The initial investment costs and the "annual revenues less expenses" of threeinvestment alternatives (X, Y and Z) are given below. The useful life of eachalternative is four years. They have no salvage value. The MARR is 10%.Capital investment ($)Annual revenues less expenses (S) 11,00036,50013,10030,00035,00012,000Find the best alternative bya) present worth analysisb) (incremental) internal rate of return analysis

business and for engineering economics in both industrial engineering and civil engineering…

The initial investment costs and the "annual revenues less expenses" of three investment alternatives (X, Y and Z) are given below. The useful life of each alternative is four years. They have no salvage value. The MARR is 10%. Y Capital investment (S) Annual revenues less expenses (S) 11,000 30,000 36,500 35,000 12,000 13,100 Find the best alternative by a) present worth analysis b) (incremental) internal rate of return analysis

The initial investment costs and the "annual revenues less expenses" of three investment alternatives (X, Y and Z) are given below. The useful life of each alternative is four years. They have no salvage value. The MARR is 10%. Y Capital investment (S) Annual revenues less expenses (S) 11,000 30,000 36,500 35,000 12,000 13,100 Find the best alternative by a) present worth analysis b) (incremental) internal rate of return analysis

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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Compare the alternatives shown on the basis of their capitalized costs using a MARR of 10% per year. Alternative First cost, $ Annual operating cost, $ per year Salvage value, $ Life, years M -155,000 -30,000 8,000 5 N -875,000 -11,000 1,000,000 00 The capitalized cost for alternative M is $- The best alternative selected on the basis of their capitalized costs is alternative M and the capitalized cost for alternative N is $-
b) National Homebuilders, Inc., plans to purchase new cut-and-finish equipment. The details of the 2 alternative options are summarized in the table below. The interest rate is 10% per year. First Cost, S Annual Operating Cost, S Salvage Value, S Equipment Life (years) Consider only machine A now. Alternative A 15,000 3,500 1,000 4 Alternative B 18,000 3,100 2,000 8 a) Calculate the AW of Machine A for one life cycle. b) Calculate the AW of Machine A for LCM = 8 years (2 life cycles). c) Demonstrate the equivalence of AW over 2 life cycles and AW over one life cycle.
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
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An economic evaluation of the following mutually exclusive alternative equipment is being conducted. Since the equipment have different service lives, repeatability is assumed using an analysis period of 30 years. The MARR is 6% per year. Capital investment Annual revenues generated Useful life Market (salvage) value The net Annual worth of Alternative B over the analysis perio A. $39,267 More Info Alternative A B $30,000 $90,000 $20,000 $30,000 10 years None 15 years None Discrete Compounding; 1=6% Single Payment OB. $23,462 Compound OC. $16,923 Amount Present Compound Amount Present Factor Worth Factor OD. $41,466 To Find F To Find P Factor To Find F Worth Factor Uniform Series Sinking Fund Factor Capital Recovery Factor To Find P OE. $20,733 Given P Given F Given A Given A Given F To Find A To Find A Given P N F/P P/F F/A P/A A/F A/P 1 1.0600 0.9434 1.0000 0.9434 1.0000 1.0600 The net Present worth of Alternative B over the analysis perio 2 1.1236 0.8900 2.0600 1.8334 0.4854 0.5454…
A consulting engineering firm is considering two models of SUVs for the company principals. A GM model will have a first cost of $36,000, an operating cost of $4000, and a salvage value of $15,000 after 3 years. A Ford model will have a first cost of $32,000, an operating cost of $3100, and also have a $15,000 resale value, but after 4 years. (a) At an interest rate of 15% per year, which model should the consulting firm buy? Conduct an annual worth analysis. (b) What are the PW values for each vehicle?
Compare the alternatives C and D on the basis of a present worth analysis using an interest rate of 15.00% per year and a study period of 10 years. (Include a minus sign if necessary.) 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 $-50000 $-8000 $-1500 $14000 10 $-21000 $-9000 $-200 $1500 5 and that of alternative D is $
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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 $-
A consulting engineering firm is considering two models of SUVs for the company principals. A GM model will have a first cost of $36,000, an operating cost of $4000, and a salvage value of $15,000 after 3 years. A Ford model will have a first cost of $32,000, an operating cost of $3100, and also have a $15,000 resale value, but after 4 years. (a) At an interest rate of 15% per year, which model should the consulting firm buy? Conduct an annual worth analysis. (b) What are the PW values for each vehicle? 10 16 -22)