【文章內(nèi)容簡介】 L S LP P M T o t a lP P M U S LP P M L S LP P M T o t a lP P M U S LP P M L S LP p kP P LP P UP pC p mC p kC P LC P UC pS t D e v ( O v e r a l l )S t D e v ( W i t h i n )S a m p l e NM e a nL S LT a r g e tU S L1 3 5 6 5 0 . 6 7 7 4 8 5 6 . 3 4 6 0 7 9 4 . 3 21 0 4 4 3 8 . 2 5 5 8 4 4 8 . 0 9 4 5 9 9 0 . 1 6 5 0 0 0 0 . 0 0 1 0 0 0 0 . 0 0 4 0 0 0 0 . 0 00 . 4 80 . 5 20 . 4 80 . 5 00 . 5 00 . 5 20 . 5 60 . 5 20 . 5 41 . 3 3 8 3 81 . 2 2 9 6 41 0 06 0 0 . 0 7 25 9 8 . 0 0 06 0 0 . 0 0 06 0 2 . 0 0 0E x p . O v e r a l l P e r f o r m a n c eE x p . W i t h i n P e r f o r m a n c eO b s e r v e d P e r f o r m a n c eO v e r a l l C a p a b i l i t yP o t e n t i a l ( W i t h i n ) C a p a b i l i t yP r o c e s s D a t aW i t h i nO v e r a l lExample 3A Histogram of camshaft length suggests mixed populations. Further investigation revealed 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 remendations? Example 3A Stat ? Quality Tools ? Capability Sixpack(Normal) Example 3A 2 01 006 0 0 . 56 0 0 . 05 9 9 . 55 9 9 . 0X b a r a n d R C h a r tS u b g rMeanM e a n = 5 9 9 . 5U C L = 6 0 0 . 3L C L = 5 9 8 . 83210RangeR = 1 . 3 4 1U C L = 2 . 8 3 5L C L = 02 01 00L a s t 2 0 S u b g r o u p s6 0 16 0 05 9 95 9 8S u b g r o u p N u m b e rValues6 0 2T5 9 8C a p a b i l i t y P l o tP r o c e s s T o l e r a n c eIIIIIIIIIS p e c i f i c a t i o n sW i t h i nO v e r a l l6 0 1 . 05 9 9 . 55 9 8 . 0N o r m a l P r o b P l o t6 0 1 . 05 9 9 . 55 9 8 . 0C a p a b i l i t y H i s t o g r a mW i t h i nS t D e v :C p :C p k :0 . 5 7 6 4 2 91 . 1 60 . 9 0O v e r a l lS t D e v :P p :P p k :C p m :0 . 6 2 0 8 6 51 . 0 70 . 8 30 . 8 7P r o c e s s C a p a b i l i t y f o r C a m s h a f t L e n g t h ( S u p p l i e r A )Example 3A 2 01 006 0 46 0 26 0 05 9 8X b a r a n d R C h a r tS u b g rMean11M e a n = 6 0 0 . 2U C L = 6 0 2 . 5L C L = 5 9 8 . 07 . 55 . 02 . 50 . 0RangeR = 3 . 8 9 0U C L = 8 . 2 2 5L C L = 02 01 00L a s t 2 0 S u b g r o u p s6 0 3 . 56 0 1 . 05 9 8 . 55 9 6 . 0S u b g r o u p N u m b e rValues6 0 25 9 8C a p a b i l i t y P l o tP r o c e s s T o l e r a n c eIIIIIIIIS p e c i f i c a t i o n sW i t h i nO v e r a l l6 0 56 0 05 9 5N o r m a l P r o b P l o t6 0 56 0 05 9 5C a p a b i l i t y H i s t o g r a mW i t h i nS t D e v :C p :C p k :1 . 6 7 2 3 10 . 4 00 . 3 5O v e r a l lS t D e v :P p :P p k :1 . 8 7 8 6 10 . 3 50 . 3 1P r o c e s s C a p a b i l i t y f o r C a m s h a f t L e n g t h ( S u p p l i e r B )What’s Six Sigma Quality — Then Original Definition by Motorola: ? if the specification limits are at least 177。 6? away from the process mean ?, . Cp ? 2, ? and the process shifts by less than ?, . Cpk ? , ? then the process will yield less than dppm rejects. 6? 6? Shift ? ? What’s Six Sigma Quality — Now Mikel J Harry claims that the process mean between lots will vary, with an average process shift of ?. k? = z? + ? k? = z? + ? Shift ? z? Note: Sigma Capability = ?(dpmo) ? ?(dppm) Process Capability for NonNormal Data Not every measured characteristic is normally distributed. Characteristic Distribution Cycle Time Exponential Reject Rate Binomial Defect Rate Poisson Process Capability for Cycle Time The Weibull Distribution is a general family of distribution with where scale parameter ? is the value at which CDF=%, and shape parameter ? determines the shape of the PDF. ? ? ??????? ??? ??????? x1 ex,。xWeibullProcess Capability for Cycle Time At ?=1, the Weibull Distribution is reduced to For an Exponential Distribution, The Exponential Distribution is thus a Weibull Distribution with ?=1. ????xe1Weibull (x。 ?=1, ?) ???xExponential (x。 ?) 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 plaint. The goal for this index is to close a plaint within one calendar week. Performance over the last 400 plaints was reviewed. Example 4 Stat ? Quality Tools ? Capability Analysis (Weibull) Example 4 2 52 01 51 050U S LP r o c e s s C a p a b i l i t y f o r C o m p l a i n t C l o s u r eC a l c u l a t i o n s B a s e d o n E x p o n e n t i a l D i s t r i b u t i o n M o d e lP P M T o t a lP P M U S LP P M L S LP P M T o t a lP P M U S LP P M L S LP p kP P LP P UP pS c a l eS h a p eS a m p l e NM e a nL S LT a r g e tU S L1 2 2 9 7 0 . 8 01 2 2 9 7 0 . 8 0 * 7 5 0 0 0 . 0 0 7 5 0 0 0 . 0 0 *0 . 3 9 *0 . 3 9 *3 . 3 41 . 0 04 0 03 . 3 4 * *7 . 0 0E x p e c t e d L T P e r f o r m a n c eO b s e r v e d