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IEN255 Chapter 4 - Present Worth Analysis

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?. Measures of investment worth. Payback period

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IEN255 Chapter 4 - Present Worth Analysis

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  1. 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?

  2. 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)

  3. Loan vs Project cash flow • Figure 4.1

  4. 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

  5. Payback period • How long does it take to recoup investment? • Most common measure • Used for initial screening

  6. Example 4.2

  7. Example 4.3

  8. 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

  9. Two competing projects • Table 4.1

  10. Present worth analysis • MARR = minimum acceptable rate of return • MARR is a management decision • estimate • service life • 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

  11. Good or bad? • If PW(i) > 0, accept • If PW(i) = 0, indifferent • If PW(i) < 0, reject

  12. Example 4.5

  13. 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

  14. Variations (future worth) • NFW = net future worth • If FW(i) > 0, accept • If FW(i) = 0, indifferent • If FW(i) < 0, reject

  15. Example 4.6

  16. 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

  17. Mutually exclusive alternatives • buying vs leasing • is a single alternative mutually exclusive? (do nothing) • revenue vs service projects • analysis period • figure 4.11

  18. Analysis period equals project lives • table solution on pg 212

  19. Analysis period differs from project lives • life is longer than analysis period • figure 4.12 • solution pg 215

  20. Project’s life is shorter than analysis period • what to do at tend? • replacement projects • fig 4.13

  21. Analysis period coincides with longest project life • fig 4.14

  22. Lowest common multiple of project lives • figure 4.15

  23. Note table 4.3

  24. IEN255 Summer’99Chapter 3, 4 & 5HW#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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