1、统计应用案例统计应用案例-报童模型报童模型 在你面前有三个门，其中两个门里面是山羊，另外一个是汽车。你当然想得到那辆汽车，而不是臭气轰轰的山羊。主持人要求你选中一个，但是不能打开。此时主持人打开另外一个，是山羊。现在主持人问你，你要不要改成第三个门？3如果产品是 贺卡:利润=$3.00成本=$0.20？设想有俩个经理设想有俩个经理 管理同样的一产品（超管理同样的一产品（超10个品种）的个品种）的采购。假定条件相同。在季节结束，采购。假定条件相同。在季节结束，张经理张经理 基本基本 没有没有 剩余剩余；王经理多个品种有剩余。王经理多个品种有剩余。销售份数 天数 1 10 2 20 3 40 4
2、20 5 10l南大北门的报亭在过去100天里，某报纸的销售记录如左表。报纸销售价1元，进价0.3元。问：报亭老板每天订几份报纸合适？lA.2 B.3 C.45l 一位物理学家，工程师和数学家在东马徒步旅行，看到一头黑山羊在山坡上吃草。物理学家 首先发表高见:“所有马国的山羊都是黑色的。”l“不对，西蒙。有些马国的山羊是黑色,”工程师纠正到。l数学家最后发言：“从所看到的，我们只能说，在东马，至少有一只山羊至少身体的 一面是黑色的。”边际成本边际成本 =边际收入边际收入7定义：期末库存余量为正时的单位成本（滞销成本）需求未满足导致的成本（机会成本）周期一开始购买的商品数量需求量D的概率密度函数
3、需求量D的概率分布函数ocucQ)(xf)(xF)()(*QFcccQFuoulPandora皮衣：单位利润皮衣：单位利润14.50$，因滞销降价，因滞销降价而带来损失而带来损失5.00$/件件.成员成员CorolnLauraTomKenyWallyWendy平均平均标准差标准差预测预测12001150125013001100120012001309从A-1看出，对于面积=0.74，z=0.65。因此74.00.55.1450.14)(*QF285,113065.01200*Q*Q面积面积=0.74需求量，需求量，Xf(x)1301200 NovDecJanFebMar AprMayJunJu
4、lAugGenerate forecast of demand and submit an order to TECReceive order from TEC at the end of the monthSpring selling seasonLeft overunits arediscountedEconomics:Each suit sells for p=$180TEC charges c=$110 per suitDiscounted suits sell for v=$90lThe“too much/too little problem”:Order too much and
5、inventory is left over at the end of the seasonOrder too little and sales are lost.lMarketings forecast for sales is 3200 unitslGather economic inputs:Selling price,production/procurement cost,salvage value of inventorylGenerate a demand model:Use empirical demand distribution or choose a standard d
6、istribution function to represent demand,e.g.the normal distribution,the Poisson distribution.lChoose an objective:e.g.maximize expected profit or satisfy a fill rate constraint.lChoose a quantity to order.0100020003000400050006000700001000200030004000500060007000ForecastActual demand .Forecasts and
7、 actual demand for surf wet-suits from the previous seasonEmpirical distribution function for the historical A/F ratios.0%10%20%30%40%50%60%70%80%90%100%0.000.250.500.751.001.251.501.75A/F ratioProbability lAll normal distributions are characterized by two parameters,mean=and standard deviation=lAll
8、 normal distributions are related to the standard normal that has mean=0 and standard deviation=1.lFor example:Let Q be the order quantity,and(,)the parameters of the normal demand forecast.Probdemand is Q or lower=Probthe outcome of a standard normal is z or lower,where(The above are two ways to wr
9、ite the same equation,the first allows you to calculate z from Q and the second lets you calculate Q from z.)Look up Probthe outcome of a standard normal is z or lower in the Standard Normal Distribution Function Table.orQzQz Start with=100,=25,Q=125Center the distribution over 0 by subtracting the
10、meanRescale the x and y axes by dividing by the standard deviation125100125QzlStart with an initial forecast generated from hunches,guesses,etc.ONeills initial forecast for the Hammer 3/2=3200 units.lEvaluate the A/F ratios of the historical data:lSet the mean of the normal distribution tolSet the s
11、tandard deviation of the normal distribution toForecastdemand Actual ratio A/FForecast ratio A/F Expected demand actual ExpectedForecast ratios A/F of deviation Standard demand actual of deviation Standard3192320099750.demand actual Expected118132003690.demand actual of deviation StandardlONeill sho
12、uld choose a normal distribution with mean 3192 and standard deviation 1181 to represent demand for the Hammer 3/2 during the Spring season.Product descriptionForecast Actual demandErrorA/F RatioJR ZEN FL 3/2 90140-501.5556EPIC 5/3 W/HD12083370.6917JR ZEN 3/2 140143-31.0214WMS ZEN-ZIP 4/3 170156140.
13、9176ZEN 3/2 3190119519950.3746ZEN-ZIP 4/3 381032895210.8633WMS HAMMER 3/2 FULL 6490367328170.5659Average0.9975Standard deviation0.36900.000.100.200.300.400.500.600.700.800.901.000100020003000400050006000QuantityProbability .Empirical distribution function(diamonds)and normal distribution function wi
14、thmean 3192 and standard deviation 1181(solid line)lCu=underage costlCo=overage costThe cost of ordering one more unit than what you would have ordered had you known demand.In other words,suppose you had left over inventory(i.e.,you over ordered).Co is the increase in profit you would have enjoyed h
15、ad you ordered one fewer unit.For the Hammer 3/2 Co=Cost Salvage value=c v=110 90=20The cost of ordering one fewer unit than what you would have ordered had you known demand.In other words,suppose you had lost sales(i.e.,you under ordered).Cu is the increase in profit you would have enjoyed had you
16、ordered one more unit.For the Hammer 3/2 Cu=Price Cost=p c=180 110=70lOrdering one more unit increases the chance of overage Expected loss on the Qth unit=Co x F(Q)F(Q)=Distribution function of demand=ProbDemand lThe ratio Cu /(Co+Cu)is called the critical ratio.lHence,to maximize profit,choose Q su
17、ch that we dont have lost sales(i.e.,demand is Q or lower)with a probability that equals the critical ratio QFCQFCuo1)(uouCCCQF)(Product descriptionForecastActual demandA/F Ratio RankPercentileHEATWAVE 3/2 1702121.252472.7%HEAT 3/2 5006351.272575.8%HAMMER 3/2 130016961.302678.8%lInputs:Empirical dis
18、tribution function table;p=180;c=110;v=90;Cu=180-110=70;Co=110-90=20lEvaluate the critical ratio:lLookup 0.7778 in the empirical distribution function table If the critical ratio falls between two values in the table,choose the one that leads to the greater order quantity(choose 0.788 which correspo
19、nds to A/F ratio 1.3)lConvert A/F ratio into the order quantity 7778.0702070uouCCC*/3200*1.34160.QForecastA F25A/F Ratio=实际需求/预测需求33 种产品26如果某款新设计的产品预测需求为3,200Q=需求量lInputs:p=180;c=110;v=90;Cu=180-110=70;Co=110-90=20;critical ratio=0.7778;mean=3192;standard deviation=1181 lLook up critical ratio in th
20、e Standard Normal Distribution Function Table:If the critical ratio falls between two values in the table,choose the greater z-statisticChoose z=0.77lConvert the z-statistic into an order quantity:4101118177.03192zQz00.010.020.030.040.050.060.070.080.090.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.70880.
21、71230.7157 0.7190 0.72240.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.74220.74540.7486 0.7517 0.75490.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.77340.77640.7794 0.7823 0.78520.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.80230.80510.8078 0.8106 0.81330.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.82890.83150.8340 0.8365 0.8389
