1. MER 160 - Design of Thermal Fluid Systems Econ Lecture 2 Professor Bruno Winter Term 2002

2. 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. Present Value @ time zero 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

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

4. P G /G and A G /G

5. Gradients 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%

6. 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)

7. Unknown Interest Rates and Years -Use tables 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 = ?

8. 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?

9. 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? 1 2 3 4 5 6 7 8 9 10 11 12 13 $100 P = ? years

10. 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 P 3 = A(P/A,I,10) and P 0 = P 3 (P/F,i,3) 1 2 3 4 5 6 7 8 9 10 11 12 13 $100 P = ? years

11. 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.

12. Use of 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 ?

13. Eng Econ Practice Problems Check Website for Practice Problems…Remember you ALL have a quiz on Engineering Econ Next Friday, not just the economists. You can see me (as Prof. Bruno) for free help on engineering econ!