ENGR.ECONOMIC ANALYSIS
14th Edition
ISBN: 9780190931919
Author: NEWNAN
Publisher: Oxford University Press
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Chapter 6, Problem 48P
To determine
The Equivalent Annual Worth among the given three route 105,205,305.
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The cash flows for three different alternatives are given in table below. MARR =10%.
Alt. A Alt. B Alt. C
Initial cost $5,000 9,000 7,500
Annual benefits $1,457 2,518 2,133
RoR 14% 13% 12.4%
Life in years 5
1. ΔRoR for the first increment (Alt. C-Alt. A) is ___________________.
A.10.12% B. 9.38% C. 11.85% D. 11.00%
2. ΔRoR for the second increment is ___________________.
A. 10.12% B. 9.38% C. 8.94% D. 9.87%
3. The best alternative for a MARR of 10% using the incremental rate of return analysis is ____________.
A. Alt. C B. Alt. A C. Alt. B D. Do nothing
Chapter 6 Solutions
ENGR.ECONOMIC ANALYSIS
Ch. 6 - Prob. 1QTCCh. 6 - Prob. 2QTCCh. 6 - Prob. 3QTCCh. 6 - Prob. 4QTCCh. 6 - Prob. 5QTCCh. 6 - Prob. 1PCh. 6 - Prob. 2PCh. 6 - Prob. 3PCh. 6 - Prob. 4PCh. 6 - Prob. 5P
Ch. 6 - Prob. 6PCh. 6 - Prob. 7PCh. 6 - Prob. 8PCh. 6 - Prob. 9PCh. 6 - Prob. 10PCh. 6 - Prob. 11PCh. 6 - Prob. 12PCh. 6 - Prob. 13PCh. 6 - Prob. 14PCh. 6 - Prob. 15PCh. 6 - Prob. 16PCh. 6 - Prob. 17PCh. 6 - Prob. 18PCh. 6 - Prob. 19PCh. 6 - Prob. 20PCh. 6 - Prob. 21PCh. 6 - Prob. 22PCh. 6 - Prob. 23PCh. 6 - Prob. 24PCh. 6 - Prob. 25PCh. 6 - Prob. 26PCh. 6 - Prob. 27PCh. 6 - Prob. 28PCh. 6 - Prob. 29PCh. 6 - Prob. 30PCh. 6 - Prob. 31PCh. 6 - Prob. 32PCh. 6 - Prob. 33PCh. 6 - Prob. 34PCh. 6 - Prob. 35PCh. 6 - Prob. 36PCh. 6 - Prob. 37PCh. 6 - Prob. 38PCh. 6 - Prob. 39PCh. 6 - Prob. 40PCh. 6 - Prob. 41PCh. 6 - Prob. 42PCh. 6 - Prob. 43PCh. 6 - Prob. 44PCh. 6 - Prob. 45PCh. 6 - Prob. 46PCh. 6 - Prob. 47PCh. 6 - Prob. 48PCh. 6 - Prob. 49PCh. 6 - Prob. 50PCh. 6 - Prob. 51PCh. 6 - Prob. 52PCh. 6 - Prob. 53PCh. 6 - Prob. 54PCh. 6 - Prob. 55PCh. 6 - Prob. 56PCh. 6 - Prob. 57PCh. 6 - Prob. 58PCh. 6 - Prob. 59PCh. 6 - Prob. 60PCh. 6 - Prob. 61PCh. 6 - Prob. 62PCh. 6 - Prob. 63PCh. 6 - Prob. 64PCh. 6 - Prob. 65PCh. 6 - Prob. 66PCh. 6 - Prob. 67PCh. 6 - Prob. 68PCh. 6 - Prob. 69PCh. 6 - Prob. 70PCh. 6 - Prob. 71PCh. 6 - Prob. 72PCh. 6 - Prob. 73PCh. 6 - Prob. 74PCh. 6 - Prob. 75PCh. 6 - Prob. 76PCh. 6 - Prob. 77P
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- Any help would be appreciated! Given the data for three different alternatives in the table below, determine the best alternative using the incremental rate of return (∆RoR) analysis. MARR =9%. A B C First cost $15,000 $25,000 $20,000 O &M Cost/ year 1,600 400 900 Benefit/year 8,000 13,000 9,000 Salvage value 3,000 6,000 4,600 Life in years 4 4 4 1. The better alternative between the first increment is ________________. A. Alt. A or Alt. B B. Alt. A C. Alt.C D. Alt. B 2. The better alternative between the second increment is ___________________. A. Alt. B or Alt. C B. Alt. B C. Alt. C D. Alt. Aarrow_forwardIt is proposed to place a cable on an existing pole line along the shore of a lake to connect two points on opposite sides. Which is more economical? Use future worth method and present worth method.arrow_forwardDetermine the better of the two alternatives using the present worth analysis. Use an interest rate of 10%. Alt. X Alt.Y Initial cost $12,500 $8,900 Annual benefit $6,800 $2,000 Salvage value $5,000 $8,900 Life in years 2 years 3 Years MARR 10%arrow_forward
- Determine which alternative, if any, should be chosen based on Annual Worth method using 15% MARR. Use Repeatability Method. Alternative A B First Cost (Investment Cost) $ 5,000 $10,200 Uniform Annual Benefit $1,100 $2,300 Useful Life 5 years 10 years a. The Annual Worth of Alternative A is = $ Blank 1 b. The Annual Worth of Alternative B is = $ Blank 2 c. Choose Alternative (Type only A or B) = Blank 3 Note: Show final answer to the nearest WHOLE NUMBER. No need to write the Unit of Measure. Blank 1 Add your answer Blank 2 Add your answer Blank 3 Add your answerarrow_forwardThe following data is available for three different alternatives. Assume an interest rate of 10% per year, compounded annually. Initial Cost Annual Benefit Useful Life (yrs) Alternative A 10,000 1,450 infinite OO Alternative B 11,000 1,930 10 Alternatives B and C are replaced at the end of their useful lives with identical replacements. Using present worth analysis, find the best alternative. Choose Alternative A because its net present worth is postive Choose Alternative A because it lasts the longest Alternative 20,000 6,707 5 Choose Alternative C because its net present worth is $9,807.17 more than its nearest competitor Choose Alterntive C because it has the highest annual benefitarrow_forwardEvaluate the two alternatives A and B and decide the economic justified alternative using: Present worth method 3 Annual worth method, Future worth method E.R.R Method , E.R.R.R method I.R.R method M.A.R.R=15%, the details of alternatives are shown in the table below Alternatives A B Investments $120,000 $155,000 Useful life (years) 15 20 Annual disbursements $25,000 $35,000 Annual revenues $45,000 $60,000 Salvage values $25,000 $30,000arrow_forward
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