Question#5: An engineer needs to decide on whether it is economically feasible to buy an equipment or not. Use the present worth method to determine the maximum purchasing cost of the equipment for it to be economically feasible if the annual MARR=15% Item Purchasing Cost (S) Annual Revenue (S) Operating Costs ($) Salvage Value (S) Useful Life (years) Value to be determined 6,000 2,000 in the first year, increasing annually by 10% thereafter (i.e. 2,200 in the 2nd year, 2,420 in the 3rd year, 2,662 in the 4th year.... etc.) 5,000 10

Question#5: An engineer needs to decide on whether it is economically feasible to buy an equipment or not. Use the present worth method to determine the maximum purchasing cost of the equipment for it to be economically feasible if the annual MARR=15% Item Purchasing Cost (S) Annual Revenue (S) Operating Costs ($) Salvage Value (S) Useful Life (years) Value to be determined 6,000 2,000 in the first year, increasing annually by 10% thereafter (i.e. 2,200 in the 2nd year, 2,420 in the 3rd year, 2,662 in the 4th year.... etc.) 5,000 10

Managerial Accounting

15th Edition

ISBN:9781337912020

Author:Carl Warren, Ph.d. Cma William B. Tayler

Publisher:Carl Warren, Ph.d. Cma William B. Tayler

Chapter12: Capital Investment Analysis

Section: Chapter Questions

Problem 7E

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From the given values below, which machine should be selected using (a) Present worth Method and (b) Future worth Method if interest rate is 10% per year? Machine A B Initial Cost (Php) Annual Operating Cost 146,000 15,000 80,000 10,000 3 220,000 10,000 75,000 25,000 Annual Revenue Salvage Value Useful Life (years) 6
From the given values below, which machine should be selected using (a) Present worth Method and (b) Future worth Method if interest rate is 10% per year? Machine Initial Cost (Php) Annual Operating Cost A В 146,000 15,000 220,000 10,000 75,000 25,000 Annual Revenue 80,000 10,000 Salvage Value Useful Life (years) 6
1. Two alternatives are being considered for installation. Which should be selected based on a present worth comparison at an interest rate of 6% per year. Alternative A Alternative B First Cost, $ 17,650 22,900 Maintenance Cost, $/year 1700 1400 Salvage value, $ 10,500 17,000 Life, years 4 3
You have been asked to evaluate two alternatives, X and Y, that may increase plant capacity for manufacturing high-pressure hydraulic hoses. The parameters associated with each alternative have been estimated. Which one should be selected on the basis of a present worth comparison at an interest rate of 13% per year? Why is yours the correct choice? Alternative First Cost X $-45,000 Maintenance cost, per $-9000 Year Salvage Value $1,000 Life 5 years Y $-55,000 $-4000 $6,500 5 years The present worth of alternative X is $ 13888 and that of alternative Y is $ 44459.4 Alternative X is selected by the company.
Reference: Case Study S Dunn Manufacturing is considering the following two alternatives. The cost information for the two proposals for replacing an equipment are provided are in table below. Initial cost Benefits/year Machine X $120,000 $20,000 for the first 10 years and $9,000 for the next 10 years Life Salvage value $40,000 MARR 5.2. The NPW of machine X is A) $35,158 B) $48,192 C) $50,752 Machine Y $96,000 $12,000 per year for 20 years. 20 years 8% $20,000
Solve and show the solution.
1
A Company is considering two alternatives with regards to an equipment which it needs. The alternatives are as follows: Alternative A; Purchase Cost of equipment Salvage value Daily operating cost Economic life, years P700,000 100,000 500 10 Alternative B: Rental at P1,500 per day At 18% interest, how many days per year must the equipment be in use if Alternative A is to be chosen.
You have been asked to evaluate two alternatives, X and Y, that may increase plant capacity for manufacturing high-pressure hydraulic hoses. The parameters associated with each alternative have been estimated. Which one should be selected on the basis of a present worth comparison at an interest rate of 10% per year? Why is yours the correct choice? Alternative First Cost Maintenance cost, per Year Salvage Value Life X $-25,000 $-8000 $1,000 5 years Y $-55,000 $-2000 $2,000 5 years and that of alternative Y is $1 The present worth of alternative X is $. Alternativ (Click to select) is selected by the company.
2. You are asked to evaluate two alternatives listed in the table below based on their net present value using an interest rate of 9%. Altern. Initial Cost Annual O&M Overhaul Cost (year 8) Salvage Value Life (years) A B $150,000 $350,000 $38,000 $18,000 $85,000 $40,000 0 4 12
QUESTION 3 If the first cost of an asset is $15,000, and an annual operating cost of $5000 with a salvage value of $3000 after 5 years. What will be the Annual worth and the Capital recovery. If i- 10% O AW -S-8465 and CR - $+4508 O AW = S-8465 and CR = $-4508 AW = S+4508 and CR = $-8465 O AW S-4508 and CR S-8465
2. It is rëquired to select one of the three machines. Which machine should be selected, based on the present worth comparison at an interest rate of 12% per year, if the costs are shown below? Туре X 7,650 1,200 Туре Y 12,900 900 2,000 Туре Z 17,000 750 First cost, $ Maintenance cost,.( $ lyear) Salvage value, ( $) Economic (years) 4,000 12 4