Compare three alternatives on the basis of their capitalized costs at /= 11.00% per year and select the best alternative. (Include a minussign if necessary.)AlternativeFirst CostAOC, per YearSalvage ValueLife, YearsE$75000$-55000$190002The capitalized cost of alternative E is $The best alternative is (Click to select)F$-315000$16000$690004. alternative F is $G$ 815000$-4000$40000090and alternative G is $

Capitalized cost is an expenses that is added to the cost basis of the fixed asset on a company…

Answered: Compare three alternatives on the basis…

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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:[A] { > Incremental analysis ([ B Alternative], [B wins ]): C A company considering 2 different machines at MARR at 12% Both life spans = 10 years Initial Cost Annual Operating lost Benefits per yin ar Salvage Value \table MM If you are and of frying investment to company decide More than 2 alternatives if the additional increment is worth while, compare Alternative: A Incremental analysis (Alternative) pairs then B C A MARR Company at Considering 2 different machines at 12% Both life spans = 10 years. M/C X м/с у Initial Cost 160000 285000 Annual Operating Cost Benefits per year Salvage Value 45 000 90000 45000 105000 20000 40000
(a) Reference the cash flows shown below. Pick the best alternative using the ERR method. MARR = 15%. e= 8%. Alternative EOY A C -$ 300 -$ 200 -$ 250 1 -10 -60 -90 2 -10 -60 -90 3 -10 -60 -90
Calculate the total capital cost and the annual worth for the following cash flow diagram (the time is infinity) 2 4 8 10 12 14 20 26 Year Infinity $2000 $3000 S50,000 $1500 $1500 $1500 Only one рayment Cost repeated every 7 years i = 5% per year $90,000
What is the best alternative using incremental Analysis? Use MARR = 15% A B C Capital Investment $ 2,000 7,000 4,200 Annual Revenues 3,200 8,000 6,000 Аппиal Costs 2, 100 5, 100 4,000 Market Value at the end of useful life 100 600 420 Useful Life (in years) 10 10 10 The correct ranking of Alternative is Blank 1 Select Alternative Blank 2 Note: Do not put comma, unit of measure and limit your answer to two decimal places. Ex: A-B-C
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 The better alternative between the second increment (B – C) is:. Group of answer choices Alt. B Alt. C Alt. B or Alt. C Alt. A
Compare the alternatives C and D on the basis of a present worth analysis using an interest rate of 13% per year and a study period of 10 years. Alternative C $-46,000 $4,000 $-1,300 $6,000 10 First Cost AOC, per Year Annual Increase in Operating Cost, per Year Salvage Value Life. Years The present worth of alternative C is $1 (Click to select) offers the lower present worth. HELIME BIEN D $-20,000 $-7,500 $-100 $1,300 5 and that of alternative D is $ 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
O b. 3425 O c. 2369 O d. 3825 Clear my choice Alternative X has a first cost of 33000 an annuàl operating cost of 6300 , and a'salvage value of 9525 after 2 year. Alternative Y has a first cost of 34000 an annual operating cost of 6400, and a salvage value of 13400 after 2 year. If MARR of 2% per year, approximately what is the PW of each alternative? Select one: O a. PW of X= -36438 and PW of Y= -33881 O b. PW of X= -36802 and PW of Y= -34220 O c. PW of X= -36077 and PW of Y= -33546 O d. PW of X= -37170 and PW of Y= -34562 Next pa US ACTIVITY 2:23 AM Ca ENG 4/21/202 hp
A Cost $4000 $2000 $6000 $1000 $9000 Annual benefit 639 410 761 117 750 Useful life, in years 20 20 20 20 20 i* 15% 20% 11.1% 9.9% 5.45% Perform incremental ROR analysis to select best one. MARR=6%; (P/A, 6%, 20) = 11.47 %3D
The following data is available for three different alternatives. Assume an interest rate of 2% per year, compounded annually. Alternative A Alternative B 7,000 6,600 975 2,148 infinite 10 Initial Cost Annual Benefit Useful Life (yrs) Alternative C 14,000 6,027 5 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 it lasts the longest Choose Alternative A because its net present worth is postive O Choose Alternative C because its net present worth is $82,185.21 more than its nearest competitor Choose Alterntive C because it has the highest annual benefit
Street lighting fixtures and their sodium vapor bulbs for a two-block area of a large city need to be installed at a first cost (investment cost) of $130,000. Annual maintenance expenses are expected to be$6,500for the first8years and$8,500 each year thereafter upto 25years. With an interest rate of 9% per year, what is the present worth cost of this project? Choose the closest answer below. A) $202,422 B $238,597 C) $249,468 D) $277,339
Solve this problem in Excel. The following two alternatives are mutually exclusive. (a) Which one will you select if your MARR is 15% per year? Use rate of return analysis. (b) Graph interest rate vs. present value for both option 1 and 2. ( c) Discuss how your selection between option 1 and 2 will change when MARR is changed and at which interest rate the two options are the same economically. Alternative Initial Cost Annual Benefit Life, years Ordinary (O) $38,000 $9,400 5 Above average (AA) $35,000 $9,000 6

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