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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 =?
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/25/2011 for the course ECONS 2104 taught by Professor Prof during the Spring '08 term at National University of Singapore.

Page1 / 4

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online