$, use $B_{k-1}$, the balance immediately before that payment.

Calculate interest.
Multiply the opening balance by the effective rate per payment period: $I_k=i_pB_{k-1}$.

Find principal repaid.
Subtract interest from the installment: $C_k=R-I_k$.

Update the balance.
Subtract the principal portion from the opening balance: $B_k=B_{k-1}-C_k$. Check that the installment equals interest plus principal.

Worked Example: Splitting Three Annual Payments

A hypothetical loan of $976 is repaid with three equal annual end-of-year payments of $500. The effective annual interest rate is 25%, and there are no fees or additional payments. Find the interest, principal, and closing balance for each year, then reconcile the totals.

Method. Work forward through the schedule, applying the rate to the balance at the start of each year. The opening balance for Year 1 is $B_0=\$976.00$.

Year 1. Interest is $I_1=0.25(\$976.00)=\$244.00$. Principal repaid is $C_1=\$500.00-\$244.00=\$256.00$. The balance after payment is $B_1=\$976.00-\$256.00=\$720.00$. The check is $\$244.00+\$256.00=\$500.00$.

Year 2. Interest is $I_2=0.25(\$720.00)=\$180.00$. Principal repaid is $C_2=\$500.00-\$180.00=\$320.00$. The balance after payment is $B_2=\$720.00-\$320.00=\$400.00$. The check is $\$180.00+\$320.00=\$500.00$.

Year 3. Interest is $I_3=0.25(\$400.00)=\$100.00$. Principal repaid is $C_3=\$500.00-\$100.00=\$400.00$. The balance after payment is $B_3=\$400.00-\$400.00=\$0.00$. The check is $\$100.00+\$400.00=\$500.00$.

Reconciliation. Total principal repaid is $\$256.00+\$320.00+\$400.00=\$976.00$, the original loan amount. Total interest is $\$244.00+\$180.00+\$100.00=\$524.00$, and total payments are $3(\$500.00)=\$1{,}500.00$. The totals agree because $\$976.00+\$524.00=\$1{,}500.00$. The common mistake this method avoids is charging interest each year on the original $976 rather than on the declining balance.

Why the Principal Portion Changes

With a fixed installment and a constant periodic rate, the principal portions have a simple relationship. The next period begins with the balance left after the current principal repayment. Therefore, the next interest charge is based on a balance lower by exactly $C_k$.

\begin{aligned} I_{k+1} &= i_p B_k = i_p(B_{k-1}-C_k) = I_k-i_pC_k,\\ C_{k+1} &= R-I_{k+1} = R-I_k+i_pC_k = (1+i_p)C_k. \end{aligned}

The last equality uses $C_k=R-I_k$. Thus, the change in the next principal portion is $C_{k+1}-C_k=i_pC_k$. If $i_p>0$ and $C_k>0$, then $C_{k+1}=(1+i_p)C_k>C_k$: the next principal portion is larger. If $i_p=0$, consecutive principal portions are equal. If $-1<i_p<0$ and $C_k>0$, then $C_{k+1}<C_k$. The claim that principal necessarily increases therefore requires a strictly positive rate and a positive current principal portion.

This relationship also explains the changing interest portions: when $i_p>0$ and principal is repaid, the balance falls, so the following interest charge falls. Because the installment stays level, that reduction in interest is matched by an increase in the principal portion. The result assumes a constant rate and an unchanged installment; a rate reset or a different payment changes the calculation.

Worked Example: Checking the Principal-portion Relationship

A hypothetical loan has a balance of $331 immediately before its first annual payment. It has a fixed effective annual rate of 10% and three equal end-of-year payments of $133.10. Calculate the schedule and check the relationship between consecutive principal portions.

First payment. Interest is $I_1=0.10(\$331.00)=\$33.10$. Principal is $C_1=\$133.10-\$33.10=\$100.00$. The new balance is $B_1=\$331.00-\$100.00=\$231.00$.

Second payment. Interest is $I_2=0.10(\$231.00)=\$23.10$. Principal is $C_2=\$133.10-\$23.10=\$110.00$. The new balance is $B_2=\$231.00-\$110.00=\$121.00$.

Third payment. Interest is $I_3=0.10(\$121.00)=\$12.10$. Principal is $C_3=\$133.10-\$12.10=\$121.00$. The new balance is $B_3=\$121.00-\$121.00=\$0.00$.

The relationship checks: $C_2=(1.10)(\$100.00)=\$110.00$, and $C_3=(1.10)(\$110.00)=\$121.00$. Total principal is $\$100.00+\$110.00+\$121.00=\$331.00$; total interest is $\$33.10+\$23.10+\$12.10=\$68.30$. The three installments total $\$399.30$, which equals the principal plus interest: $\$331.00+\$68.30=\$399.30$. The common mistake avoided here is to interpret an increasing principal portion as an increasing installment; the installment remains $133.10 throughout.

What to Check in an Amortization Schedule

A schedule should be consistent both row by row and over the whole loan. For each payment, interest plus principal must equal the installment. The closing balance must equal the opening balance minus principal repaid. Across a fully amortizing schedule, the principal portions sum to the original loan amount, subject to any stated rounding convention.

For a positive rate and positive principal portions, the principal amounts rise from one payment to the next, but the balance falls. These statements describe different quantities: principal is the amount repaid in a period, while the balance is the amount still owed afterward. With zero interest, a level-payment loan instead has equal principal portions. If a stated schedule does not satisfy these checks, first verify the rate period and the balance on which interest was calculated.

Worked Example: Quarterly Payments and a Smaller Interest Charge

A hypothetical loan of $30,301 is repaid by three equal quarterly end-of-period installments of $10,303.01. The effective rate is 1% per quarter, fixed for all three periods. Find each payment’s split and the total interest. No rounding is needed beyond cents.

Quarter 1. Interest is $I_1=0.01(\$30{,}301.00)=\$303.01$. Principal is $C_1=\$10{,}303.01-\$303.01=\$10{,}000.00$. The closing balance is $B_1=\$30{,}301.00-\$10{,}000.00=\$20{,}301.00$.

Quarter 2. Interest is $I_2=0.01(\$20{,}301.00)=\$203.01$. Principal is $C_2=\$10{,}303.01-\$203.01=\$10{,}100.00$. The closing balance is $B_2=\$20{,}301.00-\$10{,}100.00=\$10{,}201.00$.

Quarter 3. Interest is $I_3=0.01(\$10{,}201.00)=\$102.01$. Principal is $C_3=\$10{,}303.01-\$102.01=\$10{,}201.00$. The closing balance is $B_3=\$10{,}201.00-\$10{,}201.00=\$0.00$.

The principal portions follow the expected pattern: $1.01(\$10{,}000.00)=\$10{,}100.00$, then $1.01(\$10{,}100.00)=\$10{,}201.00$. Total interest is $\$303.01+\$203.01+\$102.01=\$608.03$. Total payments are $3(\$10{,}303.01)=\$30{,}909.03$, which agrees with $\$30{,}301.00+\$608.03=\$30{,}909.03$. This check also confirms that the stated 1% rate was applied per quarter, the period in which the installments are made.

Common Mistakes and Exam Checks

The most frequent errors are using the wrong balance, the wrong rate period, or the wrong interpretation of the principal sequence. Keep the following distinctions explicit when calculating or explaining a split:

Interest is based on the opening balance. For payment $k$, calculate $i_pB_{k-1}$, not $i_pL$ unless $k=1$.
The rate must match the payment interval. A quarterly installment requires the effective quarterly rate. Applying an annual rate directly to each quarter misstates interest.
The installment is not the principal portion. Principal is $R-I_k$, so a fixed installment can have changing principal amounts.
State conditions when describing the trend. Principal increases by the factor $1+i_p$ only in the comparative sense when $i_p>0$ and the current principal portion is positive. At zero interest, the portions are equal.
Check the final balance and totals. A fully amortizing schedule should end at zero, and total principal repaid should equal the starting balance.

Key takeaway: In each period, interest equals the periodic rate times the opening balance, principal equals the fixed installment minus that interest, and the balance falls by the principal repaid. With a constant positive rate and positive principal portions, consecutive principal portions satisfy $C_{k+1}=(1+i_p)C_k>C_k$.

Check Your Understanding

For each problem, use the stated rate for the stated payment period and check that the split adds to the installment.

A hypothetical loan has an opening balance of $2,000, a 5% effective annual rate, and an annual installment of $600. Find the first payment’s interest and principal portions and the balance after payment. (Answer: interest $100; principal $500; closing balance $1,500.)
Immediately before a quarterly payment, a loan balance is $4,200. The effective quarterly rate is 2%, and the installment is $1,000. Find the interest, principal, and closing balance for this payment. (Answer: interest $84; principal $916; closing balance $3,284.)
In a level-payment loan with a constant 5% rate per payment period, the first principal portion is $240. Find the second principal portion and state why it is larger. (Answer: $252, since $1.05(\$240)=\$252$ and the rate and first principal portion are positive.)
A hypothetical loan of $331 is repaid by three equal annual payments of $133.10 at an effective annual rate of 10%. Find total interest over the loan term using the payment total and the original principal. (Answer: total payments $399.30; total interest $68.30.)
A fixed-payment loan has a constant rate of $-2\%$ per payment period, and its current principal portion is positive at $500. Use the consecutive-principal relationship to find the next principal portion and determine whether it is larger or smaller. (Answer: $0.98(\$500)=\$490$, which is smaller by $10.)

What You'll Learn