1. Two alternative structures are being considered for a specific service. Compare the present worth of the cost of 24 years of service, using an interest of 5%, when neither structure has a realizable salvage value. The pertinent data are as follows: Alternative First cost Estimated life Structure I $4,500 12 years $1,000 Structure II $10,000 24 years $720 Annual disbursements

1. Two alternative structures are being considered for a specific service. Compare the present worth of the cost of 24 years of service, using an interest of 5%, when neither structure has a realizable salvage value. The pertinent data are as follows: Alternative First cost Estimated life Structure I $4,500 12 years $1,000 Structure II $10,000 24 years $720 Annual disbursements

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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I need to solve the question in handwriting and in clarity please The following are the cost of a system and the useful life is five years. The salvage value for depreciation purpose is equal to 25% of the hardware cost Item Hardware Training Installation Cost $170,000 $18,000 $18,000 a) What is the book value of the system at the end of year three if Straight line method is used? b) Suppose that after depreciation of the system for two years with straight line, the company decides to switch to the decline method for the reminder of the system life (depreciation rate 30%). What is the system Book Value at the end of four years?
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
An asset has the estimated salvage values for various lives, shown in the table below. For each possible life from 1 to 6 by 1, determine the capital recovery cost for MARR of 8%/year. EOY 0 1 2 3 4 5 6 NCF -$150,000 $30,000 $30,000 $30,000 $30,000 $30,000 $30,000 Estimated SV $150,000 $130,000 $110,000 $100,000 $90,000 $85,000 $82,000 Capital recovery cost with 1-year life $ Capital recovery cost with 2 years life $ Capital recovery cost with 3 years life $ Capital recovery cost with 4 years life $ Capital recovery cost with 5 years life $ Capital recovery cost with 6 years life $ Carry all interim calculations to 5 decimal places and then round your final answer to the nearest dollar. The tolerance is ±50.
Consider the following data on an asset: • Cost of the asset = $38,000 • Useful life = 6 years • Salvage value = $5,000 Compute the annual depreciation allowances and the resulting book values using a 150% declining balance method. Make sure that the last year's book value equals to the salvage value.
Compute the EUAC for Pump A. Ideally this value was determined to make the decision of which pump to select. Pump A, with a useful life of 6 years: Initial cost = $3500 & Salvage value at end of life of $500 Enter the EUAC for Pump A.
Consider a piece of property, plant, and equipment (PP&E) that costs $50,000, with an estimated useful life of 5 years and a $5000 salvage value. Calculate the depreciated amount after the end of useful life using the double-declining balance depreciation technique
Question Six: Two alternatives A and B are evaluated to select the most beneficial one. The details of them are shown in Figure (3). Figure (3) Alternatives Investments Useful life (years) Annual disbursements Annual revenues Salvage values Present Worth A B $60,000 $75,000 10 5 $25,000 $35,000 $45,000 $60,000 $5,000 $10,000 - $42,847 - $20,625
An existing asset that cost $16,000 twoyears ago has a market value of $12,000 today, anexpected salvage value of $2,000 at the end of itsremaining useful life of six more years, and annualoperating costs of $4,000. A new asset under consideration as a replacement has an initial cost of$10,000, an expected salvage value of $4,000 at theend of its economic life of three years, and annualoperating costs of $2,000. It is assumed that thisnew asset could be replaced by another one identical in every respect after three years at a salvagevalue of $4,000, if desired. Use a MARR of 11%,a six-year study period, and PW calculations todecide whether the existing asset should be replacedby the new one
Project A and project B have the same salvage value. The life of project A is 10 years while the life of project B is 8 years. Which project will yield the higher present value for the salvage value?a. Project Ab. Project Bc. Indeterminated. depends on the rate to be used
1- Which Project will be selected? 2- please attach your handwritten answer.
The following relates to a proposed equipment purchase: Cost $126,900 Salvage value $ 3,100 Estimated useful life 4 years Annual net cash flows $ 45,200 Depreciation method Straight-line The annual average investment amount used to calculate the accounting rate of return is: Multiple Choice $40,850 $63,450 $31,725 $61,900
A m a n h a s a n i n v e s t m e n t o f P 13, 760.00 i n c a s h f o r e q u i p m e n t . His estimates of the yearreceipts and disbursements for all purposes are shown below. He estimates that his equipment has a salvage value of P 2,000.00 at the end of its useful life.Find the rate of return of the investment. please don't use excel. answer should be 10.11%

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