Snip & Sketch ▼ ♥ Sales price Equipment cost Overhead cost Operating and maintenance cost Production time per 1,000 units Study period (planning horizon) MARR New V 5 C 14 $12.50 per unit $200,000 $50,000 per year $25 per operating hour 100 hours 5 years 15% per year Q D 16 X An electronics firm is planning to manufacture a new handheld gaming device for the preteen market. The data have been estimated for the product. Assuming a negligible market (salvag value for the equipment at the end of five years, determine the breakeven annual sales volume for this product. Click the icon to view the data estimated for the product. Click the icon to view the interest and annuity table for discrete compounding when i= 15% per year. ... The breakeven annual sales volume is units. (Round to the

Snip & Sketch ▼ ♥ Sales price Equipment cost Overhead cost Operating and maintenance cost Production time per 1,000 units Study period (planning horizon) MARR New V 5 C 14 $12.50 per unit $200,000 $50,000 per year $25 per operating hour 100 hours 5 years 15% per year Q D 16 X An electronics firm is planning to manufacture a new handheld gaming device for the preteen market. The data have been estimated for the product. Assuming a negligible market (salvag value for the equipment at the end of five years, determine the breakeven annual sales volume for this product. Click the icon to view the data estimated for the product. Click the icon to view the interest and annuity table for discrete compounding when i= 15% per year. ... The breakeven annual sales volume is units. (Round to the

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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Following cash flows for alternatives X and Y at an interest rate of 10% per year. Machine X Machine Y -146,000 -220,000 Initial cost, $ AOC, $/year Annual revenue, $/year Salvage value, $ Life, years -15,000 80,000 10,000 3 -10,000 75,000 25,000 6 In comparing the alternatives on a present worth basis, the PW of Machine X is closest to
Select the best alternative through internal rate of return analysis. Assume that the minimum attractive rate of return is 10% per year. Machine X |Machine YInitial cost $ 400,000.00 $ 500,000.00Annual operating cost $ 15,000.00 |$ 25,000.00 Annual profit $ 20,000.00 |$ 40,000.00Residual value $ 85,000.00 |$ 100,000.00Project life 15 years | none Engineering Economic Analysis
Compare the alternatives C and D on the basis of a present worth analysis using an interest rate of 14% per year and a study period of 10 years. Alternative First Cost AOC, per Year Annual Increase in Operating Cost, per Year Salvage Value Life, Years с $-48,000 $-9,000 $-1,400 $8,000 10 The present worth of alternative C is $ -117974.14 Alternative D offers the lower present worth. D $-34,000 $-9,000 $-1,500 $1,400 5 and that of alternative D is $ -73129.34 x
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 $-
ENGINEERING ECONOMY RATE WILL BE GIVEN. WRITE THE COMPLETE SOLUTIONS/EXPLANATION LEGIBLY OR TYPEWRITTEN. GIVE STRAIGHT TO THE POINT EXPLANATION. A tire manufacturer produces tires at a variable cost of $25 per unit. The plant has annual sales of $480,000 at a price of $40 per unit. Find the breakeven point in units of production. The annual fixed costs for the plant are $150,000.
Q1 Table Q1 shows the engineering project with minimum attractive rate of return (MARR) at 15% per year. Table Q1 Proposal A Investment Cost RM 10,000 Expected Life 5 years Market (Salvage Value) -RM 1,000 Annual Receipts RM 8,000 Annual Expenses RM 4,000 (a) Determine the present worth (PW), future worth (FW) and annual worth (AW) based on details given in Table Q1.
Solve the Engineering Economics Problem: Project Feasibility Indicator Using Annual Worth Analysis, which alternative is more economical? i=10% Method A First Cost: -Php250,000 Annual Cost: -Php 90,000 Salvage Value: Php 30,000 Life: 3years Method B First Cost: -Php130,000 Annual Cost: -Php 25,000 Salvage Value: Php150,000 Life: infinity
Complete the following analysis of cost alternatives and select the preferred alternative. The s A D Сapital investment $15,000 S16,000 $13,000 $18,000 Annual costs Market value 250 300 500 100 1,000 1,300 1,750 2,000 at EOY 10 FW(12%) -$49,975 –$53,658 ??? –$55,660 tudy period is 10 years and the MARR = 12% per %3D year.
Seawater contains 2.1 pounds of magnesium per ton. By using the processingmethod A, 85% of the metal can be recovered at a cost of 3.25 per ton of water pumped andprocessed. If process B is used, 70% of the available metal is recovered, at a cost of only 2.60per ton of water pumped and processed. The two processes are substantially equal as toinvestment costs and time requirements. (a) If the extracted metal can be sold for 2.40 perpound, which processing method should be used? (b) At what selling price for the metal wouldbe two processes be equally economical?Answer with complete solution brief explanation. Thank you.
Wall’s Pharmacy will have to sell a new product that has an estimated revenue of $5,100 per month and costs of $1,000 per month with an initial purchase of $28,000. How long will Wall's Pharmacy have to sell a new product if the MARR is 3% per month? Wall's Pharmacy will have to sell a new product for __ months.
CASE STUDY ANNUAL WORTH ANALYSIS THEN AND NOW Background and Information Harry, owner of an automobile battery distributorship in Atlanta, Georgia, performed an economic analysis 3 years ago when he decided to place surge protectors in-line for all his major pieces of testing equipment. The estimates used and the annual worth analysis at MARR = 15% are summarized below. Two different manufacturers' protectors were compared. Cost and installation, $ Annual maintenance cost, S per year Salvage value, $ Equipment repair savings. $ Useful life, years. O 1 2 3 4 5 6 7 8 9 10 A MARR= Year 0 1 234567 11 12 13 14 15 16 17 AW element 18 Total AW 899 10 PowrUp -26,000 -800 2,000 25.000 B 15% Investment and salvage -26,000 ...... 2,000 -6,642 6 Lloyd's -36,000 -300 C 3,000 35.000 10 PowrUp Annual maintenance 0 -800 -800 -800 -800 -800 -800 -800 D Case Study Exercises 1. Plot a graph of the newly estimated maintenance costs and repair savings projections, assuming the protectors last for seven…
A contractor can buy trucks for 800 000 usd each or rent them for 1,200 usd per truck per day. The truck has a salvage value of 100, 000 usd at the end of its useful life for 5 years. The annual maintenance cost is 20 000 usd per truck. Using the annual worth method and 14 % interest rate, determine the minimum number of days per year that each truck must be used to warrant its purchase. Show full solution tnx. xx engineering econ