Interest Formulas (Gradient Series) Lecture No.8 Chapter 3 Contemporary Engineering Economics Copyright © 2006 Contemporary Engineering Economics, 4th edition, © 2007 Linear Gradient Series L i (1 i ) iN 1 O P GM P N i (1 i ) Q N 2 G ( P / G , i, N ) P Contemporary Engineering Economics, 4th edition, © 2007 N Gradient Series as a Composite Series of a Uniform Series of N Payments of A1 and the Gradient Series of Increments of Constant Amount G. Contemporary Engineering Economics, 4th edition, © 2007 Example – Present value calculation for a gradient $2,000 series $1,250 $1,500 $1,750 $1,000 0 1 P =? 2 3 4 5 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? Contemporary Engineering Economics, 4th edition, © 2007 Method 1: Using the (P/F, i, N) Factor $2,000 $1,250 $1,500 $1,750 $1,000 0 1 P =? 2 3 4 5 $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 Contemporary Engineering Economics, 4th edition, © 2007 Method 2: Using the Gradient Factor P1 $1, 000 ( P / A ,12% ,5 ) $3, 604 .80 P2 $250( P / G ,12% ,5 ) $1, 599 .20 P $3, 604 .08 $1, 599 .20 $5,204 Contemporary Engineering Economics, 4th edition, © 2007 Gradient-to-Equal-Payment Series Conversion Factor, (A/G, i, N) Contemporary Engineering Economics, 4th edition, © 2007 Example 3.21 – Find the Equivalent Uniform Deposit Plan Contemporary Engineering Economics, 4th edition, © 2007 Solution: Given : A1 $1, 000, G $300, i 10% , and, N 6 Find : A A $1, 000 $300( A / G ,10% , 6) $1, 000 $300(2.22236) $1, 667.08 Contemporary Engineering Economics, 4th edition, © 2007 Example 3.22 Declining Linear Gradient Series Contemporary Engineering Economics, 4th edition, © 2007 Solution: F F1 F2 Equivalent Present W orth at n = 0 A1 ( F / A ,10% , 5) $200( P / G ,10% , 5) ( F / P ,10% , 5) $1, 200(6.105) $200(6.862)(1.611) $5,115 Contemporary Engineering Economics, 4th edition, © 2007 Types of Geometric Gradient Series g 0 Contemporary Engineering Economics, 4th edition, © 2007 Present Worth Factor R L 1 (1 g ) (1 i ) O | AM , if i g P PS ig N Q |TN A / (1 i ), if i g N N 1 1 Contemporary Engineering Economics, 4th edition, © 2007 Example 3.23 Annual Power Cost if Repair is Not Performed Contemporary Engineering Economics, 4th edition, © 2007 Solution – Adopt the new compressed-air system PO ld 1 (1 0.07 ) 5 (1 0.12) 5 $54, 440 0.12 0.07 $222, 283 PN ew $54, 440(1 0.23)( P / A ,12% , 5) $41, 918.80(3.6048) $151,109 Contemporary Engineering Economics, 4th edition, © 2007 Example 3.24 Jimmy Carpenter’s Retirement Plan – Save $1 Million Contemporary Engineering Economics, 4th edition, © 2007 What Should be the Size of his first Deposit (A1)? 1 (1 0 .0 6 ) (1 0 .0 8) A1 0 .0 8 0 .0 6 20 F2 0 A1 (7 2 .6 9 1 1) $ 1, 0 0 0, 0 0 0 A1 $ 1, 0 0 0, 0 0 0 7 2 .6 9 1 1 $ 1 3, 7 5 7 Contemporary Engineering Economics, 4th edition, © 2007 20