MER 439 Design of Thermal Fluid Systems Introduction to Engineering Economics 2. Factors

1. MER 439 Design of Thermal Fluid Systems Introduction to Engineering Economics 2. Factors Professor Wilk Winter 2007

2. Factor Notation Single Payment Compound Amount Factor (F/P, i%, n): F/P = (1 + ieff)n Single Payment Present Worth Factor (P/F, i%, n): P/F = (1 + ieff)-n n is in years if the ieff is used

3. Example 5 - Factors How much inheritance, to be received 20 years from now, is equivalent to receiving $10,000 now? The interest rate is 4% per year. Use factors.

4. Factor Tables Fortunately these factors are tabulated… (And Excel has nice built in functions to calculate them too)

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

6. 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 (P/F,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 = ?

7. Use of Multiple Factors Example 7: Someone deposited $700 per year (at the end of each year) at an effective interest rate of 6%. At the beginning of the tenth year she increased her deposits to $1200 per year. How much money did she have in her account immediately after she made her last deposit in the beginning of year 14?

8. 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. 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

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

10. PG/G and AG/G

11. Example 8: Gradients 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 = 12%

12. 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)

13. 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) • i.e. A = $100, P = $2000, i = 2%  n = ?

14. Unknown interest example Example 9 • 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?

15. Doubling Time • The time it will take for an amount to double, given an annual growth rate i%. (Rule of 72) n  72/i% Alternatively, i%  72/n

16. References • Newnan, D.G. and Lavelle, J.P., “Essentials of Engineering Economic Analysis,” Engineering Press, 1998. • Lindeburg, M.R., “Engineer-in-Training Reference Manual,” 8th ed., Professional Publications, 1992. • Jaluria, Y., “Design and Optimization of Thermal Systems,” McGraw-Hill, 1998. • National Council of Examiners for Engineering and Surveying, “Fundamentals of Engineering Discipline Specific Handbook,”NCEES, 1996.