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FE Review – Engineering Economics

FE Review – Engineering Economics. Valerie Thomas Industrial and Systems Engineering October 23 2013. With reference to ISyE 3025. Lectures and slides are available at http://www.isye.gatech.edu/engecon/lectures.php Cycle books from your 3025 course, one for each learning cycle.

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FE Review – Engineering Economics

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  1. FE Review – Engineering Economics Valerie Thomas Industrial and Systems Engineering October 23 2013

  2. With reference to ISyE 3025 • Lectures and slides are available at • http://www.isye.gatech.edu/engecon/lectures.php • Cycle books from your 3025 course, one for each learning cycle. And with reference to the Public FE Handbook

  3. Morning session • Engineering Economics 8% • Discounted cash flow (e.g., equivalence, PW, equivalent annual FW, rate of return) • Cost (e.g., incremental, average, sunk, estimating) • Analyses (e.g., breakeven, benefit-cost) • Uncertainty (e.g., expected value and risk)

  4. Afternoon Session • Engineering Economics 15% • Discounted cash flows (equivalence, PW, EAC, FW, IRR, loan amortization) • Types and breakdown of costs (e.g., fixed, variable, direct and indirect labor, material, capitalized) • Analyses (e.g., benefit-cost, breakeven, minimum cost, overhead, risk, incremental, life cycle) • Accounting (financial statements and overhead cost allocation) • Cost estimating • Depreciation and taxes • Capital budgeting

  5. Single Cash Flow ModelFuture Amount and Present Amount • F=P(1+i)N • F=P(F/P,i,N) • P=F(1+i)-N • P=F(P/F,i,N) FE Handbook page 114

  6. Uniform Cash Flow SeriesFuture Value • This is a sum of single cash flow payments, each of amount A (as in “Annual” payment). • F=A[(1+i)N-1]/i= A(F/A,i,N) • (F/A,i,N)=[(1+i)N-1]/i FE Handbook page 114

  7. Uniform Cash Flow SeriesGiven Future Value what is A? FE Handbook page 114

  8. Uniform Cash Flow SeriesPresent Value FE Handbook page 114

  9. Arithmetic Gradient Series (G) FE Handbook page 114

  10. Geometric gradient series (g)(Apparently this is NOT in the FE)Good for when there is inflation or rising costs (g)

  11. Capitalized CostThis can be on the FE but may not have been explicitly mentioned in ISyE 3025 (although it is in Cycle 1). • Capitalized costs generally refers to infinitely repeating costs. • For non-infinitely repeating costs: • For infinitely repeating costs A, in the limit as n goes to infinity, the above equation goes to FE Handbook page 115

  12. Time scale conversions for interest rates Compare equation from 3025, above, where M1 and M2 are the number of payment periods per year (M1 is generally 1) and iM2 is the interest rate for each M2 period, with equation from FE reference handbook, below, where r is the nominal annual interest rate and m is the number of compounding periods per year: FE Handbook page 115

  13. Time value of money calculations A deposit of $3000 is made in a savings account that pays 7.5% interest compounded annually. How much money will be available to the depositor at the end of 16 years? A. $8,877 B. $10,258 C. $9,542 D. $943 White et al. Fundamentals of Engineering Economic Analysis, p. 67.

  14. Time value of money calculations A deposit of $3000 is made in a svaings account that pays 7.5% interest compounded annually. How much money will be available to the depositor at the end of 16 years. A. 8.877 B. 10,258 C. 9,542 D. 943 Answer: • F=P(1+i)n=P(F/P,i,n)=$3,000(1.075)16 = $9,542. • This formula is on page 114 of the public FE reference handbook • A 7.5% table is not included, so table approach won’t work on this problem, but could in general be used. White et al. Fundamentals of Engineering Economic Analysis, p. 67.

  15. Interest Rate Time Conversions What is the effective annual interest rate if the nominal annual interest rate is 24% per year compounded monthly? A. 2% B. 24% C. 26.82% D. 27.12% White et al. Fundamentals of Engineering Economic Analysis,p. 67.

  16. Interest Rate Time Conversions What is the effective annual interest rate if the nominal annual interest rate is 24% per year compounded monthly? A. 2% B. 24% C. 26.82% D. 27.12% Answer – First it’s a good practice to figure out about what the answer should look like – which answer is mostly likely to be correct, and why? White et al. Fundamentals of Engineering Economic Analysis,p. 67.

  17. Interest Rate Time Conversions What is the effective annual interest rate if the nominal annual interest rate is 24% per year compounded monthly? A. 2% B. 24% C. 26.82% D. 27.12% Answer: White et al. Fundamentals of Engineering Economic Analysis, p. 67.

  18. Loan amortization You borrow $5000 at 10% per year and will pay off the load in 3 equal annual payments starting one year after the loan is made. The end-of-year payments are $2010.57. Which of the following is true for your payment at the end of year 2? • Interest is $500.00 and principal is $1510.57 • Interest is $450.00 and principal is $1560.57 • Interest is $348.94 and principal is $1661.63 • Interest is $182.78 and principal is $1827.79 White et al. Fundamentals of Engineering Economic Analysis, p. 115.

  19. Loan amortization You borrow $5000 at 10% per year and will pay off the load in 3 equal annual payments starting one year after the loan is made. The end-of-year payments are $2010.57. Which of the following is true for your payment at the end of year 2? Answer: The first thing paid in repaying a loan is the accumulated interest over that time period. First you can check the payments: A=P(A/P,0.1, 3)=$5000 x 0.4021(1) =$2010.5 (close enough) (1) from the interest table in the Public FE Handbook

  20. Loans You borrow $5000 at 10% per year and will pay off the load in 3 equal annual payments starting one year after the loan is made. The end-of-year payments are $2010.57. Which of the following is try for your payment at the end of year 2? Answer: At the end of one year, the interest accrued will be $500. So the amount of principal paid at the end of one year will be $2010.57-$500 = $1510.57 Therefore the remaining principal will be $5000 – $1510 = $3489.43. Therefore the interest at the end of the second year will be $345.94. Therefore the principal paid will be $2010.57-$345.94 = $1661.63 And the correct answer is c.

  21. Breakeven analysis

  22. Internal rate of return • ISyE 3025 cycle 2. • How to calculate the internal rate of return? • Use present worth or future worth, set present worth or future worth to zero, solve for interest rate i.

  23. Overhead cost allocationnot covered in ISyE 3025 • Overhead – all indirect costs, not included in direct material and direct labor costs. • Allocation by labor hours • Allocation by labor dollars • Example – labor hours: • http://www.dummies.com/how-to/content/how-to-calculate-overhead-allocation.html

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