Training Financial Mathematics Loan Amortization
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Loan Amortization

23 min Financial Mathematics

Loan Amortization

Amortization

Repaying a loan with equal periodic payments. Each payment covers interest plus principal reduction.

$$R = P \cdot \frac{i}{1-(1+i)^{-n}}$$

$P$ = loan amount, $i$ = rate per period, $n$ = total payments.

Amortization Schedule

For payment $k$:

  • Interest portion: $I_k = B_{k-1} \cdot i$
  • Principal portion: $P_k = R - I_k$
  • New balance: $B_k = B_{k-1} - P_k$
Example 1

Monthly payment on a $\$200{,}000$ mortgage at $6\%$ for 30 years.

$i = 0.005$, $n = 360$. $R = 200000 \cdot \frac{0.005}{1-1.005^{-360}} \approx \$1{,}199.10$.

Example 2

Total interest paid on the loan above.

Total paid $= 360 \times 1199.10 = \$431{,}676$. Interest $= 431{,}676 - 200{,}000 = \$231{,}676$.

Example 3

First two rows of an amortization schedule for a $\$10{,}000$ loan at $12\%$ monthly over 4 years.

$i = 0.01$, $n = 48$. $R \approx \$263.34$.

Month 1: Interest $= 100$, Principal $= 163.34$, Balance $= 9{,}836.66$.

Month 2: Interest $= 98.37$, Principal $= 164.97$, Balance $= 9{,}671.69$.

Practice Problems

1. Monthly payment: $\$25{,}000$ car loan, $5\%$, 5 years.
2. Total interest on the car loan above.
3. $\$150{,}000$ mortgage, $4.5\%$, 15 years. Monthly payment?
4. How much principal is paid in payment 1 of Problem 3?
5. Remaining balance after 12 payments on a $\$10{,}000$ loan at $6\%$/yr monthly, 3-year term.
6. Compare total interest: 15-year vs. 30-year mortgage on $\$200{,}000$ at $5\%$.
7. What fraction of the first payment is interest? $\$300{,}000$, $6\%$, 30 yrs.
8. If you pay an extra $\$100$/month on a $\$200{,}000$, $6\%$, 30-yr loan, how much interest do you save?
9. Bi-weekly payments: how does paying every 2 weeks change payoff time?
10. Find the outstanding balance formula after $k$ payments.
11. $\$5{,}000$ loan, $8\%$ annually, 2 years, annual payments. Full schedule.
12. APR vs. nominal rate: when are they different?
Show Answer Key

1. $R \approx \$471.78$

2. $60 \times 471.78 - 25000 = \$3{,}306.80$

3. $R \approx \$1{,}147.49$

4. Interest = $150000(0.045/12) = \$562.50$; Principal = $1147.49 - 562.50 = \$584.99$

5. $B_{12} = R \cdot \frac{1-(1+i)^{-(n-12)}}{i} \approx \$6{,}844.08$

6. 15-yr total interest $\approx \$84{,}686$; 30-yr $\approx \$186{,}512$; savings $\approx \$101{,}826$

7. $R \approx \$1{,}798.65$; Interest = $300000(0.005)= \$1{,}500$; fraction $\approx 83.4\%$

8. $\approx \$52{,}000+$ saved, payoff $\approx 25$ years

9. Effectively 13 monthly payments/year; shaves $\approx 5$ years off a 30-yr loan

10. $B_k = R \cdot \frac{1-(1+i)^{-(n-k)}}{i}$

11. $R = 5000 \cdot 0.08/(1-1.08^{-2}) \approx \$2{,}805.77$. Yr 1: Int $400$, Princ $2405.77$, Bal $2594.23$. Yr 2: Int $207.54$, Princ $2598.23$, Bal $0$.

12. When fees/costs are included; APR reflects true cost

Annuities Placement Test Practice — Financial Math