10.3 Simple & Compound Interest & Borrowing Costs
Key Takeaways
- Simple interest earns interest strictly on the initial principal amount using I = Prt and A = P(1+rt), resulting in linear growth over time.
- Compound interest earns interest on both the original principal and accumulated prior interest using A = P(1+i)^n or A = P(1+r/n)^(nt), generating exponential growth.
- Compounding frequency n (annual, semi-annual, quarterly, monthly, daily) directly impacts total growth; higher compounding frequencies yield higher effective annual rates and accumulated balances.
- Total cost of borrowing measures the complete monetary expense of a loan or credit card, calculated as Total Cost of Borrowing = Total Repayments - Principal Borrowed.
- Financial products vary in compounding rules: credit cards compound daily at high annual rates (APR), while Canadian mortgages are legally required to compound semi-annually.
10.3 Simple & Compound Interest & Borrowing Costs
Quick Summary: Interest represents either the cost of borrowing money or the yield earned on invested capital. Simple interest ($I = Prt$) is calculated solely on the principal amount, generating linear growth. Compound interest ($A = P(1+i)^n$) earns interest on both principal and accumulated prior interest, producing exponential growth. On the Ontario MPT, candidates must compare growth models, calculate total debt repayments, and analyze borrowing mechanisms including loans, credit cards, and mortgages.
Simple Interest ($I = Prt$ and $A = P(1+rt)$)
Simple interest is earned or charged strictly on the initial principal $P$ over a time period $t$. The interest rate $r$ must always match the time unit of $t$ (typically per annum / annually).
Primary Formulas
Where:
- $P$ = Principal amount (initial principal in dollars).
- $r$ = Annual interest rate expressed as a decimal (e.g., $5% = 0.05$).
- $t$ = Time in years.
Converting Non-Year Time Intervals
When time is given in months or days, convert $t$ into fractional years before substituting into $I = Prt$:
- Time in Months ($m$): $t = \frac{m}{12}$ (e.g., 9 months $\implies t = \frac{9}{12} = 0.75$ years).
- Time in Days ($d$): $t = \frac{d}{365}$ (e.g., 146 days $\implies t = \frac{146}{365} = 0.40$ years).
Compound Interest ($A = P(1+i)^n$)
Compound interest adds earned interest back into the principal balance at defined compounding intervals. Subsequent interest is then calculated on the new, larger balance.
Primary Formulas
Where:
- $P$ = Initial principal balance.
- $r$ = Nominal annual interest rate (decimal).
- $n$ = Number of compounding periods per year.
- $t$ = Time in years.
- $i = \frac{r}{n}$ = Interest rate per compounding period (periodic interest rate).
- $N = n \times t$ = Total number of compounding periods over the term.
Common Compounding Frequencies ($n$)
| Compounding Term | Periods per Year ($n$) | Periodic Rate Formula ($i$) | Total Periods for 5 Years ($N$) |
|---|---|---|---|
| Annually | $n = 1$ | $i = r / 1$ | $N = 1 \times 5 = 5$ |
| Semi-Annually | $n = 2$ | $i = r / 2$ | $N = 2 \times 5 = 10$ |
| Quarterly | $n = 4$ | $i = r / 4$ | $N = 4 \times 5 = 20$ |
| Monthly | $n = 12$ | $i = r / 12$ | $N = 12 \times 5 = 60$ |
| Daily | $n = 365$ | $i = r / 365$ | $N = 365 \times 5 = 1,825$ |
Comparing Simple vs. Compound Growth
Because simple interest grows linearly while compound interest grows exponentially, the gap between simple and compound accumulation widens dramatically over time.
graph LR
START["Principal: $1,000 at 10%"] --> S1["Year 1: Simple = $1,100 | Compound = $1,100.00"]
S1 --> S5["Year 5: Simple = $1,500 | Compound = $1,610.51"]
S5 --> S10["Year 10: Simple = $2,000 | Compound = $2,593.74"]
S10 --> S20["Year 20: Simple = $3,000 | Compound = $6,727.50"]
Growth Comparison Table ($10,000 Principal at 6.0% Annual Rate)
| Investment Horizon ($t$) | Simple Interest ($A = P(1+rt)$) | Compound Annual ($n=1$) | Compound Monthly ($n=12$) | Difference (Monthly vs. Simple) |
|---|---|---|---|---|
| 1 Year | $$10,600.00$ | $$10,600.00$ | $$10,616.78$ | $+$16.78$ |
| 5 Years | $$13,000.00$ | $$13,382.26$ | $$13,488.50$ | $+$488.50$ |
| 10 Years | $$16,000.00$ | $$17,908.48$ | $$18,193.97$ | $+$2,193.97$ |
| 20 Years | $$22,000.00$ | $$32,071.35$ | $$33,102.04$ | $+$11,102.04$ |
The Rule of 72
The Rule of 72 is a quick estimation shortcut for finding the approximate number of years required for an investment to double at a given annual compound interest rate $r%$:
Example: An investment earning $6%$ compound interest doubles in approximately $\frac{72}{6} = 12$ years.
Total Cost of Borrowing & Credit Mechanics
Borrowing money through loans, credit cards, or mortgages incurs interest costs. On the MPT, calculating the total cost of borrowing requires measuring total money repaid versus the original borrowed amount.
Primary Borrowing Cost Equation
For amortized loans with fixed monthly payments:
Financial Borrowing Vehicles
- Credit Cards: High Annual Percentage Rate (APR), typically ranging from $19.99%$ to $22.99%$. Interest compounds daily if the full statement balance is not paid by the due date. Paying only the minimum monthly payment extends debt repayment over decades and results in interest charges multiple times the original balance.
- Personal / Auto Loans: Fixed amortized term loans (e.g., 48 or 60 months) with equal monthly payments comprising principal reduction and interest.
- Canadian Mortgages: In Canada, mortgage interest rates are legally capped at semi-annual compounding ($n=2$) by the federal Interest Act, even when mortgage payments are made monthly or bi-weekly.
Worked Step-by-Step Problem Walkthroughs
Problem 1: Simple vs. Compound Interest Comparison
Problem Statement: An investor deposits $6,000 for 4 years into two different options:
- Option A: Pays simple interest at 5.0% per annum.
- Option B: Pays compound interest at 5.0% per annum compounded monthly.
- Calculate the final balance and interest earned for Option A.
- Calculate the final balance and interest earned for Option B.
- Determine how much additional interest Option B yields compared to Option A.
Step-by-Step Solution:
-
Calculate Option A (Simple Interest):
-
Calculate Option B (Compound Monthly):
- $P = $6,000$, $r = 0.05$, $n = 12$, $t = 4$
- Periodic rate $i = \frac{0.05}{12} \approx 0.0041667$
- Total periods $N = 12 \times 4 = 48$
-
Compare Yields: Conclusion: Option B yields $125.37 more interest than Option A over 4 years.
Problem 2: Total Borrowing Cost for an Auto Loan
Problem Statement: A customer buys a used vehicle for $18,000. She pays a $3,000 cash down payment and finances the remaining balance with a 48-month auto loan. The lender sets monthly payments at $365.00.
- Calculate the principal amount financed.
- Calculate the total amount repaid to the lender over 48 months.
- Compute the total cost of borrowing (interest paid).
Step-by-Step Solution:
-
Calculate Principal Financed:
-
Calculate Total Amount Repaid:
-
Compute Total Borrowing Cost: Conclusion: The borrower pays $2,520.00 in total borrowing costs (interest) over the 4-year loan term.
An educator invests $5,000 in a short-term Ontario Guaranteed Investment Certificate (GIC) that pays a simple interest rate of 4.5% per annum for a term of 18 months. What is the total interest earned and the final account balance at maturity?
A savings account has an initial principal of $8,000 compounding monthly at an annual interest rate of 6% (0.5% per month). What is the total accumulated balance in the account after 2 years (24 months)?
A teacher finances a $24,000 vehicle purchase with a 5-year auto loan. The loan agreement requires fixed monthly payments of $460.00 for 60 months. What is the total cost of borrowing (total interest paid) over the lifespan of the loan?