Lecture No.8 Chapter 3 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering...
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Transcript of Lecture No.8 Chapter 3 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering...
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Lecture No.8Chapter 3
Contemporary Engineering EconomicsCopyright © 2010
Contemporary Engineering Economics, 5th edition, © 2010
![Page 2: Lecture No.8 Chapter 3 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.](https://reader036.fdocuments.net/reader036/viewer/2022062408/56649eb55503460f94bbde44/html5/thumbnails/2.jpg)
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
![Page 3: Lecture No.8 Chapter 3 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.](https://reader036.fdocuments.net/reader036/viewer/2022062408/56649eb55503460f94bbde44/html5/thumbnails/3.jpg)
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
![Page 4: Lecture No.8 Chapter 3 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.](https://reader036.fdocuments.net/reader036/viewer/2022062408/56649eb55503460f94bbde44/html5/thumbnails/4.jpg)
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
![Page 5: Lecture No.8 Chapter 3 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.](https://reader036.fdocuments.net/reader036/viewer/2022062408/56649eb55503460f94bbde44/html5/thumbnails/5.jpg)
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
![Page 6: Lecture No.8 Chapter 3 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.](https://reader036.fdocuments.net/reader036/viewer/2022062408/56649eb55503460f94bbde44/html5/thumbnails/6.jpg)
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
![Page 7: Lecture No.8 Chapter 3 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.](https://reader036.fdocuments.net/reader036/viewer/2022062408/56649eb55503460f94bbde44/html5/thumbnails/7.jpg)
Present Worth of Geometric Gradient Series Formula:
Factor Notation:
Contemporary Engineering Economics, 5th edition, © 2010
![Page 8: Lecture No.8 Chapter 3 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.](https://reader036.fdocuments.net/reader036/viewer/2022062408/56649eb55503460f94bbde44/html5/thumbnails/8.jpg)
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
![Page 9: Lecture No.8 Chapter 3 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.](https://reader036.fdocuments.net/reader036/viewer/2022062408/56649eb55503460f94bbde44/html5/thumbnails/9.jpg)
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