CHAPTER 4 MORE INTEREST FORMULAS



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CHAPTER 4 MORE INTEREST FORMULAS After Completing This Chapter the student should be able. Solve problems modeled by Uniform Series (very common) Use arithmetic and geometric gradients Apply nominal and effective interest rates Use discrete and continuous compounding Use spreadsheets and financial functions to solve problems Questions to Consider Without a calculator, what annual rate of return do you suppose it would take to turn $5,000 into $20 Million in 51 years? 100% 200%? With a calculator, what was Anne Scheiber s actual average annual rate of return during the years she was investing? How does this rate compare to the overall performance of the stock market during the period from 1944-1995. Anne Schreiber s Bonanza Anne Scheiber died in 1995 at age 101 with an estate >$20M Left entire estate to Yeshiva University in New York She Was a retired IRS auditor, not an heiress but invested wisely Started investing in 1944 with $5,000 he life savings to that point Portfolio was garden variety, with Coca Cola and Exxon Extremely frugal, reinvested her dividends, what motivated her? Jewish woman who didn t receive promotions, wanted to give scholarships and interest free loans to help woman advance Chapter 3 presented the fundamental components or building blocks of engineering economics, most problems are more complex and requires a deeper understanding of different types of cash flows and interest calculations. Uniform Series Compound Interest Formulas Extremely common form of cash flow, automobile loans, house payments and many other commercial loans vehicles are based on a uniform payment series. Based on A = An end of year period cash receipt or disbursement in a uniform series continuing for n periods, the entire series equivalent to P or F at interest rate i. In Chapter 3 we saw that a sum P at one point in time would increase to F in n periods at an interest rate i, represented by F = P(1+i) n. Otherwise single payment Formulas. We will use this relationship to derive the uniform series equation. Page 1

Arithmetic Gradient It frequently happens that the cash flow series is not of a constant amount A. Instead, there is a uniformly increasing series as shown: Cash flows of this form may be resolved into two components: Page 2

Derivation of Arithmetic Gradient Factors The derivation of the formulas is well laid out on Page 99/100 of the text, essentially an evolution and factoring of previous formulas. Geometric Gradient As opposed to arithmetic gradient where the period-by-period change is uniform amount, there are other situations where the period change is a uniform rate, g. Nominal & Effective Interest Nominal Interest per year, r, is the annual interest rate without considering the effects of compounding. 2.5% interest every 6 months, the nominal rate per year is 2.5 x 2 = 5% Effective Interest Rate per year, i a, is the annual interest rate taking into account the effect of compounding interest on interest during the year. r i i a m Nominal interest rate per interest period (typically one year). Effective interest rate per interest period Effective interest rate per year Number of compounding sub periods per time period Effective Interest Rate per year = i a = (1+r/m) m - 1 w/ i=(r/m) Effective Interest Rate per year = i a = (1+i) m - 1 Page 3

Important to remember that in chapter three i was referred to simply as the interest rate per interest period. We were describing the effective interest rate without concerning ourselves with the more precise definition and increased complexity. Continuous Compounding Effective Interest Rate per year = i a = e r - 1 Not too concerned about this at this point in time. Spreadsheets for Economic Analysis Spreadsheets are used in most real world applications and include the following tasks: 1. Constructing cash flow tables 2. Using preprogrammed annuity functions to calculate P,F,A,n, or i. 3. Using block functions to calculate present worth or internal rate of return 4. Making graphs for analysis and convincing presentations 5. Calculating what if s for different assumptions or variables. Spreadsheet Annuity Functions Spreadsheet Block Functions Spreadsheets for Basic Graphing Problems: Text Examples: 4-1, 4-2, 4-3, 4-7, 4-12 Study Guide: 4-1, 4-7, 4-10, 4-11, 4-12, 4-17, 4-39 Page 4

=20,000 [.01(1+.01) 48 ] (1+.01) 48-1 =20,000[.01(1.6122)/1.6122-1] =20,000(.0161/.6122) =20000(.0263) A = $526.70 where.0263 = (A/P,1%,48) Page 5