1. Engineering Economic Analysis Canadian Edition Chapter 4: More Interest Formulas

2. 4-2 Chapter 4 … Examines uniform series compound interest formulas. Uses arithmetic and geometric gradients to solve problems. Evaluates non-standard series: begin-period payments; different payment and compounding periods; perpetual annuities.

3. 4-3 Chapter Assumptions in Solving Economic Analysis Problems End-of-period convention (simplifies calculations) Viewpoint of the firm (generally) Sunk costs (past has no bearing on current decisions) Owner provided capital (no debt capital) Stable prices No depreciation No income taxes

4. 4-4 Components of Engineering Economic Analysis Calculation of P, A, and F are fundamental. Some problems are more complex and require an understanding of added components: uniform series arithmetic or geometric gradients non-standard series

5. 4-5 Uniform Series Also called ordinary annuities Equidistant and equal-valued cash flows during a period of time Cash flows toward the firm; positive; inflows or Cash flows away from the firm; negative; outflows. Matching frequencies: cash flows and interest compounding Conversion of series to single sum equivalents present value or future value

6. 4-6 Economic Criteria Projects are judged against an economic criterion. SituationCriterion Fixed inputMaximize output Fixed outputMinimize input Neither fixedMaximize difference (output  input)

7. 4-7 UNIFORM SERIES COMPOUND AMOUNT (FV) SERIES DISCOUNT AMOUNT (PV) Transformations Transformation of Uniform Series

8. 4-8 Uniform Series: Future Value 0 12345 $1000 ≡ 0 FV = ?? 5 EQUIVALENCE What is the value in five years of five end-of-year $1000 deposits beginning one year from today if interest is 10% compounded annually? (Answer: $6105.10)

9. 4-9 Uniform Series: Future Value … From the cash flow diagram, we see that F = $1000(1.10 4 + 1.10 3 + 1.10 2 + 1.10 + 1) = $6105.10. In general, we can use the series formula: which in this case gives:

10. 4-10 Uniform Series: Future Value … Find the balance in ten years of annual deposits of $1500 into a fund that pays interest of 8% compounded annually. All things being equal, $1500 at the end of each year for ten years is equivalent to $21,729.84 ten years from today.

11. 4-11 Importance of Interest Income

12. 4-12 Uniform Series: Present Value 12345 $1000 ≡ PV = ?? 00 What is the value today of five end-of-year $1000 deposits beginning one year from today if interest is 10% compounded annually? (Answer: $3790.79)

13. 4-13 Uniform Series: Present Value … From the cash flow diagram, we see that F = $1000(1.10  1 + 1.10  2 + 1.10  3 + 1.10  4 + 1.10  5 ) = $3790.79. In general, we can use the series formula: which in this case gives:

14. 4-14 Uniform Series: Present Value … Example: A = $140/month, i = 1%/month (usually specified as 12% compounded monthly), n = 60 (five years). Should you pay $6320 today for these payments? The value today is less than the $6320 price, therefore, you should not accept the offer.

15. 4-15 Uniform Series: Present Value … Determine the appropriate purchase price for an energy-saving device with a five-year life (and no salvage value) if the device will provide annual savings of $2500. Assume end-of-year savings and an interest rate of 12½% compounded annually.

16. 4-16 Uniform Series: Present Value … A lender has offered you $10,000 today if you make monthly payments of $241.79 for four years. Determine the interest rate the lender is charging, expressed as a nominal monthly compounded rate). By trial and error, or interpolation, solve for i = 0.6250175% (periodic rate). Nominal rate = 12  0.6250175% = 7.500% compounded monthly.

17. 4-17 Uniform Series: More Examples You want to purchase a $50,000 car in four years. Calculate how much you must deposit in a bank account at the end of every three months in order to pay cash for the car if the rate of interest on your deposits is 5¾% compounded quarterly. i = 5¾%/4 = 1.4375%, n = 4  4 = 16. Solving for A = $2801.70.

18. 4-18 Uniform Series: More Examples … Five years ago, a couple purchased an RV (recreational vehicle) for $100,000. The RV has a market value of $25,000 today. If the market interest rate was 10% compounded annually during the last five years, find the annual equivalent cost of owning the RV. Solving for A = $22,284.81.

19. 4-19 Arithmetic Gradient Series Payment grows by a constant amount, G: The series consists of two payments: A and which is the equivalent payment for the gradient series: 0, G, 2G, 3G, … (n  1)G. 1234567 AA+GA+2GA+3GA+4GA+5GA+6G 0

20. 4-20 Arithmetic Gradient Series … A company has maintenance costs that will be $1500 six months from today and will grow by $75 every six months after that. Find the value today of the maintenance costs over a ten-year period if the rate of interest is 11¼% compounded semiannually. G = $75; i = 0.1125/2 = 0.05625; n = 10  2 = 20.

21. 4-21 Arithmetic Gradient Series … You will save for a vacation by depositing $200 in one month then $5 less each month for two years. Determine the amount you will have saved after two years if the interest rate is 4½% compounded monthly. G =  $5; i = 0.045/12 = 0.00375; n = 2  12 = 24.

22. 4-22 Geometric Gradient Series Payment grows by a constant rate, g: General formulas: 123456 A1A1 A 1 (1+g)A 1 (1+g) 2 A 1 (1+g) 3 A 1 (1+g) 4 0

23. 4-23 Geometric Gradient Series … A company has maintenance costs that will be $1500 six months from today and will grow by 4% every six months after that. Find the value today of the maintenance costs over a ten-year period if the rate of interest is 11¼% compounded semiannually. g = 0.04; i = 0.1125/2 = 0.05625; n = 10  2 = 20.

24. 4-24 Geometric Gradient Series … You will save for a vacation by depositing $200 in one month then 3% less each month for two years. Determine the amount you will have saved after two years if the interest rate is 4½% compounded monthly. g =  0.03; i = 0.045/12 = 0.00375; n = 2  12 = 24. Challenge: derive the formulas if i = g.

25. 4-25 Non-standard series: Payments at the beginning of the period For some series, the payments occur at the beginning of each payment period. These are also called annuities due. The formulas are:

26. 4-26 Annuities Due You want to lease a vehicle that is worth $42,500 by making monthly payments in advance for four years at an interest rate of 2¾% compounded monthly. Calculate the payment required. Extra: find the payment if the vehicle’s buyout value is $17,000 at the end of the lease.

27. 4-27 Non-standard series: Payment period  compounding period These are also called general annuities. Convert the nominal interest rate to the equivalent rate for the payment period. p = number of payment periods per year, and c = number of compounding periods per year.

28. 4-28 General Annuities You arrange a mortgage loan for $295,000 that requires monthly payments for 25 years at an interest rate of 5.35% compounded semiannually. Find the amount of the monthly payment.

29. 4-29 Non-standard series: Perpetual payments Series with perpetual payments are also called perpetual annuities or perpetuities. We only consider the present value of perpetual annuities.

30. 4-30 Perpetual Annuities A family wants to establish a scholarship in their name at a university. They want $2500 to be awarded annually, starting immediately. The scholarship fund has an interest rate of 6¼% compounded annually. Determine the size of the endowment the family must give. Extra: find the endowment if the award grows by 2% per year after the first award of $2500.

31. 4-31 Suggested Problems 4-25, 34, 37, 43, 51, 53, 56, 62, 63, 73, 75, 77, 78, 80, 81, 84, 86, 89, 102, 116, 124, 131.