six sigma测度

Click to edit Master text styles,Second Level,Third Level,Fourth Level,Fifth Level,Click to edit Master title style,Proprietary to Samsung Electronics Company,Six Sigma,测度,-,26,Rev. 2.0,Six Sigma,测度,(,Metrics),Measure,Define,Analyze,Improve,Control,方法论,Measure,概要,Project Y,基础统计,测定,System,分析,Six Sigma,测度,工程能力分析,Process Map,&,特性要因图,FDM,Six Sigma,测度,学习目标,理解,Six Sigma,的主要测度,DPU, DPO, DPMO,FTY, RTY, Y,NOR,理解,水准计算方法,DPU,与,DPMO,DPO (,Defects per Opportunity),每机会缺陷数,DPU (,Defects per Unit),每单位缺陷数,为了消除缺陷的基准,工厂单位改善中使用,DPMO (Defects per Million Opportunities),每百万机会缺陷数,考虑制品复杂度的比较基准,企业内或企业之间,Benchmarking,手段,Opportunity,Defects,DPO,=,Unit,Defects,DPO,=,制品,部品,2,部品,1,组装品,A,组装品,B,特性,B,特性,A,机会,(Opportunity),有可能发生缺陷的检查或试验对象,发生缺陷的机会在特性、部品、组装品等制品任何阶段 都可能存在。,复杂度,(,Complexity ),总机会数与制品或,Process,复杂度正比。,部品,特性,材料,机械,工具,次序,制品,Process,复杂度,相关部品、特性或材料越多，,成为越复杂的制品。,相关机械、工具或次序越多，,成为越复杂的,Process.,DPU/DPO/DPMO,以下,Drilling,加工的例中计算,DPU,与,DPMO,.,Data,类型是,?,Unit,?,Defect,?,Opportunity,?,DPU,?,DPMO,?,DPU,计算,总生产单位数,总缺陷数,=,DPU,DPU,不能考虑一个单位里有多少缺陷机会。,DPU,=,=,=,DPMO,总缺陷发生机会数,总缺陷数,1,000,000,=,DPMO,=,DPMO,计算中心,一般表示,6,水准时，比不良率,3.4,ppm,更恰当的是,3.4,DPMO,DPU/DPO/DPMO ,事例,使用互相不同的,Process、,制品之 间或制造领域和非领域之间比较时更 恰当。,DPMO,计算例,缺陷数,某个,PCB,以,800,个焊接点和,200,个部品构成。在这,PCB,中发现焊接不良,6,处和不良部品,2,个。,DPMO,是,?,DPMO = (6+2)/(800+200),*,百万,= 8,000,DP(M)O,计算时注意事项,例,),某部品的生产工程中不良发生机会数是,100,000,次，但正常作业,过程中只对其中,1,000,次机会进行评价，结果一个部品里发现,10,个,缺陷，以下计算中哪个正确？,DPO = 10 / 100,000,DPO = 10 / 1,000,机会是有缺陷发生的可能性，以检查及试验对象被评价时,Count,Yield,在工程的各阶段中包括再作业或部品的废弃等不良的管理指标，,是良品率的概念。,初期数率,(,FTY, First Time Yield) :,决定各别工程的品质水准时使用。再作业,/,不修理的,Process,中适用。,累计数率,(,RTY, Rolled Throughput Yield) :,表现全体工程的品质水准时使用的指标中的一个。,表现为初期数率的倍。,标准化数率,(,Y,Nor,Normalized Yield) :,表现全体工程的品质水准时使用的指标中的一个，,在,Process,中初期数率适用几何平均概念。,Yield(,数率,),的种类,Yield (,数率,),45,000,ppm,浪费,30,000,ppm,浪费,56,000,ppm,浪费,95.5%,数率,97%,数率,94.4%,数率,RTY = 0.955*0.97*0.944 = 87.4%,不但是最终阶段，其它各阶段的能力,RTY,都重视。,Hidden,Factory :,再作业,/,废弃,Yield,RTY,计算,已知的,DATA,是缺陷,DATA,时,是意味着泊松分布中,PX = 0 (,没有一个缺陷的概率,).,-,e,DPU,已知的,DATA,是不良率或数率,DATA,时,Yield,RTY = FTY,1,FTY,2,FTY,3,Y,Nor,计算,Y,Nor,= FTY,1,FTY,2,FTY,n,n,RTY,计算的检讨,生产一个单位部品所需,3,个工程，各工程的初期数率为,FTY,i,.,各单位工程中发生的缺陷数，跟随着拥有以下概率函数的泊松分布。,DPU,dpu,dpu,dpu,e,e,-,+,+,-,=,=,),(,3,2,1,-,但,为,单位工程,i,的,平均缺陷数,因此各单位工程的数率是在工程中一个缺陷都不发生的概率，如下。,i,dpu,e,X,P,-,=,=,=,0,按以上的结果，可以确认下面的等式。,3,RTY,=,FTY,i,i,= 1,Yield,想想下面的工程。,收入检查,100%,数率,SMD,检查,95.5%,数率,PBA,检查,97.0%,数率,ICT,94.4%,数率,RTY = 1.00, 0.955 0.970 0.944,= 0.874 = 87.4 %,Y,%,7,.,96,967,.,0,874,.,0,RTY,4,1,4,Nor,=,=,=,=,Yield (,数率,),Yield,例,1),现场某一位机士对特定类型,Defect,观察结果,346,个生产单位中发现,1,个。对这特性,(,Characteristic),的,RTY,?,0.00289,346,1,=,=,DPU,00289,.,0,= 0.99711,=,=,-,-,e,e,RTY,DPU,计算,),例,2),有,3,个工程组成的生产,Line，,各工程通过数率为,98%,时,Line,的,RTY,?,计算,),0.98,0.98,0.98,RTY = 0.98, 0.98 0.98 = 0.941192,RTY,计算 例,Yield,水准,水准,:,规格中心和规格上限或下限距离是标准偏差的几倍？,水准的概念,品质特性值随正态分布时，从规格中心到规格界限为止距离相当于标准偏差几倍的测度，,Process,散布越小，,水准越大。,m,1,s,3,s,平均,规格界限,测定,Process,可以生产多少均一品质制品能力的测度。,1,s,m,1,s,发生不良限度,3,s,平均,记号前数值,(,Z),越大，不良发生,的概率越小。,6,s,规格界限,3,水准的,稳定工程,6,水准的,稳定工程,由于作业者倒班、原材料,LOT,变化、设备的保,养、模具交换等现实性的变化要素而规格中心和,P,rocess,平均常时间维持一致状态是很难。,根据经验随着时间过,去，一般,Process,平均离中心约,1.5,左右变动,。,Process dynamics,LSL,USL,起点,1,起点,2,起点,3,起点,4,Process,固有能力,长期,Process,能力,目标值,水准和,DPMO,考虑,Process,现实变化，计算对应,水准的缺陷率或不良率时，,Process,平均离规格中心标准偏差的,1.5,倍变动为前提计算。,在,6,水准,Process,中，,100,万次机会中能发生,3.4,次左右缺陷，这相当于,3.4,DPMO,.,USL,LSL,规格中心,3.4,DPMO,4.5,.st,1.5,.st,999,999.6,999,995,999,991,999,987,999,979,999,968,999,952,999,928,999,892,999,841,999,767,999,663,999,517,999,313,999,032,998,650,998,134,997,445,996,533,995,339,993,790,991,802,989,276,良品数,3.4,5,9,13,21,32,48,72,108,159,233,337,483,687,968,1,350,1,866,2,555,3,467,4,661,6,210,8,198,10,724,6.0,5.9,5.8,5.7,5.6,5.5,5.4,5.3,5.2,5.1,5.0,4.9,4.8,4.7,4.6,4.5,4.4,4.3,4.2,4.1,4.0,3.9,3.8,.,st,4.5,4.4,4.3,4.2,4.1,4.0,3.9,3.8,3.7,3.6,3.5,3.4,3.3,3.2,3.1,3.0,2.9,2.8,2.7,2.6,2.5,2.4,2.3,.,lt,986,097,977,250,971,284,964,070,955,435,945,201,933,193,919,243,903,199,884,930,864,334,841,345,815,940,788,145,758,036,725,747,691,462,655,422,617,911,579,260,539,828,500,000,460,172,良品数,13,903,22,750,28,716,35,930,44,565,54,799,66,807,80,757,96,801,115,070,135,666,158,655,184,060,211,855,241,964,274,253,308,538,344,578,382,089,420,740,460,172,500,000,539,828,3.7,3.5,3.4,3.3,3.2,3.1,3.0,2.9,2.8,2.7,2.6,2.5,2.4,2.3,2.2,2.1,2.0,1.9,1.8,1.7,1.6,1.5,1.4,.,st,2.2,2.0,1.9,1.8,1.7,1.6,1.5,1.4,1.3,1.2,1.1,1.0,0.9,0.8,0.7,0.6,0.5,0.4,0.3,0.2,0.1,0.0,-0.1,.,lt,420,740,382,089,344,578,308,538,274,253,241,964,211,855,184,060,158,655,135,666,115,070,96,801,80,757,66,807,54,799,44,565,35,930,28,716,22,750,17,864,13,903,10,724,8,198,良品数,579,260,617,911,655,422,691,462,725,747,758,036,788,145,815,940,841,345,864,334,884,930,903,199,919,243,933,193,945,201,955,435,964,070,971,284,977,250,982,136,986,097,989,276,991,802,1.3,1.2,1.1,1.0,0.9,0.8,0.7,0.6,0.5,0.4,0.3,0.2,0.1,0.0,-0.1,-0.2,-0.3,-0.4,-0.5,-0.6,-0.7,-0.8,-0.9,.,st,-0.2,-0.3,-0.4,-0.5,-0.6,-0.7,-0.8,-0.9,-1.0,-1.1,-1.2,-1.3,-1.4,-1.5,-1.6,-1.7,-1.8,-1.9,-2.0,-2.1,-2.2,-2.3,-2.4,.,lt,6,210,4,661,3,467,2,555,1,866,1,350,968,687,483,337,233,159,108,72,48,32,21,13,9,5,3,良品数,993,790,995,339,996,533,997,445,998,134,998,650,999,032,999,313,999,517,999,663,999,767,999,841,999,892,999,928,999,952,999,968,999,979,999,987,999,991,999,995,999,997,-1.0,-1.1,-1.2,-1.3,-1.4,-1.5,-1.6,-1.7,-1.8,-1.9,-2.0,-2.1,-2.2,-2.3,-2.4,-2.5,-2.6,-2.7,-2.8,-2.9,-3.0,.,st,-2.5,-2.6,-2.7,-2.8,-2.9,-3.0,-3.1,-3.2,-3.3,-3.4,-3.5,-3.6,-3.7,-3.8,-3.9,-4.0,-4.1,-4.2,-4.3,-4.4,-4.5,.,lt,水准与,DPMO,表,水准计算,缺点数,DATA,时,-,求,DPMO,-,表中找出对应,DPMO,的,Z.st,值或利用,Minitab,求。,DATA,是不良率时,-,以不良率可以计算出,PPM,.,PPM = (,不良率,)1,000,000,-,从表中找出与,PPM,值一样的对应,DPMO,的,Z.st,值或利用,Minitab,求。,DATA,是数率时,-,Scrap,、,再作业等为不良计算数率。,-,从表中找出对应每百万个良品数,= (,数率,)1,000,000,Z.st,值或利用,Minitab,求。,水准的计算,例,例题,1.,对某工程生产出货的制品特征类型缺陷，,经过长期调查结果,346,个制品中发现,1,个缺陷。,计算在这工程的缺陷相关的,水准。,计算, 1,000,000,346,1,DPMO,= 2,890,=,从,表中能查出对应,DPMO=2,890,的,Z.st,值,4.26,左右,利用,Minita0b,计算,:,实习,例题,2,某一个工程组成,A, B, C, D, E 5,个作业，各作业数率为,0.99, 0.95, 0.90, 0.90, 0.95.,A,B,C,D,E,0.99,0.95,0.90,0.90,0.95,水准是每100万个良品数,= 937,375,对应的,Z.st,值,3.03,左右。,5,),95,.,0,)(,90,.,0,)(,90,.,0,)(,95,.,0,)(,99,.,0,(,=,作业的标准化数率,937375,.,0,=,计算,利用,Minitab,的计算,:,实习,例,3,有一个给顾客提供信用情报的公司，顾客要求事项分为,情报内容、迅速性、正确性、最新性、接触容易性，,对这些调查重要度和顾客满足度结果如下。,表,项目别重要度及顾客满足度,项目,最新性,接触容易性,计,/,平均,0.60,情报内容,迅速性,正确性,满足,(%),重要度,0.05,0.15,0.15,0.05,1.00,80,90,90,90,95,84.25,计算,:,确认,水准,= 2.5,左右。,利用,Minitab,的计算：,实习,例题,4,某部品销售价格,$120，,制造原价目标值定于,$100，,但各种外部及内部原因长期预算时，原价比销售价格高的,概率是,5%,左右。,计算,:,对应不良率,5% (,PPM = 50,000),的,水准为,3.15,100,120,5%,利用,Minitab,的计算,:,实习,