ENGINEERING ECONOMY - EXAMPLE 7.1 : In engineering economy, the future value of first year is FV = PV (1 + i). For second year it is FV = PV (1 + i) (1 + i). For third year it is FV = PV (1 + i) (1 + i)(1 + i) where FV = future value, PV = present value, i = interest rate per period, n = the number of compounding periods. By induction, what is the future value of $1000 for 5 years at the interest rate of 6 %?
ENGINEERING ECONOMY - ANSWER 7.1 : By induction, the future value for 5 year = FV = PV (1 + i) (1 + i) (1 + i) (1 + i)(1 + i) = $1000 (1.06) (1.06) (1.06) (1.06)(1.06) = $1338.2256. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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Question 53 - In the purchase of a machine with a period n = 8.5 years, the minimum attractive rate of return, i = 12 %, the cost P = $55000, F = $4000 is the salvage, annual maintenance A = $3500. The return of the investment or equivalent uniform annual benefit is $15000. The equivalent uniform annual cost is P (A / P, i, n) + A - F (A / F, i, n). The investment is considered acceptable only when equivalent uniform annual benefit is greater than the equivalent uniform annual cost. From the compound interest table, (A / P, i = 12 %, n = 8 years) = 0.2013, (A / P, i = 12 %, n = 9 years) = 0.1877, (A / F, i = 12 %, n = 8 years) = 0.0813, (A / F, i = 12 %, n = 9 years) = 0.0677. Prove by calculations whether the investment above is acceptable.
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