ECON 225 Engineering Economics Module 5 Chapter 11 Connect Homework
pdf
keyboard_arrow_up
School
Embry-Riddle Aeronautical University *
*We aren’t endorsed by this school
Course
225
Subject
Industrial Engineering
Date
Feb 20, 2024
Type
Pages
7
Uploaded by JusticeUniverse3669
Module 5
Chapter 11 Connect Homework
A piece of equipment has a first cost of $150,000, a maximum useful life of 7 years, and a market (salvage) value described by the relation S = 120,000 – 11,000k, where k is the number of years since it was purchased. The salvage value cannot go below zero. The AOC series is estimated using AOC = 60,000 + 14,000k. The interest rate is 11% per year. Determine the economic service life and the respective AW.
The economic service life is
2 and the AW value is $
-121,780.
Explanation Year Salvage Value, $ AOC, $ per Year 1 109,000 74,000 2 98,000 88,000 3 87,000 102,000 4 76,000 116,000 5 65,000 130,000 6 54,000 144,000 7 43,000 158,000 AW1 = -150,000(A/P,11%,1) - 74,000 + 109,000(A/F,11%,1)
AW1
=
$-131,500
AW2
=
-150,000(A/P,11%,2)
-
(74,000
+
14,000(A/G,11%,2))
+
98,000(A/F,11%,2)
AW2
=
$-121,780
AW3
=
-150,000(A/P,11%,3)
-
(74,000
+
14,000(A/G,11%,3))
+
87,000(A/F,11%,3)
AW3
=
$-122,378
AW4
=
-150,000(A/P,11%,4)
-
(74,000
+
14,000(A/G,11%,4))
+
76,000(A/F,11%,4)
AW4
=
$-125,391
AW5
=
-150,000(A/P,11%,5)
-
(74,000
+
14,000(A/G,11%,5))
+
65,000(A/F,11%,5)
AW5
=
$-129,240
AW6
=
-150,000(A/P,11%,6)
-
(74,000
+
14,000(A/G,11%,6))
+
54,000(A/F,11%,6)
AW6
=
$-133,399
AW7
=
-150,000(A/P,11%,7)
-
(74,000
+
14,000(A/G,11%,7))
+
43,000(A/F,11%,7)
AW7
=
$-137,645
The economic service life is 2 year(s) and the AW value is $-121,780.
An engineer with Calahan Technologies calculated the AW of cost values shown for a presently owned machine using estimates she obtained from the vendor and company records. A challenger has an economic service life of 7 years with an AW of $−86,000 per year. Assume that all future costs remain as estimated and the challenger’s technology will definitely replace that of the defender within 5 years. When should the company purchase the challenger?
Retention Period, Years AW of Costs, $ per Year 1 −92,000
2 −81,000
3 −87,000
4 −89,000
5 −95,000
The company should purchase the challenger at
year
2.
Explanation The company should purchase the challenger at the end of year 2, when its AW will be lower than that of the defender. A presently owned machine can last 3 more years if properly maintained at a cost of $16,000 per year. Its AOC is $32,000 per year. After 3 years, it can be sold for an estimated $10,000. A replacement costs $82,000 with a $10,000 salvage value after 3 years and an operating cost of $17,000 per year. Different vendors have offered $10,000 and $24,000, respectively, for the current system as a trade-in for the replacement machine. At i
= 9% per year, perform a replacement study and determine whether the defender should be retained or replaced. According to the replacement study, select challenger. Explanation AWC = -(82,000(A/P,9%,3)) - 17,000 + 10,000(A/F,9%,3) AWC = −
(82,000(0.3951)) - 17,000 + 10,000(0.3051) AWC = $-46,344 For the defender, the higher trade-in value applies, because that represents the best market value estimate. AWD = -(24,000(A/P,9%,3)) - 16,000 -32,000 + 10,000(A/F,9%,3) AWD = −
(24,000(0.3951)) - 16,000 -32,000 + 10,000(0.3051)
AWD = $-54,431 Since AW
C
is less than AW
D
, select challenger. State-of-the-art digital imaging equipment purchased 2 years ago for $50,000 had an expected useful life of 5 years and a $5000 salvage value. After its poor installation performance, it was upgraded for $20,000 1 year ago. Increased demand now requires another upgrade for an additional $27,000 so that it can be used for three more years. Its new annual operating cost will be $27,000 with a $11,500 salvage after the 3 years. Alternatively, it can be replaced with new equipment costing $68,500, an estimated AOC of $16,000, and an expected salvage of $25,000 after 3 years. If replaced now, the existing equipment can be traded for only $7,000. Use a MARR of 12% per year. Calculate the AWs and determine whether the company should retain or replace the defender now. The company should replace defender with challenger. Explanation AWD = -(7,000 + 27,000)(A/P,12%,3) - 27,000 + 11,500(A/F,12%,3) AWD = -(7,000 + 27,000)(0.4163) - 27,000 + 11,500(0.2963) AWD = $-37,748 AWC = -(68,500)(A/P,12%,3) - 16,000 + (25,000(A/F,12%,3)) AWC = -(68,500)(0.4163) - 16,000 + (25,000(0.2963)) AWC = $-37,111 The company should replace defender with challenger. State-of-the-art digital imaging equipment purchased 2 years ago for $50,000 had an expected useful life of 5 years and a $5000 salvage value. After its poor installation performance, it was upgraded for $20,000 1 year ago. Increased demand now requires another upgrade for an additional $27,000 so that it can be used for three more years. Its new annual operating cost will be $27,000 with a $11,500 salvage after the 3 years. Alternatively, it can be replaced with new equipment costing $68,500, an estimated AOC of $16,000, and an expected salvage of $25,000 after 3 years. If replaced now, the existing equipment can be traded for only $7,000. Use a MARR of 12% per year. Based on the poor experience with the current equipment, assume the person conducting the analysis decides the challenger may be kept for only 2 years, not three, with the same AOC and salvage estimates for the 2 years. What is the decision?
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
The company should retain defender. Explanation AWC = -(68,500)(A/P,12%,2) - 16,000 + (25,000(A/F,12%,2)) AWC = -(68,500)(0.5917) - 16,000 + (25,000(0.4717)) AWC = $-44,739 Since AW
C
is greater than AW
D
, retain defender. The estimated future market values and M&O costs for an in-place backup generator at MediCare Hospital and a possible replacement are shown below. Mrs. Jamison, hospital director, told you that she is interested only in what happens over the next 3 years. If the current generator is to be replaced, it must be done now or kept in place for the 3 years. Using an interest rate of 11% per year and a 3-year study period, determine whether or not the replacement is economically advantageous now. In-Place Replacement Year Market Value, $ M&O Cost, $ Market Value, $ M&O Cost, $ 0 36,000 73,000 1 32,000 −55,000
65,000 −37,000
2 23,000 −55,000
39,000 −37,000
3 11,000 −55,000
20,000 −37,000
4 19,000 −38,000
5 11,000 −39,000
The replacement is
economically advantageous.
Explanation Compare the cost of replacement (challenger) with the cost of keeping defender (in-place) for three more years. AWC
=
-(73,000(A/P,11%,3))
-
37,000
+
20,000(A/F,11%,3)
AWC
=
-(73,000(0.4092))
-
37,000
+
20,000(0.2992)
AWC
=
$-60,888
AWD
=
−36,000(A/P,11%,3)
−
55,000
+
11,000(A/F,11%,3)
AWD
=
−36,000(0.4092)
−
55,000
+
11,000(0.2992)
AWD
=
$-66,440
Replace the in-place with the challenger now since AW
C
is less than AW
D
.
A machine purchased 3 years ago for $140,000 is now too slow to satisfy the demand of the customers. It can be upgraded now for $76,000 or sold to a smaller company internationally for $37,000. The upgraded machine will have an annual operating cost of $92,000 per year and a $27,000 salvage value in 3 years. If upgraded, the presently owned machine will be retained for
only 3 more years, then replaced with a machine to be used in the manufacture of several other product lines. The replacement machine, which will serve the company now and for a maximum of 8 years, costs $207,000. Its salvage value will be $60,000 for years 1 through 5; $20,000 after 6 years; and $10,000 thereafter. It will have an estimated operating cost of $45,000 per year. Perform an economic analysis at 11% per year using a specified 3-year planning horizon. a) Determine if the current machine should be replaced now or 3 years from now. b) Once decided, determine the equivalent AW for the next three years. a) The current machine should be replaced
now.
b) The equivalent AW for the next three years is $
-111,754.
Explanation a) The two options are
(1) upgrade and retain defender for three years, or
(2) buy challenger now
AW1
=
-(37,000
+
76,000)(A/P,11%,3)
-
92,000
+
27,000(A/F,11%,3)
AW1
=
-(37,000
+
76,000)(0.40921)
-
92,000
+
27,000(0.29921)
AW1
=
$-130,162
AW2
=
-207,000(A/P,11%,3)
-
45,000
+
60,000(A/F,11%,3)
AW2
=
$-111,754
The defender should be replaced now.
The equivalent annual worth for the next three years will be $-111,754.
With the estimates shown below, Sarah needs to determine the trade-in (replacement) value of machine X that will render its AW equal to that of machine Y at an interest rate of 8% per year. Determine the replacement value.
The replacement value is $
13,399.
Explanation -RV(A/P,8%,3)
-
62,000
+
15,500(A/F,8%,3)
=
-
85,000(A/P,8%,5)
-
(40,000
+
2000(A/G,8%,5))
+
15,000(A/F,8%,5)
-RV(0.38803)
-
62,000
+
15,500(0.30803)
=
-
85,000(0.25046)
-
(40,000
+
2000(1.84647))
+
15,000(0.17046)
RV
=
$13,399
Machine X Machine Y Market Value, $ ? 85,000 Annual Cost, $ per Year −62,000
−40,000 for year 1,increasing by 2000 per year thereafter. Salvage Value 15,500 15,000 Life, Years 3 5
A machine purchased a year ago for $85,000 costs more to operate than anticipated. At the time of the purchase, it was expected to be used for 10 years with annual maintenance costs of $22,000 and a salvage value of $10,000. However, last year, it incurred a cost $38,000, which is expected to escalate to $39,400 this year and increase by $1,400 each year thereafter. The market value is now estimated to be $70,000–$9,000
k
, where
k
is the number of years since the machine was purchased. It is now estimated that the machine will be useful for a maximum of 7 more years. A replacement study is to be performed. Determine the values of
P
, n, AOC for year 7 and
S
of this defender.
The value of
P
is $
61,000.
The value of n is
7
years.
The AOC for year 7 is
49,200.
The value of
S
is $
-2,000.
Explanation P
=
market
value
=
70,000
-
9,000(1)
P
=
$61,000
n=
7
years
AOC
=
$39,400
+
$1,400(7)
AOC
=
$49,200
S
=
70,000
-
9,000(8)
S
=
$-2,000
Equipment purchased 2 years ago by Newport Corporation to make pneumatic vibration isolators cost $112,500. It has a market value that can be described by the relation $112,500 –
$8,600
k
, where k
is the years from time of purchase. The operating cost for the first 5 years is $58,000 per year, after which it increases by $5,700 per year. The asset’s salvage value was originally estimated to be $7000 after a predicted 10-year useful life. Determine the values of P
, S
, and AOC if (a) a replacement study is done now and it is assumed that the equipment will be kept a maximum of one more year, and (b) a replacement study is done 5 years from now and it is assumed that the equipment will be kept a maximum of only one more year after that. a) The value of P
is $ 95,300. The value of S is $ 86,700. The value of AOC is $ 58,000. b) The value of P
is $ 52,300. The value of S is $ 43,700.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
The value of AOC is $ 69,400. Explanation a) P = 112,500 - 8,600(2) P = $95,300 S = 112,500 - 8,600(3) S = $86,700 AOC = $58,000 b) P = 112,500 - 8,600(7) P = $52,300 S = 112,500 - 8,600(8) S = $43,700 AOC = $69,400