For the mutually exclusive alternatives shown, the one(s) that should be selected are:(a) Only C(b) Only AAlternativePW, $A- 25,000(c) C and DB-12,000(d) Only D10,000(e) Any of the aboveD15,000abd

Only one alternative could be selected when the alternatives are being exclusive mutually as only…

Answered: For the mutually exclusive alternatives…

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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Problem 05.023 Alternative Comparison - Different Lives Compare the alternatives C and D on the basis of a present worth analysis using an interest rate of 10% per year and a study period of 10 years. Alternative First Cost AOC, per Year Annual Increase in Operating Cost, per Year Salvage Value Life, Years The present worth of alternative C is $ с $-40,000 $-5,000 $-1,000 $15,000 10 D $-30,000 $-7,000 $-1,100 $1,000 5 and that of alternative D is $ Alternative C offers the lower present worth.
Q1 Table Q1 shows the engineering project with minimum attractive rate of return (MARR) at 15% per year. Table Q1 Proposal A Investment Cost RM 10,000 Expected Life 5 years Market (Salvage Value) -RM 1,000 Annual Receipts RM 8,000 Annual Expenses RM 4,000 (a) Determine the present worth (PW), future worth (FW) and annual worth (AW) based on details given in Table Q1.
Given the following pertinent data for four alternatives, what is the best alternative using the incremental ROR method given MARR=10% Initial Cost Annual CF, $ Life, years 30 O Alt. A Alt. B O Alt. C A O Alt. D B +22,000 +35,000 с -200,000 -275,000 -190,000 -331,350 30 D +19500 +42,000 30 30
Solve the Engineering Economics Problem: Project Feasibility Indicator Using Annual Worth Analysis, which alternative is more economical? i=10% Method A First Cost: -Php250,000 Annual Cost: -Php 90,000 Salvage Value: Php 30,000 Life: 3years Method B First Cost: -Php130,000 Annual Cost: -Php 25,000 Salvage Value: Php150,000 Life: infinity
A dyeing and finishing plant is interested in acquiring a dyeing machine for the production of a new product. Three alternatives are being considered assummarized below. Which alternative should be recommended if the plant’sMARR (hurdle rate) is 15% per year using (a) the IRR method and (b) the annual worth method?
When faced with different alternatives, there are many criteria that can be used to identify the best one. What is the tangible criterion that is usually used in an engineering economic analysis?
What is the Present worth difference of Alternative A and B. Capital investment Annual Cash inflow Life in years Salvage value Interest is 10% per year O a. 1452 O b. 1234 OC. 998 O d. 1106 Alternative A 3500 1255 4 0 B 5000 1480 6 0
Compare three mutually exclusive alternatives on the basis of their capitalized costs at i =10% per year. Alternative |First cost, $ AOC, $/year Salvage value, $| Life, years B2 300,000 |-10,000 |70,000 A1 -50,000 C3 |-30,000 5,000 2 900,000| 3,000 |200,000
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
please answer all with complete solution and draw the cash flow diagram,,, do not use excel please. thank you....
What is FW of Design A, Design B, and Design C. Which alternative should be chosen. Do not round off in the solution.
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For the mutually exclusive alternatives shown, the one(s) that should be selected are: (a) Only C (b) Only A (c) C and D (d) Only D (e) Any of the above Alternative PW, $ A -25,000 -12,000 10,000 15,000 B C D a b d