
(a):
Compare the projects based on the present worth.
(a):

Explanation of Solution
Table -1 shows the cash flow of different projects.
Table -1
Projects | B | C |
First cost (C) | 203,000 | 396,000 |
Operating cost (MO) per year | 85,000 | 119,000 |
Replacement cost (O) per year | 5,500 | |
Sludge hauling cost (S) per year less | 37,000 | |
Salvage value (SV) | 10% at C | 10% at C |
Time period (n) | 5 | 10 |
MARR (i) is 6%.
The time period for project B should equate with project time period C. Thus, all the cash flows are repeated for other five years. The time period (n1) is 10 years and time period 2 (n2) is five years.
The present worth of project B (PWA) can be calculated as follows:
The present worth of project B is -$994,275.85.
The present worth of the project C (PWC) can be calculated as follows:
The present worth of project C is -$977,141.13. Since the present worth of project C is greater than project B, select project C.
(b):
Compare the projects based on the present worth.
(b):

Explanation of Solution
The time period for project B should equate with project time period C. Thus, all the cash flows are repeated for other three years. The time period (n1) is eight years and time period 2 (n2) is five years.
The present worth of the project B (PWA) can be calculated as follows:
The present worth of project B is -$888,773.2513.
Time period (n) is eight years. The present worth of the project C (PWC) can be calculated as follows:
The present worth of project C is -$880.356.92. Since the present worth of project C is greater than project B, select project C.
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Chapter 5 Solutions
Engineering Economy
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