1 / 19

Engineering economics

Engineering economics. Interest Time value of money Evaluating economic alternatives Breakeven economics. Interest . Interest = total amount owed – principal amount (12.1). (12.2). Interest = (principal)(number of interest periods)(interest rate). (12.3). = Pni . F = ?. P. period.

yuri
Download Presentation

Engineering economics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Engineering economics • Interest • Time value of money • Evaluating economic alternatives • Breakeven economics

  2. Interest Interest = total amount owed – principal amount (12.1) (12.2) Interest = (principal)(number of interest periods)(interest rate) (12.3) = Pni

  3. F = ? P period 0 1 2 3 4 n-2 n-1 n Single payment compound amount (12.4) (12.5)

  4. F P=? period 0 1 2 3 4 n-2 n-1 n Single payment present worth (12.6)

  5. Uniform series compound amount P = ? A A A A A A A period n-2 n-1 n 0 1 2 3 4 (12.7)

  6. P A=? A A A A A A period n-2 n-1 n 0 1 2 3 4 Capital recovery (12.8)

  7. Nominal interest i = effective interest rate r = nominal interest rate m = interest periods per year y = number of years (12.9) (12.10)

  8. F = ? A A A A A A A period 0 1 2 3 4 n-2 n-1 n Uniform series compound amount (12.11) (12.12)

  9. F A=? A A A A A A period 0 1 2 3 4 n-2 n-1 n Uniform series sinking fund (12.13)

  10. A+(n-1)G A+(n-2)G A+(n-3)G P = ? A+4G A+3G A+2G A period n-2 n n-1 0 1 2 3 4 Gradient series (12.14) (12.15) (12.16)

  11. Present worth method (12.17) (12.18)

  12. years 0 1 2 3 4 5 6 $45,000 $45,000 years 0 1 2 3 4 5 6 least common multiple $30,000 $30,000 $30,000 Least common multiple

  13. Future worth method (12.19) (12.20) (12.21)

  14. Equivalent annual worth method (12.22) (12.23)

  15. Rate of return method (12.24)

  16. Simple payback period (12.25)

  17. Discounted payback period (12.26)

  18. R Cost Profit TC breakeven point Loss FC q q* Breakeven quantity Breakeven analysis

  19. Summary • Engineering economic analyses should consider the time value of money • Interest factor formulas and tables are useful in evaluating alternatives • The PW and FW methods can be used to evaluate alternatives having different lives by using the least common multiple of years. • The EUAW method is preferred because it has the advantage of not requiring the use of the least common multiple. • The breakeven point is that level of production (and sales) that results in a zero profit

More Related