Helwan University, Faculty of Engineering El-Mataria, 2013-2014 (Second Term) 2nd year Automotive and Locomotive Engineering 4th year Mechanical Power Engineering Engineering Economy and Management – Assignment # 3 Dr. Araby Ibrahim 1) Handheld fiber-optic meters with white light polarization interferometry are useful for measuring temperature, pressure, and strain in electrically noisy environments. The fixed costs associated with manufacturing are $800,000 per year. If a base unit sells for $2950 and its variable cost is $2075, (a) how many units must be sold each year for breakeven ( analytically and graphically) (b) what will the profit be for sales of 3000 units per year? Solution (a) XBE = the break even quantity (units /year) r = selling price per unit v = variable cost per unit XBE = F / (r – v) = 800,000 / (2950 – 2075) = 914 units per year (b) Profit (or loss) = TR – TC = r X – (F+ v X) = (2950 * 3000) – (800,000 + 2075 *3000) = $1,825,000 per year profit since it is positive (b) What will the profit be for sales of 7000 gallons per month? Solution The same as number 1 3) Consider the accompanying breakeven graph for an investment. (a) Determine the monthly sales volume needed to break even (analytically and graphically).To Draw X 0 1000 2000 TR 0 2950000 5900000 TC 800000 2875000 4950000 From the graph: Xbe = 914 2) Nicholea Water LLC dispenses its product Nature’s Pure Water via vending machines with most current locations at food markets and pharmacy or chemist stores.30. and answer the following questions as they pertain to the graph. The average monthly fixed cost per site is $900.18 to purify and sells for $0. . while each gallon costs $0. 000 From the graph F = 100.400000 350000 Total Revenue 300000 250000 Dollars 200000 150000 100000 50000 0 0 250 500 750 1000 1250 1500 1750 2000 Output (Units / Year) a) b) c) d) Give the equation to describe total revenue for X units per year.000 per year profit since it is positive .000 = 100000 + 1000 * v v = 100$/unit TC = F + 100X c) XBE = F / (r – v) = 100.000 + 100 *1600) = $60.000 / (200 – 100) = 1000 units per year d) At X=1600 units /year Profit (or loss) = TR – TC = r X – (F+ v X) = (200 * 1600) – (100.000 Then TR = 200. What is the "breakeven" level of X in terms of costs and revenues? If you sell 1600 units this year.000 TC = 200.000 = 1000*r r = 200 $/unit TR = 200 X b) At breakeven Xbe = 1000 units/year TC = TR = 200. Give the equation to describe total costs for X units per year. will you have a profit or loss? How much? Solution a) At breakeven Xbe = 1000 units/year TC = TR = 200. A high-use component (expected usage is 5000 units per year) can be purchased for $25 per unit with delivery promised within a week.000 4 150. and they have the following fixed and variable costs: Alternative Annual Fixed Cost ($) Variable Cost per unit ($) $120. if equipment costing $150. Over what production volume would each design (A or B or C) be chosen? Solve by drawing 6) Samsung Electronics is trying to reduce supply chain risk by making more responsible make-buy decisions through improved cost estimation.000 600.000 is purchased. Three alternatives have been identified.5 Q 10. Each machine design has unique total costs (fixed and variable) based on the annual production rate of boxes of these crackers.5 Determine the ranges of production (units produced per year) over which each alternative would be recommended for implementation by Quatro Hermanas A B C Solve by drawing 5) Three alternative designs have been created by Snakisco engineers for a new machine that spreads cheese between the crackers in a Snakisco snack. Labor and other operating costs are estimated to be $35. Salvage is estimated at 10% of first cost and i =12% per year.000 5. Neglect the element of availability (a) to determine the breakeven quantity and (b) to recommend making or buying at the expected usage level.000 boxes of crackers per year. Inc.000 Annual Variable Cost ($) 20.000 350. Design A B C Fixed Cost ($) 100. Samsung can make the component in-house and have it readily available at a cost of $5 per unit. Management is interested in the production interval of 0-150.5 Q 8Q Graphically.000 $17 220. . is investigating implementing some new production machinery as part of its operations. Alternatively. The costs for the three designs are given (where Q is the annual production rate of boxes of cheese crackers). determine which of the machine designs would be recommended for different levels of annual production of boxes of snack crackers.4) Quatro Hermanas.000 per year over the study period of 5 years. 500(A/P.500(0.Solution (a) Solve the relation AWbuy = AWmake for Q = number of units per year. the two workers can prepare 0.04 mile of ditch in 1 hour. causing the water to heat up. are expected to be $18.15 mile of ditch can be prepared.000(A/P. The single-pass system. Determine the number of miles of ditch per year the contractor would have to service for the two options to break even.5) + 15. license.5) – 35. -25Q = -150. select the make option. 7) An irrigation canal contractor wants to determine whether he should purchase a used Caterpillar mini excavator or a Toro powered rotary tiller for servicing irrigation ditches in an agricultural area of California.5) = -1200(0.000 – 5Q -20Q = -150. The water can be cooled using one of two systems: a single-pass heat exchanger or a closed-loop heat exchange system.10%.12%.748 – 106.000(0.500 with a $9000 salvage value after 10 years. The tiller costs $1200 and has a useful life of 5 years with no salvage value.67x = -316.20]/0. requires .10%.000(A/F.15 = $106. The contractor’s MARR is 10% per year.26380) = $-316.10%.250/-20 = 3713 units per year (b) Since 5000 > 3713.06275) = $-21.000(0.04 = $580 per mile FCexcavator = -26.20 per hour.56 – 580x x = 45. The initial cost of the excavator is $26.12%.000 + 9.56 per year Equate the AW relations and let x = breakeven miles per year -21.748 per year FCtiller = -1200(A/P.10) – 18. Solution VCexcavator = (15 + 1)/0. good for 3 years. Fixed costs for insurance.3 miles per year 8) An effective method to recover water used for regeneration of ion exchange resins is to use a reverse osmosis system in a batch treatment mode.16275) – 18.000 Q = -74.10) = -26. 0. It has the smaller slope of 5 versus 25 for the buy option. Alternatively. In 1 hour. the contractor can purchase a tiller and hire 2 workers at $11 per hour each.000(0.000 per year.000(A/F. Its operating cost is expected to be $1.15741) – 35. Such a system involves recirculation of the partially treated water back into the feed tank.27741) + 15. and with the tiller. etc.67 per mile VCtiller = [2(11) + 1.000 + 9. The excavator will require one operator at $15 per hour and maintenance at $1 per hour. Solution -(920 + 360)(A/P. valves.3) – 3.10 per hour. will have a useful life of 5 years.28 per hour to operate.10%. and will cost $1. etc. electricity.28x 1.93 x = 275 hours per year . The closed-loop system will cost $3850 to buy.5) – 1. What is the minimum number of hours per year that the cooling system must be used in order to justify purchase of the closed-loop system? The MARR is 10% per year.26380) – 1.a small chiller costing $920 plus stainless steel tubing.40211) – 3. costing $360.10x = -3850(A/P.10%. etc. connectors. treatment charges.82x = 500. and the salvage values are negligible.28x -1280(0.10x = -3850(0. The cost of water. will be $3.