6sigma黑带培训教材英文

1、Process Capability Analysis (Measure Phase) Information, the way you want it Scope of Module ? Process Variation ? Process Capability Specification, Process and Control Limits Process Potential vs Process Performance ? Short-Term vs Long-Term Process Capability ? Process Capability for Non-Normal Da

2、ta Cycle-Time Reject Rate Defect Rate (Exponential Distribution) (Binomial Distribution) (Poisson Distribution) Process Variation Process Variationis the inevitable differences among individual measurements or units produced by a process. Sources of Variation ? ? ? ? ? ? within unit between units be

3、tween lots between lines across time measurement error (positional variation) (unit-unit variation) (lot-lot variation) (line-line variation) (time-time variation) (repeatability ;?Min?0.5 3? ? 3? ? 3 ? 4?3?4? Example 1B Cp? USL? LSL16? 4 ?1.0 6? 6?2? C pk ?16?10 10 ? 4? ?USL ? ? ? ? LSL? ? Min? ;?

4、Min? ? ?1.0 3? ? 3? ? 3 ? 2?3?2? Example 1C Cp? USL? LSL16? 4 ?1.0 6? 6?2? C pk ?16 ? 7 7? 4? ?USL ? ? ? ? LSL? ? Min? ;? Min? ? ?0.5 3? ? 3? ? 3 ? 2?3?2? Example 1D Cp? USL? LSL 6? 16? 4 ?2.0 6?1? Cpk ?16 ?13 13? 4? ?USL ? ? ? ? LSL? ? Min? ;? Min? ? ?1.0 3? ?3?1? 3? ? 3 ? 1? Process Capability For

5、 a normally distributed characteristic, the defective rate F(x)may be estimated via the following: LSL F?x?Pr?x ? LSL? ? Pr ? x ? USL? ?LSL? ?USL? ? ? ? ? 1? ? ? ? ? ? ? ?ZLSL? 1? ? ? ZUSL? For characteristics with only one specification limit: F?x?Pr?x ? LSL?ZLSL?a) LSL only F?x?Pr?x ? USL? 1? ? ?

6、Z USL ? b) USL only USL Example 2 Specification Limits: Machine (a) (b) (c) (d) Mean 10 10 7 13 4 to 16 g Std Dev 4 2 2 1 Determine the defective rate for each machine. Example 2 Mean Std Dev ZLSL ZUSL F(xUSL) F(x) 10 10 7 4 2 2 -1.5 -3.0 -1.5 1.5 3.0 4.5 66,807 1,350 66,807 1,350 133,614 2,700 66,8

7、1166,807 3 131-9.03.001,3501,350 Lower Spec Limit = 4 g Upper Spec Limit = 16 g Process Potential vs Process Performance (a) Poor Process Potential LSLUSL (b) Poor Process Performance LSLUSL Experimental Design ?to reduce variation Experimental Design ?to center mean ?to reduce variation Process Pot

8、ential vs Process Performance Cpk 1.0 1.2 1.4 1.6 1.0 Process Potential Index (Cp) 1.21.41.61.8 318.3159.9 26.7 159.1 13.4 1.6 159.1 13.4 0.8 2.0 159.1 13.4 0.8 2,699.9 1,363.3 1,350.0 1,350.0 1,350.0 1,350.0 1.8 2.0 0.10.0 0.0 Defective Rate (measured in dppm) is dependent on the actual combination

9、 of Cpand Cpk. Process Potential vs Process Performance a) Cp= 2 Cpk= 2 b) Cp= 2 Cpk= 1 c) Cp= 2 Cpk 1 Cp Cpk ? Missed Opportunity Alternative Process Performance Index Process capability statistics measure process variation relative to specification limits. The Cpstatistic compares the engineering

10、tolerance against the processs natural variation. The Cpkstatistic takes into account the location of the process relative to the midpoint between specifications. If the process target is not centered between specifications, the Cpmstatistic is preferred. 2 ? ? ?T ? ? C pm ?C p 1? ? ? Process Stabil

11、ity A process is stable if the distribution of measurements made on the given feature is consistent over time. Stable Process ucl Time lcl Unstable Process ucl Time lcl Within vs Overall Capability Within Capability(previously called short-term capability) shows the inherent variability of a machine

12、/process operating within a brief period of time. Overall Capability(previously called long-term capability) shows the variability of a machine/process operating over a period of time. It includes sources of variation in addition to the short-term variability. Within vs Overall Capability Within Sam

13、ple Size Number of Lots Period of Time 30 50 units single lot hours or days Overall ? 100 units several lots weeks or months Number of Operators Process Potential Process Performance single operator Cp Cpk different operators Pp Ppk Within vs Overall Capability Within Capability C p Cpl Cpu Overall

14、Capability Pp Ppl ? USL ? LSL 6? Within ? ? USL? LSL 6? Overall LSL ? 3? Overall LSL? ? 3? Within USL ? ? ? 3? Within Ppu? USL? 3? Overall NSL ? ? NSL ? ?C pk ? Ppk? 3? Within 3? Overall The key difference between the two sets of indices lies in the estimates for ? Within and ? Overall . Estimating

15、? Within and ? Overall Consider the following observations from a Control Chart: S/NX1X2 X k MeanRangeStd Dev 1 2 : m x1,1 x1,2 : x1,m x2,1 x2,2 : x2,m xk,1 xk,2 : xk,m X1 X2 : Xm 2 R1 R2 : Rm S1 S2 : Sm The overall variation ? Overallis estimated by ? ? Overall ? 1 c 4 ? i?1?j?1? x ij ? mk N ?1 ? 1

16、 c4 ? i?1?j?1? x ij ? x ? mk 2 N ?1 Estimating ? Within and ? Overall The within variation ? Withinmay be estimated by one of the following: (a) R-bar Method ? ? Within ? R S d 2 whered2is a Shewhart constant = ?(k) (b) S-bar Method ? ? Within ? c 4 wherec4is a Shewhart constant = ?(k) (c) Pooled St

17、andard Deviation Method 222 1 ? n1 ?1? S1 ? n 2 ?1? S2 ? ? n m ?1? Sm ? ? Within ? ? n1 ?1 ? n 2 ?1 ? ? n m ?1 ? c 4 In MiniTab, the Pooled Standard Deviationis the default method. Estimating ? Within and ? Overall In cases where there is only 1 observation per sub-group (i.e. k=1), the Moving Range

18、 Methodis used, where MR i ? X i ? X i?1 . The within variation ?Withinis then estimated using either a) the Average Moving Range: ? ? Within ? b) the Median Moving Range: ? ? Within ? MR d 2 MR d 2 Example 3 The length of a camshaft for an automobile engine is specified at 600 2 mm. Control of the

19、length of the camshaft is critical to avoid scrap/rework. The camshaft is provided by an external supplier. Assess the process capability for this supplier. The data is available in Process Capability Analysis.MTW. Example 3 Stat ? Quality Tools ? Capability Analysis (Normal) Example 3 Process Capab

20、ility for Camshaft Length Process Data USL602.000 Target600.000 LSL598.000 Mean600.072 Sample N100 StDev (Within)1.22964 StDev (Overall) 1.33838 LSLTargetUSL Within Overall Potential (Within) Capability Cp0.54 CPU0.52 CPL0.56 Cpk0.52 Cpm Overall Capability Pp PPU PPL Ppk 0.50 0.48 0.52 0.48 0.50 596

21、597598599600601602603604 Observed Performance PPM USL 10000.00 PPM Total 50000.00 Exp. Within Performance PPM USL 58448.09 PPM Total104438.25 Exp. Overall Performance PPM USL 74856.34 PPM Total135650.67 Example 3A Histogram of camshaft length suggests mixed populations. Further investigation reveale

22、d that there are two suppliers for the camshaft. Data was collected over camshafts from both sources. Are the two suppliers similar in performance? If not, what are your recommendations? Example 3A Stat ? Quality Tools ? Capability Sixpack(Normal) Example 3A Process Capability for Camshaft Length (S

23、upplier A) 600.5 600.0 Xbar and R Chart UCL=600.3 Capability Histogram M e a n 599.5 599.0 Mean=599.5 LCL=598.8 Subgr0 3 2 1 0 R=1.341 598.0599.5601.0 1020 UCL=2.835 Normal Prob Plot R a n g e LCL=0 598.0599.5601.0 Last 20 Subgroups 601 V a lu e s 600 599 598 01020 Within StDev: 0.576429 Cp:1.16 Cpk

24、:0.90 Overall StDev: 0.620865 Pp:1.07 Ppk:0.83 I Capability Plot I Process Tolerance Within Overall I II II I I Specifications T598602 Subgroup Number Cpm:0.87 Example 3A Process Capability for Camshaft Length (Supplier B) 604 1 602 600 598 Subgr0 7.5 1020 UCL=8.225 Xbar and R Chart 1 UCL=602.5 Capa

25、bility Histogram M e a n Mean=600.2 LCL=598.0 595600605 Normal Prob Plot R a n g e 5.0 R=3.890 2.5 0.0LCL=0 595600605 Last 20 Subgroups 603.5 V a l u e s 601.0 598.5 596.0 01020 Within StDev:1.67231 Cp:0.40 Cpk:0.35 Overall StDev:1.87861 Pp:0.35 Ppk:0.31 I Capability Plot I Process Tolerance Within

26、Overall I II I Specifications II 598602 Subgroup Number Whats Six Sigma Quality Then Original Definition by Motorola: ? if the specification limits are at least 6? away from the process mean ?, i.e. Cp ? 2, ? and the process shifts by less than 1.5?, i.e. Cpk ? 1.5, ? then the process will yield les

27、s than 3.4 dppm rejects. 6?6? 4.5? Shift 1.5? Whats Six Sigma Quality Now Mikel J Harry claims that the process mean between lots will vary, with an average process shift of 1.5 ?. Note: Sigma Capability = ?(dpmo) ? ?(dppm) k?= z?+ 1.5?k?= z?+ 1.5? z? Shift 1.5? Process Capability for Non-Normal Dat

28、a Not every measured characteristic is normally distributed. Characteristic Cycle Time Reject Rate Defect Rate Distribution Exponential Binomial Poisson Process Capability for Cycle Time The Weibull Distributionis a general family of distribution with Weibull? x;?,? ? ?x ?e ? ? ?1?x ? ? ? ? ? wheres

29、cale parameter? is the value at which CDF=68.17%, andshape parameter? determines the shape of the PDF. Process Capability for Cycle Time At ?=1,the Weibull Distributionis reduced to 1 ? x e ? Weibull (x; ?=1, ?) ? ? For an Exponential Distribution, Exponential (x; ?) 1 ?e ? ? x ? The Exponential Dis

30、tributionis thus a Weibull Distribution with ?=1. Example 4 A customer service manager wants to determine the process capability for his department. A primary performance index is the time taken to close a customer complaint. The goal for this index is to close a complaint within one calendar week.

31、Performance over the last 400 complaints was reviewed. Example 4 Stat ? Quality Tools ? Capability Analysis (Weibull) Example 4 Process Data USL Target LSL Mean Sample N Shape Scale 7.00 * * 3.34 400 1.00 3.34 Process Capability for Complaint Closure Calculations Based on Exponential Distribution Mo

32、del USL Overall (LT) Capability Pp * PPU PPL Ppk 0.39 * 0.39 Observed LT Performance PPM USL 75000.00 PPM Total 75000.00 Expected LT Performance PPM USL 122970.80 PPM Total122970.80 0510152025 Example 4A Stat ? Quality Tools ? Capability Sixpack (Weibull) Example 4A Process Capability for Complaint

33、Closure 24 Individual and MR Chart Capability Histogram I n d i v i d u a l V a lu e 16 8 0 LCL=-3.779 Obser. 0 24 01020 UCL=10.46 Mean=3.34 100200300400 Weibull Prob Plot M o v . R a n g e 16 8 0 UCL=8.746 R=2.677 LCL=0 0.010.101.0010.00 Last 25 Observations 9 Capability Plot Process Tolerance Over

34、all (LT) Shape: 1.00 Scale: 3.34 Pp: * Ppk: 0.39 II I I V a lu e s 6 3 0 380390400 7 Specifications Observation Number Process Capability for Reject Rate For a Normal Distribution, the proportion of parts produced beyond a specification limit is Reject Rate?Pr?X?USL? USL? ? Pr?Z? ? USL ? ? ? ? 1?Pr?

35、Z? ? ? 1?F(Z) ? Process Capability for Reject Rate Thus, for every reject rate there is an accompanying Z-Score, Spec Limit? whereZ?Score? ? NSL? Ppk ? Recall that 3? Z?Score Ppk?Hence 3 Process Capability for Reject Rate Estimation of P pk for Reject Rate ? Determine the long-term reject rate(p) ?

36、Determine the inverse cumulative probability for p, using Calc ? Probability Distribution ? Normal ? Z-Score is the magnitude of the returned value ? P pkis one-third of the Z-Score Example 5 A sales manager plans to assess the process capability of his telephone sales departments handling of incomi

37、ng calls. The following data was collected over a period of 20 days: ? number of incoming calls per day ? number of unanswered calls per days Example 5 Stat ? Quality Tools ? Capability Analysis (Binomial) Example 5 Process Capability for Telephone Sales P Chart 0.26 0.25 UCL=0.2555 25 0.24 0.23 0.2

38、2 0.21 0.20 0.19 01020 LCL=0.1973 P=0.2264 26 Rate of Defectives % D e f e c t iv e P r o p o r t i o n24 23 22 21 20 185019502050 Sample NumberSample Size 23.5 Cumulative TfectiveSummary Stats (denotes 95% C.I.) Average P: 0.226427 (0.2222, 0.2307) Dist of Tfective % D e f e c ti v e 22.5 Tfective:

39、22.643 (22.22, 23.07) Target:0 PPM Def.: 226427 (222241, 230654) 21.5 1020 Sample Number Process Z: 0.751 (0.737, 0.765) 20222426 Ppk= 0.25 Process Capability for Defect Rate Other applications, approximating a Poisson Distribution: ? error rates ? particle count ? chemical concentration Process Cap

40、ability for Defect Rate Estimation of Y tp for Defect Rate ? Define size of an inspection unit ? Determine the long-term defects per unit(DPU ) DPU = Total Defects ? Total Units ? Determine the throughput yield (Y tp ) Ytp= expDPU Process Capability for Defect Rate Estimation of Sigma-Capability for

41、 Defect Rate ? Determine the opportunities per unit ? Determine the long-term defects per opportunity(d) d = defects per unit ?opportunities per unit ? Determine the inverse cumulative probability for d, using Calc ? Probability Distribution ? Normal ? Z-Score is the magnitude of the returned value

42、? Sigma-Capability = Z-Score + 1.5 Example 6 The process manager for a wire manufacturer is concerned about the effectiveness of the wire insulation process. Random lengths of electrical wiring are taken and tested for weak spots in their insulation by means of a test voltage. The number of weak spo

43、ts and the length of each piece of wire are recorded. Example 6 Stat ? Quality Tools ? Capability Analysis (Poisson) Example 6 Process Capability for Wire Insulation 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0.00 0102030405060708090100 LCL=0 U=0.02652 U Chart UCL=0.06904 0.08 0.07 0.06 0.05 Defect Rat

44、e S a m p l e C o u n t D P U 0.04 0.03 0.02 0.01 0.00 100110120130140150 Defects per Unit = 0.0265194 Throughput Yield = expDPU = exp0.0265194 = 0.9738 c.f. First-Time Yield = 2 / 100 = 0.02 Sample NumberSample Size 0.030 Cumulative DPUSummary Stats (denotes 95% C.I.) Mean DPU:0.0265194 (0.0237309,

45、 0.0295455) Min DPU:0 Max DPU: 0.0753425 Targ DPU: 0 Target Dist of DPU 0.025 D P U 0.020 0.015 1020304050607080901000.0000.0250.0500.075 Sample Number Example 6 Boxplot of Length Define 1 Inspection Unit i.e.Units = 125 unit length of wire = Length ?125 100110120130140150 Length Example 6A Stat ? Q

46、uality Tools ? Capability Analysis (Poisson) Example 6A Process Capability for Wire Insulation 10 U Chart UCL=8.630 5 U=3.315 0 0102030405060708090100 LCL=0 10 9 8 7 6 5 4 3 2 1 0 0.8 Defect Rate S a m p le C o u n t D P U 0.91.01.11.2 Defects per Unit = 3.31493 Throughput Yield = expDPU = exp3.31493 = 0.0363 c.f. First-Time Yield = 2 / 100 = 0.02 Sample NumberSample Size Cumulative DPU 3.5 Summary Stats (denotes 95% C.I.) Mean DPU: 3.31493 (2.96637, 3.69319) Target Dist of DPU 3.0