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A foundation generate, create code 3 of 9 none on .net projects Visual Studio 2010 Example 4 Gerald Pugh needs 1200 ea ch year to endow a sports scholarship. The highest interest rate he can nd will pay 7.5% per year with interest compounded monthly.

Gerald invests 16 000. Is this suf cient to pay for the scholarship . Solution The rate and the duration must be written in the unit of the compounding period, in this case one month. An annual interest rate of 7.5% gives an annual rate (decimal) of 7.

5 = 0.075. We will take this to imply that the rate R per month is given 100 by R = 0.

075 = 0.00625. 12 In one year, there are 12 compounding periods, so T = 12.

Using ( ): with R and T measured relative to the compounding period of one month, we have: Amount = P(1 + R)T Amount at the end of one year = 16 000(1 + 0.00625)12 = 17 242.12 So Gerald receives interest of 1242.

12 each year. This will pay for the sports scholarship and leave Gerald an extra 42.12 each year.

Perhaps this could pay for an annual Pugh Cup.. Example 5 Simone invests 500 at 4.5 % per year for four years. Interest is compounded quarterly.

Find the value of Simone s investment after the four years.. Solution 4.5 1. Annual rate = 100 = 0.

045 2. Rate R per quarter = 0.045 = 0.

01125 4 3. There are 4 4 = 16 quarters (compounding periods) in four years. So T = 16.

4. Using with three months as the unit time period, Amount = 500 (1 + 0.01125)16 = 598.

01 Generally, there are four important points to observe.. Financial Products 1. The interest rate (e.g.

6.5%) will be an annual interest rate. This will sometimes be given as a rate (decimal) = percentage 100 2.

Interest rates will always be given relative to a speci c compounding period. It is vital that the compounding period be properly identi ed and used as the basic unit in interest rate calculations. 3.

The interest rate (rate) over the time of the compounding period is the relevant fraction of the annual interest rate (rate). 4. The duration of the investment (loan) T is expressed as a multiple of the compounding period.

We can now give a general result: P is invested (or borrowed) at r% per year compounded m times per year, for T years r R = 100 = rate per year with interest compounded m times a year R = rate per unit of compounding period m mT = number of compounding periods in T years Then, assuming no money is withdrawn and no money (other than interest) is paid in, Amount after T years = P 1 + R m. Interest = P 1 + Observe that if m = 1, we have annual compounding. Hence, [3] is a special case of [2]..

Example 6 Rocky borrows $6000 for a holiday. He pays interest at 5.75% per year compounded semi-annually.

Assuming that Rocky makes no repayments during the rst six years, how much would he owe after six years . Solution P = $6000 Rate (per year) = 0.0575 Rate (per half year) = 0.0575 = 0.

02875 2 Number of compounding periods = 2 6 = 12 2 6 = 6 000(1 + 0.02875)12 = 8430.81 Amount (after six years) = P 1 + R 2 In Figure 1.

1 we show an Excel spreadsheet giving the value of an investment (or the amount of a debt) after a number of years. This spreadsheet can be used to see how any investment, or debt, with known compounding period can grow. The principal ( 100) is entered in A6 and the rate (8%) in B6.

These are given the cell names, principal and rate, respectively. In C6.
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