0.3.5 Case study: simple interest and break-even

Simple interest

Simple interest pays the same amount every year, worked out on the original sum only. Interest does not earn interest; that is compounding, which module 1.4 covers. Money markets use simple interest for loans and deposits shorter than a year.

Rule: Simple interest

\begin{gathered} I = P\,r\,t \\ A = P + I = P(1 + rt) \end{gathered}

$P$ is the principal, the sum borrowed or deposited. $r$ is the yearly rate as a decimal and $t$ the time in years. $I$ is the interest and $A$ the amount repaid at the end.

For the loan in the opening: $I = 5000 \times 0.06 \times 3 = 900$ , so $A = 5900$ .

One formula, four questions

$A = P(1 + rt)$ links four quantities. Know any three and you can find the fourth by making it the subject. Here is the rate, using the gather-and-divide method from lesson 0.3.2:

1
$A = P + Prt$
2
$A - P = Prt$
3
$r = \frac{A - P}{Pt}$

The same moves give the other two. Dividing $A = P(1 + rt)$ by $1 + rt$ gives the present value, what a future sum is worth today. Dividing $A - P = Prt$ by $Pr$ gives the time.

\begin{gathered} P = \frac{A}{1 + rt} \qquad r = \frac{A - P}{Pt} \\[6pt] t = \frac{A - P}{Pr} \end{gathered}

1 + rt

import sympy as sp A, P, r, t = sp.symbols("A P r t") formula = sp.Eq(A, P * (1 + r * t)) for letter in (P, r, t): print(letter, "=", sp.solve(formula, letter)[0]) print(5900 / (1 + 0.06 * 3)) print((5600 - 5000) / (5000 * 2))

⌘/Ctrl + Enter runs the cell

Check yourself

At 4% simple interest, how many years does it take $5,000 to grow to $7,000?

\dfrac{7000}{5000 \times 0.04} = 35

\dfrac{7000 - 5000}{0.04} = 50{,}000

\dfrac{7000 - 5000}{5000 \times 0.04} = 10

Exercise

Simple interest, rearranged

Write three functions, one for each rearrangement of $A = P(1 + rt)$ :

present_value(amount, rate, years) returns $P$ .
implied_rate(principal, amount, years) returns $r$ as a decimal.
years_needed(principal, amount, rate) returns $t$ .

Work out each formula on paper first, or check them with SymPy in the cell above.

See the tests

def test_present_value():
    """$5,900 due in 3 years at 6% is worth $5,000 today"""
    expect_close(present_value(5900, 0.06, 3), 5000.0, what="present_value(5900, 0.06, 3)")


def test_present_value_part_year():
    """$10,125 due in a quarter of a year at 5% is worth $10,000 today"""
    expect_close(present_value(10125, 0.05, 0.25), 10000.0, what="present_value(10125, 0.05, 0.25)")


def test_implied_rate():
    """$5,000 that grows to $5,600 in 2 years earned 6% a year"""
    expect_close(implied_rate(5000, 5600, 2), 0.06, what="implied_rate(5000, 5600, 2)")


def test_years_needed():
    """$5,000 takes 10 years to reach $7,000 at 4%"""
    expect_close(years_needed(5000, 7000, 0.04), 10.0, what="years_needed(5000, 7000, 0.04)")


def test_round_trip():
    """Each answer leads back to the amount: P(1 + rt) = A"""
    p, a, t = 2500, 2800, 1.5
    r = implied_rate(p, a, t)
    expect_close(p * (1 + r * t), a, what="P(1 + rt) with your implied_rate")
    expect_close(present_value(a, r, t), p, what="present_value with that rate")
    expect_close(years_needed(p, a, r), t, what="years_needed with that rate")

Break-even

Now the coffee cart. Each cup sells for a price $p$ and costs $v$ to make: coffee, milk and the cup. This is the variable cost, because it grows with sales. The cart also has a fixed cost $F$ that must be paid whatever it sells: the licence, the cart itself and insurance.

Selling $q$ cups brings in $pq$ and costs $F + vq$ , so the profit is

\begin{aligned} \text{profit}(q) &= pq - (F + vq) \\ &= (p - v)\,q - F \end{aligned}

Each cup adds $p - v$ to profit. That amount, the contribution margin, chips away at the fixed cost. For the cart, $p = 3.50$ and $v = 1.70$ , so each cup contributes $1.80.

The cart breaks even when profit is no longer negative. That is an inequality, solved as in lesson 0.3.3:

\begin{aligned} (p - v)\,q - F &\ge 0 \\ (p - v)\,q &\ge F \\ q &\ge \frac{F}{p - v} \end{aligned}

The last step divides by $p - v$ . That is positive when the price is above the variable cost, so the sign does not flip. If $p \le v$ , every cup loses money and the cart never breaks even.

F / (p - v) = 12000 / 1.80 \approx 6666.67

import math F, p, v = 12_000, 3.50, 1.70 def profit(q): return (p - v) * q - F q_star = F / (p - v) print(q_star, math.ceil(q_star)) print(round(profit(6666), 2), round(profit(6667), 2))

⌘/Ctrl + Enter runs the cell

pq

import numpy as np import matplotlib.pyplot as plt F, p, v = 12_000, 3.50, 1.70 q = np.linspace(0, 12_000, 200) plt.plot(q, p * q, label="revenue") plt.plot(q, F + v * q, label="cost") plt.axvline(F / (p - v), color="gray", linestyle="--", linewidth=0.8) plt.xlabel("cups sold in a year") plt.ylabel("dollars") plt.legend() plt.show() print((F + 6_000) / (p - v))

⌘/Ctrl + Enter runs the cell

Check yourself

The cart raises its price to $3.90 a cup. Costs stay the same. How many cups does it need to break even now?

12000 / 3.90

, rounded up: 3,077

12000 / 2.20

, rounded up: 5,455

12000 / 1.80

, rounded up: 6,667

Exercise

Break-even units

Write break_even_units(fixed, price, variable) that returns the smallest whole number of units whose profit is zero or more.

Use $F / (p - v)$ and round up with math.ceil.
If price <= variable, the business can never break even. Raise ValueError.

The tests check that profit is non-negative at your answer and negative one unit below it.

See the tests

def _profit(q, fixed, price, variable):
    return (price - variable) * q - fixed


def _raises_value_error(*args):
    try:
        break_even_units(*args)
    except ValueError:
        return True
    return False


def test_coffee_cart():
    """The coffee cart (fixed $12,000, price $3.50, cost $1.70) needs 6,667 cups"""
    expect_equal(break_even_units(12_000, 3.50, 1.70), 6667, what="break_even_units(12000, 3.50, 1.70)")


def test_exact_division():
    """When the margin divides the fixed cost exactly, no rounding: 1000 / (15 − 5) = 100"""
    expect_equal(break_even_units(1000, 15, 5), 100, what="break_even_units(1000, 15, 5)")


def test_profit_either_side():
    """Profit is at least 0 at the answer and below 0 one unit earlier"""
    args = (7_500, 24.0, 13.0)
    q = break_even_units(*args)
    assert _profit(q, *args) >= 0, f"profit at your answer {q} is {_profit(q, *args):.2f}, which is negative"
    assert _profit(q - 1, *args) < 0, f"profit at {q - 1} units is already {_profit(q - 1, *args):.2f}, so {q} is not the smallest"


def test_whole_number():
    """The answer is a whole number of units"""
    q = break_even_units(12_000, 3.50, 1.70)
    assert isinstance(q, int), f"expected a whole number (int), got {q!r}"


def test_no_fixed_cost():
    """With no fixed cost, zero units already break even"""
    expect_equal(break_even_units(0, 5, 2), 0, what="break_even_units(0, 5, 2)")


def test_never_breaks_even():
    """A price at or below the variable cost never breaks even, so raise ValueError"""
    assert _raises_value_error(1000, 2.0, 2.5), "expected ValueError when price < variable"
    assert _raises_value_error(1000, 3.0, 3.0), "expected ValueError when price == variable"

I = Prt

Case study: simple interest and break-even

Simple interest

Rule: Simple interest

One formula, four questions

Simple interest, rearranged

Break-even

Break-even units

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