Given the data below for two alternatives, find the better alternative using incremental IRR ifMARR is 15%. Alternatives can be replaced at the end of their useful lives.Alt. AAlt. BInitial Cost$10,000$25,000Maintenance Cost / Year15001200Uniform Annual Benefit480012000Salvage Value16005000Life12 Years8 Years

Incremental internal rate of return, also known as the incremental internal rate of return, is a…

Answered: Given the data below for two…

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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M12
please answer all with complete solution. do not use excel please. thank you...
Hinson’s Homegrown Farms needs a new irrigation system. System one will cost $145,000, have annual maintenance costs of $10,000, and need an overhaul at the end of year six costing $30,000. System two will have first-year maintenance costs of $5000 with increases of $500 each year thereafter. System two would not require an overhaul. Both systems will have no salvage value after 12 years. If Hinson’s cost of capital is 4%, using annual worth analysis determine the maximum Hinson’s should be willing to pay for system two.
Consider the four mutually exclusive alternatives. Each alternative has 10 years useful life and no salvage value. Which alternative should be selected based on the payback period. Note: You don't have to conside time value of money (i%). Cost Uniform Annual Benefit A $25 $4.8 3 $60 $16.8 C $90 $23.8 Focus D $120 $33.4
Fill in the missing values: QP TC TB Tprofit MC MB Mprofit ATC AFC AVC 16 500-50 ΝΑ ΝΑ ΝΑ ΝΑ ΝΑ ΝΑ 16 132 16 16 156 16 186 0 5 6 7 8 16 9 16 272 10 16 332 16 224
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Two mutually exclusive alternatives are available. If the interest is 15%, select the best alternative according to the VP method. The cash flow profiles for the alternatives are as follows. draw the diagram for those alternatives
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3
Use Present Worth Analysis to determine whether Alternative A or B should be chosen. Items are identically replaced at the end of their useful lives. Assume an interest rate of 6% per year, compounded annually. Alternative A Alternative B Initial Cost Annual Benefit Salvage Value Useful Life (yrs) 400 40 170 2 1,000 48 257 3 O Alternative A, because it costs $736.29 less than Alternative B, in terms of present worth Alternative B, because it costs $736.29 more than Alternative A, in terms of present worth Alternative A, because its present worth is positive Alternative B, because it only incurs the initial cost once every three years instead of every two years
Littrell's Nursery needs a new irrigation system. System one will cost $145,000, have annual maintenance costs of $10,000, and need an overhaul at the end of year six costing $30,000. System two will have first year maintenance costs of $5,000 with increases of $500 each year thereafter. System two would not require an overhaul. Both systems will have no salvage value after 12 years. If Littrell's cost of capital is 4%, using annual worth analysis determine the maximum Littrell's should be willing to pay for system two.
27) For the following alternatives, (a) find the maximum cost you can spend on alternative A to break even with alternative B if the interest rate is 10% per year. (b) repeat A following expected lives, 5, 8, 10, 12, 15 years using Spreadsheet and Goal seek tool.

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