lecture no.8 chapter 3 contemporary engineering economics copyright © 2010 contemporary engineering...

Lecture No.8Chapter 3

Contemporary Engineering EconomicsCopyright © 2010

Contemporary Engineering Economics, 5th edition, © 2010

Linear Gradient SeriesA Strict Gradient Series

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, 5th edition, © 2010

Example3.21 Linear Gradient: Find P, Given A1, G, N, and i

Given: A1 = $1,000, G = $250, N = 5 years, and i = 12% per year Find: P

Excel Solution:

Contemporary Engineering Economics, 5th edition, © 2010

Gradient-to-Equal-Payment Series Conversion Factor, (A/G, i, N)

Given: G = $1,000, N = 10 years, i = 12%

Find: A

Solution:

Contemporary Engineering Economics, 5th edition, © 2010

Cash Flow Series

Factor Notation

Example 3.22 – Linear Gradient: Find A, Given A1, G, i, and N

Given: A1 = $1,000, G = $300, N = 6 years, and i = 10% per year Find: A

Contemporary Engineering Economics, 5th edition, © 2010

Example 3.23 Declining Linear Gradient Series

Given: A1 = $1,200, G = -$200, N = 5 years, and i = 10% per year Find: F 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.

Contemporary Engineering Economics, 5th edition, © 2010

Present Worth of Geometric Gradient Series Formula:

Factor Notation:

Contemporary Engineering Economics, 5th edition, © 2010

Example 3.24 –Geometric Gradient Series

Given: A1 = $54,600, g = 7%, N = 5 years, and i = 12% per year Find: P

Contemporary Engineering Economics, 5th edition, © 2010

Example 3.25 Retirement Plan – Saving $1 Million

Given: F = $1,000,000, g = 6%, i = 8%, and N = 20

Find: A1

Solution:

Contemporary Engineering Economics, 5th edition, © 2010