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ENGINEERING ECONOMY - EXAMPLE 7.2 : 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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ENGINEERING ECONOMY - ANSWER 7.2 : (A / P, i = 12 %, n = 8.5 years) = (0.2013 + 0.1877) / 2 = 0.1945. (A / F, i = 12 %, n = 8.5 years) = (0.0813 + 0.0677) / 2 = 0.0745. Equivalent uniform annual cost = P (A / P, i = 12 %, n = 8.5 years) + A - F (A / F, i = 12 %, n = 8.5 years) = $55000 (0.1945) + $3500 - $4000 (0.0745) = $13899.5 < $15000 (equivalent uniform annual benefit). The investment is proven acceptable. 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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