1. www.izmirekonomi.edu.tr L6: Interest Formulas (Gradient Series) ECON 320 Engineering Economics Mahmut Ali GOKCE Industrial Systems Engineering Computer Sciences

2. www.izmirekonomi.edu.tr Linear Gradient Series P

3. www.izmirekonomi.edu.tr Gradient Series as a Composite Series

4. www.izmirekonomi.edu.tr $1,000 $1,250 $1,500 $1,750 $2,000 1 2 3 4 5 0 P =? How much do you have to deposit now in a savings account that earns a 12% annual interest, if you want to withdraw the annual series as shown in the figure? Example – Present value calculation for a gradient series

5. www.izmirekonomi.edu.tr Method 1: $1,000 $1,250 $1,500 $1,750 $2,000 1 2 3 4 5 0 P =? $1,000(P/F, 12%, 1) = $892.86 $1,250(P/F, 12%, 2) = $996.49 $1,500(P/F, 12%, 3) = $1,067.67 $1,750(P/F, 12%, 4) = $1,112.16 $2,000(P/F, 12%, 5) = $1,134.85 $5,204.03

6. www.izmirekonomi.edu.tr Method 2:

7. www.izmirekonomi.edu.tr 0 1 2 3 4 5 6 7 25 26 $3.44 million 0 1 2 3 4 5 6 7 25 26 Cash Option $175,000 $189,000 $357,000 $196,000 G = $7,000 Annual Payment Option Example 2.16 Supper Lottery

8. www.izmirekonomi.edu.tr Equivalent Present Value of Annual Payment Option at 4.5%

9. www.izmirekonomi.edu.tr Excel Solution

10. www.izmirekonomi.edu.tr Geometric Gradient Series

11. www.izmirekonomi.edu.tr Present Worth Factor

12. www.izmirekonomi.edu.tr Alternate Way of Calculating P

13. www.izmirekonomi.edu.tr Example 2.17: Find P, Given A 1,g,i,N Given: g = 5% i = 7% N = 25 years A 1 = $50,000 Find: P

14. www.izmirekonomi.edu.tr Required Additional Savings