9.6 Loans

In the last section, you learned about payout annuities.

In this section, you will learn about conventional loans (also called amortized loans or installment loans). Examples include auto loans and home mortgages. These techniques do not apply to payday loans, add-on loans, or other loan types where the interest is calculated up front.

One great thing about loans is that they use exactly the same formula as a payout annuity. To see why, imagine that you had $10,000 invested at a bank, and started taking out payments while earning interest as part of a payout annuity, and after 5 years your balance was zero. Flip that around, and imagine that you are acting as the bank, and a car lender is acting as you. The car lender invests $10,000 in you. Since you’re acting as the bank, you pay interest. The car lender takes payments until the balance is zero.

Note

Loans Formula

$P_{0} = \frac{d (1 - {(1 + \frac{r}{k})}^{- N k})}{(\frac{r}{k})}$

$P sub 0$ is the balance in the account at the beginning (the principal, or amount of the loan).

$d$ is your loan payment (your monthly payment, annual payment, etc)

$r$ is the annual interest rate in decimal form.

$k$ is the number of compounding periods in one year.

$N$ is the length of the loan, in years

Like before, the compounding frequency is not always explicitly given, but is determined by how often you make payments.

Example 1

You can afford $200 per month as a car payment. If you can get an auto loan at 3% interest for 60 months (5 years), how expensive of a car can you afford? In other words, what amount loan can you pay off with $200 per month?

Show solution

In this example,

$\begin{matrix} d = $ 200 & the monthly loan payment \\ r = 0.03 & 3 % annual rate \\ k = 12 & since we’re doing monthly payments, we’ll compound monthly \\ N = 5 & since we’re making monthly payments for 5 years \end{matrix}$

We’re looking for $P sub 0$ , the starting amount of the loan.

$P_{0} = \frac{200 (1 - {(1 + \frac{0.03}{12})}^{- 5 (12)})}{(\frac{0.03}{12})}$

$P_{0} = \frac{200 (1 - (1.0025)^{- 60})}{(0.0025)}$

$P_{0} = \frac{200 (1 - 0.861)}{(0.0025)} = $ 11, 120$

You can afford a $$ 11, 120$ loan.

You will pay a total of $12,000 ($200 per month for 60 months) to the loan company. The difference between the amount you pay and the amount of the loan is the interest paid. In this case, you’re paying $$ 12,000 - $ 11,120 = $ 880$ interest total.

Example 2

You want to take out a $140,000 mortgage (home loan). The interest rate on the loan is 6%, and the loan is for 30 years. How much will your monthly payments be?

# Ch 9.6 - a loan is a payout annuity where the BANK is the one being paid.
# This cell shows where each payment actually goes.
def payment(P0, r, k, N):
    i = r / k
    return P0 * i / (1 - (1 + i) ** (-N * k))

LOAN, APR, K, YEARS = 140000, 0.06, 12, 30      # the book's mortgage
# TRY IT: YEARS = 15. The payment goes up, the total interest collapses.
# TRY IT: APR = 0.03, then 0.09.

i = APR / K
d = payment(LOAN, APR, K, YEARS)
total = d * K * YEARS
print(f"${LOAN:,} at {100 * APR:g}% for {YEARS} years")
print(f"   monthly payment ${d:,.2f}")
print(f"   you hand the bank ${total:,.2f} in total")
print(f"   ${total - LOAN:,.2f} of that is interest "
      f"({100 * (total - LOAN) / LOAN:.0f}% of what you borrowed)\n")

print("The first three payments and the last three:")
print("payment    interest    principal    balance after")
balance = LOAN
rows = []
for m in range(1, YEARS * K + 1):
    interest = balance * i
    principal = d - interest
    balance -= principal
    rows.append((m, interest, principal, balance))
for m, interest, principal, balance in rows[:3] + rows[-3:]:
    print(f"{m:>7}    {interest:>8,.2f}    {principal:>9,.2f}    {balance:>13,.2f}")
print(f"In payment 1, ${rows[0][1]:,.2f} of your ${d:,.2f} is interest and only "
      f"${rows[0][2]:,.2f} pays down the loan.")
print(f"In the last payment it is the other way round.\n")

print("year   interest paid   principal paid   balance at year end")
for y in range(1, YEARS + 1):
    chunk = rows[(y - 1) * K: y * K]
    if y <= 3 or y % 5 == 0 or y == YEARS:
        print(f"{y:>4}   {sum(r[1] for r in chunk):>13,.2f}   "
              f"{sum(r[2] for r in chunk):>14,.2f}   {chunk[-1][3]:>19,.2f}")
half = next(m for m, _, _, bal in rows if bal <= LOAN / 2)
print(f"\nYou do not owe half the loan until payment {half} of {YEARS * K}, "
      f"i.e. year {half / K:.1f} of {YEARS}.")

Show solution

In this example,

We’re looking for $d$ .

$\begin{matrix} r = 0.06 & 6 % annual rate \\ k = 12 & since we’re doing monthly payments, we’ll compound monthly \\ N = 30 & since we’re making monthly payments for 30 years \\ P_{0} = $ 140, 000 & the starting loan amount \end{matrix}$

In this case, we’re going to have to set up the equation, and solve for $d$ .

$140, 000 = \frac{d (1 - {(1 + \frac{0.06}{12})}^{- 30 (12)})}{(\frac{0.06}{12})}$

$140, 000 = \frac{d (1 - (1.005)^{- 360})}{(0.005)}$

$140, 000 equals d open parenthesis 166.792 close parenthesis$

$d = \frac{140, 000}{166.792} = $ 839.37$

You will make payments of $839.37 per month for 30 years.

You're paying a total of $$ 302, 173.20$ to the loan company: $$ 839.37$ per month for 360 months. You are paying a total of
$$ 302,173.20 - $ 140,000 = $ 162,173.20$ in interest over the life of the loan.

Your Turn

Try it Now 4

Janine bought $3,000 of new furniture on credit. Because her credit score isn’t very good, the store is charging her a fairly high interest rate on the loan: 16%. If she agreed to pay off the furniture over 2 years, how much will she have to pay each month?

Answer

$\begin{matrix} d = unknown \\ r = 0.16 & 16 % annual rate \\ k = 12 & since we’re doing monthly payments, we’ll compound monthly \\ N = 2 & 2 year to repay \\ P_{0} = 3, 000 & the starting loan amount $3,000 loan \end{matrix}$

$3, 000 = \frac{d (1 - {(1 + \frac{0.16}{12})}^{- 2 \times 12})}{\frac{0.16}{12}}$

Solving for $d$ gives $$ 146.89$ as monthly payments.

In total, she will pay $$ 3, 525.36$ to the store, meaning she will pay $$ 525.36$ in interest over the two years.

Adapted from Math in Society by David Lippman, hosted on LibreTexts (math.libretexts.org) and licensed under CC BY-SA 3.0. Changes were made. License: CC-BY-SA-3.0.