ie2140-09-soln-03 - IE 2140 Engineering Economy Solutions...

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IE2140 (2009) soln-3-1 IE 2140 Engineering Economy Solutions to Tutorial #3 1. (Problem 6-1, pp 304) Mutually Exclusive Alternative I II III IV Capital Investment 115,000 157,000 200,000 225,000 Annual Revenues less expenses 20,200 36,900 36,900 43,000 Market Value (end of useful life) 15,000 0 18,000 25,000 Useful Life (years) 10 10 10 10 ( a ) Acceptable alternatives are those having a PW (15%) 0. Alt I: PW (15%) = – 115,000 + 20,200 [ P / A ,15%,10] + 15,000 [ P / F ,15%,10] = – $ 9,913.10 Alt II: PW (15%) = – 157,000 + 36,900 [ P / A ,15%,10] = $ 28,192.56 Alt III: PW (15%) = –200,000 + 36,900 [ P / A ,15%,10] + 18,000 [ P / F ,15%,10] = – $ 10,358.11 Alt IV: PW (15%) = – 225,000 + 43,000 [ P / A ,15%,10] + 25,000 [ P / F ,15%,10] = – $ 3,013.33 Alternatives I, III and IV are economically infeasible. Alternative II should be selected because it has a positive PW . ( b ) If total investment capital is limited to $195,000, Alternative II should be selected since it is within the budget and is also economically feasible. ( c ) Rule 1; the net annual revenues vary among the alternatives.
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IE2140 (2009) soln-3-2 2. (Problem 6-2, pp 304) D1 D2 D3 D4 Capital Investment 650,000 810,000 1,290,000 1,650,000 Annual Expenses: Power 75,000 75,000 140,000 146,000 Labor 50,000 55,000 75,000 60,000 Maintenance 710,000 650,000 470,000 420,000 Tax & Insurances 14,000 17,000 27,000 30,000 MARR =10%. Present Worth Method : PW D1 (10%) = – 650,000 – (75,000 + 50,000 + 710,000 + 14,000) [ P / A ,10%,10] = – $ 5,866,737.47 PW D2 (10%) = – 810,000 – (75,000 + 55,000 + 650,000 + 17,000) [ P / A ,10%,10] = – $ 5,707,219.98 PW D3 (10%) = –1,290,000 – (140,000 + 75,000 + 470,000 + 27,000) [ P / A ,10%,10] = – $ 5,664,931.78 PW D4 (10%) = –1,650,000 – (146,000 + 60,000 + 420,000 + 30,000) [ P / A ,10%,10] = – $ 5,680,836.02 Select D3 to minimize the present worth of cost. Future Worth Method
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ie2140-09-soln-03 - IE 2140 Engineering Economy Solutions...

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