IB Mathematics Analysis and Approaches Topic SL…

From Simple Interest To Compound Interest

Simple interest pays the same fixed amount each year, calculated only on the original sum, so the balance grows in equal steps like an arithmetic sequence.
Compound interest is different: each year's interest is added to the balance, so the next year earns interest on a bigger amount.
That means each year you multiply the balance by the same factor, which is exactly a geometric sequence (SL 1.3), and it is why savings grow faster over time.

The Compound Interest Formula

$$ FV = PV\left(1 + \frac{r}{100k}\right)^{kn} $$

$PV$ is the present value, the amount you start with, and $FV$ is the future value, the amount at the end.
$r$ is the nominal annual interest rate as a percentage, $n$ is the number of years, and $k$ is how many times interest is added per year.
The piece $\dfrac{r}{100k}$ is the interest rate for one period: divide the annual percentage $r$ by $100$ to make it a decimal, then by $k$ to share it across the $k$ periods in a year.
The exponent $kn$ is the total number of compounding periods: $k$ per year for $n$ years.

Note

So $k = 1$ is yearly, $k = 2$ half-yearly, $k = 4$ quarterly and $k = 12$ monthly, and a larger $k$ compounds slightly faster.

Example question

$\$1000$ is invested at a nominal annual rate of $5\%$. Find its value after 3 years if interest is compounded (a) annually and (b) monthly.

Solution

(a) $k = 1$: $FV = 1000\left(1 + \tfrac{5}{100}\right)^{3} = 1000(1.05)^3 = \$1157.63$.
(b) $k = 12$: $FV = 1000\left(1 + \tfrac{5}{1200}\right)^{36} = \$1161.47$, slightly more because it compounds more often.

Exam technique

The formula booklet (SL 1.4) gives this formula, and you are never asked to derive it.
The single most common error is leaving $k = 1$ when the interest is not annual, so read the compounding frequency carefully.
The built-in finance solver on your GDC is allowed on Paper 2 and is often faster.

Depreciation: The Same Idea Running Backwards

Depreciation is when something loses a fixed percentage of its value each year, which is geometric decay with a ratio below 1.
Losing $p\%$ means keeping $(100 - p)\%$, so each year you multiply by the factor $1 - \tfrac{p}{100}$.

$$ \text{value after } n \text{ years} = PV\left(1 - \frac{p}{100}\right)^{n} $$

Example

A $\$20\,000$ car depreciates $15\%$ per year, so it keeps $85\%$ and the factor is $0.85$.
After 4 years its value is $20\,000(0.85)^4 = \$10\,440.13$.

Real Value: Adjusting For Inflation

The nominal value is the number of dollars in the account; the real value is what those dollars can actually buy.
Inflation shrinks buying power, so to find the real value you grow the money by the interest rate but discount it by the inflation rate.
If the interest rate only matches inflation, the real value does not change even though the nominal balance rises.

Note

As the compounding frequency $k$ grows without bound, the factor $\left(1 + \tfrac{r}{100k}\right)^{kn}$ approaches continuous compounding and the number $e$.
This link to $e$ is enrichment only and is never examined in this sub-topic.

Active recall

Why does compound interest grow faster than simple interest?
In the formula, why is the per-period rate $\tfrac{r}{100k}$ and the exponent $kn$?
What value of $k$ do you use for quarterly compounding?
Write the multiplying factor for a value depreciating 15% per year.
Why can a rising nominal balance still mean no real gain?

From Simple Interest To Compound Interest

Simple interest pays the same fixed amount each year, calculated only on the original sum, so the balance grows in equal steps like an arithmetic sequence.
Compound interest is different: each year's interest is added to the balance, so the next year earns interest on a bigger amount.
That means each year you multiply the balance by the same factor, which is exactly a geometric sequence (SL 1.3), and it is why savings grow faster over time.

The Compound Interest Formula

$$ FV = PV\left(1 + \frac{r}{100k}\right)^{kn} $$

$PV$ is the present value, the amount you start with, and $FV$ is the future value, the amount at the end.
$r$ is the nominal annual interest rate as a percentage, $n$ is the number of years, and $k$ is how many times interest is added per year.
The piece $\dfrac{r}{100k}$ is the interest rate for one period: divide the annual percentage $r$ by $100$ to make it a decimal, then by $k$ to share it across the $k$ periods in a year.
The exponent $kn$ is the total number of compounding periods: $k$ per year for $n$ years.

Note

So $k = 1$ is yearly, $k = 2$ half-yearly, $k = 4$ quarterly and $k = 12$ monthly, and a larger $k$ compounds slightly faster.

Example question

$\$1000$ is invested at a nominal annual rate of $5\%$. Find its value after 3 years if interest is compounded (a) annually and (b) monthly.

Solution

(a) $k = 1$: $FV = 1000\left(1 + \tfrac{5}{100}\right)^{3} = 1000(1.05)^3 = \$1157.63$.
(b) $k = 12$: $FV = 1000\left(1 + \tfrac{5}{1200}\right)^{36} = \$1161.47$, slightly more because it compounds more often.

Exam technique

The formula booklet (SL 1.4) gives this formula, and you are never asked to derive it.
The single most common error is leaving $k = 1$ when the interest is not annual, so read the compounding frequency carefully.
The built-in finance solver on your GDC is allowed on Paper 2 and is often faster.

Depreciation: The Same Idea Running Backwards

Depreciation is when something loses a fixed percentage of its value each year, which is geometric decay with a ratio below 1.
Losing $p\%$ means keeping $(100 - p)\%$, so each year you multiply by the factor $1 - \tfrac{p}{100}$.

$$ \text{value after } n \text{ years} = PV\left(1 - \frac{p}{100}\right)^{n} $$

Example

A $\$20\,000$ car depreciates $15\%$ per year, so it keeps $85\%$ and the factor is $0.85$.
After 4 years its value is $20\,000(0.85)^4 = \$10\,440.13$.

Real Value: Adjusting For Inflation

The nominal value is the number of dollars in the account; the real value is what those dollars can actually buy.
Inflation shrinks buying power, so to find the real value you grow the money by the interest rate but discount it by the inflation rate.
If the interest rate only matches inflation, the real value does not change even though the nominal balance rises.

Note

As the compounding frequency $k$ grows without bound, the factor $\left(1 + \tfrac{r}{100k}\right)^{kn}$ approaches continuous compounding and the number $e$.
This link to $e$ is enrichment only and is never examined in this sub-topic.

Active recall

Why does compound interest grow faster than simple interest?
In the formula, why is the per-period rate $\tfrac{r}{100k}$ and the exponent $kn$?
What value of $k$ do you use for quarterly compounding?
Write the multiplying factor for a value depreciating 15% per year.
Why can a rising nominal balance still mean no real gain?

IB Mathematics Analysis and Approaches SL 1.4—financial Apps – Compound Interest, Annual Depreciation Notes

IB Mathematics Analysis and Approaches SL 1.4—financial Apps – Compound Interest, Annual Depreciation Notes

From Simple Interest To Compound Interest

The Compound Interest Formula

Depreciation: The Same Idea Running Backwards

Real Value: Adjusting For Inflation

From Simple Interest To Compound Interest

The Compound Interest Formula

Depreciation: The Same Idea Running Backwards

Real Value: Adjusting For Inflation

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

IB Mathematics Analysis and Approaches SL 1.4—financial Apps – Compound Interest, Annual Depreciation Notes

IB Mathematics Analysis and Approaches SL 1.4—financial Apps – Compound Interest, Annual Depreciation Notes

From Simple Interest To Compound Interest

The Compound Interest Formula

Depreciation: The Same Idea Running Backwards

Real Value: Adjusting For Inflation

From Simple Interest To Compound Interest

The Compound Interest Formula

Depreciation: The Same Idea Running Backwards

Real Value: Adjusting For Inflation