分析教案講稿17 sa marcordwell

decisionquestionsisconsiderable,butthisarticlewillfocusononlyone:linearprogramming.Theideaspresentedinthisarticlearebasedonasimpleexample.Supposeaprofit-seekingfirmhastwoconstraints:labour,limitedto16,000hours,andmaterials,limitedto15,000kg.Thefirmmanufacturesandsellstwoproducts,XandY.TomakeX,thefirmuses3kgofmaterialandfourhoursoflabour,whereastomakeY,thefirmusescontributionsmadebyeachproductare$30forXand$40forY.Thecostofmaterialsisperblemistoproducetheequationsforwillbe3X+5Y≤15,000,andthelabourconstraintwillbe4X+4Y≤16,000.Youshouldnotforgetthenon-negativityconstraint,ifneeded,ofX,Y≥0.Thecontributionfunctionis30X+40Y=

theoptimalsolutonwillbeatpointBTheoptimalpontisX=2,500and=1,500,whichgenerates$135,000incontribution.Checkthisforyourself(seeequationsisassumedinthisarticle.orderto profit.However,thingscanchangeand,inparticular,constraintscanrelaxortighten.Managementneedstoknowthefinancialbepaidforthem?Andhowmuchshouldbebought?Thesedynamicsareimportant.thismean?Ifmanagementisofferedmorethan$5perk erthenormalprice.Payinglessthan$13($5+$8)perkgtoobtainmorematerialswillmakethefirmbetteroffrenderitworseoffintermsofcontributionbuy‘expensive’extramaterials(thosecostingcostingmorethan13per13perkg).ItFIGURE1:OPTIMALNY4A3210C122.5 66studentaccountantMarchenablethebusinesstosatisfythedemandsofanimportantcustomerwhomight,inbejustifiableifitwillpreventapenaltyonthecontract.Thecostofthisisrarely perkg,isnotattractive.Qualityisafactor,recognisethat‘price’isnoteverything.materials,thenthatconstraintrelaxesandsoitslineonthegraphmovesoutwardsandlinewillbetotallyoutsidethelabourlineononthegraph).Labourwouldth onlyconstraint.Weneedtofindouthowmanymaterialsatwhich4,000unitsofYareproduced.Tomake4,000unitsofYweneed20,000kgof-15,000).Note:Althoughinterpretationisimportantatthislevel,therewillstillbemarks

availableforthebasiccalculations. Working1intersectionof:3X+5Y=150004X+4Y=secondbythreeweget:12X+20Y=12X+12Y==12,000,orY=SubstitutingY=1,500inanyoftheaboveequationswillgiveustheXvalue:3X+5(1,500)=3X=X=Thecontributiongainedis(2,500x30)+(1,500x40)=$135,0001kgandresolveasfollows:3X+5Y=15,0014X+4Y=DecisionmakingisanimportantaspectofthePaperF5syllabus.Thefirststepinanylinearprogrammingproblemistoproducetheequationsforconstraintsandthecontributionfunction.Thisshould

andbythreeforthesecondproduces:12X+20Y=12X+12Y=8Y=Y=SubstitutingY=1,500.5inanyoftheaboveequationswillgiveusX:3X+5(1,500.5)=

3X=X=x40)=Thenewlevelofcontributi(2,499.5x40)= optimalistheshadowprice:142,505-142,500=$5perGeoffCordwell aminerforPaperMarch2008studentaccountant16,000小時(shí)；材料，限制為15,000公斤。該公司生產(chǎn)和銷售兩種產(chǎn)品，X和Y。為了制造X，34Y54X30Y408103X5Y15,0004X4Y16,000。如果需要的話，您不應(yīng)忘記X,Y≥0的非負(fù)約束。30X40YC（1，最優(yōu)生產(chǎn)計(jì)劃）將表明，通過推出貢獻(xiàn)函數(shù)，最優(yōu)解將位于B點(diǎn)——材料和勞動(dòng)力約束的交點(diǎn)。X2,500，Y1,500135,000（5（2）。重要的一點(diǎn)是，這意味著什么？如果向管理層提供更多材料，則應(yīng)準(zhǔn)備好支付不超過正常價(jià)格每公斤5美元的費(fèi)用13（58）當(dāng)然，可能有充分的理由購(gòu)買“昂貴”的額外材料（13）1：最000s421.D4X4Y16,000（人工）3X+5Y=15,000（材料）12345X000662008313識(shí)到“價(jià)格”并不是一切。更多材料的吸引力的點(diǎn)（圖表上的D點(diǎn)）。勞動(dòng)力將成為唯一的限制。D（4,000Y）4,000位的Y，我們需要20,000公斤的材料。因此，所需額外材料的最大量為5,000公斤（20,000B3X5Y15,0004X4Y16,00012X20Y60,00012X12Y兩個(gè)方程的差異為：8Y12,000YY1,500X3X5(1,500)15,0003X7,500X2,500x301,500x40)工作2：材料的影子價(jià)格為了找到這一點(diǎn)，我們將材料約束放寬1kg，并解析如下：3X+5