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Question 101 - (a) Let | A > = (Aa Ab Ac), | B > = (Ba Bb Bc), | C > = (Ca Cb Cc). Find | A > + | C > - | B > in term of Aa, Ab, Ac, Ba, Bb, Bc, Ca, Cb and Cc. (b) Let d | E > = d (Ea Eb Ec) = (d Ea d Eb d Ec). If | E > = (6 7 8), find the value of 10 | E >.

Question 101 - (a) Let | A > = (Aa Ab Ac), | B > = (Ba Bb Bc), | C > = (Ca Cb C..

Answer 101 - (a) | A > + | C > - | B > = (Aa + Ca - Ba Ab + Cb - Bb Ac + Cc - Bc). (b) Let d = 10, Ea = 6, Eb = 7, Ec = 8. Then 10 | E > = (10 x Ea 10 x Eb 10 x Ec) = (10 x 6 10 x 7 10 x 8) = (60 70 80). 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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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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