Basics Of Engineering Economy
2nd Edition
ISBN: 9780073376356
Author: Leland Blank, Anthony Tarquin
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Question
Chapter 5, Problem 32APQ
To determine
Comparing alternatives.
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Evaluate the two alternatives A and B and decide the economic justified
alternative using:
Present worth method
Annual worth method
Future worth method
I.R.R method
E.R.R Method
E.R.R.R method
M.A.R.R = 15%
the details of alternatives are shown in the table below
Alternatives
A
Investments
$6,000 $7,500
Useful life (years)
10
Annual disbursements
$2,500 $3,500
Annual revenues
$4,500 $6,000
Salvage values
$500
$1,000
TRUE OR FALSE
4. If there is a past estimation error, there is a need to replace the property immediately.5. The physical life is always greater than all the other ālifeā factors under replacement analysis.6. If breakeven analysis is conducted with PW analysis for different MEAs, it doesnāt guarantee that the fastestalternative to reach its breakeven point has also the highest equivalent worth.
Evaluate the two alternatives A and B and decide the economic justified
alternative using:
Present worth method, Annual worth method, Future worth method
I.R.R method
, E.R.R Method
.E.R.R.R method
M.A.R.R=15%, the details of alternatives are shown in the table below
Alternatives
A
B
Investments
$60,000 $75,000
Useful life (years)
5
15
Annual disbursements
$25,000 $35,000
Annual revenues
$45,000 $60,000
Salvage values
$5,000 $10,000
Chapter 5 Solutions
Basics Of Engineering Economy
Ch. 5 - Prob. 1PCh. 5 - Prob. 2PCh. 5 - Prob. 3PCh. 5 - Prob. 4PCh. 5 - Prob. 5PCh. 5 - Prob. 6PCh. 5 - Prob. 7PCh. 5 - Prob. 8PCh. 5 - Prob. 9PCh. 5 - Prob. 10P
Ch. 5 - Two machines with the following cost estimates are...Ch. 5 - Prob. 12PCh. 5 - Prob. 13PCh. 5 - Prob. 14PCh. 5 - Prob. 15PCh. 5 - Prob. 16PCh. 5 - Prob. 17PCh. 5 - Prob. 18PCh. 5 - Estimates have been presented to Holly Farms,...Ch. 5 - Prob. 20PCh. 5 - Prob. 21PCh. 5 - Prob. 22PCh. 5 - Prob. 23PCh. 5 - Prob. 24PCh. 5 - Prob. 25PCh. 5 - Prob. 26PCh. 5 - A major repair on the suspension system of Janes...Ch. 5 - Prob. 28PCh. 5 - Prob. 29PCh. 5 - Prob. 30PCh. 5 - Prob. 31PCh. 5 - Prob. 32APQCh. 5 - Prob. 33APQCh. 5 - Prob. 34APQCh. 5 - Prob. 35APQCh. 5 - Prob. 36APQCh. 5 - The AW values of three revenue alternatives are ...Ch. 5 - Prob. 38APQCh. 5 - Prob. 39APQCh. 5 - Use an interest rate of 10% per year. The...Ch. 5 - Prob. 41APQCh. 5 - Prob. 42APQ
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Similar questions
- If a project is evaluated using annual worth method, then we know that: Select one: a.Ā Present worth will always be smaller than annual worth b.Ā Present worth will have the opposite sign of annual worth c.Ā Present worth will have the same sign as annual worth d.Ā Present worth will always be larger than annual wortharrow_forward5. The physical life is always greater than all the other "life" factors under replacement analysis. 6. If breakeven analysis is conducted with PW analysis for different MEAS, it doesn't guarantee that the fastest alternative to reach its breakeven point has also the highest equivalent worth.arrow_forwarda company is considering the purchase of a new Lathe, where there are 3 alternative Lathes with brands D, E and F with economic value for each Lathe as follows; Ā Lathe Machine D Lathe Machine E Lathe Machine F Initial Cost (Million Rp) 530 540 480 Maintenance Cost (MillionRp) 90 80 100 Residual Value (Million Rp) 60 130 80 Operating Life 10 10 10 Ā Question: Do a Present Worth Analysis to be able to determine the Alternative chosen from the three Generators with MARR = 13%?arrow_forward
- If you have the present worth of an alternative with a 5-year life, you can obtain its annual worth by:a. multiplying the PW by i.b. multiplying the PW by (A/F,i,5).c. multiplying the PW by (P/A,i,5).d. multiplying the PW by (A/P,i,5).arrow_forwardNeed answers ASAP... The annual worth can be calculated from the alternativeās: a. either ( a) or ( b) b. future worth by multiplying by ( F/A, i, n) c. all of the above d. present worth by multiplying by ( A/P, i, n)arrow_forwardIf the MARR=10%, compute the value ofĀ XĀ that makes the alternatives equally desirable.Ā Do not use spreadsheets.Ā Ā Alternatives Machine A Machine B First cost $12,000 $20,000 Annual Operating cost Ā $1,400/year Ā X Salvage value $2,000 $3,000 Life 4 years 8 yearsarrow_forward
- 1. A piece of construction equipment was bought 3 years ago for $ 500,000, expected life of 8 years and a salvage value of $20,000. The annual operating cost for this equipment is $58,000. It now can be sold for $200,000. An alternative piece of equipment can now be bought for $ 600,000, a salvage value of $150,000 and an expected life of 10 years. The annual operating cost for this equipment is $15,000. At MARR= 10% should we replace the old equipment? Use both EAC and P.W. Replace O Not replacearrow_forwardFor a certain potential investment project, we have the following estimates: Capital investment I Annual revenues R $100,000 $30.000 Annual expenses $9.000 Market value $10,000 Useful life 10 years MARR 15% a. Find the breakeven market value. b. Find the breakeven annual expense. c. Is the decision more sensitive to changes in market value or annual expense? How do you know?arrow_forwardQuestions: A.) Future worth of Alternative A, B, C & D B.) Using the future worth method, which Alternative should be chosen? C.) Annual worth of Alternative A, B, C, & D D.) Using Annual worth method, which alternative should be chosen?arrow_forward
- TRUE OR FALSE The future worth of a perpetuity from one (1) to ? years is undefined because as ? approaches infinity, the valuebecomes 0. The breakeven point always maximizes the profit.arrow_forwardWhen comparing mutually exclusive alternatives that have different lives by the present worth method, it is necessary to: (a) Always compare them over a period equal to the life of the longer-lived alternative. (b) Always compare them over a time period of equal service. (c) Always compare them over a period equal to the life of the shorter-lived alternative. (d) Find the present worth over one life cycle of each alternative. (e) None of the above a C darrow_forwardPrecision Engineering Factory consumesĀ 50,000Ā units of a component per year. The ordering, receiving and handling costs are RsĀ 3Ā per order while the trucking costs are RsĀ 12Ā per order. Further details are as follows: deterioration and obsolescence cost RsĀ 0.004Ā per unit per year; interest cost ReĀ 0.06Ā per unit per year; storage cost RsĀ 1,000Ā per year forĀ 50,000Ā units. Calculate the economic order quality.arrow_forward
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