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Financial Maths Siew Wei & Andrea Phan. Exercise 6D: Compound Interest.

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Presentation on theme: "Financial Maths Siew Wei & Andrea Phan. Exercise 6D: Compound Interest."— Presentation transcript:

1 Financial Maths Siew Wei & Andrea Phan

2 Exercise 6D: Compound Interest

3 Compound Interest  More common form of interest  Compound Interest is calculated each time period on a sum of money to which the previous amount of interest has been added to the next time period. Example:

4  If $250 has been invested at 10% per annum, where interest is added to the account each year: First Year: Interest = $250 x 10% x 1 = $25.00 Money in Account = $250 + $25 = $275.00 Second Year: Money in Account = $275 + $27.50 = $302.50 Third Year: Money in Account = $302.50 + $30.25 = $322.75

5 Compound Interest Formulas  Formula for calculating the amount of an investment earning compound interest:  Formula for finding the amount of interest earned:

6 Compound Interest Formulas  However, the interest can be added at different time periods, not just annually, therefore, there is another formula used to take into account situations where the interest isn’t added annually:

7 EXAMPLE 5. Calculate the amount of money that would be invested at 6.75% per annum compound interest to achieve a sum of $36000 in 4 years. P= ?, t= 4, n= 1, r= 6.75, A= 36000 P= A/(1+((r/n)/100))^nt = 36000/(1+((6.75/1)/100))^1x4 = 27722.40896 = $27722.41

8 EXAMPLE 2 6. Calculate the interest paid on a loan of $2750 at 11% per annum, compounded quarterly for 4 years. Firstly, determine the amount of investment, $A A = P x (1+((r/n)/100)^(n x t) = 2750 x (1+((11/4)/100)^(4x4) = $4244.65 Then, subtract the principal from the amount of investment. I = A – P = 4244.65 – 2750 = $1494.65

9 Exercise 6E: Flat Rate Depreciation

10 Depreciation It is when something decrease in value or… DEPRECIATES The longer something last, the lower the depreciation rate will be e.g. - Cars and Phone - Office equipment

11 Depreciation is an estimate of the annual reduction in the value of items. The book value of an item is its value at any given time:  Equation: Book value = purchase price − depreciation The two most common way of calculating depreciation: - flat rate depreciation - reducing balance depreciation.

12 Flat rate Depreciation The item will depreciated in value by a fixed amount each year, normally a percentage of the original value. Then to calculate the book value of a product, you will have to subtract the depreciation from the purchase price.

13 Example 8. A developer installs a new air-conditioning system into an office block she is building. The cost is $122 870, and its value depreciates by 12.5% per year by the flat rate method. a Calculate the annual depreciation of the system. D=Prt/100  D = (122870 x 12.5 x 1) /100  D = 15358.75 c Draw a graph to show book value against time. This means the book value will decrease by $15358.75 annually b Calculate the book value of the system after 4 years. V= P – (Prt/100)  V = 122870 – ((122870 x 12.5 x 4)/100) V= 122870 – 61435 V= 61435 d Estimate the number of years it would take for the book value of the system to be zero. 0= 122870 – ((122870 x 12.5 x t)/100)  -122870= -(122870 x 12.5 x t )/100  12287000 = -122870 x -12.5 x- t t= 8 years

14 Exercise 6F: Reducing Balance Depreciaiton

15 Reducing Balance Depreciation Instead of finding out an annual fixed amount of depreciation it is calculated by the % of book value decrease annually. ^_^  The Book value will decrease most in the first year  Following years the percentage of depreciation will decrease

16 To find the depreciation the book value is subtracted from the purchased price:

17 Example 7 A new computer system, purchased by an engineering company, has an initial value of $75 000. a Calculate the value of the system after 3 years if the annual depreciation rate is 30% using the reducing balance method. V = P x (1 – (r/100))^t  V = 75000 x (1 – (30/100))^3 V= $25725 b Calculate the amount of depreciation in 3 years. D = P – (P x (1- (r/100))^t)  D = 75000 – 25725 D= 49275

18 Exercise 6G: Hire- Purchase & Flat interest rate

19 Hire-Purchase  A method of purchasing goods when having an insufficient amount of cash.  Purchaser agrees to hire the item from the seller and makes periodic payments of an agreed amount.  The item will be owned by the purchaser at the end of the period of the agreement.  If the purchaser stops making payments any time throughout the agreement, no money will be refunded to the purchaser and the item is returned to the vendor.

20 Flat interest rate  Annual rate of the total interest paid as a proportion of the original debt.  Same as simple interest rate, but in the context of hire-purchase.

21 EXAMPLE 5. Exercise gym equipment, which normally costs $750, can be bought through hire-purchase with a $200 deposit and $26.40 a month for 30 months. Calculate: a. The amount of interest being charged Total paid = 200 + (26.40 x 30) = $992 Interest paid = Total paid – Purchase Price = 992 – 750 = $242 b. The flat rate of interest per annum r f = 100I/Pt = (100 x 242)/(550 x 2.5) = 17.6 %

22 EXAMPLE 2 6. A microwave oven is advertised with a marked price of $576 and the opportunity to buy it on hire-purchase, with no deposit and an interest rate of 10% repayable over a year with four equal instalments. Calculate: a. The amount of interest Interest rate = 10% Interest = 0.10 x 576 = $57.60 b. The total amount to be repaid Original amount + Interest = 576 + 57.6 = $633.60 c. The amount of each instalment = 644.60/4 = $158.40


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