2. Why Engineering Economy Chapter 1

3. WHAT IS ECONOMICS ? The study of how limited resources are used to satisfy unlimited human wants. The study of how individuals and societies choose to use scarce resources that nature and previous generations have provided.

4. Engineers must work within the realm of economics and justification of engineering projects Work with limited funds (capital) Capital is not unlimited – rationed Capital does not belong to the firm Belongs to the Owners of the firm Capital is not “free”…it has a “cost”

5. Section 1.1 Definition ENGINEERING ECONOMY IS INVOLVED WITH THE APPLICATION OF DEFINED MATHEMATICAL RELATIONSHIPS THAT AID IN THE COMPARISON OF ECONOMIC ALTERNATIVES

6. Problem Solving Approach  Understand the Problem  Collect all relevant data/information  Identify the criteria for decision making  Define the feasible alternatives  Evaluate each alternative using a common perspective.  Select the “best” alternative  Implement and monitor

7. Economic Issues to be Answered before deciding on an alternative  How much does the option cost  How much will the option save  How do we get the money to pay for it  What are the tax effects  What is the criteria to be used to decide on the option  What are the assumptions used in the estimates  How dependent is a decision on the assumptions-sensitivity analysis

8. Engineering Economy Study Approach  The parameters associated with an alternative include: 1. First Cost (Initial outlay) 2. Estimated Useful Life 3. Estimated Annual Income or Revenue 4. Estimated Annual Expenses or Costs 5. Salvage Value 6. Interest Rate 7. Tax Effects

9. Time Value of Money  The time value of money is the change in the amount of money over a given time period.  This is the most important concept in Engineering Economy.

10. Cash Flows  The parameters listed make up the cash flows associated with an alternative.  Cash flows are said to be positive when they flow into the firm (i.e. revenues)  Cash flows are said to be negative when they flow out of the firm (i.e. expenses)

11. Alternatives  In addition to list of generated alternatives, there is the do nothing alternative. (status quo)  This is the alternative to choose when none of the generated alternatives achieve the chosen decision criteria.  Alternatives can be either independent or mutually exclusive

12. Interest  Interest is a rental for the use of money.  It is what establishes equivalent values for different periods of time  It is the difference between a beginning amount and an ending amount

13. Two Interest Perspectives  Interest Earned – this is the perspective of a person who either saves, invests, or loans a sum of money out, and at a later time receives a larger sum. Interest = total amount now – original amount  Interest paid – this is the perspective of a person who borrows a sum of money, and at a later time repays a larger sum. Interest = Amount owed now – original amount

14. Interest Rate  The interest rate is the amount of interest accrued for a period of time divided by the original amount The time unit for interest payments is called the interest period. Often the interest period is a year. The above expression is for a single period.

15. Single Period Interest Paid  Example 1.3  You borrow $10,000 for one full year  Must pay back $10,700 at the end of one year. Interest Amount? Interest Rate?  Interest Amount (I) = $10,700 - $10,000  Interest Amount = $700 for the year. The $700 represents the return to the lender for this use of his/her funds for one year  Interest rate (i) = 700/$10,000 = 7%/Yr. 7% is the return earned by the lender

16. Single Period Interest Paid  Example 1.4  Borrow $20,000 for 1 year at 9% interest per year  Interest? Total Due?  i = 0.09 per year and N = 1 Year  Interest (I) = (0.09)($20,000) = $1,800  Pay $20,000 + (0.09)($20,000) at end of 1 year  Total amt. paid one year hence $20,000 + $1,800 = $21,800

17. Example 1.4  Note the following  Total Amount Due one year hence is ($20,000) + 0.09($20,000) ($20,000) + 0.09($20,000) =$20,000(1+.09) = $21,800 =$20,000(1+.09) = $21,800 The (1.09) factor accounts for the repayment of the $20,000 and the interest amount The (1.09) factor accounts for the repayment of the $20,000 and the interest amount This (1+i) factor will be one of the important interest factors to be seen later This (1+i) factor will be one of the important interest factors to be seen later

18. Interest Earned and Rate of Return (Ex 1.4 data)  Assume you invest $20,000 for one year in a venture that will return to you, 9% per year. Original $20,000 back Original $20,000 back Plus…….. Plus…….. The 9% return on $20,000 = $1,800 The 9% return on $20,000 = $1,800 We say that you earned 9%/year on the investment! This is your RATE of RETURN (ROR) on the investmen

19. Interest rate  ROI or Return on Investment is another term for rate of return used in settings where the original amount invested is provided by capital funds  The calculation for determining interest rate, rate of return, and return on investment are identical.  The time unit for which the amount of interest accrues is the interest period.

20. Economic Equivalence  Economic Equivalence  Two sums of money at two different points in time can be economically equivalent if: We consider an interest rate and, We consider an interest rate and, No. of time periods between the two sums No. of time periods between the two sums This illustrates the time value of money concept This illustrates the time value of money concept

21. Equivalence Example  Return to Example 1.4 i = 9%  Diagram the loan (Cash Flow Diagram)  The company’s perspective is shown T=0t = 1 Yr $20,000 is received here $21,800 paid back here + _

22. Equivalence Example $20,000 now is economically equivalent to $21,800 one year from now IF the interest rate is set to equal 9%/year T=0t = 1 Yr $20,000 is received here $21,800 paid back here

23. The Cash Flow Diagram:  Extremely valuable analysis tool  First step in the solution process  Graphical Representation on a time scale  Does not have to be drawn “to exact scale” But, should be neat and properly labeled But, should be neat and properly labeled Will be helpful on most in-class exams Will be helpful on most in-class exams

24. Important TERMS  CASH INFLOWS Money flowing INTO the firm from outside Money flowing INTO the firm from outside Revenues, Savings, Salvage Values, etc. Revenues, Savings, Salvage Values, etc.  CASH OUTFLOWS Disbursements Disbursements First costs of assets, labor, salaries, taxes paid, utilities, rents, interest, etc. First costs of assets, labor, salaries, taxes paid, utilities, rents, interest, etc.

25. Net Cash Flows  A NET CASH FLOW is Cash Inflows – Cash Outflows Cash Inflows – Cash Outflows (for a given time period) (for a given time period)  We normally assume that all cash flows occur: At the END of a given time period. This is the end-of-period convention At the END of a given time period. This is the end-of-period convention

26. Cash Flow diagrams - timeline Cash Flow diagrams - timeline  Assume a 5-year problem  The basic time line is shown below The present is time 0, generally.

27. Displaying Cash Flows  A sign convention is applied Positive cash flows (Inflows) are normally drawn upward from the time line Positive cash flows (Inflows) are normally drawn upward from the time line Negative cash flows (Outflows) are normally drawn downward from the time line Negative cash flows (Outflows) are normally drawn downward from the time line

28. Sample CF Diagram Positive CF at t = 1 Negative CF’s at t = 2 & 3

29. Example 1.17  A father wants to deposit an unknown lump ‑ sum amount into an investment opportunity 2 years from now that is large enough to withdraw $4,000 per year for state university tuition for 5 years starting 3 years from now.  Construct the cash flow diagram, assume the rate of return is estimated to be 15.5% per year.

30. Example 1.17 CF Diagram Using PV(.155,5,4000), P = $13,251.40

31. Multi-Period Simple and Compound Interest  Prior discussion on interest and interest rate were for one interest period  For more than one interest period there are two “types” of interest calculations Simple Interest Simple Interest Compound Interest Compound Interest  Compound Interest is more common worldwide and applies to most analysis situations

32. Simple Interest Over Time  Simple Interest  Calculated on the principal amount only  Easy (simple) to calculate  Simple Interest is: (principal)*(interest rate)*(number of periods) (principal)*(interest rate)*(number of periods) $I = (P)*(i)*(n) i = interest rate I = interest $I = (P)*(i)*(n) i = interest rate I = interest

33. Simple Interest Over Time  Ex 1.7  An engineer borrows $1,000 for 3 years at 5% per year, simple interest  Let “P” = the principal sum ($1,000)  i = the interest rate (5%/year)  Let n = number of years (3)

34. Simple Interest Over Time  Simple Interest Definition  I = P(i)(N)  For Ex. 1.7: I = $1,000(0.05)(3) = $150.00 I = $1,000(0.05)(3) = $150.00 Total Interest over 3 Years Total Interest over 3 Years

35. Simple Interest Over Time  Year-by-Year Analysis: Simple Interest  Year 1 I 1 = $1,000(0.05) = $50.00 accrues I 1 = $1,000(0.05) = $50.00 accrues 1 2 3 P=$1,000 I 1 =$50.00 $50 interest accrues but is not paid 0

36. Simple Interest Over Time  Year 2 I 2 = $1,000(0.05) = $50.00 accrues I 2 = $1,000(0.05) = $50.00 accrues 1 2 3 I 1 =$50.00 P=$1,000 I 2 =$50.00 Another $50.00 interest accrues but is not paid 0

37. Simple Interest Over Time  Year 3 I 3 = $1,000(0.05) = $50.00 accrues I 3 = $1,000(0.05) = $50.00 accrues I 2 =$50.00 1 2 3 I 1 =$50.00 P=$1,000 I 3 =$50.00 Pay back $1,000 + $150 of interest The unpaid interest did not earn interest over the 3-year period 0

38. Simple Interest Summary  In a multi-period situation with simple interest: The accrued interest does not earn interest during the succeeding time period. The accrued interest does not earn interest during the succeeding time period. Normally, the total sum borrowed (lent) is paid back at the end of the agreed time period PLUS the accrued (owed but not paid) interest. Normally, the total sum borrowed (lent) is paid back at the end of the agreed time period PLUS the accrued (owed but not paid) interest.

39. Compound Interest Over Time  Compound Interest is different  For compound interest the interest accrued for each interest period is calculated on the principal plus the total amount of interest accumulated in all prior periods. This accrued interest is then added to the prior balance to form a new principal balance.  Interest then “earns interest”  Compound interest is the interest type used to calculate the time value of money for Engineering Economy Analysis

40. Compound Interest Over Time  Unlike simple interest, compound interest does not have a formula for calculating the total amount of interest over several interest periods.  Compound interest has to be calculated each period to determine the principal plus accumulated interest that the interest rate is applied to in the next period. Interest = (interest rate) X [principal + accrued interest]

41. Compound Interest Example 1.8  Here an engineer borrows $1000 @ 5% per year compound interest. How much is due after 3 years. P = $1,000 P = $1,000 i = 5% per year compounded annually (C.A.) i = 5% per year compounded annually (C.A.) N = 3 years N = 3 years

42. Example 1.8 Cash Flow I 2 =$52.50 1 2 3 P=$1,000 I 3 =$55.13 I 1 =$50.00 Owe at t = 3 years: $1,000 + 50.00 + 52.50 + 55.13 = $1,157.63 Owe at t = 3 years: $1,000 + 50.00 + 52.50 + 55.13 = $1,157.63 $1157.63 For 3 yrs of compound interest, accrued not paid: i = 5% 0 + -

43. Compound interest  For the example:  P 0 = +$1,000  I 1 = $1,000(0.05) = $50.00  Owe P 1 = $1,000 + 50 = $1,050 (but we don’t pay yet!)  New Principal sum at end of t = 1: = $1,050.00

44. Compound Interest: t = 2  Principal at end of year 1: $1,050.00  I 2 = $1,050(0.05) = $52.50 (owed but not paid)  Add to the current unpaid balance yields: $1,050 + 52.50 = $1,102.50 $1,050 + 52.50 = $1,102.50 New unpaid balance or New Principal Amount New unpaid balance or New Principal Amount  Now, go to year 3…….

45. Compound Interest: t = 3  New Principal sum: $1,102.50  I 3 = $1102.50(0.05) = $55.125 = $55.13  Add to the beginning of year principal yields: $1102.50 + 55.13 = $1157.63 $1102.50 + 55.13 = $1157.63 This is the loan payoff at the end of 3 years This is the loan payoff at the end of 3 years  Note how the interest amounts were added to form a new principal sum with interest calculated on that new amount

46. Comparison of simple and compound interest, Ex 1.7 and 1.8

47. Example 1.9  Five plans are shown that will pay off a loan of $5,000 over 5 years with interest at 8% per year. We illustrate differences in equivalence depending upon interest type, interest timing, and method for repaying principal.  Plan 1. Simple Interest, pay all at the end  Plan 2. Compound Interest, pay all at the end  Plan 3. Simple interest, pay interest at end of each year. Pay the principal at the end of N = 5  Plan 4. Compound Interest, pay interest and part of the principal each year (pay 20% of the Prin. Amt.)

48. Example 1.9 Plan 5. Compound Interest, make equal payments of the compound interest and principal reduction over 5 years with end-of-year payments. Plan 5. Compound Interest, make equal payments of the compound interest and principal reduction over 5 years with end-of-year payments. Note: The following tables will show the five approaches. For now, do not try to understand how all of the numbers are determined (that will come later!). Focus on the plans and how these tables illustrate economic equivalence.

49. Plan 1: @ 8% Simple Interest  Simple Interest: Pay all at end on $5,000 Loan

50. Plan 2: Compound Interest @ 8%/yr Compound interest: Pay all at the End of 5 Years Compound interest: Pay all at the End of 5 Years

51. Plan 3: Simple Interest Pd. Annually  Principal Paid at the End (balloon Note)

52. Plan 4: Compound Interest  20% of Principal and accrued interest Paid back annually

53. Plan 5: Equal Repayment Plan with Compound Interest Equal Annual Payments (Part Principal and Part Interest Equal Annual Payments (Part Principal and Part Interest

54. Comparison Plan How Paid Total Paid 1 Simple interest Paid at end $7,000 2 Compound interest Paid at end $7,346.64 3 Interest only paid annually $7,000 4 Interest and 20% Prin. Paid annually $6,200 5 Equal annual payments $6,261.41

55. Analysis  Note that the amounts of the annual payments are different for each repayment schedule and that the total amounts repaid for most plans are different, even though each repayment plan requires exactly 5 years.  The difference in the total amounts repaid can be explained (1) by the time value of money, (2) by simple or compound interest, and (3) by the partial repayment of principal prior to year 5.

56. Terminology and Symbols  Specific symbols and their respective definitions have been developed for use in engineering economy.  Symbols tend to be standard in most engineering economy texts world-wide.  Mastery of the symbols and their respective meanings is most important in understanding the subsequent material!

57. Terminology and Symbols  P = value or amount of money at a time designated as the present or time 0. A single dollar amount.  In an equivalence context, P is referred to as present worth (PW), present value (PV), net present value (NPV), discounted cash flow (DCF), and capitalized cost (CC); dollars.

58. Terminology and Symbols  F = value or amount of money at some future time. A single dollar amount.  Also, F is called future worth (FW) and future value (FV); dollars

59. Terminology and Symbols A = series of consecutive, equal, end ‑ of ‑ period amounts of money. A = series of consecutive, equal, end ‑ of ‑ period amounts of money. Also, A is called the annual worth (AW), annuity, equivalent uniform annual worth (EUAW); dollars per year, dollars per month Also, A is called the annual worth (AW), annuity, equivalent uniform annual worth (EUAW); dollars per year, dollars per month n = number of relevant interest periods; years, months, days, n = number of relevant interest periods; years, months, days,

60. Terminology and Symbols  i = interest rate or rate of return per time period; percent per year, percent per month  t = time, stated in periods; years, months, days, etc

61. P and F  The symbols P and F represent one- time occurrences:  Specifically: $P $F t = n 0 1 2 … … n-1 n +-+-

62. P and F: It should be clear that a present value P represents a single sum of money at some time prior to a future value F It should be clear that a present value P represents a single sum of money at some time prior to a future value F

63. Annual Amounts It is important to note that the symbol A always represents a uniform amount (i.e., the same amount each period) that extends through consecutive interest periods. It is important to note that the symbol A always represents a uniform amount (i.e., the same amount each period) that extends through consecutive interest periods. Cash Flow diagram for annual amounts might look like the following: Cash Flow diagram for annual amounts might look like the following: 0 1 2 3.. N-1 n $A A = equal, end of period cash flow amounts +-+-

64. Interest Rate – i% per period  The interest rate i is assumed to be a compound rate, unless specifically stated  as “simple interest”  The effective rate i is expressed in percent per interest period; for example, 12% per year.

65. Terminology and Symbols  Most engineering economy problems: Involve the dimension of time Involve the dimension of time At least 4 of the symbols { P, F, A, i% and n } will be used, and At least 4 of the symbols { P, F, A, i% and n } will be used, and At least 3 of 4 have a known estimated value. At least 3 of 4 have a known estimated value.

66. Computer Solutions  Use of a spreadsheet similar to Microsoft’s Excel is fundamental to the analysis of engineering economy problems.  Appendix A of the text presents a primer on spreadsheet use.  All engineers are expected by training to know how to manipulate data, macros, and the various built-in functions common to spreadsheets.

67. Spreadsheets  Excel supports (among many others) six built-in functions to assist in time value of money analysis  Master each on your own and set up a variety of the homework problems (on your own)

68. Excel’s Financial Functions  To find the present value P, when there is either a single future payment and/or an annuity: PV (i%, n, A, F)  To find the present value P of any series:  NPV (i%, second_cell:last_cell) + first_cell  To find the equal, periodic value A, when there is a single P and /or F payment : PMT(i%, n, P, F)  To find the future value F, when there is either a single P or and/or an annuity: FV(i%, n, A, P)  Nesting of 1 function for the argument of another is allowed.

69. Financial Functions  To find the number of periods n: NPER (i%, A, P, F)  To find the compound interest rate i: RATE (n, A, P, F)  To find the compound interest rate i: IRR (first_ cell:last_ cell)

70. Minimum Attractive Rate of Return  An investment is a commitment of funds and resources in a project with the expectation of earning a return over and above the worth of the resources that were committed.  A firm’s financial managers set a minimum interest rate that that all accepted projects must meet or exceed.  The rate, once established by the firm is termed the Minimum Attractive Rate of Return (MARR).

71. The MARR  The MARR is expressed as a percent per year.  Numerous models exist to aid a firm’s financial managers in estimating what this rate should be in a given time period.  In some circles, the MARR is termed the Hurdle Rate.  Capital (investment funds) is not free.  It costs the firm money to raise capital or to use the owners of the firm’s capital.  This cost is often expresses as a % per year.

72. Costs of Capital Firms raise capital from the following sources:  EQUITY – using the owner’s funds (retained earnings, cash on hand– belongs to the owners, or new stock) Owners expect a return on their money and hence, there is a cost to the firm Owners expect a return on their money and hence, there is a cost to the firm  DEBT – the firm borrows from outside the firm and pays an interest rate on the borrowed funds

73. Costing Capital  Financial models exist that will approximate the costs of the components making up the weighted average cost of capital for a given time period.  Once this “cost” is approximated, then new projects up for funding MUST return at least the cost of the funds used in the project PLUS some additional percent return.  Costs of capital are expressed as a % per year just like an interest rate.

74. Setting a MARR  If we start with the WACC…  Add a buffer percent (?? Varies from firm to firm)  This yields an approximation to a reasonable MARR  This becomes the Hurdle Rate that all prospective projects should earn in order to be considered for funding.

75. Graphical Presentation: MARR 0% MARR - % Safe Investment WACC - % Acceptable range for new projects ROR - %

76. The MARR as an Opportunity Forgone  Assume a firm’s MARR = 12%  Two independent projects, A and B  A costs $400,000 and presents an estimated 13% Rate of Return per year.  B cost $100,000 with an estimated Rate of Return of 14.5%

77. Opportunity Forgone  What if the firm has a budget of, say, $150,000?  A cannot be funded – not sufficient funds!  B is funded and earns 14.5% return or more  A is not funded, hence the firm loses the OPPORTUNITY to earn 13%  This often happens!

78. Opportunity Forgone  In the event a MARR is unknown, and there are insufficient funds for all projects, the de facto MARR becomes the rate of return of the first unfunded project.

79. Problem Perspectives  Before solving, one must decide upon the perspective of the problem  Most problems will present two perspectives  Assume a borrowing situation; for example: Perspective 1: From the lender’s view Perspective 1: From the lender’s view Perspective 2: From the borrower’s view Perspective 2: From the borrower’s view Impact upon the sign convention Impact upon the sign convention

80. Lending – Borrowing Example  Assume $5,000 is borrowed and payments are $1,100 per year.  Draw the cash flow diagram for this First, whose perspective will be used? First, whose perspective will be used? Lender’s or the Borrower’s ? ? ? Lender’s or the Borrower’s ? ? ? Problem will “infer” or you must decide…. Problem will “infer” or you must decide….

81. Lending – Borrowing  From the lenders perspective: 0 1 2 3 4 5 -$5,000 A = +$1,100/yr

82. Lending - Borrowing  From the borrower’s perspective: 0 1 2 3 4 5 P = +$5,000 A = -$1,100/yr

83. Spreadsheet Applications  Example 1.18 a – For the series of positive cash flows given in the problem, find the equivalent future value for the end of year 4. Obtain answers for both simple interest and compound interest. a – For the series of positive cash flows given in the problem, find the equivalent future value for the end of year 4. Obtain answers for both simple interest and compound interest.

84. Spreadsheet Applications  Example 1.18 b – For the revised series of positive cash flows given in the problem, (changing $300,000 in years 3 and 4 to $600,000), find the equivalent future value for the end of year 4. Obtain answers for both simple interest and compound interest. b – For the revised series of positive cash flows given in the problem, (changing $300,000 in years 3 and 4 to $600,000), find the equivalent future value for the end of year 4. Obtain answers for both simple interest and compound interest.