简化版英文运筹学案例

一、LP1.Example2.3-1(UrbanRenewalModel)ThecityofErstvilleisfacedwithaseverebudgetshortage.Seekingalong-termsolution,thecitycouncilvotestoimprovethetaxbasebycondemninganinner-cityhousingareaandreplacingitwithamoderndevelopment.Theprojectinvolvestwophases:(1)demolishingsubstandardhousestoprovidelandforthenewdevelopment,and(2)buildingthenewdevelopment.Thefollowingisasummaryofthesituation.1.Asmanyas300substandardhousescanbedemolished.Eachhouseoccupiesa.25-acrelot.Thecostofdemolishingacondemnedhouseis$2000.2.Lotsizesfornewsingle-,double-,triple-,andquadruple-familyhomes(units)are.18,.28,.4,and.5acre,respectively.Streets,openspace,andutilityeasementsaccountfor15%ofavailableacreage.3.Inthenewdevelopmentthetripleandquadrupleunitesaccountforatleast25%ofthetotal.Singleunitsmustbeatleast20%ofallunitsanddoubleunitsatleast10%.4.Thetaxleviedperunitforsingle,double,triple,andquadrupleunitsis$1000,$1900,$2700,and$3400,respectively.5.Theconstructioncostperunitforsingle-,double-,triple-,andquadruple-familyhomesis$50,000,$70,000,$130,000,and$160,000,respectively.Financingthroughalocalbankcanamounttoamaximumof$15million.Howmanyunitsofeachtypeshouldbeconstructedtomaximizetaxcollection?Mathematicalmodel:x1=Numberofunitsofsingle-familyhomesx2=Numberofunitsofdouble-familyhomesx3=Numberofunitsoftriple-familyhomesx4=Numberofunitsofquadruple-familyhomesx5=NumberofoldhomestobedemonishedThecompletemodelthusbecomesMaxz=1000x1+1900x2+2700x3+3400x4Subjectto.18x1+.28x2+.4x3+.5x4-.2125x50x53000,,,,1500021601307050075.75.25.25.01.1.9.1.02.2.2.8.5432154321432143214321xxxxxxxxxxxxxxxxxxxxxxSolution:Totaltaxcollection=z=$343,965Numberofsinglehomes=x1=35.8336unitsNumberofdoublehomes=x2=98.5399unitsNumberoftriplehomes=x3=44.7945unitsNumberofquadruplehomes=x4=0unitsNumberofhomesdemonished=x5=244.49245units1.Example2.3-2(CurrencyArbitrageModel)Supposethatacompanyhasatotalof5milliondollarsthatcanbeexchangedforeuros(€)，Britishpounds(£)，yen(￥),andKuwaitidinars(KD).Currencydealerssetthefollowinglimitsontheamountofanysingletransaction:5milliondollars,3millioneuros,3.5millionpounds,100millionyen,and2.8millionKDs.Thetablebelowprovidestypicalspotexchangerates.Thebottomdiagonalratesarethereciprocalofthetopdiagonalrates.Forexample,rate(€$)=1/rate($€)=1/.769=1.30.$€£￥KD$1.769.625105.342€1/.7691.813137.445£1/.6251/.8131169.543￥1/1051/1371/1691.0032KD1/.3421/.4451/.5431/.00321Isitpossibletoincreasethedollarholdings(abovetheinitial$5million)bycirculatingcurrenciesthroughthecurrencymarket?MathematicalModel:Forthepurposeofdevelopingthemodelandsimplifyingthenotation,thefollowingnumericcodeisusedtorepresentthecurrencies.Currency$€£￥KDCode12345DefineijxAmountincurrencyiconvertedtocurrencyj,iandj=1,2,…,5ThecompletemodelisnowgivenasMaximizez=ySubjectto0)0032.543.445.342(.0)0032.1169137105(0)543.11691813.625(.0)445.11371813.1769(.5)342.11051625.1769.1(45352515545352515434241445434241534323133534323152423212252423215141312115141312xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxy,04,3,2,1,8.25,3,2,1,1005,4,2,1,5.35,4,3,1,35,4,3,2,554321ijjjjjjxjxjxjxjxjxSolution:Theoptimumsolutionis:SolutionInterpretationy=5.09032Finalholdings=$5,090,320Netdollargain=$90,320,whichrepresentsa1.8064%rateofreturnx12=1.46206Buy$1,462,060worthofeurosx15=5Buy$5,000,000worthofKDx25=3Buy€3,000,000worthofKDx31=3.5Buy£3,500,000worthofdollarsx32=0.931495Buy£931,495worthofeurosx41=100Buy￥100,000,000worthofdollarsx42=100Buy￥100,000,000worthofeurosx43=100Buy￥100,000,000worthofpoundsx53=2.085BuyKD2,085,000worthofpoundsx54=.96BuyKD960,000worthofyen3.Example2.3-7(CrudeOilRefiningandGasolineBlending)ShaleOil,locatedontheislandofAruba,hasacapacityof1,500,000bblofcrudeoilperday.Thefinalproductsfromtherefineryincludethreetypesofunleadedgasolinewithdifferentoctanenumbers(ON):regularwithON=87,premiumwithON=89,andsuperwithON=92.Therefiningprocessencompassesthreestages:(1)adistillationtowerthatproducesfeedstock(ON=82)attherateof.2bblperbblofcrudeoil,(2)acrackerunitthatproducesgasolinestock(ON=98)byusingaportionofthefeedstockproducedfromthedistillationtowerattherateof.5bblperbbloffeedstock,and(3)ablenderunitthatblendsthegasolinestockfromthecrackerunitandthefeedstockfromthedistillationtower.Thecompanyestimatesthenetprofitperbarrelofthethreetypesofgasolinetobe$6.70,$7.20,and$8.10,respectively.Theinputcapacityofthecrackerunitis200,000barrelsoffeedstockaday.Thedemandlimitsforregular,premium,andsupergasolineare50,000,30,000,and40,000barrelsperday.Developamodelfordeterminingtheoptimumproductionschedulefortherefinery.MathematicalModel:Letijx=bbl/dayofinputstreamiusedtoblendfinalproductj,i=1,2;j=1,2,3ThecompletemodelisthussummarizedasMaximize)(10.8)(20.7)(70.6231322122111xxxxxxzSubjectto0,,,,,)(929882)(899882)(879882000,40000,30000,50000,200)(2000,500,1)(10)(5232221131211231323132212221221112111231322122111232221232221131211xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxSolution:Theoptimumsolutionisz=1,482,000,x11=20,625,x21=9375,x12=16,