ie2140-09-soln-02 - IE 2140 Engineering Economy Solutions...

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IE2140 (2009) soln-2-1 IE 2140 Engineering Economy Solutions to Tutorial #2 1. (Problem 5-4, pp 244) Machine XYZ Investment Cost $18,000 Useful life 20 years Market value $5,000 Annual operating expenses $250 Overhaul cost – EoY 7 $500 Overhaul cost – EoY 14 $800 MARR 10% PW (10%) = – 18,000 + 5,000 [ P / F ,10%,20] 250 [ P / A ,10%,20] – 500 [ P / F ,10%,7] – 800 [ P / F ,10%,14] = – 18,000 + 5,000 (0.148644) – 250 (8.513564) – 500 (0.513158) – 800 (0.263331) = – $19,852.42 2. (Problem 5-21, pp 246) Proposal A Investment Cost $15,000 Expected life 6 years Market (salvage) value − $2,000 Annual receipts $11,000 Annual expenses $6,000 MARR 12% F 6 = –15,000 [ F / P ,12%,6] + (11,000 - 6,000) [ F / A ,12%,6] – 2,000 = –15,000 (1.973823) + (5,000) (8.115189) 2,000 = $ 8,968.60
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IE2140 (2009) soln-2-2 3. (Problem 5-26, pp 246) AW (18%) = −15,000 [ A / P , 18%, 2] + 10,000 − 3,000 + 10,000 [ A / F , 18%, 2] = 15,000 (0.638716) + 10,000 − 3,000
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ie2140-09-soln-02 - IE 2140 Engineering Economy Solutions...

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