報(bào)童問題模型課件完整版

報(bào)童問題模型1ppt課件1、報(bào)童問題的提出2、報(bào)童問題所屬范疇3、報(bào)童模型的建立與求解4、報(bào)童模型的推廣與應(yīng)用2ppt課件1、報(bào)童問題的提出

在日常生活中，經(jīng)常會(huì)碰到一些季節(jié)性強(qiáng)、更新快、不易保存等特點(diǎn)的物品，如海產(chǎn)、山貨、時(shí)裝、生鮮食品和報(bào)紙等，當(dāng)商店購進(jìn)這些商品時(shí)，買的數(shù)量越多，價(jià)格越便宜獲利越大。但買得太多也可能賣不出去，需要削價(jià)處理，人力物力都受損；如果進(jìn)貨太少，又可能發(fā)生缺貨現(xiàn)象，失去銷售機(jī)會(huì)而減少利潤。3ppt課件

這就產(chǎn)生一個(gè)問題：訂貨量過多，出現(xiàn)過剩，會(huì)造成損失；訂貨量少，又可能會(huì)失去銷售機(jī)會(huì)，影響利潤，那么應(yīng)該如何確定訂貨策略呢？將這一現(xiàn)象具體到報(bào)童銷售報(bào)紙上，就引發(fā)了報(bào)童問題：報(bào)童問題：報(bào)童每天需訂購多少份報(bào)紙？4ppt課件問題報(bào)童售報(bào)：(零售價(jià))a>(購進(jìn)價(jià))b>(退回價(jià))c售出一份賺a-b；退回一份賠b-c

每天購進(jìn)多少份使收入最大？分析購進(jìn)太多賣不完退回賠錢購進(jìn)太少不夠銷售賺錢少應(yīng)根據(jù)需求確定購進(jìn)量每天需求量是隨機(jī)的優(yōu)化問題的目標(biāo)函數(shù)應(yīng)是長期的日平均收入每天收入是隨機(jī)的存在一個(gè)合適的購進(jìn)量等于每天收入的期望5ppt課件2、報(bào)童問題所屬范疇

單周期隨機(jī)型存貯模型這種單周期購入—售出（報(bào)紙、日歷、雜志，各種季節(jié)性貨物、時(shí)裝），并且超出該購入—售出周期商品就會(huì)嚴(yán)重貶值的存貯問題，存貯論中統(tǒng)稱為賣報(bào)童問題。這類問題的庫存控制策略是以利潤期望最大為目標(biāo)，確定一次購入的經(jīng)濟(jì)訂貨批量。6ppt課件3、模型的建立與求解建模

設(shè)每天購進(jìn)n份，日平均收入為G(n)調(diào)查需求量的隨機(jī)規(guī)律——每天需求量為r的概率f(r),r=0,1,2…準(zhǔn)備求n使G(n)最大

已知售出一份賺a-b；退回一份賠b-c7ppt課件求解將r視為連續(xù)變量8ppt課件結(jié)果解釋取n使P1~賣不完的概率，a-b~售出一份賺的錢P2~賣超的概率，b-c~退回一份賠的錢0nrp(r)P1P29ppt課件4、報(bào)童問題的推廣與應(yīng)用在科學(xué)的管理方法和手段在管理實(shí)踐中運(yùn)用越來越多的今天，管理者同樣需要考慮，怎樣改進(jìn)粗放的管理模式，才能提高企業(yè)的管理水平，從而提高企業(yè)的效益。在管理實(shí)踐中，我們會(huì)發(fā)現(xiàn)，與報(bào)童問題類似的問題非常多，這樣我們就可以將報(bào)童問題的研究方法運(yùn)用到實(shí)踐中，通過科學(xué)的調(diào)查、計(jì)算，把過去經(jīng)驗(yàn)的管理方法，上升到科學(xué)的管理方法。10ppt課件報(bào)童問題的推廣與應(yīng)用：多產(chǎn)品報(bào)童問題；考慮風(fēng)險(xiǎn)偏好的報(bào)童問題；基于需求預(yù)測的報(bào)童問題；考慮采購提前期的報(bào)童問題；11ppt課件ProductAvailability:TradeoffsHighavailability=>responsivetocustomersattractincreasedsaleshigherrevenueHighavailability=>largerinventoryhighercostsriskofobsolescence12ppt課件NewsboyModelsingleperiodmodel(onesellingseason) (one-timeorder,e.g.forquantitydiscount)demanduncertaintyorderplaced(anddelivered)beforedemandisknownunmetdemandislostunsoldinventoryattheendoftheperiodisdiscard(orsalvagedatlowervalue)Howmuchtoorder?NewsvendorModel13ppt課件FactorsaffectingavailabilityDemanduncertaintyOverstockingcost

C0=lossincurredwhenaunitunsoldatendofsellingseasonUnderstockingcost

Cu=profitmarginlostduetolostsale(becausenoinventoryonhand)Customer/Cycleservicelevel

CSL=levelofproductavailability=Prob(Demand<stocklevel)14ppt課件DeterminingOptimalLevelofProductAvailabilitySingleperiodPossiblescenariosSeasonalitemswithasingleorderinaseasonOne-timeordersinthepresenceofquantitydiscountsContinuouslystockeditemsDemandduringstockoutisbackloggedDemandduringstockoutislostExample:SellingparkasatLLBeanCostperparka=c=$45Salepriceperparka=p=$100Inventoryholding(untilseasonend)andtransportationcost(tooutletstore)perparka=$10Discountpriceperparka(seasonendsales)=$50Salvagevalueperparka=$50-$10=$40=sCostofoverstocking=Co=$45+$10-$50=c-s=$5Marginalprofitfromsellingparka=costofunderstocking=Cu

=$100-$45=p-c=$5516ppt課件L.L.BeanExample–DemandDistributionDemandDi

(inhundreds)ProbabilitypiCumulativeProbabilityofDemandBeingDiorLess(Pi)ProbabilityofDemandBeingGreaterthanDi40.010.010.9950.020.030.9760.040.070.9370.080.150.8580.090.240.7690.110.350.65100.160.510.49110.200.710.29120.110.820.18130.100.920.08140.040.960.04150.020.980.02160.010.990.01170.011.000.00Table13-1LLBean:ExpectedProfitExpecteddemandExpectedprofitiforder10Expectedprofitiforderk

18ppt課件LLBean:Expectedprofit19ppt課件Newsvendor:MarginalAnalysisStockoneunitif…Stock2units(insteadof1unit)if...Stock1 Stock2 Stock3D=0D=1D=2D=320ppt課件Increaseorderfromktok+1ifProb(Demand<k)<

Cu

Co+CuOrderk+1insteadofkif

Pr(D>k)Cu

+Pr(D<k)(-Co)

>0or Pr(D<k)(-Co)+[1-Pr(D<k)]Cu

>0orderk+1keepordersizeatkinsteadofk1moreunsold1fewerlostsale0CuPk1-Pk-CoAdditionalcontribution21ppt課件LLBean22ppt課件L.L.BeanExampleAdditionalHundredsExpectedMarginalBenefitExpectedMarginalCostExpectedMarginalContribution11th5,500x0.49=2,695500x0.51=2552,695–255=2,44012th5,500x0.29=1,595500x0.71=3551,595–355=1,24013th5,500x0.18=990500x0.82=410990–410=58014th5,500x0.08=440500x0.92=460440–460=–2015th5,500x0.04=220500x0.96=480220–480=–26016th5,500x0.02=110500x0.98=490110–490=–38017th5,500x0.01=55500x0.99=49555–495=–440Table13-223ppt課件NewsvendorModel-

DemandDistributionContinuousOrderysuchthatCSL*=Prob(Demand<

y)=

Cu

Co+CuyCriticalratioCriticalfractileOptimalCycleSe