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ie2140-09-soln-01

# ie2140-09-soln-01 - IE 2140 Engineering Economy Solutions...

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IE2140 (2009) soln-1-1 IE 2140 Engineering Economy Solutions to Tutorial #1 1. (Problem 4-34, pp 197) The maximum amount the company can spent on the equipment is the present equivalent of a uniform series of \$22,000 per year, at an interest rate of 15% per year. P? \$22,000 1 2 3 4 0 5 P = 22,000 [ P / A , 15%, 5] = 22,000 (3.352155) = \$73,748.41 The company can justify spending up to \$73,748.41 for this piece of equipment. 2 . (Problem 4-80, pp 202) 1,000 800 2003 2004 2005 2002 600 400 F =? The cash flows can be decomposed into 2 components: 1. A uniform annual cash flow ( A =1,000, N =4) with time zero at EoY 2000. 2. A uniform gradient annual cash flow ( G =-200, N =4) with time zero at EoY 2000. The future equivalent value at EoY 2005 is F = 1,000 [ F / A , 8%, 4] [ F / P , 8%, 1] 200 [ P / G , 8%, 4] [ F / P , 8%, 5] = 1,000 (4.506112) (1.08) 200(4.650093) (1.4693) = \$ 3,500.10 EoY Total = CF 1 + CF 2 2000 0 0 0 2001 1,000 1,000 0 2002 800 1,000 -200 2003 600 1,000 -400 2004 400 1,000 -600 2005 F =?

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IE2140 (2009) soln-1-2 3 . (Problem 4-85, pp 203) Note that N = 7 (six years in addition to the first) The cash flow diagram is as follows: \$200k 0 f =6% 1 2 3 4 5 6 7 i = 0.15, f = 0.06 Method 1: Let 0.08490566 1 06 . 1 15 . 1 1 1 1 f i i CR
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