Chapter 3: Combining Factors

1. Chapter 3: Combining Factors Developed by Dr. Don Smith, P.E., Texas A&M University & Dr. Michael Bennett, P.Eng., PMP, UOIT, Oshawa, Ontario

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 “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 P3 P0 P of this series is at t = 2 (P3 or F3) P3 = $500(P/A,i%,4) or, could refer to as F3 P0 = P3(P/F,i%,2) or, F3(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 P3 P0 F for this series is at t = 6 F6 = A(F/A,i%,4)

7. Figure 3.3, 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. Example 3.3 An engineering company in British Columbia that owns 50 hectares of valuable land has decided to lease the mineral rights to a mining company. The primary objective is to obtain long-term income to finance ongoing projects 6 and 16 year from the present time. The engineering company makes a proposal to the mining company that it pay $20,000 per year fro 10 years, beginning 1 year from now, plus $10,000 six years from now and $15,000 16 years from now. If the mining company wants to pay off its lease immediately, how much should it pay now, if the investment should make 16% per year?

11. 7 19 18 16 15 4 17 5 3 2 1 20 6 First Step - Cash Flow $15,000 $10,000 A = $20,000 P = ?

12. Solution P = 20000(P / A,16%,20) + 10000(P / F,16%,6) + 15000(P / F, 16%,16) = $124,075

13. 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

14. 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

15. 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

16. 3.2 Write the Equivalence Statement P = $500(P/A,10%,3)(P/F,10%,2) + $300(P/F,10%,4) - 400(P/F,10%,5) Substituting the factor values into the equivalence expression and solving…

17. 3.2 Substitute the factors and solve $1,027.58 P = $500( 2.4869 )( 0.8264 ) + $300( 0.6830 ) - 400( 0.6209 ) = $831.06 $204.90 $248.36

18. Section 3.3 Calculations for Shifted Gradients

19. 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

20. 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.

21. 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.

22. 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.

23. 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: $500/ yr increasing by $100/year through year 10; i = 10%; Find the PW at t = 0

24. 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

25. 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

26. 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

27. 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!

28. 0 1 2 3 … … … n Present worth point is at t = 2 for this example 3.3 Shifted Geometric Gradient • Conventional Geometric Gradient A1

29. 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

30. 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

31. 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

32. 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

33. 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

34. Section 3.4 Shifted Decreasing Arithmetic Gradients

35. 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

36. 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

37. 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

38. 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

39. 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.

40. 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

41. Section 3.5 Spreadsheet Applications

42. 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

43. 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.

44. 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

45. Mc Mc Mc Graw Graw Graw Hill Hill Hill End of Slide Set