ii) Select which alternative using AW and i= (10) 12% - per year. First Cost, $ -200,000 80,000 Annual operating cost, S/year 7,000 -30,000 Salvage value, $ 5,000 Life, years
Homework Answers
(i) Calculate the Annual Worth AW of Material X:-
First Cost = $15000
Maintenance Cost = $9000
Salvage = $2000
Life = 5 Years
PVAF @15% for 5 years = 3.3522
PVIF @15% for 5 years = 0.4972
[-First cost – Maintainance cost per year (PVAF @15% for 5 years) + Salvage (PVIF @15% for 5 years)]/ PVAF @15% for 5 years
[-$15000 - $9000(3.3522) + $2000(0.4972)]/3.3522
[-$15000 – $30169.8 + $994.4]/3.3522
-$44175.4/3.3522
= -$13178
Calculate the Annual Worth AW of Material Y:-
First Cost = $35000
Maintenance Cost = $7000
Salvage = $20000
Life = 5 Years
PVAF @15% for 5 years = 3.3522
PVIF @15% for 5 years = 0.4972
[-First cost – Maintainance cost per year (PVAF @15% for 5 years) + Salvage (PVIF @15% for 5 years)]/ PVAF @15% for 5 years
[-$35000 - $7000(3.3522) + $20000(0.4972)]/3.3522
[-$35000 – $23465.4 + $9944]/3.3522
-$48521.4/3.3522
= -$14474
On the basis of above analysis, Material X should be selected
(ii) Calculate the Annual Worth AW of Alternative C:-
First Cost = $80000
Annual Operating Cost = $30000
Salvage = $5000
Life = 8 Years
PVAF @12% for 8 years = 4.9676
PVIF @12% for 8 years = 0.4039
[-First cost – Annual Operating cost per year (PVAF @12% for 8 years) + Salvage (PVIF @12% for 8 years)]/ PVAF @12% for 8 years
[-$80000 - $30000(4.9676) + $5000(0.4039)]/4.9676
[-$80000 – $149028 + $2019.5]/4.9676
-$227008.5/4.9676
= -$45698
Calculate the Annual Worth AW of Alternative D:-
First Cost = $200000
Annual Operating cost = $7000
(-First cost * Int Rate) – Maintainance cost per year
(-$200000 * 0.12) - $7000
= - $31000
On the basis of above analysis, Alternative D should be selected
Add Answer to:
QUESTION 1 i) A metallurgical engineer is considering two materials for use in a space vehicle. All estimates are made....
An industrial engineer is considering two robots for improving efficiency in a fibre-optic manufacturing company. Robot...
An industrial engineer is considering two robots for improving efficiency in a fibre-optic manufacturing company. Robot X will have a first cost of RM88000, an annual maintenance and operation (M&O) cost of RM32000, and a RM35000 salvage value after its useful life of 6 years. A more sophisticated Robot Y will have a first cost of RM157000, an annual M&O cost of RM29800, and a RM55000 salvage value after its 6 year life. Which should be selected on the basis...
Question 11 For the cash flows provided hereunder, answer the questions and determine which alternative is best at an interest rate of 10% Alternative 350,000 150,000 0,000 450,000 First cost, S Main...
Question 11 For the cash flows provided hereunder, answer the questions and determine which alternative is best at an interest rate of 10% Alternative 350,000 150,000 0,000 450,000 First cost, S Maintenance costs, $ 20.000 15,000 ear Overhaul cots in vear 5 10.000 Salvage values, S 8.000 20,000 200,000 Life, years Match the closest correct answers for the below questions: Calculate the Present worth PW of A A. -86,748.66 B. -97,743.33] # Calculate the Annual worth AW of B Calculate...
Question 11 For the cash flows provided hereunder, answer the questions and determine which alternative is best at an interest rate of 10% Alternative 350,000 150,000 0,000 450,000 First cost, S Main...
Question 11 For the cash flows provided hereunder, answer the questions and determine which alternative is best at an interest rate of 10% Alternative 350,000 150,000 0,000 450,000 First cost, S Maintenance costs, $ 20.000 15,000 ear Overhaul cots in vear 5 10.000 Salvage values, S 8.000 20,000 200,000 Life, years Match the closest correct answers for the below questions: Calculate the Present worth PW of A A. -86,748.66 B. -97,743.33] # Calculate the Annual worth AW of B Calculate...
Date Table 2 (MARR-10%) First cost, S Annual cost, S per year Salvage value, S Life, years -40,00...
Date Table 2 (MARR-10%) First cost, S Annual cost, S per year Salvage value, S Life, years -40,000 -25,000 20,000 10 -75,000 15,000 7,000 a) Conduct PW analysis b) Conduct AW analysis c) Calculate capitalized cost for N d) Calculate capital recovery for MN
Date Table 2 (MARR-10%) First cost, S Annual cost, S per year Salvage value, S Life, years -40,000 -25,000 20,000 10 -75,000 15,000 7,000 a) Conduct PW analysis b) Conduct AW analysis c) Calculate capitalized cost...
U 15% PowrUp Investment Annual and salvage maintenance Repair -300 NGOM 300 OWN 300 Lloyd's Investment...
U 15% PowrUp Investment Annual and salvage maintenance Repair -300 NGOM 300 OWN 300 Lloyd's Investment Annual Repair savings and salvage maintenance Savings 36,000 -800 25,000 35,000 -800 25,000 35,000 -800 25,000 300 35,000 -800 25,000 35,000 -800 25,000 -300 35,000 2.000 -800 25,000 -300 35,000 -300 35,000 -300 35,000 15 -300 35,000 16 -300 35,000 17 AW element -6,642 -800 25,000 -7,025 -300 35,000 18 Total AW $ 17,558 $ 27,675 Figure 6-9 Annual worth analysis of surge protector...
03) In order to improve the recovery of highly brackish groundwater, the engineer at Clean Water ...
03) In order to improve the recovery of highly brackish groundwater, the engineer at Clean Water Engineering has established three altermatives for this project Assess all of the following projects, using MARR of 9% per year 129冫. Alternative A 500,000 70,000 Alternative B 250,000 80,000 155,000 in first year and decreases by $8000 thereafter First Cost, S Annual operating cost.s Alternative C 150,000 55,000 Annual income, $ 175,000 140,000 50,000 every 10 sears Major overall. S Salvage. S Life. Year...
Problem 06.022 Evaluating Alternatives by Annual Worth Analysis An environmental engineer wants to evaluate three different...
Problem 06.022 Evaluating Alternatives by Annual Worth Analysis An environmental engineer wants to evaluate three different methods for disposing of nonhazardous chemical waste: land application, fluidized-bed incineration, and private disposal contract. Use the estimates below to help her determine which has the least cost at i= 11% per year on the basis of an annual worth evaluation. 1 T First Cost AOC per Year Salvage Value Life Land $-150,000 $-97,000 $29,000 4 years Incineration -720,000 $-62,000 $370,000 6 years Contract...
Question 9: Giving the below data, which machine should be selected using i-5% per year A)...
Question 9: Giving the below data, which machine should be selected using i-5% per year A) Using Presnet Worth Analsys B) Using Annual Worth Analsys Machine X First Cost $20,000 operating cost $9,000 per year $5,000 Salvage value Years $35,000 $4,000 per year $7,000 Lombing of money Different nterest Question 6: An industrial engineer made a modification to a chip manufacturing process that will save her company $10,000 per year. At an interest rate of 8% per year, how much...
Required information Problem 14.056 The two machines shown are being considered for a chip manufacturing operation....
Required information Problem 14.056 The two machines shown are being considered for a chip manufacturing operation. Assume the MARR is a real return of 14% per year and that the inflation rate is 5.2% per year. -780.000 Machine First Cost. $ M&O. $ per year Salvage Value, $ Life, years -145,000 - 70.000 40,000 -5,000 200,000 Problem 14.056.b: Compare two alternatives based on their AW values with inflation consideration Which machine should be selected on the basis of an annual...
A delivery car had a first cost of $34,000, an annual operating cost of $15,000, and...
A delivery car had a first cost of $34,000, an annual operating cost of $15,000, and an estimated $5000 salvage value after its 6-year life. Due to an economic slowdown, the car will be retained for only 2 years and must be sold now as a used vehicle. At an interest rate of 8% per year, what must the market value of the used vehicle be in order for its AW value to be the same as the AW if...
An industrial engineer is considering two robots for improving efficiency in a fibre-optic manufacturing company. Robot X will have a first cost of RM88000, an annual maintenance and operation (M&O) cost of RM32000, and a RM35000 salvage value after its useful life of 6 years. A more sophisticated Robot Y will have a first cost of RM157000, an annual M&O cost of RM29800, and a RM55000 salvage value after its 6 year life. Which should be selected on the basis...
Question 11 For the cash flows provided hereunder, answer the questions and determine which alternative is best at an interest rate of 10% Alternative 350,000 150,000 0,000 450,000 First cost, S Maintenance costs, $ 20.000 15,000 ear Overhaul cots in vear 5 10.000 Salvage values, S 8.000 20,000 200,000 Life, years Match the closest correct answers for the below questions: Calculate the Present worth PW of A A. -86,748.66 B. -97,743.33] # Calculate the Annual worth AW of B Calculate...
Question 11 For the cash flows provided hereunder, answer the questions and determine which alternative is best at an interest rate of 10% Alternative 350,000 150,000 0,000 450,000 First cost, S Maintenance costs, $ 20.000 15,000 ear Overhaul cots in vear 5 10.000 Salvage values, S 8.000 20,000 200,000 Life, years Match the closest correct answers for the below questions: Calculate the Present worth PW of A A. -86,748.66 B. -97,743.33] # Calculate the Annual worth AW of B Calculate...
Date Table 2 (MARR-10%) First cost, S Annual cost, S per year Salvage value, S Life, years -40,000 -25,000 20,000 10 -75,000 15,000 7,000 a) Conduct PW analysis b) Conduct AW analysis c) Calculate capitalized cost for N d) Calculate capital recovery for MN
Date Table 2 (MARR-10%) First cost, S Annual cost, S per year Salvage value, S Life, years -40,000 -25,000 20,000 10 -75,000 15,000 7,000 a) Conduct PW analysis b) Conduct AW analysis c) Calculate capitalized cost...
U 15% PowrUp Investment Annual and salvage maintenance Repair -300 NGOM 300 OWN 300 Lloyd's Investment Annual Repair savings and salvage maintenance Savings 36,000 -800 25,000 35,000 -800 25,000 35,000 -800 25,000 300 35,000 -800 25,000 35,000 -800 25,000 -300 35,000 2.000 -800 25,000 -300 35,000 -300 35,000 -300 35,000 15 -300 35,000 16 -300 35,000 17 AW element -6,642 -800 25,000 -7,025 -300 35,000 18 Total AW $ 17,558 $ 27,675 Figure 6-9 Annual worth analysis of surge protector...
03) In order to improve the recovery of highly brackish groundwater, the engineer at Clean Water Engineering has established three altermatives for this project Assess all of the following projects, using MARR of 9% per year 129冫. Alternative A 500,000 70,000 Alternative B 250,000 80,000 155,000 in first year and decreases by $8000 thereafter First Cost, S Annual operating cost.s Alternative C 150,000 55,000 Annual income, $ 175,000 140,000 50,000 every 10 sears Major overall. S Salvage. S Life. Year...
Problem 06.022 Evaluating Alternatives by Annual Worth Analysis An environmental engineer wants to evaluate three different methods for disposing of nonhazardous chemical waste: land application, fluidized-bed incineration, and private disposal contract. Use the estimates below to help her determine which has the least cost at i= 11% per year on the basis of an annual worth evaluation. 1 T First Cost AOC per Year Salvage Value Life Land $-150,000 $-97,000 $29,000 4 years Incineration -720,000 $-62,000 $370,000 6 years Contract...
Question 9: Giving the below data, which machine should be selected using i-5% per year A) Using Presnet Worth Analsys B) Using Annual Worth Analsys Machine X First Cost $20,000 operating cost $9,000 per year $5,000 Salvage value Years $35,000 $4,000 per year $7,000 Lombing of money Different nterest Question 6: An industrial engineer made a modification to a chip manufacturing process that will save her company $10,000 per year. At an interest rate of 8% per year, how much...
Required information Problem 14.056 The two machines shown are being considered for a chip manufacturing operation. Assume the MARR is a real return of 14% per year and that the inflation rate is 5.2% per year. -780.000 Machine First Cost. $ M&O. $ per year Salvage Value, $ Life, years -145,000 - 70.000 40,000 -5,000 200,000 Problem 14.056.b: Compare two alternatives based on their AW values with inflation consideration Which machine should be selected on the basis of an annual...
A delivery car had a first cost of $34,000, an annual operating cost of $15,000, and an estimated $5000 salvage value after its 6-year life. Due to an economic slowdown, the car will be retained for only 2 years and must be sold now as a used vehicle. At an interest rate of 8% per year, what must the market value of the used vehicle be in order for its AW value to be the same as the AW if...
Most questions answered within 3 hours.
-
Which complex in the electron transport chain does not
contribute to the proton gradient across the...
asked 5 seconds from now -
The EastTN Company sells computer parts through a retail store
that it operates. The firm's comparative...
asked 14 seconds ago -
Use MATLAB:
Use a conditional while loop to find the first factorial that is
greater than...
asked 3 minutes ago -
Many describe estate and gift taxation as “political football.”
Discuss arguments for and against the transfer...
asked 5 minutes ago -
Pine wood has the following composition: 50.31% C,
6.20% H2, 43.08% Oz and 0.41% ash.
(a)...
asked 17 minutes ago -
A toy cannon uses a spring to project a 5.22-g soft rubber ball.
The spring is...
asked 18 minutes ago -
If a firm has an EV of $1,840 million and EBITDA of $377
million, what is...
asked 18 minutes ago -
Candy Kane Cosmetics (CKC) produces Leslie Perfume, which
requires chemicals and labor. Two production processes are...
asked 23 minutes ago -
Determine the heat involved in the combustion of liquid
hydrazine by using the following reactions. Use...
asked 24 minutes ago -
The proportion of Canadians with green eyes is 0.36. As part of
a study of the...
asked 33 minutes ago -
3) How does Descartes’s hyperbolic doubt compare to the
skepticism that Socrates expresses in the Platonic...
asked 31 minutes ago -
How many coulombs are required to plate a layer of chromium
metal 0.27 mm thick on...
asked 44 minutes ago