1. Mc Factors: How Time and Interest Affect Money Graw Hill CHAPTER II
ENGINEERING ECONOMY Fifth Edition Mc
Graw
Hill
Blank and Tarquin
CHAPTER II
Factors: How Time and Interest Affect Money
Adopted and modified by Dr. W-.W. Li of UTEP, Fall, 2003
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

2. 2.1 Basic Derivations: F/P factor
F/P Factor To find F given P P0 Fn N
………….
To Find F given P
Compound forward in time
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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)
5/14/2019
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
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

5. What is the future value, F?
2.2 Example- F/P Analysis
Example: P= $1,000;n=3;i=10%
What is the future value, F?
P=$1,000
F = ??
i=10%/year
F3 = $1,000[F/P,10%,3] = $1,000[1.10]3
= $1,000[1.3310] = $1,331.00
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

6. 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%?
F9 = $100,000
i = 15%/yr
…………
P= ??
P0 = $100,000(P/F, 15%,9) = $100,000(1/(1.15)9)
= $100,000(0.2843) = $28,430 at time t = 0
5/14/2019
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
Annuity Cash Flow
P = ??
………….. n
n-1
$A per period
5/14/2019
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
Desire an expression for the present worth – P of a stream of equal, end of period cash flows - A
P = ??
n n
A = given
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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
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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
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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.
5/14/2019
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
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

13. 2.3 F/A and A/F Derivations
Annuity Cash Flow
………….. N
Find $A given the Future amt. - $F
$A per period
5/14/2019
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!
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

15. By substitution we see:
2.3 A/F Factor
By substitution we see:
Simplifying we have:
Which is the (A/F,i%,n) factor
5/14/2019
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
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Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

17. 2.3 F/A and A/F Derivations
Annuity Cash Flow
………….. N
Find $F given thethe $A amounts
$A per period
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

18. Formosa Plastics has major fabrication plants in Texas and Hong Kong.
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.
5/14/2019
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 = ??
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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, = million 8 years from now/
5/14/2019
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.
5/14/2019
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
5/14/2019
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
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

24. 2.5 Linear Gradient Example
A1+n-1G
A1+n-2G
Assume the following:
A1+2G
A1+G
n N
This represents a positive, increasing arithmetic gradient
5/14/2019
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
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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
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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!
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

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

29. 2 periods to the left of the “1G” point or,
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!
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

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

31. The Present Worth Point of the
2.5 Gradient Component
The Gradient Component $0
$100
$200
$300
$400
$500
$600 X
The Present Worth Point of the
Gradient
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

32. The Present Worth Point of the
PW of the Base Annuity is at t = 0
PWBASE Annuity=$100(P/A,i%,7)
Base Annuity – A = $100 X
The Present Worth Point of the
Gradient
5/14/2019
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!
5/14/2019
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.
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

35. 2.5 Present Worth: Gradient Component
General CF Diagram – Gradient Part Only 1G 2G 3G
(n-2)G
(n-1)G 0G
We want the PW at time t = 0 (2 periods to the left of 1G)
……….. n n
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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
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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
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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)
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

39. Convert G to an equivalent A
2.5 The A/G Factor
Convert G to an equivalent A
A/G,i,n =
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

40. Present Worth Point is here!
2.5 Gradient Example
Consider the following cash flow
$100
$200
$300
$400
$500
Present Worth Point is here!
And the G amt. = $100/period
Find the present worth if i = 10%/yr; n = 5 yrs
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

41. 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 $ amount
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

42. 2.5 The Gradient Component $0
$100
$200
$300
$400
= 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.
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

43. 2.5 PW of the Gradient Component
= 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) = $ for the gradient PW
5/14/2019
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%) = $ $686.18
Equals $
Note: The two sums occur at t =0 and can be added together – concept of equivalence
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

45. 2.5 Example Summarized This Cash Flow… $500 $400 $300 $200 $100
$100
$200
$300
$400
$500
Is equivalent to $ at time 0 if the interest rate is 10% per year!
5/14/2019
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
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

47. 2.6 Geometric Gradients: Increasing
Typical Geometric Gradient Profile
Let A1 = the first cash flow in the series
…… n n A1
A1(1+g)
A1(1+g)2
A1(1+g)3
A1(1+g)n-1
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

48. 2.6 Geometric Gradients: Decreasing
Typical Geometric Gradient Profile
Let A1 = the first cash flow in the series
…… n n A1
A1(1-g)
A1(1-g)2
A1(1-g)3
A1(1-g)n-1
5/14/2019
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!
5/14/2019
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
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

51. 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 A1 value and rewrite as..
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

52. 2.6 Geometric Gradients Subtract (1) from (2) and the result is….. (2)
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Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

53. 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.
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

54. 2.6 Geometric Gradient: i = g Case
For the case i = g
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Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

55. 2.6 Geometric Gradients: Summary
Pg = A1(P/A,g,i,n)
g not = to i
Case: g = i
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

56. 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.
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

57. 2.6 Geometric Gradient Example (+g)
g = +11% per period; A1 = $1700; i = 8%/yr
PW(8%) = ??
$1700
$1700(1.11)1
$1700(1.11)2
$1700(1.11)3
$1700(1.11)5
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

58. 2.6 Geometric Gradient ( -g )
Consider the following problem with a negative growth rate – g.
g = -10%/yr; i = 8%; n = 4
A1 = $1000
$900
$810
$729
P0=??
We simply apply a “g” value = -0.10
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

59. 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
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

60. 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
F = P(1+i)n
5,000 = 3,000(1+i)5
(1+i)5 = 5,000/3000 =
$3,000
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

61. 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?
Draw the cash flow diagram as….
Fn = $2000
i = 5%/year; n is unknown!
…… n
P = $1,000
5/14/2019
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…..
…… n
P = $1,000
Fn = $2000
Fn=? = 1000(F/P,5%,x): = 1000(1.05)x
Solve for “x” in closed form……
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

63. 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 = / = yrs
With discrete compounding it will take 15 years to amass $2,000 (have a little more that $2,000)
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University

64. 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.
5/14/2019
Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas A&M University