For the following table, if the MARR is 10% per year and a useful life for each alternativeof six years that equals the study period. The rank order of alternatives from least capitalinvestment to greatest capital investment isDo nothingFill the table with the missing values.> ACвDo nothing> A? (2)? (4)A capital investmentA annual revenuesA annual costsA market valueA IRR-15,000-2000-30004,000900460-150-1,0006,000-100-2,2203,350? (5)(6)12.7%10.9%Alternative chosen(1)(3)1. write the alternative chosen2. write the two alternatives compared3. write the alternative chosen4. write the two alternatives compared5. what is IRR incremental (final answer)6. write the alternative chosen

The internal rate of return procedure is one among the fluctuated strategies accessible for…

Answered: For the following table, if the MARR is…

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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Given the following information about possible investments and a MARR of 15% per year. Alternative A Investment Cost Annual Benefits Useful Life P250,000 P60,000 5 years a.) Which alternative is preferred, based on the LCM method? b.) Which alternative is preferred using a five-year study period. Alternative B P400,000 P40,000 P15 years
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
A company plans to finance a large project that will last 5 years. The company will borrow $480,000 now to finance the project. How much will the company have to receive in annual payments for 5 years, provided it will also receive a final lump sum payment after 10 years in the amount of $50,000. The MARR is 20% per year on this investment. Your Answer: Answer units
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
If the MARR=10%, compute the value of X that makes the alternatives equally desirable. Do not use spreadsheets. Alternatives Machine A Machine B First cost $12,000 $20,000 Annual Operating cost $1,400/year X Salvage value $2,000 $3,000 Life 4 years 8 years
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?
Please help.
Complete the following analysis of cost alternatives and select the preferred alternative. The s A D Сapital investment $15,000 S16,000 $13,000 $18,000 Annual costs Market value 250 300 500 100 1,000 1,300 1,750 2,000 at EOY 10 FW(12%) -$49,975 –$53,658 ??? –$55,660 tudy period is 10 years and the MARR = 12% per %3D year.
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
Q4

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