Compound Interest & Compounding Frequencies
- The Compound Interest Formula (Formula Booklet):
$$FV = PV \left( 1 + \frac{r}{100k} \right)^{kn}$$
where:
- $FV$: Future Value (final accumulated balance)
- $PV$: Present Value (initial principal invested)
- $r$: Nominal annual interest rate in percent (e.g. for $6\%$, $r = 6$, not $0.06$)
- $k$: Number of compounding periods per year
- $n$: Number of years of investment
- $kn$: Total number of compounding periods
- Standard Compounding Frequencies ($k$):
- Annually: $k = 1$
- Semi-annually (Half-yearly): $k = 2$
- Quarterly: $k = 4$
- Monthly: $k = 12$
- Daily: $k = 365$
- Total Interest Earned: $$\text{Interest Earned} = FV - PV$$
Example 1
Comparing Compounding Frequencies
An investor deposits $\$8000$ into a savings bond paying a nominal annual interest rate of $4.8\%$ for $5$ years. Calculate the future value if interest is compounded:
(a) Annually ($k = 1$)
(b) Monthly ($k = 12$)
Give your answers to the nearest dollar.
- Part (a): Annual Compounding ($k = 1$): $$FV = 8000 \left( 1 + \frac{4.8}{100 \times 1} \right)^{1 \times 5} = 8000(1 + 0.048)^5 = 8000(1.048)^5$$ $$FV = 8000(1.26417) \approx \$10\,113.38 \implies \$10\,113$$
- Part (b): Monthly Compounding ($k = 12$): $$FV = 8000 \left( 1 + \frac{4.8}{100 \times 12} \right)^{12 \times 5} = 8000 \left( 1 + \frac{4.8}{1200} \right)^{60}$$ $$FV = 8000 (1 + 0.004)^{60} = 8000 (1.004)^{60}$$ $$FV = 8000(1.27064) \approx \$10\,165.13 \implies \$10\,165$$
- Comparison: Monthly compounding yields $\$10\,165 - \$10\,113 = \$52$ more than annual compounding due to interest earning interest more frequently.
Answer: (a) $\$10\,113$; (b) $\$10\,165$
Annual Depreciation (Reducing Balance Method)
- Depreciation Definition: The reduction in the market value of a physical asset (such as machinery, vehicles, or technology) over time at a constant percentage rate.
- Reducing Balance Depreciation Formula: $$FV = PV (1 - r)^n$$ where $PV$ is the initial purchase price, $r$ is the annual rate of depreciation expressed as a decimal (e.g. for $15\%$, $r = 0.15$), and $n$ is the number of years.
- Connection to Geometric Sequences: Depreciation is modeled by a geometric sequence where the common ratio is $R = 1 - r < 1$: $$u_{n+1} = u_1 (1 - r)^n$$ The balance decays exponentially toward zero over time.
Example 2
Vehicle Depreciation Modeling
A company buys a transport truck for $\$45\,000$. The truck depreciates at a rate of $14\%$ per year.
(a) Find the value of the truck after $4$ years.
(b) Calculate the total loss in value of the truck over this $4$-year period.
Give your answers to the nearest dollar.
- Part (a): Apply the depreciation formula with $PV = 45\,000, r = 0.14, n = 4$: $$FV = 45\,000(1 - 0.14)^4 = 45\,000(0.86)^4$$ $$(0.86)^4 \approx 0.547008$$ $$FV = 45\,000 \times 0.547008 \approx \$24\,615.36 \implies \$24\,615$$
- Part (b): Calculate total loss in value: $$\text{Loss} = PV - FV = \$45\,000 - \$24\,615 = \$20\,385$$
Answer: (a) $\$24\,615$; (b) $\$20\,385$
Solving for Investment Duration $n$ Using Logarithms
- Problem Type: Finding how many years (or compounding periods) are required for an investment to reach or double a specified target value.
- Step-by-Step Algebraic Derivation:
Given $FV = PV \left( 1 + \frac{r}{100k} \right)^{kn}$:
- Divide by $PV$: $$\frac{FV}{PV} = \left( 1 + \frac{r}{100k} \right)^{kn}$$
- Take natural logarithms ($\ln$) of both sides: $$\ln\left(\frac{FV}{PV}\right) = \ln\left[\left( 1 + \frac{r}{100k} \right)^{kn}\right] = kn \ln\left( 1 + \frac{r}{100k} \right)$$
- Isolate the number of years $n$: $$n = \frac{\ln(FV / PV)}{k \ln\left( 1 + \frac{r}{100k} \right)}$$
- Rounding in Financial Contexts: If interest is compounded periodically (e.g. annually or quarterly), round up to the next full compounding period so the target threshold is actually reached.
Example 3
Solving for Doubling Time
An investor deposits $\$6000$ at a nominal annual interest rate of $5.4\%$ compounded quarterly. Find the minimum number of full quarters required for the investment to double in value ($FV \ge \$12\,000$).
- Set up the compound interest inequality with $PV = 6000, FV = 12\,000, r = 5.4, k = 4$: $$6000 \left( 1 + \frac{5.4}{400} \right)^{4n} \ge 12\,000$$ $$6000 (1 + 0.0135)^{4n} \ge 12\,000 \implies (1.0135)^{4n} \ge 2$$
- Let $N = 4n$ be the total number of compounding quarters: $$(1.0135)^N \ge 2$$
- Take natural logarithms on both sides: $$\ln(1.0135^N) \ge \ln 2 \implies N \ln(1.0135) \ge \ln 2$$ $$N \ge \frac{\ln 2}{\ln(1.0135)} \approx \frac{0.693147}{0.0134098} \approx 51.69\text{ quarters}$$
- Round up to the next full quarter: $N = 52$ quarters ($52 \div 4 = 13$ years).
Answer: $52$ quarters ($13$ full years)
Example 4
Solving for Investment Horizon via Natural Logarithms
Maria deposits $\$6000$ in a high-yield account offering $4.2\%$ nominal annual interest, compounded monthly.
(a) Calculate the balance in the account after $8$ years. [3 marks]
(b) Find the minimum number of complete years required for the investment to grow to at least $\$15\,000$. [4 marks]
- Part (a): Monthly compounding parameters: $$PV = 6000, \quad r = 4.2, \quad k = 12, \quad n = 8 \implies kn = 96$$ $$FV = 6000\left(1 + \frac{4.2}{1200}\right)^{96} = 6000(1 + 0.0035)^{96} = 6000(1.0035)^{96} \quad [M1]$$ $$FV = 6000(1.39868) \approx \$8392.08 \implies \$8392 \quad [A1]$$
- Part (b): Set up growth inequality: $$6000(1.0035)^{12n} \ge 15\,000 \implies (1.0035)^{12n} \ge \frac{15\,000}{6000} = 2.5 \quad [M1]$$
- Take natural logarithms of both sides: $$\ln\left[(1.0035)^{12n}\right] \ge \ln(2.5) \implies 12n \ln(1.0035) \ge \ln(2.5) \quad [M1]$$ $$n \ge \frac{\ln(2.5)}{12\ln(1.0035)} = \frac{0.91629}{12(0.0034939)} = \frac{0.91629}{0.041926} \approx 21.855 \text{ years} \quad [A1]$$
- State minimum complete years: Since interest compounds monthly and complete years are requested: $$n = 22 \text{ complete years} \quad [A1]$$
Answers: (a) $\$8392$; (b) $22$ complete years
Example 5
Asset Depreciation & Sinking Fund Replacement
A printing press bought for $\$65\,000$ depreciates at an annual reducing rate of $15\%$. Concurrently, a replacement sinking fund is opened with an initial deposit of $\$12\,000$ earning $6\%$ annual interest compounded annually.
(a) Find the value of the press after $5$ years. [2 marks]
(b) Determine the number of years $t$ when the accumulated sinking fund first exceeds the residual value of the press. [5 marks]
- Part (a): Evaluate depreciation after 5 years: $$V(5) = 65\,000(1 - 0.15)^5 = 65\,000(0.85)^5 = 65\,000(0.443705) \approx \$28\,840.84 \implies \$28\,841 \quad [M1][A1]$$
- Part (b): Equate fund growth and asset depreciation: $$\text{Sinking fund: } S(t) = 12\,000(1 + 0.06)^t = 12\,000(1.06)^t \quad [M1]$$ $$\text{Asset value: } V(t) = 65\,000(0.85)^t$$ We require $S(t) > V(t)$: $$12\,000(1.06)^t > 65\,000(0.85)^t \implies \frac{(1.06)^t}{(0.85)^t} > \frac{65\,000}{12\,000} \quad [M1]$$ $$\left(\frac{1.06}{0.85}\right)^t > \frac{65}{12} \implies (1.247059)^t > 5.41667 \quad [A1]$$
- Solve for $t$ using logarithms: $$t \ln(1.247059) > \ln(5.41667) \implies t > \frac{\ln(5.41667)}{\ln(1.247059)} = \frac{1.68944}{0.22079} \approx 7.65 \text{ years} \quad [M1]$$ Thus, after $8$ years, the fund first exceeds the residual value [A1].
Answers: (a) $\$28\,841$; (b) $8$ years
Examiner Commentary: Combining the powers into $\left(\frac{1.06}{0.85} \right)^t$ simplifies the algebra dramatically, avoiding the need to solve multiple separate logarithm expressions.
$\$5000$ is invested at $4.5\%$ compound interest per annum, compounded annually. Calculate the value of the investment after $6$ years, to the nearest dollar.
View Complete Solution
- Identify parameters: $PV = 5000, r = 4.5, k = 1, n = 6$.
- Apply formula: $$FV = 5000\left(1 + \frac{4.5}{100}\right)^6 = 5000(1.045)^6$$ $$FV = 5000(1.30226) \approx \$6511.30 \implies \$6511$$
Final Answer: $\$6511$
$\$10\,000$ is deposited into an account earning $5.2\%$ nominal annual interest, compounded quarterly. Find the balance after $4$ years, to the nearest dollar.
View Complete Solution
- Identify parameters: $PV = 10\,000, r = 5.2, k = 4, n = 4 \implies kn = 16$.
- Evaluate quarterly rate: $$\frac{r}{100k} = \frac{5.2}{400} = 0.013$$
- Calculate future value: $$FV = 10\,000(1 + 0.013)^{16} = 10\,000(1.013)^{16}$$ $$FV = 10\,000(1.22949) \approx \$12\,294.94 \implies \$12\,295$$
Final Answer: $\$12\,295$
A laptop is purchased for $\$2000$. It depreciates at a rate of $20\%$ per year. Without a calculator, find the value of the laptop after $3$ years.
View Complete Solution
- Depreciation multiplier: $1 - 0.20 = 0.80 = \frac{4}{5}$.
- Evaluate after 3 years: $$FV = 2000 \times (0.80)^3 = 2000 \times \left(\frac{4}{5}\right)^3 = 2000 \times \frac{64}{125}$$ $$\frac{2000}{125} = 16 \implies 16 \times 64 = \$1024$$
Final Answer: $\$1024$
An initial sum of $\$7500$ grows to $\$9600$ in $5$ years under annual compound interest. Find the annual interest rate $r$, correct to $3$ significant figures.
View Complete Solution
- Set up the equation: $$9600 = 7500(1 + r)^5 \implies (1 + r)^5 = \frac{9600}{7500} = 1.28$$
- Take the 5th root: $$1 + r = (1.28)^{1/5} \approx 1.05063$$ $$r \approx 0.05063 \implies 5.06\%$$
Final Answer: $5.06\%$
Sophie has $\$15\,000$ to invest for $4$ years. Bank A offers $6.0\%$ compounded annually. Bank B offers $5.85\%$ compounded monthly. Which bank yields the greater return, and by how much?
View Complete Solution
- Evaluate Bank A (Annually, $k = 1$): $$FV_A = 15\,000(1 + 0.06)^4 = 15\,000(1.06)^4 \approx 15\,000(1.26248) \approx \$18\,937.15$$
- Evaluate Bank B (Monthly, $k = 12, kn = 48$): $$FV_B = 15\,000\left(1 + \frac{5.85}{1200}\right)^{48} = 15\,000(1 + 0.004875)^{48}$$ $$FV_B \approx 15\,000(1.26296) \approx \$18\,944.43$$
- Compare: Bank B yields $\$18\,944.43 - \$18\,937.15 = \$7.28$ more than Bank A.
Final Answer: Bank B yields $\$7.28$ more
A company deposits $\$12\,000$ at $4.2\%$ compounded semi-annually. Find the minimum number of complete years needed for the balance to exceed $\$20\,000$.
View Complete Solution
- Set up inequality ($k = 2$): $$12\,000\left(1 + \frac{4.2}{200}\right)^{2n} > 20\,000$$ $$(1.021)^{2n} > \frac{20\,000}{12\,000} = \frac{5}{3} \approx 1.66667$$
- Take natural logarithms: $$2n \ln(1.021) > \ln\left(\frac{5}{3}\right)$$ $$2n > \frac{\ln(1.66667)}{\ln(1.021)} \approx \frac{0.510826}{0.020783} \approx 24.58\text{ half-years}$$
- Convert to years ($n = \frac{24.58}{2} = 12.29$): To complete full years, round up to $13$ years. (After 12.5 years / 25 half-years, it exceeds $\$20\,000$; so $13$ full complete years).
Final Answer: $13$ full years (or $25$ half-year periods)
Due to inflation of $3.5\%$ per year, the purchasing power of money decreases according to $V = P(1 - 0.035)^t$. Find how many years it takes for the purchasing power of $\$1000$ to be reduced to $\$500$ (i.e. halved).
View Complete Solution
- Set up the equation: $$500 = 1000(0.965)^t \implies (0.965)^t = 0.5$$
- Solve via logarithms: $$t \ln(0.965) = \ln(0.5) \implies t = \frac{\ln(0.5)}{\ln(0.965)}$$ $$t \approx \frac{-0.693147}{-0.035626} \approx 19.455\text{ years}$$
Final Answer: $19.5$ years
A manufacturing press is purchased for $\$120\,000$. It depreciates at a constant annual rate of $16.5\%$. The machine is sold for scrap when its residual value drops below $15\%$ of its original purchase price. After how many full years of service will the machine be scrapped?
View Complete Solution
- Set up threshold condition: $$FV < 0.15 \times PV \implies PV(1 - 0.165)^n < 0.15 \times PV$$ $$(0.835)^n < 0.15$$
- Solve using natural logarithms ($\ln 0.835 < 0$, flip sign): $$n \ln(0.835) < \ln(0.15) \implies n > \frac{\ln(0.15)}{\ln(0.835)}$$ $$n > \frac{-1.89712}{-0.18032} \approx 10.52\text{ years}$$
- Round up to full years: $n = 11$ full years.
Final Answer: $11$ full years
An investment of $\$P$ grows at a nominal annual rate of $r\%$ compounded annually. After $2$ years the balance is $\$7260$, and after $3$ years the balance is $\$7986$. Without using a calculator, find $r$ and $P$.
View Complete Solution
- Set up consecutive year equations: $$P(1 + r/100)^2 = 7260 \quad \text{--- (1)}$$ $$P(1 + r/100)^3 = 7986 \quad \text{--- (2)}$$
- Divide (2) by (1): $$1 + \frac{r}{100} = \frac{7986}{7260} = \frac{798.6}{726} = 1.10$$ $$\frac{r}{100} = 0.10 \implies r = 10\%$$
- Substitute back into (1) to find $P$: $$P(1.10)^2 = 7260 \implies P(1.21) = 7260$$ $$P = \frac{7260}{1.21} = \frac{726\,000}{121} = \$6000$$
Final Answer: $r = 10\%, P = \$6000$
On his $18$th birthday, Lucas receives $\$25\,000$. He deposits the entire amount into a high-yield account earning $5.4\%$ nominal annual interest compounded monthly.
(a) Calculate the value of Lucas's investment on his $25$th birthday ($7$ years later), to the nearest dollar.
(b) Find the total interest earned over this $7$-year period.
(c) Determine the age Lucas will be when the investment first reaches $\$60\,000$.
View Complete Solution
- Part (a): $PV = 25\,000, r = 5.4, k = 12, n = 7 \implies kn = 84$: $$FV = 25\,000\left(1 + \frac{5.4}{1200}\right)^{84} = 25\,000(1 + 0.0045)^{84} = 25\,000(1.0045)^{84}$$ $$FV \approx 25\,000(1.45789) \approx \$36\,447.33 \implies \$36\,447$$
- Part (b): Interest earned: $$\text{Interest} = \$36\,447 - \$25\,000 = \$11\,447$$
- Part (c): Solve for $n$ with $FV \ge 60\,000$: $$25\,000(1.0045)^{12n} = 60\,000 \implies (1.0045)^{12n} = 2.4$$ $$12n = \frac{\ln(2.4)}{\ln(1.0045)} \approx \frac{0.875469}{0.0044899} \approx 194.99\text{ months}$$ $$n = \frac{194.99}{12} \approx 16.25\text{ years}$$ $$\text{Lucas's Age} = 18 + 16.25 \approx 34.25 \implies 34\text{ years and } 3\text{ months}$$
Final Answer: (a) $\$36\,447$; (b) $\$11\,447$; (c) Age $34$ (specifically $34$ years and $3$ months)
Compound Interest Formula
- Compound Interest: $$FV = PV \left( 1 + \frac{r}{100k} \right)^{kn}$$
- Interest Rate: $r$ is nominal annual rate in percent.
- Compounding Compaction: $k$ is compounding periods/yr; $kn$ is total periods.
Depreciation & Growth Models
- Annual Depreciation: $$FV = PV (1 - r)^n \quad (r \text{ as decimal})$$
- Simple Interest: $$I = \frac{P \times r \times t}{100}$$
- Logarithmic Duration: $$n = \frac{\ln(FV / PV)}{k \ln\left(1 + \frac{r}{100k}\right)}$$
Compounding Frequencies Table
- Annually: $k = 1$
- Half-Yearly / Semi-Annually: $k = 2$
- Quarterly: $k = 4$
- Monthly: $k = 12$
- Daily: $k = 365$
Critical IB Examination Pitfalls in Financial Math
- Double Percentage Conversion: In $FV = PV(1 + \frac{r}{100k})^{kn}$, the $100$ is already in the denominator! If $r = 5\%$, write $5$, NOT $0.05$. Writing $0.05 / 100$ divides by $100$ twice.
- Forgetting the $k$ in the Exponent: Multiplying the denominator by $k$ but forgetting to multiply $n$ by $k$ in the exponent $(kn)$ will severely underestimate compounding growth.
- Confusing Depreciation with Compound Interest: Depreciation subtracts the rate: $1 - r$. Never add when dealing with decaying asset value!
- Premature Currency Rounding: Never round intermediate compounding values to the nearest cent/dollar until the very final answer step. Store numbers in calculator memory.
- GDC TVM Sign Conventions: When using the financial TVM solver on GDCs, cash paid out (investment) must be entered as a negative value ($PV < 0$) so the received balance is positive ($FV > 0$).
Question 1
What is the value of $k$ when an interest rate is compounded quarterly?
Question 2
A car worth $\$30\,000$ depreciates by $10\%$ per year. What is its value after $2$ years?
Question 3
If $\$1000$ is invested at $6\%$ nominal interest compounded monthly, what is the interest rate applied each month?
Question 4
In the formula $(1 + r)^n = 3$, how is $n$ expressed in terms of natural logarithms?
An investor deposits $\$4000$ into an account paying a nominal annual interest rate of $20\%$ compounded semi-annually.
View Markscheme
$\text{Rate} = \frac{20\%}{2} = \mathbf{10\%} \quad (\text{or } 0.10)$ A1
(b)
Number of periods: $kn = 2 \times 1.5 = 3$ half-years M1
$FV = 4000(1 + 0.10)^3 = 4000(1.10)^3$ M1
$= 4000(1.331) = \mathbf{\$5324}$ A1
(c)
$\text{Interest} = FV - PV = \$5324 - \$4000$ M1
$= \mathbf{\$1324}$ A1
A business buys high-precision laser equipment for $\$85\,000$. The equipment depreciates at $12.5\%$ per annum.
View Markscheme
$FV = 85\,000(1 - 0.125)^5 = 85\,000(0.875)^5$ M1
$= 85\,000(0.512888) \approx \$43\,595.49 \implies \mathbf{\$43\,595}$ A1
(b)
$85\,000(0.875)^n < 25\,000 \implies (0.875)^n < \frac{25}{85} = \frac{5}{17}$ M1
$n \ln(0.875) < \ln\left(\frac{5}{17}\right)$ A1
$n > \frac{\ln(5/17)}{\ln(0.875)} = \frac{-1.22378}{-0.13353} \approx 9.165$ M1
Rounding up to complete years: $\mathbf{10\text{ years}}$ A1
Unit Overview