Compare the alternatives C and D on the basis of a present worth analysis using an interest rate of 14% per year and a study period of10 years.AlternativeFirst CostAOC, per YearAnnual Increase in OperatingCost, per YearSalvage ValueLife, Yearsс$-48,000$-9,000$-1,400$8,00010The present worth of alternative C is $ -117974.14Alternative D offers the lower present worth.D$-34,000$-9,000$-1,500$1,4005and that of alternative D is $ -73129.34 x

Present worth helps to assess the current value of money received or paid in the future, considering…

Answered: Compare the alternatives C and D on 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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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
If a project is evaluated using annual worth method, then we know that: Select one: a. Present worth will always be smaller than annual worth b. Present worth will have the opposite sign of annual worth c. Present worth will have the same sign as annual worth d. Present worth will always be larger than annual 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.
Required information One of two methods must be used to produce expansion anchors. Method A costs $65.000 initially and will have a $14.000 salvage value after 3 years. The operating cost with this method will be $23,000 per year. Method B will have a first cost of $140,000, an operating cost of $14,000 per year, and a $32.000 salvage value after its 3-year life. The interest rate for both the methods is 15%. Which method should be used on the basis of a present worth analysis? The present worth of method A Is $ and that of method B is $ Method (Cick to solect) v is selected by the company.
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 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
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 show all steps with explanation to getting the answer! Thanks
Q4
Don't answer by pen paper and don't use Excel

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