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CHAPTER 2

CHAPTER 2. Factors: How Time and Interest Affect Money. 1. Foundations: Overview. F/P and P/F Factors P/A and A/P Factors F/A and A/F Factors Interpolate Factor Values P/G and A/G Factors Geometric Gradient Calculate i Calculate “n” Spreadsheets. CHAPTER 2: Section 1.

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CHAPTER 2

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  1. CHAPTER 2 Factors: How Time and Interest Affect Money

  2. 1. Foundations: Overview • F/P and P/F Factors • P/A and A/P Factors • F/A and A/F Factors • Interpolate Factor Values • P/G and A/G Factors • Geometric Gradient • Calculate i • Calculate “n” • Spreadsheets

  3. CHAPTER 2: Section 1 F/P and P/F Factors

  4. Fn …………. n P0 2.1 Basic Derivations: F/P factor • F/P FactorTo find F given P To Find F given P Compound forward in time

  5. 2.1 Derivation by Recursion: F/P factor • F1 = P(1+i) • F2 = F1(1+i)…..but: • F2 = P(1+i)(1+i) = P(1+i)2 • F3 =F2(1+i) =P(1+i)2 (1+i) = P(1+i)3 In general: Fn = P(1+i)n Fn = P(F/P,i%,n)

  6. 2.1 Present Worth Factor from F/P • Since Fn = P(1+i)n • We solve for P in terms of FN • P = F{ 1/ (1+i)n} = F(1+i)-n • Thus: P = F(P/F,i%,n) where (P/F,i%,n) = (1+i)-n • Thus, the two factors are: • F = P(1+i)n finds the future worth of P; • P = F(1+i)-n finds the present worth from F

  7. Fn …………. n P 2.1 P/F factor – discounting back in time • Discounting back from the future P/F factor brings a single future sum back to a specific point in time.

  8. F = ?? 0 1 2 3 P=$1,000 i=10%/year Example 2.1 - F/P Analysis • Example: P= $1,000;n=3;i=10% • What is the future value, F? F3 = $1,000[F/P,10%,3] = $1,000[1.10]3 = $1,000[1.3310] = $1,331.00

  9. F9 = $100,000 ………… 0 1 2 3 8 9 P= ?? Example 2.2 – P/F Analysis • Assume F = $100,000, 9 years from now. What is the present worth of this amount now if i =15%? i = 15%/yr P0 = $100,000(P/F, 15%,9) = $100,000(1/(1.15)9) = $100,000(0.2843) = $28,430 at time t = 0

  10. CHAPTER 2: Section 2 P/A and A/P Factors

  11. P = ?? ………….. 1 2 3 .. .. n-1 n 0 2.2 Uniform Series Present Worth and Capital Recovery Factors • Annuity Cash Flow $A per period

  12. P = ?? 0 1 2 3 n-1 n A = given 2.2 Uniform Series Present Worth and Capital Recovery Factors • Desire an expression for the present worth – P of a stream of equal, end of period cash flows - A

  13. 2.2 Uniform Series Present Worth and Capital Recovery Factors • Write a Present worth expression [1] Term inside the brackets is a geometric progression. Mult. This equation by 1/(1+i) to yield a second equation

  14. 2.2 Uniform Series Present Worth and Capital Recovery Factors • The second equation [2] To isolate an expression for P in terms of A, subtract Eq [1] from Eq. [2]. Note that numerous terms will drop out.

  15. 2.2 Uniform Series Present Worth and Capital Recovery Factors • Setting up the subtraction [2] - [1] = [3]

  16. 2.2 Uniform Series Present Worth and Capital Recovery Factors • Simplifying Eq. [3] further n

  17. 2.2 Uniform Series Present Worth and Capital Recovery Factors • This expression will convert an annuity cash flow to an equivalent present worth amount one period to the left of the first annuity cash flow.

  18. 2.2 Capital Recovery Factor A/P, i%, n The present worth point of an annuity cash flow is always one period to the left of the first A amount • Given the P/A factor Solve for A in terms of P Yielding…. A/P,i%,n factor

  19. CHAPTER 2: Section 3 F/A and A/F Factors

  20. $F ………….. N 2.3 F/A and A/F Derivations • Annuity Cash Flow 0 Find $A given the Future amt. - $F $A per period

  21. 2.3 Sinking Fund and Series Compound amount factors (A/F and F/A) • Take advantage of what we already have • Recall: • Also: Substitute “P” and simplify!

  22. 2.3 A/F Factor • By substitution we see: • Simplifying we have: • Which is the (A/F,i%,n) factor

  23. 2.3 F/A factor from the A/F Factor • Given: • Solve for F in terms of A

  24. $F ………….. N 2.3 F/A and A/F Derivations • Annuity Cash Flow 0 Find $F given the $A amounts $A per period

  25. 2.3 Example 2.5 • Formosa Plastics has major fabrication plants in Texas and Hong Kong. • It is desired to know the future worth of $1,000,000 invested at the end of each year for 8 years, starting one year from now. • The interest rate is assumed to be 14% per year.

  26. 2.3 Example 2.5 • A = $1,000,000/yr; n = 8 yrs, i = 14%/yr • F8 = ??

  27. 2.3 Example 2.5 Solution: The cash flow diagram shows the annual payments starting at the end of year 1 and ending in the year the future worth is desired. Cash flows are indicated in $1000 units. The F value in 8 years is F = l000(F/A,14%,8) =1000( 13.23218) = $13,232.80 = 13.232 million 8 years from now.

  28. 2.3 Example 2.6 • How much money must Carol deposit every year starting, l year from now at 5.5%per year in order to accumulate $6000 seven years from now?

  29. 2.3 Example 2.6 • Solution • The cash How diagram from Carol's perspective fits the A/F factor. • A= $6000 (A/F,5.5%,7) = 6000(0.12096) = $725.76 per year • The A/F factor Value 0f 0.12096 was computed using the A/F factor formula

  30. CHAPTER 2: Section 4 Interpolation in Interest Tables

  31. 2.4 Interpolation of Factors • All texts on Engineering economy will provide tabulated values of the various interest factors usually at the end of the text in an appendix • Refer to the back of your text for those tables.

  32. 2.4 Interpolation of Factors • Typical Format for Tabulated Interest Tables

  33. 2.4 Interpolation (Estimation Process) • At times, a set of interest tables may not have the exact interest factor needed for an analysis • One may be forced to interpolate between two tabulated values • Linear Interpolation is not exact because: • The functional relationships of the interest factors are non-linear functions • Hence from 2-5% error may be present with interpolation.

  34. 2.4 An Example • Assume you need the value of the A/P factor for i = 7.3% and n = 10 years. • 7.3% is most likely not a tabulated value in most interest tables • So, one must work with i = 7% and i = 8% for n fixed at 10 (Tables 12 & 13) • Proceed as follows:

  35. 2.4 Basic Setup for Interpolation • Work with the following basic relationships

  36. 2. 4 i = 7.3% using the A/P factor • For 7% we would observe: (A/P,7%,10) = 0.14238

  37. 2. 4 i = 7.3% using the A/P factor • For i = 8% we observe: (A/P,8%,10) = 0.14903

  38. 2. 4 Estimating for i = 7.3% • Form the following relationships

  39. 2.4 Final Estimated Factor Value • Observe for i increasing from 7% to 8% the A/P factors also increases. • One then adds the estimated increment to the 7% known value to yield:

  40. 2.4. The Exact Value for 7.3% • Using a previously programmed spreadsheet model the exact value for 7.3% is:

  41. CHAPTER 2 Section 5 P/G and A/G Factors

  42. 2.5 Arithmetic Gradient Factors • In applications, the annuity cash flow pattern is not the only type of pattern encountered • Two other types of end of period patterns are common • The Linear or arithmetic gradient • The geometric (% per period) gradient • This section presents the Arithmetic Gradient

  43. 2.5 Arithmetic Gradient Factors • An arithmetic (linear) Gradient is a cash flow series that either increases or decreases by a constant amount over n time periods. • A linear gradient is always comprised of TWO components:

  44. 2.5 Arithmetic Gradient Factors • The Two Components are: • The Gradient component • The base annuity component • The objective is to find a closed form expression for the Present Worth of an arithmetic gradient

  45. A1+(n-1)G A1+(n-2)G A1+2G A1+G 0 1 2 3 n-1 N 2.5 Linear Gradient Example • Assume the following: This represents a positive, increasing arithmetic gradient

  46. 2.5 Example: Linear Gradient • Typical Negative, Increasing Gradient: G=$50 The Base Annuity = $1500

  47. 2.5 Example: Linear Gradient • Desire to find the Present Worth of this cash flow The Base Annuity = $1500

  48. 2.5 Arithmetic Gradient Factors • The “G” amount is the constant arithmetic change from one time period to the next. • The “G” amount may be positive or negative! • The present worth point is always one time period to the left of the first cash flow in the series or, • Two periods to the left of the first gradient cash flow!

  49. (n-1)G (n-2)G +2G G 0 1 2 3 n-1 N 2.5 Derivation: Gradient Component Only • Focus Only on the gradient Component “0” G Removed Base annuity

  50. 2.5 Present Worth Point… • The Present worth point of a linear gradient is always: • 2 periods to the left of the “1G” point or, • 1 period to the left of the very first cash flow in the gradient series. DO NOT FORGET THIS!

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