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## Factors: How Time and Interest Affect Money

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**1. **Chapter 2 Factors:
How Time and Interest Affect Money

**2. **Learning Objectives 4/15/2012 2

**3. ** Single Payment Factors(F/P and P/F) Objective:
Derive factors to determine the present or future worth of a cash flow
Cash Flow Diagram basic format

**4. **Basic Derivation: F/P factor 4/15/2012 4

**5. **Derivation: F/P factor Find F given P
F1 = P + Pi = P(1+i)
F2 = F1 + F1i = F1(1+i)..or
F2 = P(1+i) + P(1+i)i = P(1+i)(1+i) = P(1+i)2
F3 = F2+ F2 i = 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)
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**6. **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
P = F(P/F,i%,n) where
(P / F, i%, n) = (1+i)-n 4/15/2012 6

**7. **P/F factor discounting back in time Discounting back from the future 4/15/2012 7

**8. **Example- F/P Analysis Example: P= $1,000;n=3;i=10%
What is the future value, F? 4/15/2012 8

**9. **Example P/F Analysis Assume F = $100,000, 9 years from now. What is the present worth of this amount now if i =15%? 4/15/2012 9

**10. **4/15/2012 10

**11. **Uniform Series Present Worth 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. 4/15/2012 11

**12. **Capital Recovery Factor (CRF) = A/P factor Given the P/A factor 4/15/2012 12

**13. **4/15/2012 13

**14. **Modified HW Problem 2.5 A maker of microelectromechanical systems [MEMS], believes it can reduce product recalls if it purchases new software for detecting faulty parts. The cost of the new software is $225,000.
How much would the company have to save each year for 4 years to recover its investment if it uses a minimum attractive rate of return of 15% per year?
(A/P, 15%, 4) p. 745 Table 19 (interest rate i=15%) ? column: A/P, row: n=4
A/P factor = 0.35027
Company would have to save $225,000 x 0.35027 = $78,810.75 each year.
4/15/2012 14

**15. **HW Problem 2.12* Comparison of Table with equation:
V-Tek Systems is a manufacturer of vertical compactors, and it is examining its cash flow requirements for the next 5 years. The company expects to replace office machines and computer equipment at various times over the 5-year planning period. Specifically, the company expects to spend $9000 two years from now, $8000 three years from now, and $5000 five years from now. What is the present worth of the planned expenditures at an interest rate of 10% per year?
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**16. **Sinking Fund and Series Compound Amount Factors (A/F and F/A) Take advantage of what we already have
Recall:
Also:
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**17. **4/15/2012 17

**18. **Modern context of a Sinking Fund In modern finance, a sinking fund is a method enabling an organization to set aside money over time to retire its indebtedness. More specifically, it is a fund into which money can be deposited, so that over time its preferred stock, debentures or stocks can be retired. For the organization that is retiring debt, it has the benefit that the principal of the debt or at least part of it, will be available when due. For the creditors, the fund reduces the risk the organization will default when the principal is due.
In some US states, Michigan for example, school districts may ask the voters to approve a taxation for the purpose of establishing a Sinking Fund. The State Treasury Department has strict guidelines for expenditure of fund dollars with the penalty for misuse being an eternal ban on ever seeking the tax levy again.
Historical Context: A Sinking Fund was a device used in Great Britain in the 18th century to reduce national debt.
4/15/2012 18

**19. **HW Problem 2.23 Southwestern Moving and Storage wants to have enough money to purchase a new tractor-trailer in 3 years. If the unit will cost $250,000, how much should the company set aside each year if the account earns 9% per year?
(A/F, 9%, 3) or n = 3, F = $250,000, i = 9%
Using Table 14 (pg 740), the A/F = 0.30505
A= $250,000 x 0.30505 = $76262.50 4/15/2012 19

**20. **Interpolation (Estimation Process) 4/15/2012 20

**21. **Example 2.7 4/15/2012 21

**22. **Basic Setup for Interpolation 4/15/2012 22

**23. **4/15/2012 23

**24. **Arithmetic Gradient Factors 4/15/2012 24

**25. **Linear Gradient Example Assume the following: 4/15/2012 25

**26. **Example: Linear Gradient 4/15/2012 26

**27. **Arithmetic Gradient Factors 4/15/2012 27

**28. **Present Worth Point 4/15/2012 28

**29. **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. 4/15/2012 29

**30. **Present Worth: Gradient Component(Example 2.10*) Three contiguous counties in Florida have agreed to pool tax resources already designated for county-maintained bridge refurbishment. At a recent meeting, the county engineers estimated that a total of $500,000 will be deposited at the end of next year into an account for the repair of old and safety-questionable bridges throughout the three-county area. Further, they estimate that the deposits will increase by $100,000 per year for only 9 years thereafter, then cease.
Determine the equivalent (a) present worth and (b) annual series amounts if county funds earn interest at a rate of 5% per year. 4/15/2012 30

**31. **4/15/2012 31

**32. **Example 2.10 (b)* 4/15/2012 32

**33. **Equations for P/G and A/G 4/15/2012 33

**34. **Geometric Gradients 4/15/2012 34

**35. **cash flow diagrams for geometric gradient series 4/15/2012 35

**36. **Geometric Gradients: Increasing 4/15/2012 36

**37. **Geometric Gradients 4/15/2012 37

**38. **Pg /A Equation 4/15/2012 38 In summary, the engineering economy relation and factor formulas to calculate Pg in period t = 0 for a geometric gradient series starting in period 1 in the amount A1 and increasing by a constant rate of g each period are

**39. **Engineers at SeaWorld, a division of Busch Gardens, Inc., have completed an innovation on an existing water sports ride to make it more exciting. The modification costs only $8000 and is expected to last 6 years with a $1300 salvage value for the solenoid mechanisms. The maintenance cost is expected to be high at $1700 the first year, increasing by 11% per year thereafter. Determine the equivalent present worth of the modification and maintenance cost. The interest rate is 8% per year.
4/15/2012 39

**40. **4/15/2012 40 continued

**41. **Cash flow diagram 4/15/2012 41

**42. **4/15/2012 42

**43. **i rate is unknown 4/15/2012 43

**44. **Example: i unknown 4/15/2012 44

**45. **Unknown Number of Years 4/15/2012 45

**46. **Unknown Number of Years 4/15/2012 46

**47. **Formulas and factors derived and applied in this chapter perform equivalence calculations for present, future, annual, and gradient cash flows. Capability in using these formulas and their standard notation manually and with spreadsheets is critical to complete an engineering economy study.
Using these formulas and spreadsheet functions, you can convert single cash flows into uniform cash flows, gradients into present worths, and much more.
You can solve for rate of return i or time n.
A thorough understanding of how to manipulate cash flows using the material in this chapter will help you address financial questions in professional practice as well as in everyday living.
4/15/2012 47

**48. **WHAT A DIFFERENCE THE YEARS AND COMPOUND INTEREST CAN MAKE Real World Situation - Manhattan Island purchase.
It is reported that Manhattan Island in New York was purchased for the equivalent of $24 in the year 1626. In the year 2001, the 375th anniversary of the purchase of Manhattan was recognized.
F = P (1+i)n = 24 (1+ 0.06)382 = 111,443,000,000 (2008)
F = P + Pin = 24 + 24(0.06)382 = $550.08 (simple interest) 4/15/2012 48