Engineering economy lorie m. Cabanayan francisco d. Cuaresma
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COMPILED LECTURE IN ENGINEERING ECONOMY
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- Interest Problems With Compounding More Often Than Once Per Year 30 Single Amounts
Periodic Rate - The amount of interest you are charged each period, like every month.
Effective interest rate - the actual annual interest rate that accrues, after taking into consideration the effects of compounding (when compounding occurs more than once per year). The rate that you actually get charged on an annual basis. Remember you are paying interest on interest. To illustrate: Consider a principal amount of $1,000 to be invested for three years at a nominal rate of 12% compounded semi-annually. The interest earned during the first six months would be: $1,000 x (0.12/2) = $60 The interest earned during the second six months would be $1,060 x (0.12/2) = 63.60 Then total interest earned during the year is $60.00 + $63.20 = $123.60 The effective annual interest rate for the entire year is ($123.60 / $1,000) X 100 = 12.36% The relationship between effective annual interest, i, and nominal interest r, is i = (1 + r/M) M – 1 where M is the number of compounding periods per year and r is expressed in decimal. Problem: A credit card company charges an interest rate of 1.375 per month on the unpaid balance of all accounts. The annual interest rate they claim is 12 x 1.375% = 16.5%. What is the effective rate of interest per year being charged by the company? i = (1 + 0.165/12) 12 – 1 = 0.1781 or 17.81% per year Interest Problems With Compounding More Often Than Once Per Year 30 Single Amounts If a nominal interest rate is quoted and the number of compounding periods per year and number of years are known, any problem involving future amounts, annual, or present equivalent values can be calculated by straightforward use of equations F = P(1 + i) N and i = (1 + r/M) M – 1 respectively. Problem. Suppose that a $100 lump-sum amount is invested for 10 years at a nominal rate of 6% compounded quarterly. How much is it worth at the end of the tenth year? Solution: There are 4 compounding periods per year, or a total of 4 x 10 = 40 periods. The interest rate per interest period is 6%/4 = 1.5%. F = P (F/P, 1.5%, 40) = $100.00 (1.015) 40 = $100.00 (1.814) = $181.40 Other solution: Using i = (1 + r/M) M – 1 i = (1 + 6/10) 10 – 1 = 6.14% F = P (F/P, 6.14%, 10) = $100.00 (1.0614) 10 = $181.40 Problem. At a certain interest rate compounded semi-annually, P2,000 will amount to 6,500 in 10 years. What is the amount at the end of 15 years? Solution: Solve first for the interest For 10 years: F = P (1+i) N N = 10 (2) = 20 periods 6,500 = 2,000 (1+i) 20 3.25 = (1+i) 20 3.25 1/20 = 1+i 1.06 = 1+i i = 1.06 – 1 i = 0.06 or 6% For 15 years: F = 2,000 (1.06) 30 Download 436.52 Kb. Do'stlaringiz bilan baham: |
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