ch05a process capability[1]
TRANSCRIPT
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Chapter 5a Process Capability
This chapter introduces the topic of processcapability studies. The theory behind
process capability and the calculation of Cpand Cpk is presented
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Specification Limit &Process Limit
Look at indv. values and avg. values of xsIndv xs values n = 84 - considered as populationAvgs n = 21 - sample taken
= same (in this case)
Normally distributed individual xs and avg. valueshaving same mean, only the spread is different W > R elationship = popu. Std. dev. of avgs
If n = 5 = 0.45 W W = popn std dev of indiv. xsSP RE AD OF AVGS IS HALF OF SP RE AD FO R INDV.VALU E S
xXandX
xW
nx !
xW
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R elationship betweenpopulation and sample values
Assume Normal Dist.E stimate popu. std. dev.
c4 } ; n = 84
@ (c4 = 0.99699)
= 4.17
= 2.09417.4
nx!
W!W
4c
!W
3n4)1n(4
0.996994.16 !
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C entral Limit Theorem
If the population from which samples aretaken is NOT normal, the distribution of
SAMPLE AVERAGES will tend towardnormality provided that sample size, n, is atleast 4.Tendency gets better as n oStandardized normal for distribution of averages Z =
n
x
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Central Limit Theoremis one reason whycontrol chart worksNo need to worry aboutdistribution of s is notnormal, i.e. indv.values.Averages distributionwill tend to ND
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C ontrol Limits &Specifications
C ontrol limits - limits for avgs, and establishedas a func. of avgsSpecification limits -allowable variation insize as per designdocuments e.g. drawing@ for individual valuesestimated by designengineers
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C ontrol limits, Process spread, Dist of averages, & distribution of individual valuesare interdependent. determined by theprocessC . C harts C ANNOT determine process meetsspec.
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Process C apability &Tolerance
When spec. established without knowingwhether process capable of meeting it or not serious situations can resultProcess capable or not actually lookingat process spread, which is called processcapability (6 W)Lets define specification limit as tolerance(T) : T = USL -LSL
3 types of situation can resultthe value of 6 W < USL-LSLthe value of 6 W = USL - LSLthe value of 6 W > USL - LSL
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C ase I and C ase II situations
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C ase 3 situation
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Process C apabilityProcedure (s method)
1. Take subgroup size 4 for 20 subgroups2. Calculate sample s.d., s, for each subgroup3. Calculate avg. sample s.d. Ds = s/g4. Calculate est. population s.d.5. Calculate Process Capability =
R - method1. Same as 1. above2. Calculate R for each subgroup3. Calculate avg. Range, = R/g4. Calculate5. Calculate Calculate 6
4o cs !WW
R
oW2o dR !
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Process C apability (6 W) AndTolerance
Cp - Capability IndeT = U-LCp = 1 Case II 6W = TCp > 1 Case I 6W < TCp < 1 Case III 6W > TUsually Cp = 1.33 (de facto
std.)Measure of processperformanceShortfall of Cp - measurenot in terms of nominal or target value >>> must useCpk
Formulas
Cp = (T)/6 W
Cpk =3
Z(min)
Z (USL) =xUSL
WLSLx
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E xampleDetermine Cp and Cpk for aprocess with average 6.45,W = 0.030, having USL =
6.50 , LSL = 6.30 -- T = 0.2
SolutionCp= T/6 W= 0.2/6(0.03)=1.11Cpk = Z(min)/3Z(U) = (USL - Dx )/ W =
6.50-6.45)/0.03 = 1.67Z(L) = ( Dx LSL)/ W = 6.45-
6.30)/0.03 = 5.00@Cpk = 1.67/3 = 0.56
Process NOT capable sincenot centered. Cp > 1 doesntmean capable. Have tocheck Cpk
UL T
6.30 6.506.45=x
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C omments On C p, C pkCp does not change when process center (avg.) changesCp = Cpk when process is centred
Cpk e Cp always this situationCpk = 1.00 de facto standardCpk < 1.00 p process producing rejectsCp < 1.00 p process not capableCpk = 0 process center is at one of spec.limit (U or L)Cpk < 0 i.e. ve value, avg outside of limits
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E xercise
1. Find Cp, CpkDx = 129.7 (Length of radiator hose)W = 2.35Spec. 130.0 s 3.0What is the %defective?
2. Find Cp, Cpk
Spec.U = 58 mmL = 42 mmW = 2 mmWhen = 50When = 54
3. Find Cp, CpkU = 56L = 44W = 2
When = 50When = 56