Loans and Amortization

A loan payment usually has two jobs: pay the interest accrued for the period and reduce the amount still owed. Amortization is the process of reducing that principal through a sequence of payments.

A quoted annual rate and a monthly payment are not enough to reconstruct a loan. The compounding convention, payment frequency, payment timing, number of payments, fees and rounding rule also matter. The worked amounts below are hypothetical Canadian-dollar examples, not current offers or personalized borrowing recommendations.

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Start with the balance equation

Let $P$ be the amount borrowed, $i$ the interest rate per payment period, $M$ a level payment at each period’s end, and $B_k$ the balance immediately after payment $k$ . Suppose the rate stays constant, interest is charged before each payment, and there are no extra fees, missed payments or additional borrowing.

B_0=P,\qquad B_k=(1+i)B_{k-1}-M.

The period’s interest is $iB_{k-1}$ , and the principal repaid is $M-iB_{k-1}$ . Their sum is the payment. This is why the interest component usually falls as the balance decreases.

If a payment equals the period’s interest, the principal does not fall. If it is smaller, unpaid interest increases the balance in this model. For $1200 at 1% per month, a $12 end-of-month payment leaves $1200 owing; a $10 payment leaves $1202 after the first month.

Derive a payment that clears the loan

Expand the recurrence through successive periods:

B_1=P(1+i)-M,

B_2=P(1+i)^2-M(1+i)-M.

For integer $n\ge1$ and $i\ne0$ , the finite geometric sum gives

B_n=P(1+i)^n-M\frac{(1+i)^n-1}{i}.

To clear the balance immediately after payment $n$ , set $B_n=0$ and solve:

M=\frac{Pi}{1-(1+i)^{-n}}.

This formula applies here to a positive periodic rate with equal end-of-period payments. At zero interest, use $M=P/n$ . Beginning-of-period payments or a rate change require adjusted timing or a new schedule.

The same result follows by equating the amount borrowed to the discounted value of the payments:

P=\frac{M}{1+i}+\frac{M}{(1+i)^2}+\cdots+\frac{M}{(1+i)^n}.

Each payment is discounted from its own date. This connects amortization to dated cash-flow comparisons.

Calculate a short loan before rounding the ledger

Suppose $1200 is repaid in six monthly payments at exactly 1% per month. Then

M=\frac{1200(0.01)}{1-(1.01)^{-6}}\approx207.058040.

The ideal mathematical payment is about $207.058040, and its six-payment total is about $1242.348240. The interest cost in that unrounded model is about $42.35.

Money payments normally require a cent-valued amount. Rounding $M$ to $207.06 at every payment does not automatically make the last balance exactly zero. A complete payment schedule must specify how to handle that final difference.

Reconcile a cent-rounded amortization schedule

For this teaching ledger, round each monthly interest charge to cents, taking an exact half-cent upward. Pay $207.06 in each of the first five months. In month six, pay the remaining principal plus that month’s rounded interest.

Reconcile a cent-rounded amortization schedule: reference table
Payment	Opening balance ($)	Interest ($)	Payment ($)	Principal repaid ($)	Closing balance ($)
1	1200.00	12.00	207.06	195.06	1004.94
2	1004.94	10.05	207.06	197.01	807.93
3	807.93	8.08	207.06	198.98	608.95
4	608.95	6.09	207.06	200.97	407.98
5	407.98	4.08	207.06	202.98	205.00
6	205.00	2.05	207.05	205.00	0.00

For example, the second interest charge is $1004.94(0.01)=10.0494$ dollars, which rounds to $10.05. Subtracting it from $207.06$ leaves $197.01$ of principal repayment. The closing balance is $1004.94-197.01=807.93$ dollars.

The six payments total $1242.35. Principal repayments total $1200.00, and interest totals $42.35. Those three totals reconcile exactly under the stated rounding rule.

Interest and principal portions of six loan payments

The first five payments are $207.06. The final adjusted payment is $207.05. All column heights use the same zero-based scale.

Diagram explanation and exact data

For each month subtract the rounded interest from the payment to obtain principal repaid. Total principal repaid is 1200 dollars and total interest is 42.35 dollars. The last payment clears the cent-rounded balance exactly.

Interest and principal portions of six loan payments: exact reference data
Month	Interest ($)	Principal repaid ($)	Payment ($)
1	12.00	195.06	207.06
2	10.05	197.01	207.06
3	8.08	198.98	207.06
4	6.09	200.97	207.06
5	4.08	202.98	207.06
6	2.05	205.00	207.05

Actual contracts may accrue interest daily, round differently, add charges or alter payment dates. Use their ledger conventions rather than treating this instructional schedule as universal. The Financial Consumer Agency of Canada’s personal-loan guidance explains the importance of interest, fees and repayment conditions.

A smaller payment can mean more total interest

Keep the same $1200 principal and 1% monthly rate but spread the loan over twelve monthly payments. The unrounded formula gives

M_{12}=\frac{1200(0.01)}{1-(1.01)^{-12}}\approx106.618546.

The payment is lower, but twelve ideal payments total about $1279.42, including $79.42 interest. Under the six-period model, interest was about $42.35. Compare total borrowing cost as well as affordability per payment.

This comparison keeps the rate and fee assumptions fixed. A real comparison must include charges, prepayment conditions and any rate changes. An extra principal payment reduces later interest in the constant-rate model, but contractual fees can affect the overall outcome.

Match compounding to payment frequency

Consider a separate stipulated mortgage model: $300,000 principal, a 6% nominal annual rate compounded semiannually, and monthly end-of-period payments. This is a specified contract convention for the example, not a claim about every Canadian mortgage.

One half-year grows a balance by $1+0.06/2=1.03$ . Six equivalent monthly periods must produce the same factor:

(1+i)^6=1.03,\qquad i=1.03^{1/6}-1\approx0.004938622.

The equivalent monthly rate is about 0.4938622%. It is not $0.06/12=0.005$ , because that would describe a different nominal compounding convention.

With a 25-year amortization and twelve payments per year, $n=300$ . Holding the stipulated rate constant throughout the calculation gives

M=\frac{300000i}{1-(1+i)^{-300}}\approx1919.419871.

The payment is about $1919.42. Incorrectly using 0.5% per month would instead give about $1932.90. A small rate-conversion difference affects a long schedule materially.

Separate the mortgage term from the amortization period

The term is the period governed by the current mortgage agreement. The amortization period is the projected time needed to repay the loan under its payment assumptions. These are different concepts, as the Financial Consumer Agency of Canada explains.

A five-year term does not mean that a loan with a 25-year amortization is fully repaid after five years. Nor does it guarantee that the original rate will remain available for all 25 years.

Using the previous example’s unrounded monthly payment and rate, the balance after 60 payments is

B_{60}=300000(1+i)^{60}-M\frac{(1+i)^{60}-1}{i}\approx269510.11.

That is the model’s outstanding principal at five years, not its total future interest. A renewal may change the rate and payment. The projection must then be recalculated from the remaining balance and the new terms. The cent-rounded real ledger may also differ slightly from this exact-formula illustration.

Audit a payment schedule with identities

For each row, check that opening balance plus interest minus payment equals closing balance. Separately check that payment equals interest plus principal repaid. Across the whole schedule, principal repaid plus the final outstanding balance must equal the original principal, provided no additional principal or fees were added.

Do not call a monthly interest rate an annual percentage rate without an appropriate definition. Do not describe scheduled future interest as an amount already owed today. Rate quotations, interest accrual and legal disclosure measures are related but are not interchangeable labels.

Practice questions

A $600 loan has zero interest and twelve equal monthly payments. Find the payment.
A balance of $1000 accrues 1% interest and then receives a $100 payment. Find the new balance and the principal repaid.
A $5000 balance accrues 1% interest for a period, followed by a $40 payment. Does the balance decrease?

Check your answers

$M=600/12=50$ dollars. The zero-rate case does not use a quotient containing the periodic rate in the denominator.
Interest is $10. The principal repayment is $100-10=90$ , and the new balance is $1000+10-100=910$ dollars.
Interest is $50, greater than the $40 payment. The new balance is $5000+50-40=5010$ dollars. It increases by $10 under this model.