1、統(tǒng)計(jì)應(yīng)用案例-報(bào)童模型,在你面前有三個(gè)門(mén)，其中兩個(gè)門(mén)里面是山羊，另外一個(gè)是汽車(chē)。你當(dāng)然想得到那輛汽車(chē)，而不是臭氣轟轟的山羊。主持人要求你選中一個(gè)，但是不能打開(kāi)。此時(shí)主持人打開(kāi)另外一個(gè)，是山羊?，F(xiàn)在主持人問(wèn)你，你要不要改成第三個(gè)門(mén)？,3,誰(shuí)是更好的采購(gòu)經(jīng)理?,如果產(chǎn)品是 賀卡: 利潤(rùn) = $3.00 成本 = $0.20 ？,設(shè)想有倆個(gè)經(jīng)理 管理同樣的一產(chǎn)品（超10個(gè)品種）的采購(gòu)。假定條件相同。在季節(jié)結(jié)束， 張經(jīng)理 基本 沒(méi)有 剩余 ； 王經(jīng)理多個(gè)品種有剩余。,報(bào)亭老板如何決策？,銷(xiāo)售份數(shù) 天數(shù) 1 10 2 20 3 40 4 20 5 10,南大北門(mén)的報(bào)亭在過(guò)去100天里，某報(bào)紙的銷(xiāo)售記錄如

2、左表。報(bào)紙銷(xiāo)售價(jià)1元，進(jìn)價(jià)0.3元。問(wèn)：報(bào)亭老板每天訂幾份報(bào)紙合適？ A. 2 B. 3 C. 4,5,什么是分析能力？,一位物理學(xué)家，工程師和數(shù)學(xué)家在東馬徒步旅行， 看到一頭黑山羊在山坡上吃草。物理學(xué)家 首先發(fā)表高見(jiàn):“所有馬國(guó)的山羊都是黑色的?！?“不對(duì)，西蒙。有些馬國(guó)的山羊是黑色,” 工程師糾正到。 數(shù)學(xué)家最后發(fā)言：“從所看到的，我們只能說(shuō)，在東馬，至少有一只山羊至少身體的 一面是黑色的?！?決策思路,邊際成本 = 邊際收入,7,單周期最優(yōu)訂貨量-1,定義： 期末庫(kù)存余量為正時(shí)的單位成本（滯銷(xiāo)成本） 需求未滿(mǎn)足導(dǎo)致的成本（機(jī)會(huì)成本） 周期一開(kāi)始購(gòu)買(mǎi)的商品數(shù)量 需求量D的概率密度函數(shù) 需求

3、量D的概率分布函數(shù),案例：報(bào)童模型,Pandora皮衣：?jiǎn)挝焕麧?rùn)14.50$，因滯銷(xiāo)降價(jià)而帶來(lái)?yè)p失5.00$/件.,9,單周期最優(yōu)訂貨量-2,從A-1看出， 對(duì)于面積=0.74， z=0.65 。因此,面積=0.74,需求量， X,f(x),ONeills Hammer 3/2 wetsuit,Hammer 3/2 timeline and economics,Economics: Each suit sells for p = $180 TEC charges c = $110 per suit Discounted suits sell for v = $90,The “too much/

4、too little problem”: Order too much and inventory is left over at the end of the season Order too little and sales are lost. Marketings forecast for sales is 3200 units,Newsvendor model implementation steps,Gather economic inputs: Selling price, production/procurement cost, salvage value of inventor

5、y Generate a demand model: Use empirical demand distribution or choose a standard distribution function to represent demand, e.g. the normal distribution, the Poisson distribution. Choose an objective: e.g. maximize expected profit or satisfy a fill rate constraint. Choose a quantity to order.,The N

6、ewsvendor Model: Develop a Forecast,Historical forecast performance at ONeill,Forecasts and actual demand for surf wet-suits from the previous season,Empirical distribution of forecast accuracy,Normal distribution tutorial,All normal distributions are characterized by two parameters, mean = m and st

7、andard deviation = s All normal distributions are related to the standard normal that has mean = 0 and standard deviation = 1. For example: Let Q be the order quantity, and (m, s) the parameters of the normal demand forecast. Probdemand is Q or lower = Probthe outcome of a standard normal is z or lo

8、wer, where (The above are two ways to write 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.,Converting between Normal distribution

9、s,Start with = 100, = 25, Q = 125,Center the distribution over 0 by subtracting the mean,Rescale the x and y axes by dividing by the standard deviation,Start with an initial forecast generated from hunches, guesses, etc. ONeills initial forecast for the Hammer 3/2 = 3200 units. Evaluate the A/F rati

10、os of the historical data: Set the mean of the normal distribution to Set the standard deviation of the normal distribution to,Using historical A/F ratios to choose a Normal distribution for the demand forecast,ONeills Hammer 3/2 normal distribution forecast,ONeill should choose a normal distributio

11、n with mean 3192 and standard deviation 1181 to represent demand for the Hammer 3/2 during the Spring season.,Empirical vs normal demand distribution,Empirical distribution function (diamonds) and normal distribution function with mean 3192 and standard deviation 1181 (solid line),“Too much” and “to

12、o little” costs,Cu = underage cost Co = overage cost The 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 had you ordered one fewer u

13、nit. For the Hammer 3/2 Co = Cost Salvage value = c v = 110 90 = 20 The 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 ordered one

14、 more unit. For the Hammer 3/2 Cu = Price Cost = p c = 180 110 = 70,Balancing the risk and benefit of ordering a unit,Ordering 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 = Q) but the benefit/gain of orde

15、ring one more unit is the reduction in the chance of underage: Expected gain on the Qth unit = Cu x (1-F(Q),As more units are ordered, the expected benefit from ordering one unit decreases while the expected loss of ordering one more unit increases.,Newsvendor expected profit maximizing order quanti

16、ty,To maximize expected profit order Q units so that the expected loss on the Qth unit equals the expected gain on the Qth unit: Rearrange terms in the above equation - The ratio Cu / (Co + Cu) is called the critical ratio. Hence, to maximize profit, choose Q such that we dont have lost sales (i.e.,

17、 demand is Q or lower) with a probability that equals the critical ratio,Finding the Hammer 3/2s expected profit maximizing order quantity with the empirical distribution function,Inputs: Empirical distribution function table; p = 180; c = 110; v = 90; Cu = 180-110 = 70; Co = 110-90 =20 Evaluate the

18、 critical ratio: Lookup 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 corresponds to A/F ratio 1.3) Convert A/F ratio into the order quantity,25,A/F Ratio =實(shí)際需求/ 預(yù)測(cè)需求,33 種產(chǎn)品,26,用歷史數(shù)據(jù)構(gòu)建的需求分布函數(shù),如果某款新設(shè)計(jì)的產(chǎn)品預(yù)測(cè)需求為3,200,Q = 需求量,Hammer 3/2s expected profit maximizing order quantity using the normal distribution,Inputs: p = 180; c = 110; v = 90; Cu = 180-110 = 70; Co = 110-90 =20; critica