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CHAPTER III

M c. Graw. Hill. ENGINEERING ECONOMY Fifth Edition. Blank and Tarquin. CHAPTER III. COMBINING FACTORS Adopted and modified by Dr. W-.W. Li of UTEP, Fall, 2003. 3.1 Uniform Series that are SHIFTED. A shifted series is one whose present worth point in time is NOT t = 0.

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CHAPTER III

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  1. Mc Graw Hill ENGINEERING ECONOMYFifth Edition Blank and Tarquin CHAPTER III COMBINING FACTORS Adopted and modified by Dr. W-.W. Li of UTEP, Fall, 2003 Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  2. 3.1 Uniform Series that are SHIFTED • A shifted series is one whose present worth point in time is NOT t = 0. • Shifted either to the left of “0” or to the right of t = “0”. • Dealing with a uniform series: • The PW point is always one period to the left of the first series value • No matter where the series falls on the time line. Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  3. 0 1 2 3 4 5 6 7 8 3.1 Shifted Series Consider: A = -$500/year P2 P0 P of this series is at t = 2 (P2 or F2) P2 = $500(P/A,i%,4) or, could refer to as F2 P0 = P2(P/F,i%,2) or, F2 (P/F,i%,2) Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  4. P0 A = -$500/year 0 1 2 3 4 5 6 7 8 P3 F6 = ?? 3.1 Shifted Series: P and F Require F6 • F for this series is at t = 6 • F6 = A(F/A,i%,4) where n = 4 time periods forward. Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  5. 3.1 Suggested Steps • Draw and correctly label the cash flow diagram that defines the problem • Locate the present and future worth points for each series • Write the time value of money equivalence relationships • Substitute the correct factor values and solve Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  6. 3.2 Series with other single cash flows • It is common to find cash flows that are combinations of series and other single cash flows. • Solve for the series present worth values then move to t = 0. • Solve for the PW at t = 0 for the single cash flows. • Add the equivalent PW’s at t = 0. Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  7. F4 = $300 A = $500 0 1 2 3 4 5 6 7 8 F5 = -$400 3.2 Series with Other cash flows • Consider: i = 10% • Find the PW at t = 0 and FW at t = 8 for this cash flow –watch the signs! Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  8. F4 = $300 A = $500 0 1 2 3 4 5 6 7 8 i = 10% F5 = -$400 3.2 The PW Points are: 1 2 3 t = 1 is the PW point for the $500 annuity; “n” = 3 Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  9. Back 4 periods F4 = $300 A = $500 0 1 2 3 4 5 6 7 8 i = 10% Back 5 Periods F5 = -$400 3.2 The PW Points are: 1 2 3 t = 1 is the PW point for the two other single cash flows Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  10. 3.2 Write the Equivalence Statement P = $500(P/A,10%,3)(P/F,10%,1) + $300(P/F,10%,4) - 400(P/F,10%,5) Substituting the factor values into the equivalence expression and solving…. Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  11. 3.2 Substitute the Factors and Solve $1,130.30 P = $500( 2.4869 )( 0.9090 ) + $300( 0.6830 ) - 400( 0.6209 ) = $1583.56 $204.90 $248.36 Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  12. 3.3 The Linear Gradient – Revisited • The Present Worth of an arithmetic gradient (linear gradient) is always located: • One period to the left of the first cash flow in the series ( “0” gradient cash flow) or, • Two periods to the left of the “1G” cash flow Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  13. 3.3 Shifted Gradient • A Shifted Gradient is one whose present value point is removed from time t = 0. • A Conventional Gradient is one whose present worth point is t = 0. Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  14. Gradient Series ……..Base Annuity …….. n-1 n 1 2 … 0 3.3 Example of a Conventional Gradient • Consider: • This Represents a Conventional Gradient. • The present worth point is t = 0. Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  15. Gradient Series ……..Base Annuity …….. n-1 n 1 2 3 4 0 3.3 Example of a Shifted Gradient • Consider: The Present Worth Point for the Base Annuity and the Gradient would be here! This Represents a Shifted Gradient. Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  16. G = +$100 Base Annuity = $100 1 2 3 4 ……….. ……….. 9 10 0 3.3 Shifted Gradient: Example • Given: • Let Cash Flow (CF) start at t = 3: • $100/ yr increasing by $100/year through year 10; i = 10%; Find the PW at t = 0. Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  17. Base Annuity = $100 1 2 3 4 ……….. ……….. 9 10 0 3.3 Shifted Gradient: Example • PW of the Base Annuity P2 = $100(P/A,10%,8) nannuity = 8 time periods P2 = $100( P/A,10%,8 ) = $100( 5.3349 ) = $533.49 P0 = $533.49( P/F,10%,2 ) = $533.49( 0.8264 ) = $440.88 Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  18. G = +$100 1 2 3 4 ……….. ……….. 9 10 0 3.3 Gradient Present Worth • For the gradient component P2 = $100(P/G,10%,8) • PW of gradient is at t = 2: • P2 = $100( P/G,10%,8 ) = $100( 16.0287 ) = $1,602.87 • P0 = $1,602.87( P/F,10%,2 ) = $1,602.87( 0.8264 ) • = $1,324.61 Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  19. 3.3 Example: Final Solution • For the Base Annuity • P0 =$440.88 • For the Linear Gradient • P0 = $1,324.61 • Total Present Worth: • $440.88 + $1,324.61 = $1,765.49 Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  20. 0 1 2 3 … … … n 3.3 Shifted Geometric Gradient • Conventional Geometric Gradient P of G. Grad. Is at t = “0” A1 Present worth point is at t = 0 for a conventional geometric gradient! Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  21. 0 1 2 3 … … … n Present worth point is at t = 2 for this example 3.3 Shifted Geometric Gradient • Shifted Geometric Gradient A1 Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  22. 0 1 2 3 4 5 6 7 8 3.3 Geometric Gradient Example i = 10%/year A = $700/yr A1 = $400 @ t = 5 12% Increase/yr Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  23. 0 1 2 3 4 5 6 7 8 3.3 Geometric Gradient Example i = 10%/year A = $700/yr PW point for the annuity 12% Increase/yr PW point for the gradient Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  24. 5 6 7 8 Present Worth of the Gradient at t = 4 P4 = $400{ P/A1,12%,10%,4 } = 3.3 The Gradient Amounts P0 = $1,494.70( P/F,10%,4) = $1,494.70( 0. 6830 ) P0 = $1,020,88 Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  25. 0 1 2 3 4 5 6 7 8 3.3 The Annuity Present Worth • PW of the Annuity i = 10%/year A = $700/yr P0 = $700(P/A,10%,4) = $700( 3.1699 ) = $2,218.94 Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  26. 3.3 Total Present Worth • Geometric Gradient @ t = • P0 = $1,020,88 • Annuity • P0 = $2,218.94 • Total Present Worth” • $1,020.88 + $2,218.94 • = $3,239.82 Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  27. 0 1 2 3 4 5 6 7 8 3.4 Shifted Decreasing Linear Gradients • Given the following shifted, decreasing gradient: F3 = $1,000; G=-$100 i = 10%/year Find the Present Worth @ t = 0 Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  28. 0 1 2 3 4 5 6 7 8 3.4 Shifted Decreasing Linear Gradients • Given the following shifted, decreasing gradient: F3 = $1,000; G=-$100 i = 10%/year PW point @ t = 2 Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  29. 0 1 2 3 4 5 6 7 8 3.4 Shifted Decreasing Linear Gradients F3 = $1,000; G=-$100 i = 10%/year Use (P/F,10%,2) P2 or, F2: Then take back to t = 0 P0 here Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  30. i = 10%/year 0 1 2 3 4 5 6 7 8 3.4 Shifted Decreasing Linear Gradients F3 = $1,000; G=-$100 Base Annuity = $1,000 (P/F,10%,2) P2 or, F2: Then take back to t = 0 P0 here Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  31. 0 1 2 3 4 5 6 7 8 3.4 Time Periods Involved F3 = $1,000; G=-$100 i = 10%/year 1 2 3 4 5 P2 or, F2: Take back to t = 0 P0 here Dealing with n = 5 periods. Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  32. 0 1 2 3 4 5 6 7 8 3.4 Time Periods Involved F3 = $1,000; G=-$100 G = -$100/yr $1,000 i = 10%/year 1 2 3 4 5 P2 = $1,000( P/A,10%,5 ) – 100( P/G,10%.5 ) P2= $1,000( 3.7908 ) - $100( 6.8618 ) = $3,104.62 P0 = $3,104.62( P/F,10%,2 ) = $3104.62( 0 .8264 ) = $2,565.65 Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  33. 3.5 Spreadsheet Applications • Assume Excel is the spreadsheet of choice • Instructors may vary on the degree of emphasis placed on spreadsheet use • Student’s Goal: • Learn the Excel Financial Functions • Create your own spreadsheets to solve a variety of problems Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  34. 3.5 NPV Function in Excel • NPV function is basic • Requires that all cell in the range so defined have an entry. • The entry can be $0…but not blank! • Incorrect results can be generated if one or more cells in the defined range is left blank . • A “0” value must be entered. Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

  35. 3.5 Spreadsheets • It is assumed if an instructor desires to apply spreadsheets, he or she will provide examples and go over each example and the associated cell formulas. • See Appendix A for further details on Excel applications Blank & Tarquin: 5th Edition. Ch.3 Authored by Dr. Don Smith, Texas A&M University

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