MSE 604 - HW 6 (1)
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School
California State University, Northridge *
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Course
604
Subject
Industrial Engineering
Date
Dec 6, 2023
Type
Pages
6
Uploaded by ColonelWillpowerDove33
California State University, Northridge
MSE604
ENGINEERING ECONOMIC ANALYSIS
Semester: Fall 2023
Professor: Ahmad Sarfaraz
Submitted By:
Bhushan Chaudhari
Falguni Pandey
Tejal Nikam
Mayur Ahire
6-07.
Fiesta Foundry is considering a new furnace that will allow them to be more
productive. Three alternative furnaces are under consideration.
Solution:
Furnace A: -450,000 + 70,000 (P/A, 12%, 15) + 50,000(P/F, 12%, 15)
= -450,000 + 70,000(6.8109) + 50,000(0.1827)
= 35,895.33
Furnace B: -400,000 + 62,000 (P/A, 12%, 15) + 40,000 (P/F, 12%, 15)
= -400,000 + 62,000(6.8109) + 40,000(0.1827)
= 29,581.45
Furnace C: -300,000 + 57,000 (P/A, 12%, 15) + 35,000 (P/F, 12%, 15)
= -300,000 + 57,000(6.1809) + 35,000(0.1827)
= 94,613.65
Therefore, the answer is C which is the highest value.
6-14
The estimated negative cash flows for three design alternatives are shown below. The
MARR is 12% per year and the study period is seven years. Which alternative is best based
on the IRR method? Doing nothing is not an option.
Answer. There are three alternative investments.
A, B and C
Minimum acceptable rate of return is equal to 12%
Study period is equal to 7 years
Among the three alternatives, the firm will choose alternative A
Even though alternative A requires higher capital investment then alternative B and C
The annual expenses of alternative A is lower than alternative B and C
Annual expenses of alternative A = 85,6000
But its annual expenses = $7,400
Alternative B requires lower capital investment
But alternative B has annual expenses 12100 dollars which is higher than alternative A and C
Conclusion
Alternative A is the best option for capital investment
6-28.
Consider the following EOY cash flows for two mutually exclusive alternatives (one
must be chosen):
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Solution:
The MARR is 5% per year.
Answer.
A. For Lead Acid
AW = -Capital investment (A/P, i, n) - Annual expenses + Market value at the end of useful
life (A/F, i, n)
AW of Lead Acid = -6000 * (A/P, 5%,12) - 2500
= -6000 * 0.112825 - 2500
= -3176.95 ~
-3177 $
B. For Lithium Ion
AW = -Capital investment (A/P, i, n) - Annual expenses + Market value at the end of useful
life (A/F, i, n)
AW of Lithium ion = -14000 * (A/P, 5%,18) - 2400 + 2600 * (A/F, 5%,18)
= -14000 * 0.085546 - 2400 + 2600 * 0.035546
= -3505.22 ~
-3505
$
The annual worth of Lead Acid is numerically higher.
This implies that Lead Acid alternative will be less costly.
Thus,
Lead Acid alternative should be chosen
6-38.
Two mutually exclusive alternatives for office building refrigeration and air
conditioning
are being investigated. Their relevant costs and lives are summarized as follows:
If the hurdle rate (MARR) is 15% per year and the study period is 18 years, which chilling
system should be recommended?
(6.5)
Solution:
Annual Worth of AC = - P (A / P, i, n) + A + F (A/F, i, n)
= - 60,000 (A / P, 15%, 6) + 40,000 + 10,000 (A/F, 15%, 6)
= - 60,000 (0.2642) + 40,000 + 10,000 (0.1142)
= - 15852 + 40,000 + 1142
= - 15852 + 41,142
= 25290
Annual Worth of CC = - P (A / P, i, n) + A
= - 80,000 (A / P, 15%, 9) + 32,000
= - 80,000 (0.2096) + 32,000
= -16768 + 32000
= 15232
The annual worth of AC chilling system is more than CC chilling system. Therefore AC
chilling system should be recommended.
6-52
. Compare alternatives A and B with the equivalent worth method of your choice if the
MARR is 15% per year. Which one would you recommend? State all assumptions.
The two alternatives are analyzed using the present worth (PR) method. The assumption is
that the alternatives are mutually exclusive.
Solution:
PW of A = -50,000-500 (P/A,15%,20) -5000 (P/G,15%,20)-5000 (P/F,15%,10) - 5000
(P/F,15%,5)-5000 (P/F,15%,15)-5000 (P/F,15%,20) + 10,000 (P/F,15%,20)
= -5000 -5000 (((1.15)20-1)/ (0.15 (1.15)20))- 500 (((1.15)20 -1)/ (0.15)20 (1.15)20)-
20/(0.15)20 (1.15)20))
= -5000 (1.15)-5 -5000 (1.15)-10- 5000(1.15)-15-5000(1.15)-20 +10,000(1.15)-20
Hence PW of A= -102,117
Therefore, the PW of Alternative A is $-102,117.
PW of B = -20,000-10,000(P/A,15%,10)-1000(P/G,15%,10)
=-20,000-10,000(((1.15)10-1)/((0.15)(0.15)10))-1000[(((1.15)10-1)/((0.15)2(0.15)10))-
(10/(0.15)(1.15)10
=-2000 -50,187 – 16974
Hence PW of B = -87,166
Therefore, the alternative of B is $87,166.
The PW of B is less than the PW of A.
Therefore, alternative B is less costly.
Hence alternative B is recommended
.
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