# Engineering Economy IEN255 Chapter 4 - Present Worth Analysis  Do the product or not?  3 main issues  How much additional investment in plant & equipment.

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Engineering Economy IEN255 Chapter 4 - Present Worth Analysis  Do the product or not?  3 main issues  How much additional investment in plant & equipment to mfg the product?  How long to recover initial investment  Can we make a profit a \$X price?

Engineering Economy Measures of investment worth  Payback period  Cash flow equivalence  present worth  future worth  annual worth (chap 5)  rate of return (chap 6)  (tax concerns later)

Engineering Economy Loan vs Project cash flow  Figure 4.1

Engineering Economy Example 4.1  Purchase cost = \$300,000  5000 x 40% x 3 = 6000 productive hours  6,000/60% = 10,000 hours of paid time per year  Avoided cost = 10,000 hours x \$25 /hour = \$250,000/year  So, net benefits = (\$250000 - \$175000) = \$75000 per year  Fig 4.2

Engineering Economy Payback period  How long does it take to recoup investment?  Most common measure  Used for initial screening

Engineering Economy Example 4.2

Engineering Economy Example 4.3

Engineering Economy Payback period - Pros and Cons  Pro  simple  minimize further analysis (screen all projects)  Cons  no time value of money  no consideration of length of investment

Engineering Economy Two competing projects  Table 4.1

Engineering Economy Present worth analysis  MARR = minimum acceptable rate of return  MARR is a management decision  estimate ä service life ä cash flows (in and out) (if An positive net cash inflow and An is negative if net cash outflow)  determine net cash flows  find present worth of each net cash flow

Engineering Economy Good or bad?  If PW(i) > 0, accept  If PW(i) = 0, indifferent  If PW(i) < 0, reject

Engineering Economy Example 4.5

Engineering Economy Investment pool (borrowed funds)  place to get funds for projects within a company  In pool => \$75000(F/P, 15%, 3) = \$114,066  Project = \$119,470 - \$114,066 = \$5404  Bring back to present = \$3553  fig 4.5

Engineering Economy Variations (future worth)  NFW = net future worth  If FW(i) > 0, accept  If FW(i) = 0, indifferent  If FW(i) < 0, reject

Engineering Economy Example 4.6

Engineering Economy Capitalized equivalent method  Perpetual service life  capitalized cost  PW(I) = A(P/A,I,N  )= A/i (4.3)  Project’s life is extremely long

Engineering Economy Mutually exclusive alternatives  buying vs leasing  is a single alternative mutually exclusive? (do nothing)  revenue vs service projects  analysis period  figure 4.11

Engineering Economy Analysis period equals project lives  table solution on pg 212

Engineering Economy Analysis period differs from project lives  life is longer than analysis period  figure 4.12  solution pg 215

Engineering Economy Project’s life is shorter than analysis period  what to do at tend?  replacement projects  fig 4.13

Engineering Economy Analysis period coincides with longest project life  fig 4.14

Engineering Economy Lowest common multiple of project lives  figure 4.15

Engineering Economy Note table 4.3

Engineering Economy IEN255 Summer’99 Chapter 3, 4 & 5 HW#2  Homework Assignment: Chapter 3 #’s 3.66; 3.73; 3.78 Chapter 4 #’s 4.1; 4.3; 4.7; 4.22; 4.26; 4.34; 4.39; 4.48  Due together (Tues June 29) Chapter 5 - will not be collected * problems will be done in class, others will be posted. #’s 5.1;5.6*; 5.11*; 5.17; 5.20; 5.28*; 5.32; 5.34*; 5.38*; 5.42*

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