Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Interest Formulas (Gradient Series)

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Presentation transcript:

Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Interest Formulas (Gradient Series) Lecture No. 8 Chapter 3 Contemporary Engineering Economics Copyright © 2016

Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Linear Gradient Series A Strict Gradient Series Gradient Series as a Composite Series of a Uniform Series of N Payments of A 1 and the Gradient Series of Increments of Constant Amount G

Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Example 3.18: Linear Gradient: Find P, Given A 1, G, N, and i  Given: A 1 = $1,000, G = $250, N = 5 years, and i = 12% per year  Find: P

Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Solution  Excel Solution

Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Gradient-to-Equal-Payment Series Conversion Factor, (A/G, i, N) Cash Flow Series Factor Notation  Given: G = $1,000, N = 10 years, i = 12%  Find: A  Solution

Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Example 3.19: Linear Gradient: Find A, Given A 1, G, i, and N  Given: A 1 = $1,000, G = $300, N = 6 years, and i = 10% per year  Find: A

Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Solution

Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Example 3.20: Declining Linear Gradient Series: Find F, Given A 1, G, I, and N  Given: A 1 = $1,200, G = -$200, N = 5 years, and i = 10% per year  Find: F

Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Solution  Strategy: Since we have no interest formula to compute the future worth of a linear gradient series directly, we first find the equivalent present worth of the gradient series and then convert this P to its equivalent F.  Solution

Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Present Worth of Geometric Gradient Series  Formula  Factor Notation

Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Example 3.21: Geometric Gradient Series  Given: A 1 = $54,600, g = 7%, N = 5 years, and i = 12% per year  Find: P

Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Solution

Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Example 3.22: Retirement Plan: Saving $1 Million  Given: o F = $1,000,000, o g = 6%, o i = 8%, and o N = 20  Find: A 1

Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Solution