1. Engineering Orientation Engineering Economics

2. Value and Interest The value of a dollar given to you today is of greater value than that of a dollar given to you one year from today Cost of Money Simple and Compound Interest Cash Flow Diagrams Cash Flow Patterns Equivalence of Cash Flow Patterns

3. Value and Interest “Value” is not synonymous with “amount”. The value of an amount of money depends on when the amount is received or spent. First Cost is what you pay for an item when you buy it

4. Cost of Money Interest that could be earned if the amount invested in a business or security was instead invested in government bonds or in time deposit.

5. Cost of Money Buy a car for $20,000 of your own cash vs. US bonds returning 5%/yr ($1,000 forever) In effect you are paying $1,000 for ever (even after the car is a certifiable clunker destined for destruction)

6. Value and Interest The difference between the anticipated amount and its current value is called interest. At an interest rate of 10% what is the value now of the expectation of receiving $1 in one year?

7. Simple and Compound Interest You have a business project costing $100,000 You get a loan for 7.5% for 5 years at simple interest payable at the end of the loan The loan costs $7,500 for each of five years for a total interest of $37,500 Total cost over 5 years = $137,500

8. Simple and Compound Interest Is the banker really willing to lend you money for 5 years? Isn’t he also lending you $7,500 for four years, $15,000 for three years, $22,500 for two years, $30,000 for four years and $37,500 for five years?

9. Simple and Compound Interest Usually the banker does think you owe him interest on the interest (known as compound interest) He will charge you about $375 after year 1, $750 after year 2. $1,125 after year 3, $1,500 after year 4 and $1,875 after year 5 The cost of the 5 year project is thus about $142,125 (Approximation) Compound interest can be a significant part of an engineering project

10. Terms and formulae P or PV Principal is the amount borrowed N # of pay periods r Interest rate per period F or FV, Future worth, value in the future of what you have to payback Formulae: Simple interest = P(1 + Nr) ( = $137,500) Compound interest = P(1 + r) N ( = $143,563)

11. Pay periods Suppose your load is compounded quarterly, monthly or daily instead of yearly. Student loan of $25,000 at 8% for Annually for two years, Quarterly and Daily

12. Pay periods

13. Example What amount must be paid in two years to settle a current debt of $1,000 if the interest rate is 6%?

14. Cash Flow Diagrams

15. Cash Flow Patterns

17. Example A new widget twister, with a life of six years, would save $2,000 in production costs each year. Using a 12% interest rate, determine the highest price that could be justified for the machine. Although the savings occur continuously throughout each year, follow the usual practice of lumping all amounts at the ends of years.

18. Example How soon does money double if it is invested at 8% interest?

19. Example 2 Find the value in 2002 of a bond described as “Acme 8% of 2015” if the rate of return set by the market for similar bonds is 10%.

20. Example Compute the annual equivalent maintenance costs over a 5-year life of a laser printer that is warranted for two years and has estimated maintenance costs of $100 annually. Use i = 10%.

21. When Is An Investment Worth It? ‘Break Even Point’ (BEP) has a simple definition: BEP occurs when the project has earned back the cost it took to make it

22. BEP Example Cost of producing new widget is $1,000,000. If profit per widget is $1.00 and we’re selling 1,000/day when is the BEP?. Know - How: Equate cost to total money stream. Solve: 1,000 [widgets/day]  1.00 [$/widget]  D [days] = $1,000,000. Solving for D gives: D = 1,000 days = 2.74 years. Most companies require BEP of 12 -18 months to fund a new widget

23. 23 Return on Investment ROI = The ratio of annual return to the cost of the investment If an investment of $500,000 produces an income of $40,000 per year, its ROI = $40,000/$500,000 = 0.08 = 8%. Many successful large companies operate with ROI’s of 15% or more

24. 24 Return on Investment

25. Unusual Cash Flows and Interest Periods

26. PAYMENTS AT BEGINNINGS OF YEARS Using a 10% interest rate, find the future equivalent of:

27. Study Examples Compute the effective annual interest rate i e equivalent to 5% nominal annual interest compounded continuously.

28. Cost of losing one semester Two students, Frank and Mary start they Engineering Studies on the same date and they make the commitment of retiring thirty years after their forecasted graduation date (the date they would graduate if no delays are introduced). This date will not change if any delays make any of them graduate later. Calculate the difference in their earnings if for some reasons Frank is required to graduate one semester later than what was intended.

29. The Model Assumptions: We have a constant inflation You have a yearly Salary Increase greater than what you loose because of inflation Salary increases and inflation are constant They work for the same company their entire carrier

30. Study Examples Your perfectly reliable friend, Frank, asks for a loan and promises to pay back $150 two years from now. If the minimum interest rate you will accept is 8%, what is the maximum amount you will loan him? a) $119 b) $126 c) $129 d) $139 The annual amount of a series of payments to be made at the end of each of the next twelve years is $500. What is the present worth of the payments at 8% interest compounded annually? a) $500 b) $3,768 c) $6,000 d) $6,480

31. Study Examples Maintenance expenditures for a structure with a twenty- year life will come as periodic outlays of $1,000 at the end of the fifth year, $2,000 at the end of the tenth year, and $3,500 at the end of the fifteenth year. With interest at 10%, what is the equivalent uniform annual cost of maintenance for the twenty-year period? a) $200 b) $262 c) $300 d) $325 The purchase price of an instrument is $1 2,000 and its estimated maintenance costs are $500 for the first year, $1 500 for the second and $2500 for the third year. After three years of use the instrument is replaced; it has no salvage value. Compute the present equivalent cost of the instrument using 1 0% interest. a) $14,070 b) $15,570 c) $15,730 d) $16,500

32. Study Examples If $10,000 is borrowed now at 6% interest, how much will remain to be paid after a $3,000 payment is made four years from now? a) $7,000 b) $9,400 c) $9,625 d) $9,725