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CHAPTER 3

CHAPTER 3. COMBINING FACTORS. Learning Objectives. Shifted Series Shifted Series and Single Amounts Shifted Gradients Decreasing Gradients Spreadsheets. Section 3.1. Calculations For Uniform Series That Are Shifted. 1. Uniform Series that are SHIFTED.

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CHAPTER 3

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  1. CHAPTER 3 COMBINING FACTORS

  2. Learning Objectives • Shifted Series • Shifted Series and Single Amounts • Shifted Gradients • Decreasing Gradients • Spreadsheets

  3. Section 3.1 Calculations For Uniform Series That Are Shifted

  4. 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 t = “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.

  5. 0 1 2 3 4 5 6 7 8 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)

  6. F at t = 6 0 1 2 3 4 5 6 7 8 Shifted Series: P and F Consider: A = -$500/year P2 P0 F for this series is at t = 6 F6 = A(F/A,i%,4)

  7. 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

  8. Section 3.2 Uniform Series and Randomly Placed Single Amounts

  9. 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

  10. 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

  11. 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

  12. 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 = 0 is the PW point for the two other single cash flows

  13. 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….

  14. 3.2 Substitute the factors and solve $1,130.42 P = $500( 2.4869 )(0.9091 ) + $300( 0.6830 ) - 400( 0.6209 ) = $1086.96 $204.90 $248.36

  15. Section 3.3 Calculations for Shifted Gradients

  16. 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

  17. (n-1)G (n-2)G +2G G 0 1 2 3 n-1 N “0” G Removed Base annuity

  18. 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.

  19. 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.

  20. Gradient Series ……..Base Annuity …….. n-1 n 1 2 … 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.

  21. G = +$100 Base Annuity = $100 0 3.3 Shifted Gradient: Example • Given: 1 2 3 4 ……….. ……….. 9 10 Let C.F start at t = 3: $100/ yr increasing by $100/year through year 10; i = 10%; Find the PW at t = 0

  22. Base Annuity = $100 1 2 3 4 ……….. ……….. 9 10 0 3.3 Shifted Gradient: Example Nannuity = 8 time periods • PW of the Base Annuity P2 = $100( P/A,10%,8 ) = $100( 5.3349 ) = $533.49 P0 = $533.49( P/F,10%,2 ) = $533.49( 0.8264 ) = $440.88

  23. G = +$100 1 2 3 4 ……….. ……….. 9 10 0 3.3 Gradient Present Worth • For the gradient component • 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

  24. 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

  25. 0 1 2 3 … … … n 3.3 Shifted Geometric Gradient • Conventional Geometric Gradient A1 Present worth point is at t = 0 for a conventional geometric gradient!

  26. 0 1 2 3 … … … n Present worth point is at t = 2 for this example 3.3 Shifted Geometric Gradient • Conventional Geometric Gradient A1

  27. 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

  28. 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

  29. 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

  30. 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

  31. 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

  32. Section 3.4 Shifted Decreasing Arithmetic Gradients

  33. 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

  34. 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

  35. 0 1 2 3 4 5 6 7 8 3.4 Shifted Decreasing Linear Gradients F3 = $1,000; G=-$100 i = 10%/year P2 or, F2: Take back to t = 0 P0 here

  36. 0 1 2 3 4 5 6 7 8 3.4 Shifted Decreasing Linear Gradients F3 = $1,000; G=-$100 Base Annuity = $1,000 i = 10%/year P2 or, F2: Take back to t = 0 P0 here

  37. 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.

  38. 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

  39. Section 3.5 Spreadsheet Applications

  40. 3.5 Spreadsheet Applications • Assume Excel is the spreadsheet of choice • Goal: • Learn the Excel Financial Functions • Create your own spreadsheets to solve a variety of problems

  41. 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.

  42. 3.5 Spreadsheets • See Appendix A for further details on Excel applications

  43. Assignment for Chapter 3 • Problems 12, 20, 22, 29, 38, 40, 44, 48, and 54-62

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