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MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors

MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012. Some Definitions. Capital: Invested money and resources Interest: The return on capital

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MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors

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  1. MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

  2. Some Definitions Capital: Invested money and resources Interest: The return on capital Nominal IR: the interest rate per year without adjusting for the number of compounding periods Effective IR: the interest rate per year adjusting for the number of compounding periods

  3. Equivalence Different sums of money at different times can be equal in economic value. i.e. $100 today with i = 6% is equivalent to $106 in one year. Equivalence depends on the interest rate! Equivalence occurs when different cash flows at different times are equal in economic value at a given interest rate.

  4. time Salvage “Costs” Income Operating & Maintenance Costs Replacement Costs Initial Capital Cost Cash Flow Diagrams: An Important Tool - Arrows up represent “income” or “profits” or “payoffs” - Arrows down represent “costs” or “investments” or “loans” - The “x axis” represents time, most typically in years

  5. Time Value of Money If $4500 is invested today for 12 years at 15% interest rate, determine the accumulated amount. Draw this. F = P(1+i)n P =Present Value (in dollars) F = Future Value (in dollars) F n = 12, i = 15% t=0 t=12 $4500

  6. Factors Single Payment Compound Amount Factor (future worth) (F/P, i%, n) : Single Payment Present Worth Factor (P/F, i%, n): n is in years if the ieff is used.

  7. Example - Factors How much inheritance to be received 20 years from now is equivalent to receiving $10,000 now? The interest rate is 8% per year compounded each 6-months.

  8. Uniform Series (Annuity) An Annuity is a series of equal amount money transactions occurring at equal time periods Ordinary Annuity - one that occurs at the end of each time period Uniform Series Present Worth Factor Capital Recovery Factor

  9. Annuities Can Relate an Annuity to a future value: Uniform Series Compound Amount Factor Uniform Series Sinking Fund Factor

  10. Annuity Example How much money can you borrow now if you agree to repay the loan in 10 end of year payments of $3000, starting one year from now at an interest rate of 18% per year?

  11. Factors Fortunately these factors are tabulated… And Excel has nice built in functions to calculate them too….

  12. Spreadsheet Function P = PV(i,N,A,F,Type) F = FV(i,N,A,P,Type) i = RATE(N,A,P,F,Type,guess) Where, i = interest rate, N = number of interest periods, A = uniform amount, P = present sum of money, F = future sum of money, Type = 0 means end-of-period cash payments, Type = 1 means beginning-of-period payments, guess is a guess value of the interest rate

  13. Gradient Factors Engineering Economic problems frequently involve disbursements or receipts that increase or decrease each year (i.e. equipment maintenance) If the increase is the same every year this is called a uniform arithmetic gradient.

  14. The Uniform amount of increase each period is the gradient amount The amount in the initial year is called a base amount, and it doesn’t need to equal the gradient amount Present Value @ time zero Gradient Factors

  15. PG = Present worth of the gradient starting in year 2… This is what is calculated by P/G factor. PT (total) = PG+PA PA comes from using the P/A factor on an annuity equal to the base amount. Gradient Factors To get the Gradient Factors we subtract off the base amount, and start things in year (period) 2:

  16. PG/G and AG/G P/G = factor to convert a gradient series to a present worth. A/G = factor to convert a gradient series to an equivalent uniform annual series.

  17. Gradient Example Find the PW of an income series with a cash flow in Year 1 of $1200 which increases by $300 per year through year 11. Use i = 15%

  18. Review of Factors Using the tables.. Single Payment factors (P/F), (F/P) Uniform Series factors (P/A), (F/A) Gradients (A/G), (P/G)

  19. Unknown Interest Rates and Years • Unknown Interest rate: • i.e. F = $20K, P = $10K, n = 9  i = ? • Or A = $1770, n = 10, P = $10K  i =? • Unknown Years – sometimes want to determine the number of years it will take for an investment to pay off ( n is unknown) • A = $100, P = $2000, i = 2%  n = ?

  20. Unknown interest example If you would like to retire with $1million 30 years from now, and you plan to save $6000 per year every year until then, what interest rate must your savings earn in order to get you that million?

  21. Use of Multiple Factors Many cash flow situations do not fit the single factor equations. It is often necessary to combine equations Example? What is P for a series of $100 payments starting 4 years from now? years 1 2 3 4 5 6 7 8 9 10 11 12 13 $100 P = ?

  22. Use of Multiple Factors Several Methods: 1. Use P/F of each payment 2. F/P of each and then multiply by P/F 3. Get F =A (F/A, i,10), then P = F (F/P,i,13) 4. Get P3 = A(P/A,I,10) and P0 = P3(P/F,i,3) years 1 2 3 4 5 6 7 8 9 10 11 12 13 $100 P = ?

  23. Use of Multiple Factors Step for solving problems like this: 1. Draw Cash Flow Diagram. 2. Locate P or F on the diagram. 3. Determine n by renumbering if necessary. 4. use factors to convert all cash flows to equivalent values at P or F.

  24. Multiple Factors: Example A woman deposited $700 per year for 8 years. Starting in the ninth year she increased her deposits to $1200 per year for 5 more years. How much money did she have in her account immediately after she made her last deposit?

  25. Eng Econ Practice Problems Check Website for Practice Problems…Remember you ALL have a quiz on Engineering Econ on Monday, not just the economists!

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