Compare alternatives A and B with the present worth method if the MARR is 14% per year. Which one wouldyou recommend? Assume repeatability and a study period of 12 years.Capital InvestmentOperating CostsA$50,000$6,000 at end of year 1 and increasingby $600 per year thereafterOverhaul CostsLifeSalvage Value$6,000 every 3 years12 years$12,000 if just overhauledB$25,000$12,000 at end of year 1 and increasingby $1,200 per year thereafterNone6 yearsnegligibleClick the icon to view the interest and annuity table for discrete compounding when the MARR is 14% perThe PW of Alternative A is S(Round to the nearest dollar.)The PW of Alternative B is $. (Round to the nearest dollar.)Alternative A should be selected.다.

To compare alternatives A and B using the Present Worth (PW) method with a Minimum Acceptable Rate…

Answered: Compare alternatives A and B with the…

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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Economics A patch of land adjacent to the proposed international airport is likely to increase in value. The cost of land now is INR 8,00,000 and is expected to be worth INR 15,00,000 within 5 years. During this period it can be rented for small scale industry at INR 15000 per year. Annual taxes are presently is INR 8500 and likely to remain constant. What rate of return will be earned on the investment if the estimates are accurate? Draw cash flow diagram.
Three independent alternatives are given below. If MARR is 18%, what is your decision? A B C Initial Cost $4.50 $1.90 $1.20 Annual Revenues $4.00 $2.50 $3.00 Salvage Value $0.50 $0.90 $0.00 Annual Operating & Maintenance Costs $1.20 $1.90 $2.70 Estimated life, in years 3 Infinite
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. A
Would love some help on how to approach this - thanks! 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
Conduct a ROR analysis for the above information. If MARR= 18%, which project should be chosen and what is its corresponding ROR Investment cost Annual maintenance cost Salvage value Project A 20,000 2,000 5,000 Useful life (years) 20 O Project B, ROR - 19.69% Project A, ROR=20.69% O Project B, ROR=20.69% Project A, ROR = 19.69% Project B 10,000 4,000 1,000 20
MARR is 8%
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 answer
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).
Required information The tabulation of the incremental cash flows between alternatives A and B is shown. Alternative A has a 3-year life and alternative B a 6-year life. Year Incremental Cash Flows (BA), $ -22,500 5,000 5,000 11,000 5,000 5,000 5,000 NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. 0123456 If neither alternative has a salvage value, what is the first cost of alternative A? The first cost of alternative A is $-
Select the best alternative through internal rate of return analysis. Assume that the minimum attractive rate of return is 10% per year. Machine X |Machine YInitial cost $ 400,000.00 $ 500,000.00Annual operating cost $ 15,000.00 |$ 25,000.00 Annual profit $ 20,000.00 |$ 40,000.00Residual value $ 85,000.00 |$ 100,000.00Project life 15 years | none Engineering Economic Analysis
Using Incremental with EUAW analysis find the best alternative, MARR = %10. should use Excel and show your equations seperately, see below example: [A Benefit - [IC (A/P, i%, n) - Salvage (A/F, i, n)] + A Cost+ G Cost (A/G, i, n)] Data B C A First Cost $2,300,000 $2,780,000 $2,540,000 Salvage Value $82,000 $118,000 $97,000 Annual Benefit $580,000 $670,000 $650,000 M&O $65,000 $78,000 $71,000 M&O Gradient $11,000 $15,000 $12,500 Useful Life, Years 9 9 9 You - ZOOM
Please no written by hand solution Two projects are shown below. Interest rate is 10% per year with yearly compounding. Project A Project B Project Life 10 years 10 years Benefits $500 at year = 0 and $100 per year (at year = 1,2,3,…,10) $15 per year starting year 2 (at year = 2,3,4,…,10) Disbenefits $40 per year (at year = 1,2,3,…,10) $1 per year starting year 2 (at year = 2,3,…,10) Costs $600 at year = 0, $100 at year = 1 $50 at year = 0 (1) Find the conventional B/C ratio for Project A. (2) Find the conventional B/C ratio for Project B. (3) Based on B/C analysis (using conventional B/C ratio), which between A and B is more attractive?

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