1. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Recommendations – no obligation Chapter 3

2. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 2 Overview EXCEL DEMO OF CHAPTER 3 MATERIAL CHAPTER 3 ESSENTIALS –FORMULAS –EXAMPLES 3.25, 3.26, 3.27 –TABLES

3. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 3 The time value of money If you have money, you can either –Lend it and get interest –Invest it and hope you get interest and that you get it back –Invest it and hope you can resell it for more –Use it to buy something or a service In the first two instances, when you ask for it back in the future, you’ll get your money plus what it earned. In other words, it will have grown Unfortunately, during the same time period, money in general will have lost some of its value due to inflation But if you invest wisely, the net should be higher In this chapter we assume that the two effects are combined in the quoted rates

4. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 4 Graphical Representation Time Income Expense

5. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 5 Graphical Representation - Example Time Borrowed Money Repayment

6. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 6 Simple Interest At 5 % per year (per annum) Deposit $100 One Year $5

7. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 7 Compound Interest At 5 % per year (per annum), compounded annually Deposit $100 One Year $5.00105*.05=$5.25 110.25*.05=$5.5125 115.76*0.05 = $5.788

8. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 8 Equivalence We can compare different cash flows by comparing their value at the same time The choice of time is arbitrary, but is usually the present or some date in the future Equivalence is valid for only one interest rate

9. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 9 Equivalence - example @ 5% compound interest $100 held as cash for 3 years will be worth $86.38 $100 loaned today can be paid back in 3 equal annual installments of $36.72 $100 promised to be paid 2 years from now is worth $90.70 today

10. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 10 Uniform Series

11. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 11 Linear Gradient

12. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 12 Geometric Gradient

13. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 13 Irregular Series

14. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 14 Notation A n = payment or receipt occurring at the end of period n A = a constant payment or receipt at end of each period N = number of periods i = interest rate per interest period P = a sum of money at time zero = Present value or worth F = a sum of money at some future date (usually the end of the time span that you are analyzing

15. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 15 CHAPTER 3 FORMULAS

16. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 16 CHAPTER 3 FORMULAS (CONTINUED)

17. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 17 Derivations DERIVATION OF SOME FORMULAS –SINGLE CASH FLOW –UNIFORM SERIES OF PAYMENTS –LINEAR GRADIENT SERIES –GEOMETRIC GRADIENT SERIES –IRREGULAR SERIES GO OVER REMAINING CHAPTER 3 EXAMPLES

18. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 18 Formulas F – future value P – present value N – number of periods i – interest rate per period A – payment at the end of each period G – increase in payment per period starting in the second A 1 – Initial payment n – Index for periods g –percentage change from payment to payment

19. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 19 Formulas – Future value, Present value

20. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 20 Formulas – Payments –the last term in 7) should be raised to N-1

21. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 21 Formulas – Payments (continued)

22. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 22 Formulas – Gradient Series

23. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 23 Formulas -- Geometric Gradient Series Geometric Gradient Series

24. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 24 Formulas -- Geometric Gradient (continued)

25. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 25 Formulas -- Geometric Gradient (continued)

26. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 26 Tables

27. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 27

28. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 28

29. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 29 Equivalence at any point in time

30. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 30

31. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 31 Example 3.6 - solution

32. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 32

33. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 33 3.25

34. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 34 3.26

35. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 35 3.26

36. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 36 3.26

37. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 37 Example 3.27

38. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 38 3.27 – cash flows

39. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 39 3.27 – part a)

40. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 40 3.27 part b)

41. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 41 3.27 part b)

42. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 42 3.27 part b)

43. ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 43 3.27 part b)