1. ## maths in economics

• Calculate the value of £1,200 invested for 5 years at 4% interest rate compounded:
• annually
• semi-annually
• quarterly
• continuously
• How many years would it take for an amount of money invested at 4% interest rate compounded continuously to triple in value?

2. Originally Posted by daisy
• Calculate the value of £1,200 invested for 5 years at 4% interest rate compounded:
• annually
• semi-annually
• quarterly
• continuously
• How many years would it take for an amount of money invested at 4% interest rate compounded continuously to triple in value?
For the annual and quarterly compounding, go here:

Balance Roll with Interest

For start date and end date, just enter 2 dates with a 5 year difference, like 1/1/2008 and 1/1/2013.

You should be able to figure out the semi-annually from here.

3. The compound Interest Equation is as follows:

$\displaystyle P = C(1 + \frac{r}{n})^{nt}$

where:

$\displaystyle P = \text{Future Value}$

$\displaystyle C = \text{Initial Deposit}$

$\displaystyle r = \text{Interest Rate}$

$\displaystyle n = \text{Number of times invested per year}$

$\displaystyle t = \text{Number of years invested}$

Let's go ahead and put what we know:

$\displaystyle P = ?$

$\displaystyle C = 1200$

$\displaystyle r = .04$

$\displaystyle n = \text{Varies}$

$\displaystyle t = 5$

1) Annually (n=1):

$\displaystyle P = 1200(1 + .04)^{5}$

$\displaystyle P = 1459.98$

2) Semi-annually (n=2):

$\displaystyle P = 1200(1 + \frac{.04}{2})^{10}$

$\displaystyle P = 1462.79$

3) Quarterly (n=4):

$\displaystyle P = 1200(1 + \frac{.04}{4})^{20}$

$\displaystyle P = 1464.23$

4) Continuously ($\displaystyle n\to \infty$):

This has a special equation:

$\displaystyle P = Ce^{rt}$

All the variables remain the same.

$\displaystyle P = 1200e^{5(.04)}$

$\displaystyle P = 1465.68$

And there you go.

4. How many years would it take for an amount of money invested at 4% interest rate compounded continuously to triple in value?
$\displaystyle \begin{gathered} P = Ce^{rt} \hfill \\ \Leftrightarrow \hfill \\ t = \tfrac{1} {r} \cdot \ln \tfrac{P} {C} = \tfrac{1} {{.04}} \cdot \ln \tfrac{3} {1} \approx 27.46530721 \hfill \\ \end{gathered}$

or 27 years and 169.84 days