Compound Growth & Interest

What is Compounding?

Compounding means earning returns on your returns. Each period, you earn interest not just on your original amount, but also on all previously earned interest.

Simple Example: If you invest 100 EUR at 10% annual interest:

  • Year 1: 100 EUR → 110 EUR (earned 10 EUR)
  • Year 2: 110 EUR → 121 EUR (earned 11 EUR, not just 10 EUR!)
  • Year 3: 121 EUR → 133.10 EUR (earned 12.10 EUR)

Part 1: The Three Main Compounding Formulas

1. Annual/Period Compounding (Most Common)

Formula: \(A = P(1 + r)^t\)

Where:

  • \(A\) = Final amount
  • \(P\) = Principal (starting amount)
  • \(r\) = Rate per period (as decimal: 8% = 0.08)
  • \(t\) = Number of periods

When to use: When interest compounds once per period (annually, monthly, etc.)

Example: 1,000 EUR invested at 6% annual interest for 5 years

  • \(A = 1000(1.06)^5 \approx 1338.23\) EUR

2. Multiple Compounding Per Year

Formula: \(A = P\left(1 + \frac{r}{n}\right)^{nt}\)

Where:

  • \(r\) = Annual nominal rate (as decimal)
  • \(n\) = Compounding frequency per year
    • Monthly: \(n = 12\)
    • Quarterly: \(n = 4\)
    • Daily: \(n = 365\)
  • \(t\) = Time in years

When to use: When interest compounds multiple times per year

Example: 5,000 EUR at 8% annual rate, compounded monthly for 3 years

  • \(A = 5000\left(1 + \frac{0.08}{12}\right)^{12 \times 3} \approx 6351.19\) EUR

3. Continuous Compounding

Formula: \(A = Pe^{rt}\)

Where:

  • \(e\) ≈ 2.71828 (Euler’s number)
  • Other variables are basically the same as above

When to use: Theoretical maximum compounding

Example: 2,000 EUR at 7% continuously compounded for 10 years

  • \(A = 2000e^{0.07 \times 10} \approx 4027.51\) EUR

Part 2: Comparing Methods

Effective Annual Rate (EAR)

The EAR shows the actual yearly return when compounding occurs more frequently.

From periodic compounding: \(\text{EAR} = \left(1 + \frac{r}{n}\right)^n - 1\)

From continuous: \(\text{EAR} = e^r - 1\)

Example: 12% nominal rate

Compounding EAR Calculation Result
Annual 12% 12.00%
Quarterly \((1.03)^4 - 1\) 12.55%
Monthly \((1.01)^{12} - 1\) 12.68%
Continuous \(e^{0.12} - 1\) 12.75%

Part 3: General Exponential Growth

Population/Bacteria Growth

If something multiplies by factor \(k\) every \(\Delta t\) time units:

Formula: \(P(t) = P_0 \cdot k^{t/\Delta t}\)

Example: Bacteria triple every 4 hours, starting with 500

  • After 12 hours: \(P(12) = 500 \cdot 3^{12/4} = 500 \cdot 27 = 13,500\) bacteria

Converting to continuous rate: \(r = \dfrac{\ln (k)}{\Delta t}\)

  • For tripling every 4 hours: \(r = \dfrac{\ln (3)}{4} = 0.275\) per hour

That’s the important part that made the task in lecture 01-06 difficult! We know, that the bacteria triples every 4 hours. That way, we can essentially either work with continous compounding or with periodic compounding, as long as the bacteria triples every 4 hours.

  • Tripling every 4 hours: \(A = P \cdot 3^{t/4}\)
  • Equivalent continuous form: \(A = P \cdot e^{(\ln (3)/4)t}\)
  • They match because \(3^{t/4} = e^{(\ln 3/4)t}\)

Practice Problems

Basic Application

  1. You invest 3,000 EUR at 5% annual interest. What’s the value after 8 years?
  2. A savings account offers 4.8% compounded monthly. If you deposit 2,500 EUR, what’s the value after 3 years?
  3. How long does it take 1,000 EUR to grow to 2,500 EUR at 9% annual interest?
  4. A card charges 18% APR (annual percentage rate, the nominal yearly rate) compounded daily. If you owe 5,000 EUR and make no payments, what will you owe after 2 years?
  5. A city of 50,000 grows at 3% annually. What’s the population after 20 years?
  6. You need 50,000 EUR in 10 years for a down payment. If you can earn 7% annually, how much must you invest today?
  7. A bacteria culture doubles every 3 hours. If you start with 1,000 bacteria at noon, how many will there be at midnight?
  1. \(A = 3000(1.05)^8\)
  2. \(A = 2500(1 + 0.048/12)^{12 \times 3} = 2500(1.004)^{36}\)
  3. \(t = \ln(2.5)/\ln(1.09)\)
  4. \(A = 5000(1 + 0.18/365)^{365 \times 2}\)
  5. \(P = 50000(1.03)^{20}\)
  6. \(P = 50000/(1.07)^{10}\)
  7. From noon to midnight = 12 hours = 4 doubling periods, hence Bacteria = \(1000 \times 2^4\)