ENGINEERING ECONOMY Fifth Edition Blank and Tarquin Mc
ENGINEERING ECONOMY Fifth Edition Blank and Tarquin Mc Graw Hill CHAPTER II Factors: How Time and Interest Affect Money Adopted and modified by Dr. W-. W. Li of UTEP, Fall, 2003 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 1
2. 1 Basic Derivations: F/P factor F/P Factor To find F given P To Find F given P Fn …………. N P 0 12/7/2020 Compound forward in time Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 2
2. 1 Derivation by Recursion: F/P factor F 1 = P(1+i) F 2 = F 1(1+i)…. . but: F 2 = P(1+i)2 F 3 =F 2(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) 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 3
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 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 4
2. 2 Example- F/P Analysis Example: P= $1, 000; n=3; i=10% What is the future value, F? F = ? ? 0 P=$1, 000 1 2 3 i=10%/year F 3 = $1, 000[F/P, 10%, 3] = $1, 000[1. 10]3 = $1, 000[1. 3310] = $1, 331. 00 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 5
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%? F 9 = $100, 000 i = 15%/yr 0 1 2 3 ………… 8 9 P= ? ? P 0 = $100, 000(P/F, 15%, 9) = $100, 000(1/(1. 15)9) = $100, 000(0. 2843) = $28, 430 at time t = 0 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 6
2. 2 Uniform Series Present Worth and Capital Recovery Factors Annuity Cash Flow P = ? ? 1 2 3 …………. . . 0 n-1 n $A period 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 7
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 P = ? ? 0 1 2 3 n-1 n A = given 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 8
2. 2 Uniform Series Present Worth and Capital Recovery Factors 12/7/2020 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 Simplifying further 12/7/2020 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 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. 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 11
2. 2 Capital Recovery Factor A/P, i%, n Given the P/A factor The present worth point of an annuity cash flow is always one period to the left of the first A amount Solve for A in terms of P Yielding…. A/P, i%, n factor 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 12
2. 3 F/A and A/F Derivations $F Annuity Cash Flow …………. . 0 $A period 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University N Find $A given the Future amt. - $F 13
2. 3 Sinking Fund and Series Compound amount factors (A/F and F/A) Take advantage of what we already have Substitute “P” and simplify! Recall: Also: 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 14
2. 3 A/F Factor By substitution we see: Simplifying we have: Which is the (A/F, i%, n) factor 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 15
2. 3 F/A factor from the A/F Factor Given: Solve for F in terms of A 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 16
2. 3 F/A and A/F Derivations $F Annuity Cash Flow …………. . 0 $A period 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University N Find $F given thethe $A amounts 17
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 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. 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 18
2. 3 Example 2. 5 • A = $1, 000/yr; n = 8 yrs, i = 14%/yr • F 8 = ? ? 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 19
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 = l 000(F/A, 14%, 8) = 1000( 13. 23218) = $13, 232. 80 = 13. 232 million 8 years from now/ 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 20
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. 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 21
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 (% period) gradient • This section presents the Arithmetic Gradient 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 22
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 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 23
2. 5 Linear Gradient Example A 1+n-1 G Assume the following: A 1+n-2 G A 1+G 0 1 2 3 n-1 N This represents a positive, increasing arithmetic gradient 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 24
2. 5 Example: Linear Gradient • Typical Negative, Increasing Gradient: G=$50 The Base Annuity = $1500 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 25
2. 5 Example: Linear Gradient • Desire to find the Present Worth of this cash flow The Base Annuity = $1500 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 26
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! 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 27
2. 5 Derivation: Gradient Component Only Focus Only on the gradient Component A 1+n-1 G “ 0” G A 1+n-2 G A 1+G 0 12/7/2020 1 Blank & Tarquin: 5 -th Edition Ch. 1 2 3 Don Smith Texas A&M Authored by: Dr. University n-1 N 28
2. 5 Present Worth Point… The Present worth point of a linear gradient is always: 2 periods to the left of the “ 1 G” point or, n 1 period to the left of the very first cash flow in the gradient series. n DO NOT FORGET THIS! 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 29
2. 5 Present Worth Point… $700 $600 $500 $400 $300 $200 $100 X 0 1 2 3 4 5 6 7 The Present Worth Point of the Gradient 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 30
2. 5 Gradient Component • The Gradient Component $600 $500 $400 $300 $200 $100 $0 X 0 1 2 3 4 5 6 7 The Present Worth Point of the Gradient 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 31
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 X 0 1 2 3 4 5 6 7 The Present Worth Point of the Gradient 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 32
2. 5 Present Worth: Linear Gradient The present worth of a linear gradient is the present worth of the two components: n 1. The Present Worth of the Gradient Component and, n 2. The Present Worth of the Base Annuity flow n Requires 2 separate calculations! 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 33
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. 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 34
2. 5 Present Worth: Gradient Component General CF Diagram – Gradient Part Only (n-1)G 3 G 1 G (n-2)G 2 G 0 G We want the PW at time t = 0 (2 periods to the left of 1 G) 0 12/7/2020 1 2 3 4 ………. . n-1 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University n 35
2. 5 To Begin- Derivation of P/G, i%, n 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 36
2. 5 The P/G factor for i and N Subtracting the last two equations 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 37
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) 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 38
2. 5 The A/G Factor Convert G to an equivalent A A/G, i, n = 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 39
2. 5 Gradient Example • Consider the following cash flow $500 $400 $300 $200 $100 0 1 2 3 4 5 Present Worth Point is here! And the G amt. = $100/period Find the present worth if i = 10%/yr; n = 5 yrs 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 40
2. 5 Gradient Example- Base Annuity • First, The Base Annuity of $100/period A = +$100 0 1 2 3 4 5 • 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 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 41
2. 5 The Gradient Component $400 $300 $0 0 1 $100 2 $200 3 4 5 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. 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 42
2. 5 PW of the Gradient Component PG@t=0 = G(P/G, 10%, 5) = $100(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 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 43
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 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 44
2. 5 Example Summarized This Cash Flow… $500 $400 $300 $200 $100 0 1 2 3 4 5 Is equivalent to $1065. 26 at time 0 if the interest rate is 10% per year! 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 45
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 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 46
2. 6 Geometric Gradients: Increasing • Typical Geometric Gradient Profile • Let A 1 = the first cash flow in the series 0 1 A 1 12/7/2020 2 3 4 ……. . n-1 n A 1(1+g)2 A 1(1+g)3 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University A 1(1+g)n-1 47
2. 6 Geometric Gradients: Decreasing • Typical Geometric Gradient Profile • Let A 1 = the first cash flow in the series 0 1 2 3 A 1(1 -g)2 4 ……. . A 1(1 -g)3 n-1 n A 1(1 -g)n-1 A 1(1 -g) A 1 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 48
2. 6 Geometric Gradients: Derivation • First Major Point to Remember: • A 1 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 A 1 cash flow point in time • Remember: The PW point in time is one period to the left of the first cash flow – A 1! 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 49
2. 6 Geometric Gradients: Derivation • For a Geometric Gradient the following parameters are required: • The interest rate period – i • The constant rate of change – g • No. of time periods – n • The starting cash flow – A 1 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 50
2. 6 Geometric Gradients: Starting • Pg = The Aj’s time the respective (P/F, i, j) factor • Write a general present worth relationship to find Pg…. Now, factor out the A 1 value and rewrite as. . 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 51
2. 6 Geometric Gradients (1) (2) Subtract (1) from (2) and the result is…. . 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 52
2. 6 Geometric Gradients Solve for Pg and simplify to yield…. • This is the (P/A, g, i, n) factor and is valid if g not equal to i. 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 53
2. 6 Geometric Gradient: i = g Case For the case i = g 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 54
2. 6 Geometric Gradients: Summary • Pg = A 1(P/A, g, i, n) g not = to i 12/7/2020 Case: g = i Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 55
2. 6 Geometric Gradient: Example • Assume maintenance costs for a particular activity will be $1700 one year from now. • Assume an annual increase of 11% per year over a 6 -year time period. • If the interest rate is 8% per year, determine the present worth of the future expenses at time t = 0. • First, draw a cash flow diagram to represent the model. 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 56
2. 6 Geometric Gradient Example (+g) • g = +11% period; A 1 = $1700; i = 8%/yr 0 1 2 3 $1700(1. 11)1 4 5 6 7 $1700(1. 11)2 $1700(1. 11)3 PW(8%) = ? ? $1700(1. 11)5 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 57
2. 6 Geometric Gradient ( -g ) • Consider the following problem with a negative growth rate – g. g = -10%/yr; i = 8%; n = 4 A 1 = $1000 $900 0 1 2 $810 3 $729 4 P 0=? ? We simply apply a “g” value = -0. 10 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 58
2. 7 When the i – rate is unknown • A class of problems may deal with all of the parameters know except the interest rate. • For many application-type problems, this can become a difficult task • Termed, “rate of return analysis” • In some cases: • i can easily be determined • In others, trial and error must be used 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 59
2. 7 Example: i unknown • Assume on can invest $3000 now in a venture in anticipation of gaining $5, 000 in five (5) years. • If these amounts are accurate, what interest rate equates these two cash flows? $5, 000 0 1 2 3 4 5 • F = P(1+i)n $3, 000 • 5, 000 = 3, 000(1+i)5 • (1+i)5 = 5, 000/3000 = 1. 6667 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 60
2. 8 Unknown Number of Years • Some problems require knowing the number of time periods required given the other parameters • Example: • How long will it take for $1, 000 to double in value if the discount rate is 5% per year? Fn = $2000 • Draw the cash flow diagram as…. i = 5%/year; n is unknown! 0 1 P = $1, 000 12/7/2020 2 . . . ……. Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University n 61
2. 8 Unknown Number of Years • Solving we have…. . 0 1 2 Fn = $2000 . . . ……. n P = $1, 000 • Fn=? = 1000(F/P, 5%, x): 2000 = 1000(1. 05)x • Solve for “x” in closed form…… 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 62
2. 8 Unknown Number of Years • Solving we have…. . • (1. 05)x = 2000/1000 • Xln(1. 05) =ln(2. 000) • X = ln(1. 05)/ln(2. 000) • X = 0. 6931/0. 0488 = 14. 2057 yrs • With discrete compounding it will take 15 years to amass $2, 000 (have a little more that $2, 000) 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 63
2. 9 Basic Sensitivity Analysis • Sensitivity analysis is a procedure applied to a formulated problem whereby one can assess the impact of each input parameter relating to the output variable. • Sensitivity analysis is best performed using a spreadsheet model. • The procedure is to vary the input parameters within certain ranges and observe the change on the output variable. 12/7/2020 Blank & Tarquin: 5 -th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University 64
- Slides: 64
