ENGINEERING ECONOMIC ENHANCED EBOOK
14th Edition
ISBN: 9780190931940
Author: NEWNAN
Publisher: OXF
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Question
Chapter 9, Problem 73P
(a)
To determine
To calculate:The no. of years the treated part last to be the preferred alternative if installed cost of corrosion treated part is $400
(b)
To determine
To calculate: The no. of years the treated part last to be the preferred alternative if untreated part last only for 4 years.
(c)
To determine
To calculate:The no. of years the treated part last to be the preferred alternative if MARR is 12%.
(d)
To determine
To calculate: The no. of years the treated part last to be the preferred alternative if sub part (a), (b) and (c) happen simultaneously.
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Students have asked these similar questions
Two lathes are being considered in the manufacture of certain machine parts. Data is given below, all cost in peso:
LATHE A
LATHE B
First Cost
40,000
56,000
Salvage Value
5,000
7,000
Annual Maintenance
2,000
2,800
Operation, Cost/hour
4
3.5
Life, in years
10
12
Time per part (hours)
0.40
0.25
REQUIRED:
Determine the number of machine parts/year that could be produced so that 2 lathes will be equally economical if the MARR is 18%. Use AWM
If the number of parts is 10,000 units, which lathe will you recommend? Use ROR
If the number of parts is 10,000 units, which lathe will you recommend? Use PWM
If the number of parts is 10,000 units, which lathe will you recommend? Use EUAC
Problem 13
The Keiser Mining Company must deposit money in an account TODAY (n=0) to cover the anticipated cleanup
costs that will begin when the mine closes. These cleanup costs of $6000 will begin in year 5 and repeat every
2 years, as shown below. The account will pay 5% annual interest.
$6
$6
$6
$in
thousands
of
5
-
ST
10
11
=
The amount Keiser Mining Company should deposit TODAY is closest to:
a. $45,900
b. $48,200
c. $50,600
d. $58,500
Which equation below gives at10% interest the total EUAC of an asset with an initial cost of $30,000, an estimated salvage value of $12,000 after its 7 -year service lile a, O&M costs of $25,000 per year?
EUAC=($30,000−12,000)(A/P,10%,7)+($12,000)(A/F,1006,7)+$25,000
EUAC=($30,000−12,000)(A/P,10%,7)+($12,000)(0.10)+$25,000(A/F,10%,7)
EUAC=($30,000−12,000)(A/R,10%,7)+($12,000)(0.10)+$25,000
EUAC=($30,000−12,000)(A/P,10%,7)+$25,000
Chapter 9 Solutions
ENGINEERING ECONOMIC ENHANCED EBOOK
Ch. 9 - Prob. 2QTCCh. 9 - Prob. 3QTCCh. 9 - Prob. 1PCh. 9 - Prob. 2PCh. 9 - Prob. 3PCh. 9 - Prob. 4PCh. 9 - Prob. 5PCh. 9 - Prob. 6PCh. 9 - Prob. 7PCh. 9 - Prob. 8P
Ch. 9 - Prob. 9PCh. 9 - Prob. 10PCh. 9 - Prob. 11PCh. 9 - Prob. 12PCh. 9 - Prob. 13PCh. 9 - Prob. 14PCh. 9 - Prob. 15PCh. 9 - Prob. 16PCh. 9 - Prob. 17PCh. 9 - Prob. 18PCh. 9 - Prob. 19PCh. 9 - Prob. 20PCh. 9 - Prob. 21PCh. 9 - Prob. 22PCh. 9 - Prob. 23PCh. 9 - Prob. 24PCh. 9 - Prob. 25PCh. 9 - Prob. 27PCh. 9 - Prob. 28PCh. 9 - Prob. 29PCh. 9 - Prob. 30PCh. 9 - Prob. 31PCh. 9 - Prob. 32PCh. 9 - Prob. 33PCh. 9 - Prob. 34PCh. 9 - Prob. 35PCh. 9 - Prob. 36PCh. 9 - Prob. 37PCh. 9 - Prob. 38PCh. 9 - Prob. 39PCh. 9 - Prob. 40PCh. 9 - Prob. 41PCh. 9 - Prob. 42PCh. 9 - Prob. 43PCh. 9 - Prob. 44PCh. 9 - Prob. 45PCh. 9 - Prob. 46PCh. 9 - Prob. 47PCh. 9 - Prob. 48PCh. 9 - Prob. 49PCh. 9 - Prob. 50PCh. 9 - Prob. 51PCh. 9 - Prob. 52PCh. 9 - Prob. 53PCh. 9 - Prob. 54PCh. 9 - Prob. 55PCh. 9 - Prob. 56PCh. 9 - Prob. 57PCh. 9 - Prob. 58PCh. 9 - Prob. 59PCh. 9 - Prob. 60PCh. 9 - Prob. 61PCh. 9 - Prob. 62PCh. 9 - Prob. 63PCh. 9 - Prob. 64PCh. 9 - Prob. 65PCh. 9 - Prob. 66PCh. 9 - Prob. 67PCh. 9 - Prob. 68PCh. 9 - Prob. 69PCh. 9 - Prob. 70PCh. 9 - Prob. 71PCh. 9 - Prob. 72PCh. 9 - Prob. 73PCh. 9 - Prob. 74PCh. 9 - Prob. 75PCh. 9 - Prob. 77PCh. 9 - Prob. 78PCh. 9 - Prob. 79PCh. 9 - Prob. 80PCh. 9 - Prob. 81PCh. 9 - Prob. 83PCh. 9 - Prob. 84PCh. 9 - Prob. 85PCh. 9 - Prob. 86PCh. 9 - Prob. 87PCh. 9 - Prob. 88P
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