1 / 122

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.

kaden
Télécharger la présentation

CHAPTER 2

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. 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!

More Related