Application of Financial Mathematics

NoteLearning objectives

• Compute simple interest and the acquired capital.
• Compute compound interest and the acquired capital.
• Convert an interest rate from one period to another.
• Recognise which regime (simple or compound) applies to a situation.

📊 Lecture slides: View the slides for this module

1 Diagnostic Activity

Answer the following questions from memory before reading further.

1. Compute the simple interest earned on €1,000 invested at 5% per year for 2 years.
2. Is compound interest modelled by an arithmetic or geometric sequence?
3. A monthly interest rate is 0.5%. What is the equivalent annual rate under simple (proportional) conversion?

NoteCheck your answers

1. \(1{,}000 \times 0.05 \times 2 = €100.\)
2. Geometric — each period multiplies the capital by a constant factor \((1+i).\)
3. \(0.5\% \times 12 = 6\%\) per year.

Use your results to identify topics needing attention.

2 Opening Problem

You invest €5,000 at an annual rate of 4%. Bank A compounds the interest once a year; Bank B compounds it every month (same annual rate).

After 10 years, how much more does Bank B give you than Bank A? Is the difference significant?

Write down your estimate before continuing.

3 Core Concepts

Simple interest and compound interest are two different ways of computing, for example, the return on invested capital or the cost of a loan. Each has its practical uses depending on the financial and economic context. They are the direct application of the arithmetic and geometric sequences studied in Module 4.

Simple Interest

NotePrinciple

With simple interest, interest is computed only on the initial capital, without reinvestment. The interest is the same for each period, so the acquired capital follows an arithmetic sequence.

Let \(C_0\) be the initial capital, \(i\) the interest rate per period, and \(n\) the number of periods. The interest of one period is \(C_0\,i\) (constant), so:

\[ \text{Interest over } n \text{ periods} = C_0\,i\,n, \qquad C_n = C_0\,(1 + i\,n). \]

TipTypical uses of simple interest

• short-term loans;
• one-off investments without reinvestment;
• late-payment penalties and fees.

Worked example

We invest \(100{,}000\) € at a simple annual rate of \(4\%\) for \(4\) years. The yearly interest is constant: \(100{,}000 \times 0.04 = 4{,}000\) € per year. After \(4\) years the acquired capital is

\[ C_4 = 100{,}000\,(1 + 0.04 \times 4) = 100{,}000 \times 1.16 = 116{,}000 \text{ €}. \]

Converting Interest Rates

Sometimes the rate and the period are not expressed in the same unit of time. Under the simple (proportional) convention, we scale the rate proportionally to the period.

TipProportional conversion (simple regime)

An annual rate \(i_{\text{year}}\) corresponds to a monthly rate \(i_{\text{month}} = \dfrac{i_{\text{year}}}{12}\), and conversely \(i_{\text{year}} = 12\,i_{\text{month}}\).

CautionRate exercises (click for solutions)

1. Convert an annual rate of \(6\%\) to a monthly rate: \(\dfrac{6\%}{12} = 0.5\%\) per month.
2. Convert a monthly rate of \(1\%\) to an annual rate: \(12 \times 1\% = 12\%\) per year.
3. A capital of \(5{,}000\) € invested for \(4\) months earns \(50\) € of interest. Find the annual rate. The \(4\)-month rate is \(\dfrac{50}{5000} = 1\%\), so the monthly rate is \(0.25\%\) and the annual rate is \(12 \times 0.25\% = 3\%\).

Compound Interest

NotePrinciple

With compound interest, the interest earned each period is added to the capital and itself earns interest in the following periods (the “snowball” effect). The acquired capital follows a geometric sequence.

CautionDerivation — the compound-interest formula (click to expand)

At the end of one period the capital is multiplied by \((1 + i)\): the capital \(C_0\) becomes \(C_1 = C_0 + C_0 i = C_0(1 + i)\). Repeating, \(C_2 = C_1(1 + i) = C_0(1 + i)^2\), and by induction

\[ C_n = C_0\,(1 + i)^{\,n}. \]

This is exactly a geometric sequence with first term \(C_0\) and common ratio \(q = 1 + i\). The interest earned over \(n\) periods is \(C_n - C_0 = C_0\big[(1 + i)^n - 1\big]\).

Worked example

We invest \(100{,}000\) € at a compound annual rate of \(5\%\) for \(10\) years:

\[ C_{10} = 100{,}000 \times 1.05^{10} \approx 162{,}889 \text{ €}. \]

For comparison, the same capital at a simple rate of \(6\%\) for \(10\) years gives \(100{,}000(1 + 0.06 \times 10) = 160{,}000\) €: over a long horizon the compounding effect can outweigh a higher simple rate — but not always, as the final-exam case in Module 8 shows.

ImportantExam tip — equivalent compound rate

Converting a compound rate between periods is not proportional. A monthly compound rate \(i_m\) corresponds to the annual rate \(i_{\text{year}} = (1 + i_m)^{12} - 1\). For \(i_m = 0.5\%\) this gives \((1.005)^{12} - 1 \approx 6.17\%\), not \(6\%\).

Average Placement Rate

The average placement rate is the single rate that would produce the same final result (the same acquired capital) as a set of distinct investments, each with its own rate, amount, and duration. It summarises the overall performance of several combined investments in one figure.

TipPresent value

\[P = \frac{V_c}{(1 + i)^n}\]

The present value \(P\) is the initial capital that would produce a future value \(V_c\) at rate \(i\) after \(n\) periods — it lets us compare cash flows at different points in time.

4 Worked Example

Worked example — Bank A vs Bank B (the Opening Problem)

We now solve the problem posed in §2 step by step.

Bank A — annual compounding (\(k = 1\)):

\[ V_A = 5{,}000 \times (1 + 0.04)^{10} = 5{,}000 \times 1.04^{10} \approx 7{,}401.22 \text{ €}. \]

Bank B — monthly compounding (\(k = 12\)):

\[ V_B = 5{,}000 \times \!\left(1 + \frac{0.04}{12}\right)^{10 \times 12} = 5{,}000 \times (1.003\overline{3})^{120} \approx 7{,}453.71 \text{ €}. \]

Difference:

\[ 7{,}453.71 - 7{,}401.22 = 52.49 \text{ €}. \]

Monthly compounding yields €52.49 more — a modest but real gain from the compounding frequency.

5 Guided Practice

An employee places €3,000 in a savings account offering a compound annual rate of 3.5%, compounded quarterly.

1. Write the formula for the acquired capital after \(n\) years.
2. Calculate the acquired capital after 4 years.
3. How many years does it take to double the initial capital?

TipHint

With quarterly compounding, \(k = 4\). Use \(V_c = P\!\left(1 + \dfrac{i}{k}\right)^{nk}\) with \(P = 3{,}000\), \(i = 0.035\), \(k = 4\). For doubling, solve \(\left(1 + \dfrac{0.035}{4}\right)^{4n} = 2\).

CautionShow the solution

1. Formula:

\[ V_c = 3{,}000 \times \!\left(1 + \frac{0.035}{4}\right)^{4n} = 3{,}000 \times 1.00875^{4n}. \]

2. After 4 years (\(n = 4\), so the exponent is \(4 \times 4 = 16\)):

\[ V_c = 3{,}000 \times 1.00875^{16} \approx 3{,}000 \times 1.1490 \approx 3{,}447 \text{ €}. \]

3. Doubling: solve \((1.00875)^{4n} = 2\). Testing \(n = 20\): \((1.00875)^{80} \approx 2.002 \approx 2\). About 20 years.

6 Interactive Exploration

Use the controls below to explore how the principal, rate, number of years, and compounding frequency interact.

viewof principal = Inputs.range([100, 10000], {
  value: 5000,
  step: 100,
  label: "Principal P (€)"
})

viewof rate = Inputs.range([0, 20], {
  value: 4,
  step: 0.1,
  label: "Annual rate i (%)"
})

viewof periods = Inputs.range([1, 30], {
  value: 10,
  step: 1,
  label: "Number of years n"
})

viewof freq = Inputs.select([1, 2, 4, 12], {
  value: 1,
  label: "Capitalisation frequency k (per year)"
})

i = rate / 100
simpleVal = principal * (1 + i * periods)
compoundVal = principal * Math.pow(1 + i / freq, periods * freq)
cumInterest = compoundVal - principal

timeData = Array.from({length: periods + 1}, (_, t) => ({
  t,
  compound: principal * Math.pow(1 + i / freq, t * freq),
  simple: principal * (1 + i * t)
}))

growthChart = Plot.plot({
  marks: [
    Plot.line(timeData, {x: "t", y: "compound", stroke: "steelblue",
      label: "Compound"}),
    Plot.line(timeData, {x: "t", y: "simple", stroke: "orange",
      strokeDash: [4, 4], label: "Simple"}),
    Plot.dot(timeData, {x: "t", y: "compound", fill: "steelblue", r: 3}),
    Plot.dot(timeData, {x: "t", y: "simple", fill: "orange", r: 3})
  ],
  x: {label: "Year"},
  y: {label: "Capital (€)"},
  color: {legend: true}
})

html`
<div style="margin-bottom:1rem">${growthChart}</div>
<div class="metric-grid">
  <div class="metric-card">
    <span class="metric-label">Simple interest value</span>
    <strong>${simpleVal.toFixed(2)} €</strong>
  </div>
  <div class="metric-card">
    <span class="metric-label">Compound interest value</span>
    <strong>${compoundVal.toFixed(2)} €</strong>
  </div>
  <div class="metric-card">
    <span class="metric-label">Cumulative interest earned</span>
    <strong>${cumInterest.toFixed(2)} €</strong>
  </div>
  <div class="metric-card">
    <span class="metric-label">Gain over simple interest</span>
    <strong>${(compoundVal - simpleVal).toFixed(2)} €</strong>
  </div>
</div>
`

Investigation

1. Set \(k = 1\) (annual). Now set \(k = 12\) (monthly), keeping all other parameters fixed. How much extra does monthly compounding earn after 10 years?
2. Keep \(k = 1\) and increase \(n\) from 5 to 30. Does the gap between compound and simple interest grow slowly or quickly?
3. Which has more impact on the final compound value: doubling the rate or doubling the number of years? Test with the sliders.
4. Set \(i = 0\) (zero rate). What do both formulas give? Why?

7 Applied Problem

A firm needs to repay a loan of €200,000 in 8 years. Two banks compete for the business. Bank A offers a simple annual rate of 7%. Bank B offers a compound annual rate of 5.5%.

1. Calculate the total amount repaid to each bank after 8 years.
2. Which bank offers the better deal?
3. Bank A also offers to apply its rate monthly under simple (proportional) conversion. What is the monthly rate? What is the total repayment?
4. Bank B offers monthly compounding at the same annual rate. What is the effective annual compound rate? Is it higher or lower than 5.5%?

8 Common Errors

WarningCommon error — applying proportional conversion to compound rates

A student converts a monthly compound rate of 0.5% to an annual rate as \(0.5\% \times 12 = 6\%\).

This is only valid for simple (proportional) rates. For compound rates, the correct conversion is:

\[ i_{\text{year}} = (1 + i_{\text{month}})^{12} - 1 = (1.005)^{12} - 1 \approx 6.17\%. \]

Always use the compound-rate conversion formula when interest is reinvested each period.

9 Check Your Understanding

1. €8,000 invested at 3% simple interest for 5 years. What is the acquired capital?
2. What does it mean economically when the capitalisation frequency \(k\) increases while the annual rate stays fixed?
3. €10,000 invested at 4% compound (annual) for 8 years. Calculate the acquired capital.
4. A student computes the annual compound rate equivalent to a monthly rate of 1% as \(1\% \times 12 = 12\%\). What is the correct answer?
5. Which repayment is smaller: €100,000 at 6% simple for 10 years, or €100,000 at 5% compound for 10 years?

10 Summary

1. Simple interest applies the rate to the original capital only (\(V_s = P(1+in)\)); compound interest applies it to the growing capital (\(V_c = P(1+i/k)^{nk}\)) — the difference grows with time and rate.
2. Simple interest rates convert proportionally across periods; compound rates do not — converting \(i_m\) to an annual rate requires \((1+i_m)^{12} - 1\), which always gives a higher annual rate.
3. The capitalisation frequency \(k\) matters: more frequent compounding (quarterly, monthly, daily) produces a higher final value for the same nominal annual rate, because interest starts earning interest sooner.

11 Independent Work

• Complete simple-interest exercises 1–5 in the exercise bank.
• Complete compound-interest exercises 1–5.
• Work through the loan comparison problem.
• Record errors in your error log.
• Review glossary entries for: simple interest, compound interest, capitalisation frequency, present value, acquired capital.

12 Continue

Continue to Module 6 →