Engineering Economy
Engineering Economy
8th Edition
ISBN: 9780073523439
Author: Leland T Blank Professor Emeritus, Anthony Tarquin
Publisher: McGraw-Hill Education
Question
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Chapter 5, Problem 21P

(a):

To determine

Compare the projects based on the present worth.

(a):

Expert Solution
Check Mark

Explanation of Solution

Table -1 shows the cash flow of different projects.

Table -1

ProjectsBC
First cost (C)203,000396,000
Operating cost (MO) per year85,000119,000
Replacement cost (O) per year5,500 
Sludge hauling cost (S) per year less 37,000
Salvage value (SV)10% at C10% at C
Time period (n)510

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:

PWB=C(MO+O)((1+i)n11i(1+i)n1)+C+(C×0.1)(1+i)n2+(C×0.1)(1+i)n1=(203,000(85,000+5,500)((1+0.06)1010.06(1+0.06)10)+203,000+(203,000×0.1)(1+0.06)5+(203,000×0.1)(1+0.06)10)=(203,000(90,500)(1.79084810.06(1.790848))+203,000+20,3001.338226+20,3001.790848)=203,000(90,500)(0.7908480.107451)136,524.03+11,335.42=203,000(90,500)(7.36008)136,524.03+11,335.42=203,000666,087.24136,524.03+11,335.42=994,275.85

The present worth of project B is -$994,275.85.

The present worth of the project C (PWC) can be calculated as follows:

PWC=C(MOS)((1+i)n1i(1+i)n)+(C×0.1)(1+i)n=(396,000(119,00037,000)((1+0.06)1010.06(1+0.06)10)+(396,000×0.1)(1+0.06)10)=(396,00082,000(1.79084810.06(1.790848))+39,6001.790848)=396,00082,000(0.7908480.107451)+22,112.43=396,00082,000(7.36008)+22,112.43=396,000603,526.56+22,112.43=977,414.13

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):

To determine

Compare the projects based on the present worth.

(b):

Expert Solution
Check Mark

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:

PWB=C(MO+O)((1+i)n11i(1+i)n1)+C+(C×0.1)(1+i)n2+(C×0.1)(1+i)n1=(203,000(85,000+5,500)((1+0.06)810.06(1+0.06)8)+203,000+(203,000×0.1)(1+0.06)5+(203,000×0.1)(1+0.06)8)=(203,000(90,500)(1.59384810.06(1.593848))+203,000+20,3001.338226+20,3001.593848)=203,000(90,500)(0.5938480.095631)136,524.03+12,736.4717=203,000(90,500)(6.209786)136,524.03+12,736.4717=203,000561,985.633136,524.03+12,736.4717=888,773.25

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:

PWC=C(MOS)((1+i)n1i(1+i)n)+(C×0.1)(1+i)n=(396,000(119,00037,000)((1+0.06)810.06(1+0.06)8)+(396,000×0.1)(1+0.06)8)=(396,00082,000(1.59384810.06(1.593848))+39,6001.593848)=396,00082,000(1.59384810.06(1.593848))+24,845.5311=396,00082,000(6.209786)+24,845.5311=396,000509,202.452+24,845.5311=880,356.92

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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