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Time and Interest in Engineering Economy

Understand how time and interest affect money in engineering economy. Learn about present worth, future worth, and annuity calculations.

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Time and Interest in Engineering Economy

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  1. Mc Graw Hill ENGINEERING ECONOMYFifth Edition Blank and Tarquin CHAPTER II Factors: How Time and Interest Affect Money Adopted and modified by Dr. W-.W. Li of UTEP, Fall, 2003 Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  2. 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 Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  3. 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) Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  4. 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 Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  5. F = ?? 0 1 2 3 P=$1,000 i=10%/year 2.2 Example- 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 Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  6. F9 = $100,000 ………… 0 1 2 3 8 9 P= ?? 2.2 Example – 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 Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  7. P = ?? ………….. 1 2 3 .. .. n-1 n 0 2.2 Uniform Series Present Worth and Capital Recovery Factors • Annuity Cash Flow $A per period Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  8. 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 Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  9. 2.2 Uniform Series Present Worth and Capital Recovery Factors Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  10. 2.2 Uniform Series Present Worth and Capital Recovery Factors • Simplifying further Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  11. 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. Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  12. 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 Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  13. $F ………….. N 2.3 F/A and A/F Derivations • Annuity Cash Flow 0 Find $A given the Future amt. - $F $A per period Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  14. 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! Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  15. 2.3 A/F Factor • By substitution we see: • Simplifying we have: • Which is the (A/F,i%,n) factor Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  16. 2.3 F/A factor from the A/F Factor • Given: • Solve for F in terms of A Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  17. $F ………….. N 2.3 F/A and A/F Derivations • Annuity Cash Flow 0 Find $F given thethe $A amounts $A per period Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  18. 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. Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  19. 2.3 Example 2.5 • A = $1,000,000/yr; n = 8 yrs, i = 14%/yr • F8 = ?? Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  20. 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/ Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  21. 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. Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  22. 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 Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  23. 2.5 Arithmetic Gradient Factors • An arithmetic (linear) Gradient is a cash flow series that either increases or decreases by a contestant amount over n time periods. • A linear gradient is always comprised of TWO components: • The Gradient component • The base annuity component • The objective is to find a closed form expression for the Present Worth of an arithmetic gradient Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  24. A1+n-1G A1+n-2G 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 Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  25. 2.5 Example: Linear Gradient • Typical Negative, Increasing Gradient: G=$50 The Base Annuity = $1500 Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  26. 2.5 Example: Linear Gradient • Desire to find the Present Worth of this cash flow The Base Annuity = $1500 Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  27. 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! Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  28. A1+n-1G A1+n-2G A1+2G A1+G 0 1 2 3 n-1 N 2.5 Derivation: Gradient Component Only • Focus Only on the gradient Component “0” G Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  29. 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! Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  30. 0 1 2 3 4 5 6 7 2.5 Present Worth Point… $700 $600 $500 $400 $300 $200 $100 X The Present Worth Point of the Gradient Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  31. $600 $500 $400 $300 $200 $100 $0 0 1 2 3 4 5 6 7 2.5 Gradient Component • The Gradient Component X The Present Worth Point of the Gradient Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  32. X 0 1 2 3 4 5 6 7 The Present Worth Point of the Gradient 2.5 Present Worth Point… • PW of the Base Annuity is at t = 0 • PWBASE Annuity=$100(P/A,i%,7) Base Annuity – A = $100 Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  33. 2.5 Present Worth: Linear Gradient • The present worth of a linear gradient is the present worth of the two components: • 1. The Present Worth of the Gradient Component and, • 2. The Present Worth of the Base Annuity flow • Requires 2 separate calculations! Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  34. 2.5 Present Worth: Gradient Component • The PW of the Base Annuity is simply the Base Annuity –A{P/A, i%, n} factor • What is needed is a present worth expression for the gradient component cash flow. • We need to derive a closed form expression for the gradient component. Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  35. (n-1)G (n-2)G 3G 2G 1G 0G 0 1 2 3 4 ……….. n-1 n We want the PW at time t = 0 (2 periods to the left of 1G) 2.5 Present Worth: Gradient Component • General CF Diagram – Gradient Part Only Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  36. 2.5 To Begin- Derivation of P/G,i%,n Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  37. 2.5 The P/G factor for i and N Subtracting the last two equations Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  38. 2.5 Further Simplification on P/G Remember, the present worth point of any linear gradient is 2 periods to the left of the 1-G cash flow or, 1 period to the left of the “0-G” cash flow. P=G(P/G,i,n) Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  39. 2.5 The A/G Factor • Convert G to an equivalent A A/G,i,n = Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  40. $500 $400 $300 $200 $100 0 1 2 3 4 5 2.5 Gradient Example • Consider the following cash flow Present Worth Point is here! And the G amt. = $100/period Find the present worth if i = 10%/yr; n = 5 yrs Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  41. 0 1 2 3 4 5 2.5 Gradient Example- Base Annuity • First, The Base Annuity of $100/period A = +$100 • PW(10%) of the base annuity = $100(P/A,10%,5) • PWBase = $100(3.7908)= $379.08 • Not Finished: We need the PW of the gradient component and then add that value to the $379.08 amount Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  42. $400 $300 $200 $100 $0 0 1 2 3 4 5 2.5 The Gradient Component PG@t=0 = G(P/G,10%,5) = $100(P/G,10%,5) Could substitute n=5, i=10% and G = $100 into the P/G closed form to get the value of the factor. Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  43. 2.5 PW of the Gradient Component PG@t=0 = G(P/G,10%,5) = $100(P/G,10%,5) P/G,10%,5) Sub. G=$100;i=0.10;n=5 6.8618 Calculating or looking up the P/G,10%,5 factor yields the following: Pt=0 = $100(6.8618) = $686.18 for the gradient PW Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  44. 2.5 Gradient Example: Final Result • PW(10%)Base Annuity = $379.08 • PW(10%)Gradient Component= $686.18 • Total PW(10%) = $379.08 + $686.18 • Equals $1065.26 • Note: The two sums occur at t =0 and can be added together – concept of equivalence Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  45. $500 $400 $300 $200 $100 0 1 2 3 4 5 2.5 Example Summarized This Cash Flow… Is equivalent to $1065.26 at time 0 if the interest rate is 10% per year! Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  46. 2.6 Geometric Gradients • An arithmetic (linear) gradient changes by a fixed dollar amount each time period. • A GEOMETRIC gradient changes by a fixed percentage each time period. • We define a UNIFORM RATE OF CHANGE (%) for each time period • Define “g” as the constant rate of change in decimal form by which amounts increase or decrease from one period to the next Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  47. 0 1 2 3 4 …….. n-1 n 2.6 Geometric Gradients: Increasing • Typical Geometric Gradient Profile • Let A1 = the first cash flow in the series A1 A1(1+g) A1(1+g)2 A1(1+g)3 A1(1+g)n-1 Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  48. 0 1 2 3 4 …….. n-1 n A1(1-g)n-1 A1(1-g)3 A1(1-g)2 A1(1-g) A1 2.6 Geometric Gradients: Decreasing • Typical Geometric Gradient Profile • Let A1 = the first cash flow in the series Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  49. 2.6 Geometric Gradients: Derivation • First Major Point to Remember: • A1 does NOT define a Base Annuity/ • There is not BASE ANNUITY for a Geometric Gradient! • The objective is to determine the Present Worth one period to the left of the A1 cash flow point in time • Remember: The PW point in time is one period to the left of the first cash flow – A1! Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

  50. 2.6 Geometric Gradients: Derivation • For a Geometric Gradient the following parameters are required: • The interest rate per period – i • The constant rate of change – g • No. of time periods – n • The starting cash flow – A1 Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

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