Chapter 4. One-sided Process Capability Assessment in the Presence of Gauge Measurement Errors

Transcription

1 hapter One-sided Process apability Assessment in the Presence of Gauge Measurement Errors n the manufacturing industry, many product characteristics are of one-sided specifications The process capability indices and PL are often used to measure process performance (see equation ()) The index measures the capability of a smaller-the-better process with an upper specification limit SL, whereas the index PL measures the capability of a larger-the-better process with a lower specification limit LSL f the quality characteristic of manufacturing process is normally distributed, the process yield can be expressed by %yeild = Φ (3 ), () where Φ is the cumulative distribution function of the standard normal distribution, and = or PL t is clear that the relationship between the index and process yield is one-to-one Thus, the index provides an exact measure of process yield Table 8 displays some commonly used capability values of, the corresponding process yield, and nonconformity units in parts per million (NPPM) Montgomery [3] recommended some minimum quality requirements on For existing processes, the capability must be no less than 5, and for new processes, the capability must be no less than 5 For existing processes on safety, strength, or critical parameters, the capability must be no less than 5, and for new processes on safety, strength, or critical parameters, the capability must be no less than 6 sing the index, the practitioners can evaluate their process capability and make decisions Table 8 The corresponding process yield and NPPM for Process yield (%) NPPM n section, we discuss the relationship between the empirical process capability and the true process capability n section, we obtain the pdf, the expected value, the variance and the MSE of And, we compare the MSE of with that of n section 3, we use the confidence interval bounds 3

2 in Pearn & Shu [] to estimate the minimum process capability by, we show that a large measurement error results in significantly underestimating the true process capability n section, we use the critical values in Pearn & hen [35] to test whether the process capability meets the requirement, and we show that the α -risk and the power both become decrease with measurement error n section 5, we present our modified confidence interval bounds and critical values for the cases that measurement errors are unavoidable Finally in section 6, an example is presented Empirical Process apability Suppose that X ~ Normal ( μ, σ ) represents the relevant quality characteristic of a manufacturing process, Mittag [7] introduced the degree of error contamination, τ = σm / σ The relationship between the empirical process capability index and the true process capability index is =, () + τ where or PL is denoted here as Since the variation of the empirical data we observe is greater than the variation of the original data, the denominator of the index becomes larger, and we would understate the true capability of the process if we calculate the process capability based on the empirical data n Table 9, we tabulate some empirical process capabilities with τ = () for various true process capability = 5,, 33, 5, 67,, and 5 f τ =, then for = 35 the true process capability is = 5, and for = 77 the true process capability = 5 The empirical process capability diverges from the true one more when the measurement error increases t is obvious that the gauge accuracy is less important if the required process capability is only marginally capable, and becomes more critical as the true capability requirement gets more stringent Table 9 Process capability with τ = () for various τ

3 Sampling Distribution of Since the process parameters μ and σ are unknown, we therefore cannot evaluate the actual process capability But, given sample data taken from the process, we could estimate process capability Denoting { X i, i =,, n } the random sample of size n from the quality characteristics X, the natural estimators of and PL are ˆ SL X =, ˆ 3S PL X LSL =, 3S (3) n n / where X = i= Xi / n and S = [ i= ( Xi X)/( )] are conventional estimators of μ and σ Assume that X and M are stochastically independent, ~ Normal ( μ, σ = σ + σ M ) With a given sample { i, i =,, n}, the estimators of and PL are SL = b, PL = b 3S LSL, 3S () Based on the same argument used in hou & Owen [6] and Pearn & hen [35], the estimator ( or PL ) is distributed as dt ( n δ ), where d = b (3 ) n n and ( t ) n δ is a non-central t distribution with n - degrees of freedom and non-centrality parameter δ = 3 n / + τ The mean, the variance, and the mean squared error of the estimator are E( ) = + τ (5) (( n ) / ) (( n 3) / ) ( ) (( n ) / ) (( n 3) / ) Var( Γ Γ Γ Γ ) = + [ Γ(( n ) / )] + τ 9 n[ Γ(( n )/ )] (6) (( n )/) (( n 3)/) MSE( Γ Γ ) = ( ) ( ) + + τ [ Γ(( n )/)] + τ Γ(( ) / ) Γ(( n 3) / ) + 9[ n (( )/)] ( ) Γ n (7) For τ >, is a biased estimator of, and the bias (/ + τ ) decreases in τ Since Γ(( ) / ) Γ(( n 3) / ) /[ Γ(( n ) / )] is positive, then Var( ) < Var( ) To compare MSE ( ) with MSE ( ), we consider the function f (, n, τ ) = MSE ( )/ MSE ( ) By some reduction, we have f (, n, τ ) = if and only if Γ(( n )/) Γ(( )/) Γ(( n 3)/) [ Γ(( n )/)] τ = [ Γ(( n ) / )] Γ(( ) / ) Γ(( n 3) / ) (8) or τ = Denote the right side of the equal sign in the above formula as τ, we 3

4 have f (, n, τ ) > if τ > τ and f (, n, τ ) < if τ < τ exclusive of t represents that MSE ( ) > MSE ( ) if τ > τ, MSE ( ) < MSE ( ) if τ < τ exclusive of, and MSE ( ) = MSE ( ) if τ = τ or Table τ values for n = 5(5) n τ n τ n τ n τ Table lists the τ values for n = 5(5) Figures 6(a)-6(b) display the surface plots of the ratios γ = f (, n, τ ) with n = 5() and τ in [, ] for =, and 33 The value τ is greater than 5 for small n (n ), and greater than for n 5 When 5 < n, τ is between and For large n, γ is greater than for almost every value of τ, and γ increases if τ increases The maximum values of γ are 39, and 537, respectively, and the minimum values of γ are 86 (/), and 797 (/55), respectively The maximum values of γ occur at n = and τ =, and the minimum values of γ occur at n = 5 and τ = 788 The difference between MSE( ) and MSE( ) with γ > is more significant than that with γ < Figure 6(a) Surface plot of γ with n = 5() and τ in [, ] for = Figure 6(b) Surface plot of γ with n = 5() and τ in [, ] for = 33 3 onfidence Bounds Based on The lower confidence bounds present a measure on the minimum capability of the process based on the sample data Let k = 3 / bn and k = 3 PL / bn, and we have S L = X + ks and LSL = X - ks A θ % lower confidence bound for satisfies P ( ) = θ t can be 35

5 written as SL μ P( ) = P 3σ X + ks μ Z 3 n = P = P 3 σ S / σ k n Z 3 n 3 = P n = P t = n S / σ b ( ( δ 3 ) t ) = θ (9) Similarly, a θ % lower confidence bound for PL satisfies P ( PL L ) = θ t can be shown as Pt ( n ( δ L = 3 nl) t ) = θ, where Z is distributed as Normal(, ), t = - k n, and t = k n To find the exact θ % lower confidence bounds, Pearn & Shu [] provided an algorithm and a Matlab program to solve the above equations With measurement errors, we use to estimate but not Thus, t = - (3 / bn ) n and t = (3 PL / bn ) n instead of t and, are substituted into the equations to t obtain the confidence bounds Denote the bounds originated from t and t as and L The confidence coefficient by the confidence bound (denoted by θ ) we obtained is SL μ ( ) τ θ = P = P + 3σ + k S 3 / + μ Z n = = τ P P k n 3 σ + τ S / σ 3 / + 3 Z n 3 τ n n P t( δ S σ bn = P = = / + τ ) t, () where k = 3 / bn And, θ can be also obtained by the confidence bound, which can be expressed as L θ P t t () + τ 3 nl = ( δl = ) Figures 7(a)-7(b) plot θ versus τ with n = 5(5) and =, and 33, for 95% confidence intervals (because E( ) = E ( / +τ ), we consider the cases that = / + τ ) Since is smaller than in the presence of measurement errors, and (or L ) is smaller than (or L ), it is necessary that θ is always greater than θ Thus, Figures 7(a)-7(b) show that the confidence coefficients become increase because the estimated lower confidence interval bounds become decrease by measurement errors 36

6 θ versus τ Figure 7(a) Plots of with = and n = 5(5) (from top to bottom) for 95% confidence intervals θ versus τ Figure 7(b) Plots of with = 33 and n = 5(5) (from top to bottom) for 95% confidence intervals apability Testing Based on We usually use statistical testing to determine if our processes meet the capability requirement The null hypothesis is H : c (process is not capable), and the alternative hypothesis is H : > c (process is capable) of testing, where c is our required process capability The critical value is used to determine whether the null hypothesis should be rejected f the point estimator of the process capability is greater than the critical value, we reject the null hypothesis and conclude that the process is capable Otherwise, we would believe that the process is incapable Suppose that the nominal size of our statistical testing is α (type error), the critical value c can be determined by ( ) α = P c = c () Since is distributed as ( 3 dtn δ = n), where d = b (3 ) n n, we can obtain that c is b c = t, α ( δ = 3 nc ), 3 n (3) where tn, α ( δ ) is the upper α th quantile of tn ( δ ) distribution And the power of the test can be calculated as ( ) = P ( 3 n 3 ) > nc π ( ) = P > c 3 n 3 n = P 3 n > c = P b b t( δ = 3 n) > c b ( ( δ 3 ), α ( δ 3 ) ) = P t = n > t = nc () However, in the presence of measurement errors, the α -risk (denoted by α ) and the power (denoted by π ) are 37

7 ( ) ( = P 3 n 3 ) nc = c α = P c = c 3 n 3 n = P 3 n c = c = P b b t( δ = 3 n ) c = c b c = P t( δ = 3 n ) t, α ( δ = 3 nc ) + τ (5) ( ) ( ) π ( ) = P > c = P 3 n > 3 nc 3 n 3 3 = > = ( = 3 ) > n n P c P t δ n n c b b b = P t( δ = 3 n ) > t, α ( δ = 3 nc) (6) + τ Figure 8(a) Surface plot of α with n = 5(), τ in [, ] for c =, α = 5 Figure 8(b) Surface plot of α with n = 5(), τ in [, ] for c = 33, α = 5 Figure 9(a) Plots of π versus τ, with n = 5, α = 5, for c =, = () (from bottom to top) Figure 9(b) Plots of π versus τ, with n = 5, α = 5, for c = 33, = 33()33 (from bottom to top) Earlier discussions indicate that we underestimate the true process 38

8 capability using instead of The probability that is greater than c would be less than that of using Thus, the α -risk using to estimate is less than the α -risk if using to estimate The power, if using to estimate, is also less than the power using That is, we have α α and π π Figures 8(a)-8(b) are the surface plots of α with n = 5() and τ in [, ] for =, 33 and α = 5 Figures 9(a)-9(b) are the plots of π versus τ with n = 5 and α = 5 for c =, 33 and = c () c + Note that for τ =, α = α and π = π in those figures n Figures 8(a)-8(b), α decreases as τ or n increases, and the decreasing rate is more significant with large c values We find that for large τ values α is smaller than 5 n Figures 9(a)-9(b), π decreases as τ increases, but increases as n increases Decrement of π by τ is more significant for large c values Because of measurement errors, π may decrease significantly For instance, in Figure 9(a) the π values ( c =, n = 5) for = is π = 9 if there is no measurement error (τ = ) But, when τ =, π decreases to, the decrement of the power is Modified onfidence Bounds and ritical Values We showed earlier that the coefficients increase owing to the underestimating the lower confidence bounds We also showed that both the α -risk and the power of the test decrease in measurement error The probability of passing non-conforming product units decreases, but the probability of correctly judging a capable process as incapable also decreases Since the lower confidence bound of the process capability is severely underestimated, and the power becomes much weak, the producers cannot firmly state that their processes meet the capability requirement even if their processes are sufficiently capable Good product units would be incorrectly rejected in this case (rejected products are either scrapped or requiring rework) nnecessary cost may accompany those incorrect decisions to the producers mproving the gauge capability and training the operators by proper education are some advice for reducing the measurement errors Nevertheless, measurement errors may be unavoidable in most manufacturing processes Thus in this section, we adjust the confidence bounds to give a more precise estimation of process capability, and revise critical values to improve the power for testing hypothesis Suppose that the desired confidence coefficient is θ, the adjusted confidence interval of with confidence interval bound, can be established as SL μ ( ) τ θ = P = P + 3σ 39

9 + k S μ Z 3 n / + τ = P = P k n 3 σ + τ S / σ Z 3 n / + τ 3 n 3 n = P = P t( δ = ) t S / σ b (7) + τ Similarly, the adjusted confidence interval of bound L, can be established as θ = P t = + τ 3 nl ( δl ) t PL with confidence interval (8) To find the exact θ % lower confidence bounds, an S-plus program is developed to solve the equations Figures (a)-(b) are comparisons among,, and for =, 33 with n = 5, where is the 95% lower confidence bound of, is the 95% lower confidence bound of, and is the adjusted 95% lower confidence bound for Note that in this case, the probability that the interval with the bound contains the actual value is greater than that of the interval with the bound or does, while the probability that the interval with the bound or contains the actual value is just 95 From Figures (a)-(b), we see that the lower confidence bounds remained underestimated, even if we adjust the formula to calculate the bounds But, the magnitude of underestimation using adjusted confidence bounds is significantly reduced Figure (a) Plots of,, and (from top to bottom) versus τ with n = 5 and for = Figure (b) Plots of,, and (from top to bottom) versus τ with n = 5 and for = 33

10 n order to improve the power of the test, we consider the revised critical values c satisfied c < c Thus, the probability that is greater than c is greater than the probability that is greater than c Both the α -risk and the power increase when we use c as a new critical value in the testing Suppose that the α -risk using the revised critical value c is α, the revised critical c must satisfy α ( ) P ( 3 n 3 ) nc = = c = P c = c 3 n 3 n = P c = c b b 3 n = P t( δ = 3 n ) c = c b c 3 n = P t( δ = 3 n ) c τ b + (9) To ensure that the α -risk is within the preset magnitude, we let α = α, thus c can be obtained as b c c = t, α ( δ = 3 n ), 3 n () + τ and the power π is π ( ) P ( 3 n 3 ) nc = > ( ) = P > c 3 n 3 n = P > c b b 3 n = P t( δ = 3 n ) > c b c = P t( δ = 3 n ) > t, α ( δ = 3 n ) () + τ + τ Figures (a)-(b) plot π versus τ with n = 5 and α = 5 for c =, 33, and = c () c + From those figures, we see that the powers corresponding to the adjusted critical values c remain decreasing in measurement error, but the decrements originated from the new critical values c are very small We have improved a certain degree of power For instance, if we compare the π values in Figure 9(a) ( c =, n = 5) for = with the π values in Figure (a) ( c =, n = 5) for =, we see that π = and π = 885 with τ = n this case, using the adjusted critical values c the power is improved by 83 Tables -3 in Appendix provide the revised critical values for some commonly used capability requirements sing those tables, the practitioner may select the proper critical values for capability testing

11 Figure (a) Plots of π versus τ, with n = 5, α = 5, for c =, = () (from bottom to top) Figure (b) Plots of π versus τ, with n = 5, α = 5, for c = 33, = 33()33 (from bottom to top) 6 Application Example TFT-LDs (thin-film-transistor liquid crystal display) consist of a lower glass plate on which the TFT is formed, an upper glass plate on which the color filter is formed, and the injected liquid crystal between both glass plates (see Figure (a)) The TFT plays a critical role in transmitting and controlling electric signals, which determines the amount of voltage applied to the liquid crystal The liquid crystal controls light permeability using different molecular structures that vary in accordance with the voltage n this way, the desired color and image is displayed as it passes through the color filter (see figure (b)) The TFT-LD consumes less energy compared to a RT (cathode-ray tube), is slimmer and weighs less The TFT-LD emerges as the most widely used display solution, because of its high reliability, viewing quality and performance, compact size and environment- friendly features Because of the heat resistance, non-conductance, and simple processing steps Non-alkali thin film glass is the major material of manufacturing TFT-LD While manufacturing non-alkali thin film glass, flatness is one of the critical quality characteristics f the flatness of glass is not in control, the TFT-LD products may result in a certain degree of chromatic aberration Table 6 observations for flatness (unit: um)

12 Figure (a) Structure of TFT-LD (a) Figure (b) Structure of TFT-LD (b) onsider a supplier in manufacturing TFT-LD products in Taiwan, the production specifications of flatness for a particular model of non-alkali thin film glass are: SL = 5 um (5 mm), T = um A total of 6 observations were collected which are displayed in Table To determine whether the process is Satisfactory ( > 33) with unavoidable measurement errors τ =, we propose the following procedure, STEP : Determine the capability requirement c (normally chosen to, 33, 5) and the α -risk (normally set to, 5, or 5), STEP : alculate the value of the point estimator from the sample, STEP 3: heck the appropriate table listed in Tables -3 and finding the corresponding critical value c based on α, τ, and n, STEP : onclude that the process meets the capability requirement if is greater than c Otherwise, we do not have enough information to conclude that the process is capable With the proposed procedure, we first determine that c = 33 and α = 5 Based on the sample data of 6 observations, we obtain the sample mean = 93, the sample standard deviation S = 85, and the point estimator = 5 From Table, we find the critical value c = 5 based on α, τ and n Since > c, we conclude that the process is Satisfactory Moreover, by inputting, τ, n, and the desired confidence coefficient θ = 95 into the computer program we can obtain the 95% lower confidence bound of this process capability as 385 3