Compare the alternatives shown on the basis of their capitalized costs using an interest rate of 10% per yearquarterly. Alternative M Alternative N First cost, P 150,000 800,000 Annual operating cost, P per year 50,000 Salvage value, Life, years 10,000 1,000,000 8,000 25

Compare the alternatives shown on the basis of their capitalized costs using an interest rate of 10% per yearquarterly. Alternative M Alternative N First cost, P 150,000 800,000 Annual operating cost, P per year 50,000 Salvage value, Life, years 10,000 1,000,000 8,000 25

Essentials Of Investments

11th Edition

ISBN:9781260013924

Author:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.

Publisher:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.

Chapter1: Investments: Background And Issues

Section: Chapter Questions

Problem 1PS

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Question

Compare the alternatives shown on the basis of their capitalized costs using an interest rate of 10% per
yearquarterly.
Alternative M
Alternative N
First cost, P
Annual operating cost, P per year 50,000
Salvage value,
Life, years
150,000
800,000
10,000
1,000,000
8,000
5
25

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Can someone please solve this question using hand calculations only? PLEASE AND THANK YOU!!!
24
The equivalent annual worth of alternative A over an infinite time period is closest to:a. $-25,000b. $-27,200c. $-31,600d. $-37,100
The equivalent annual worth of alternative A is closest to:a. $-25,130b. $-37,100c. $-41,500d. $-42,900
Two mutually exclusive alternatives have the estimates shown below. Use annual worth analysis to determine which should be selected at an interest rate of 10% per year. Q $-44,000 R $-84,000 First Cost AOC per Year Salvage Value $-10,000 $-5,000 in year 1, increasing by $1,000 per year thereafter $4,000 $5,000 Life 2 years 4 years Alternative (Click to select) should be selected.
expense of $500. Consider a yearly Two machines are evaluated for a new project, consider a yearly interest rate of 6% and the information described in the following table. Determine EUAC for both alternatives and determine the best option. Purchase cost Salvage value Duration Machine A Machine B $10,000 $15,000 $1,250 $1,950 5 years 10 years 11 13 L
Compare three alternatives on the basis of their capitalized costs at /= 8% per year and select the best alternative. Alternative E F G $-460,000 $-30,000 $90,000 4 First Cost AOC, per Year Salvage Value Life, Years $-80,000 $-65,000 $40,000 2 The capitalized cost of alternative E is $ The best alternative is (Click to select) ▼ alternative F is $ $-960,000 $-3,000 $450,000 ∞0 and alternative G is $
Compare three alternatives on the basis of their capitalized costs at i = 13% per year and select the best alternative. Alternative E F G First Cost $-85,000 $-325,000 $-825,000 AOC, per Year $-65,000 $-20,000 $-6,000 Salvage Value $25,000 $75,000 $500,000 Life, Years 2 4 ∞ The capitalized cost of alternative E is $ , alternative F is $ , and alternative G is $ . The best alternative is E G F .
Two investment opportunities are as follows: A B First cost $150 $115 Uniform annual benefit $26 $20 End-of-useful-life salvage value $O SO Useful life, in years 10 10 If the MARR is 13%, which alternative should be selected?
Compare the machines below using present worth analysis at i10% per year and find which one should be selected Machine Y First Cost Operating Cost Savage Lfe cycle Machine X 20.000 3.000 3.000 3years Firat Cost 30.000 .000 Annuel Operating Cost Salvage Life cycle 3.000 6 yoars
Consider the following alternatives below.Based on a 7% interest rate, assuming both alternatives perform the same task and there is a continuing requirement, which alternative should be selected and what is its equivalent uniform annual cost?
A bridge that was constructed at a cost of P 75,000 is expected to last 30 years, at the end of which time its renewal cost will be P 40,000. Annual repairs and maintenance are P 3,000. What is the capitalized cost of the bridge at an interest of 6%? 133,433.00 50,000.00 125,000.00 140,625.00 75,948.67 58,433.00

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