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Unformatted text preview: Statistical Quality Control
OEM 2009 ESI 6321 Applied Probability Methods in Engineering Statistical Quality Control A collection of tools used to monitor and guide quality improvement activities. Quality of design: Quality level (performance, reliability, function) that is the result of engineering and/or management decisions Reduction of variability Elimination of defects Goal: all units are identical, and defectfree Quality of conformance: 2 Process Variability Any production process contains variability. Chance (common) causes of variability Essentially unavoidable, inherent variability in the process Assignable (special) causes of variability Avoidable variability in the process: Improperly adjusted machines Operator errors Defective raw materials 3 Control Chart Process is in control: Only chance causes of variability are present Assignable causes of variability are present Process is out of control: The most powerful of all Statistical Quality Control tools is the control chart. A control chart displays periodic measurements of some quality characteristic, and visualizes whether the process is in or out of control.
4 Control Chart A control chart consists of a center line and two control limits for the value of the quality characteristic that we are interested in. If the observed value of the quality characteristic falls beyond one of the two control limits, the process is out of control. This type of control chart is called a Shewhart control chart.
5 Variables vs. Attributes Control Chart There are two general types of control charts: Variables control charts. Attributes control charts. Variable control charts are used for quality characteristics that can be measured on a continuous scale. Attribute control charts are used for discrete quality characteristics; for instance: conforming vs. nonconforming.
6 Control Chart for Mean The most widely used variables control chart monitors the mean of a process parameter. This mean is then the quality characteristic of interest. It is often reasonable to assume that the quality characteristic is normally distributed (when the process is in control). We will base our choice of center line and control limits on this assumption. In principle, the approach can be generalized to other population distributions.
7 Designing a Control Chart Example: consider the manufacturing of automobile piston rings. A critical quality characteristic is the average realized inside ring diameter. We know that the process mean inside ring diameter is 74 mm, and its standard deviation is 0.01 mm if the process is in control. Every hour a random sample of 5 rings is taken, and their average inside ring diameter is computed and plotted in a control chart.
8 Example Control Chart
Control chart
74.0180 74.0135 74.0090 74.0045 74.0000 73.9955 73.9910 73.9865 73.9820 1 3 5 7 9 11 13 15 sample number average average 9 Designing a Control Chart Intuitively speaking, sample means far from the population mean (74 mm) should signal that the process is out of control. Denote the hourly sample (of size n) by X1,...,Xn and the corresponding sample 1 n mean by X = ! Xj n j =1 Denote the population mean and standard deviation if the process is in control by and .
10 Designing a Control Chart This means that, if the process is in control, the mean and standard deviation of the sample mean X are equal to !2 E(X) = and ! 2 = X n Recall that the sample mean is assumed normally distributed. Let Z be a standard normal random variable. Then we define z/2 as follows: Pr Z > z! /2 = ! / 2
11 ( ) Designing a Control Chart Then we have that
1 ! " = Pr !z" /2 < Z # z" /2 ( ) % ( X! = Pr ' !z" /2 < # z" /2 * & ) $/ n = Pr ! z" /2$ / n < X # + z" /2$ / n ( ) Therefore, with probability 100(1)% the sample mean falls in the interval $ ! z # / " /2 & % n , + z" / 2# / n' ) (
12 Designing a Control Chart Recall that, in the example, the incontrol population mean is = 74 mm and the standard deviation is = 0.01 mm. Thus the standard deviation of the sample mean is equal to !X = ! n = 0.01 5 = 0.0045 and the sample mean falls in the interval #74 ! 0.0045z ,74 + 0.0045z % " /2 " /2 & $
with probability 100(1)%.
13 Designing a Control Chart In control charts, we usually use z/2=3, which corresponds to =0.0027. The limits of the control interval are called the 3sigma control limits. In the example, we obtain LCL = 74 ! 0.0045z" /2 = 73.9865 UCL = 74 + 0.0045z" /2 = 74.0135
and we say that the process is in control as long as the sample means are within these limits. The mean (74 in this case) forms the center line.
14 Example Control Chart
Control chart
74.0180 74.0135 74.0090 74.0045 74.0000 73.9955 73.9910 73.9865 73.9820 1 3 5 7 9 11 13 15 sample number average average UCL CL LCL 15 Control Chart for Mean Summarizing: If the incontrol process mean and standard deviation are known, the parameters of the Shewhart control chart for the mean are given by UCL = + z! /2" / n CL = LCL = # z! /2" / n
16 Hypothesis Testing In fact, what we are doing when checking whether or not a process is in control is test a socalled statistical hypothesis on whether the mean of the actual process (population) is equal to the incontrol process mean. In the example, H0 : = 74 H1 : ! 74 The control chart tests this hypothesis after each observed sample (mean)! 17 Designing a Control Chart We need to choose: Sampling method Sample size Sample frequency 18 Rational Subgroups The observations in a sample should include only the chance variability. The most common approach is to select n observations at the beginning (or end) of each time period (snapshot approach). This method minimizes variability within a sample, and maximizes variability between samples. This approach therefore allows for the detection of process shifts.
19 Rational Subgroups In addition, observations could be grouped by Machine Workstation Operator etc. 20 Sample Size and Frequency Large samples make it easier to detect small shifts in the process. However, with large samples it is difficult to prevent variation due to assignable causes within the sample. There is also tradeoff between the size and frequency of the sample. 21 Patterns on Control Charts Usually, a process is said to be out of control if a measurement falls beyond the control limits. However, the pattern of consecutive measurements can signal an out of control process as well. For example, many consecutive points above or below the center line is cause for concern. 22 Western Electric rules A process is out of control if one of the following is satisfied: One point outside 3 control limits. Two out of three consecutive points outside 2 control limits (on the same side). Four out of five consecutive points outside 1 control limits (on the same side). Eight consecutive points on one side of the center line. 23 Example Control Chart
Control chart
74.0180 74.0135 74.0090 74.0045 74.0000 73.9955 73.9910 73.9865 73.9820 1 3 5 7 9 11 13 15 sample number average UCL +2sigma +sigma CL sigma 2sigma LCL run average 24 Control Chart for Mean If the process mean and standard deviation are unknown, we need to estimate them based on preliminary samples. Suppose we have m samples available, each of size n. We will denote these samples by (Xi1,...,Xin) for i=1,...,m Usually we should have m=20 to 25, and typically we will have n=4 to 6. 25 Control Chart for Mean We can then estimate the population mean by the overall mean, sometimes also called grand mean X = ! X , where X m
i =1 i 1 m i is the i th sample mean We will use the grand mean as an estimate of the center line on the control chart for the mean. 26 Control Chart for Mean The following estimator of the standard deviation is insensitive to shifts between samples:
S= ! S , where S m
i =1 i 1 m 2 i is the i th sample variance : Si2 = Even for a normal population, S is a biased estimator of the population standard deviation: ! (X n "1
j =1 1 n ij " X i )2 E(S) = c4!
where the value of c4, for different values of n, can be found in Table I in the handout. Correction factor important since n is small!
27 Control Chart for Mean This means that the process standard deviation can be estimated as S ! = ^ c4 The parameters of the Shewhart control control chart can then be estimated as " ^ s UCL = x + z! /2 = x + z! /2 n c4 n
CL = x LCL = x # z! /2 " ^
n = x # z! /2 s c4 n
28 Example A component part for a jet aircraft engine is manufactured by an investment casting process. The vane opening on this casting is an important functional parameter of the part. The following table shows measurements for m=20 samples of n=5 parts each. The table also shows the sample means, standard deviations, and ranges.
29 Example
obs 1 33 33 35 30 33 38 30 29 28 38 28 31 27 33 35 33 35 32 25 35 obs 2 29 31 37 31 34 37 31 39 33 33 30 35 32 33 37 33 34 33 27 35 obs 3 31 35 33 33 35 39 32 38 35 32 28 35 34 35 32 27 34 30 34 36 obs 4 32 37 34 34 33 40 34 39 36 35 32 35 35 37 35 31 30 30 27 33 obs 5 33 31 36 33 34 38 31 39 43 32 31 34 37 36 39 30 32 33 28 30 mean 31.6 33.4 35.0 32.2 33.8 38.4 31.6 36.8 35.0 34.0 29.8 34.0 33.0 34.8 35.6 30.8 33.0 31.6 28.2 33.8 33.3 S 1.673 2.608 1.581 1.643 0.837 1.140 1.517 4.382 5.431 2.550 1.789 1.732 3.808 1.789 2.608 2.490 2.000 1.517 3.421 2.387 2.345 range 4 6 4 4 2 3 4 10 15 6 4 4 10 4 7 6 5 3 9 6 5.8
30 Example: Control Chart for Mean We first estimate the process standard deviation. Based on the sample standard deviations:
! = ^
s c4 = 2.345 0.94 = 2.495 31 Example: Control Chart for Mean The 3control limits for the then:
UCL = x + z! /2 LCL = x $ z! /2 Xchart are " ^
n " ^ n = 33.3 + 3 # = 33.3 $ 3 # 2.495 5 2.495 5 = 36.67 = 29.97 32 Control Chart for Mean  alternative In practice (particularly when n is small), the sample range is often used to estimate the variability. Define the average range to be
R= ! R , where R is the i m
i =1 i i 1 m th sample range The relationship between the range and the standard deviation of a normal population is given by = E(R)/d2, where the values of d2 for different n can be found in Figure 16.8.
33 Control Chart for Mean  alternative This means that the process standard deviation can alternatively be estimated as R ! = ^ d2 The parameters of the Shewhart control control chart can then be estimated as " ^ r
UCL = x + z! /2 CL = x LCL = x # z! /2 n = x + z! /2 d2 n r d2 n
34 " ^
n = x # z! /2 Example: Control Chart for Mean alternative We first estimate the process standard deviation. Based on the sample ranges:
! = ^
r d2 = 5.8 2.326 = 2.494 (Note that this is almost identical to the estimate based on the sample standard deviations.)
35 Example: Control Chart for Mean alternative The 3control limits for the then:
UCL = x + z! /2 LCL = x $ z! /2 Xchart are " ^
n " ^ n = 10.71 + 3 # = 10.71 $ 3 # 2.494 5 2.494 5 = 36.67 = 29.97 36 Control Chart for Mean  alternative Figure 16.8 also provides 3 A2 = d2 n so that, if we use z/2=3,
UCL = x + 3 CL = x LCL = x " 3 ! ^
n = x +3 r d2 n r d2 n = x + A2 r ! ^
n = x "3 = x " A2 r
37 Example: Control Chart for Mean 38 Control Charts for Variability Another important variable control chart monitors the variability of a process parameter. Variability can be measured by Standard deviation Range We will construct Shewhart control charts for both of these. 39 Control Chart for Standard Deviation Recall that E(S) = c4! . The standard deviation of S is given by
2 ! S = ! 1 " c4 We again assume that the quality characteristic (S in this case) is (approximately) normally distributed. 40 Control Chart for Standard Deviation The parameters of the control chart for the standard deviation are therefore equal to s 2 UCL = s + z! /2" S = s + z! /2 ^ 1 # c4 c4
CL = s LCL = s # z! /2" S = s # z! /2 ^ s c4
2 1 # c4 We again estimate by
S c4 , where S = 1
m i =1 !S m i 41 Example In the example: s = 1 m ! s =2.345
i =1 i m The control chart parameters are:
UCL = s + z! /2 s c4
2 1 " c4 = 2.345 + 3 2.345 0.94 1 " 0.942 = 4.90
0 CL = s = 2.345 s 2.345 2 LCL = s " z! /2 1 " c4 = 2.345 " 3 1 " 0.942 = "0.21 c4 0.94 42 Example: Control Chart for Standard Deviation 43 Control Chart for Range For a normal population, the standard deviation of the range is related to the standard deviation of the population. This gives the following estimator of the standard deviation of the range: ! R = d3! = d3 ^ ^ r d2 where the values of d2 and d3, for different values of n, can be found in Figure 16.8 and Table II in the handout.
44 Control Chart for Range The control chart for the range then has a center line given by r , and control limits
# z d & UCL = r + z! /2" R = %1 + ! /2 3 ( r ^ % d2 ( $ ' # z d & LCL = r ) z! /2" R = %1 ) ! /2 3 ( r ^ % d2 ( $ ' Figure 16.9 provides values
" " 3d % 3d % D3 = $1 ! 3 ' , D4 = $1 + 3 ' d2 & d2 & # #
45 Example: Control Chart for Range In the example: ! R = d3! = 0.864 " 2.494 = 2.154 ^ ^
so that UCL = r + z! /2" R = 5.8 + 3 # 2.154 = 12.261 ^ LCL = r + z! /2" R = 5.8 $ 3 # 2.154 = $0.661 ^ 0 46 Example Control Chart for Range 47 Out of Control Observations Four of the twenty samples are out of control with respect to the 3control limits for the mean An additional sample is out of control with respect to the 3control limits for the standard deviation and range We should not use these samples to estimate the control limits Therefore, we reestimate the control limits with a reduced set of 15 samples 48 Example: Control Chart for Mean 49 Example: Control Chart for Standard Deviation 50 Example Control Chart for Range 51 Out of Control Observations This process can be repeated if deemed necessary. Note that it is important to investigate whether the causes for the outofcontrol observations can be traced and used to improve process performance! 52 Control Chart for Individual Measurements Suppose our sample size is equal to n=1, for instance because: Automated inspection of every unit Production rate is slow Processspecific reasons make taking multiple samples meaningless In this case, the sample variability cannot be estimated as before. We use the moving range, the difference between two successive observations, to measure process variability.
53 Control Chart for Individual Measurements The moving range is given by MRi = Xi  Xi1 Denote the sample mean of the observations by X, and the average of the moving ranges by 1 m 1 m MR = " MRi = m ! 1 " X i ! X i !1  m ! 1 i =2 i =2 An estimate of the population standard deviation is ! = mr / d2 = mr / 1.128 ^
54 Control Charts for Individual Measurements The control limits for a control chart for individual measurements are then: UCL = x + z! /2" = x + z! /2 ^ CL = x mr 1.128 mr 1.128 LCL = x # z! /2" == x # z! /2 ^ 55 Control Charts for Individual Measurements The control limits for a control chart for moving ranges are: With z/2=3 we then have # z d & UCL = mr + z! /2" R = %1 + ! /2 3 ( r ^ % d2 ( $ ' # z d & LCL = mr ) z! /2" R = %1 ) ! /2 3 ( r ^ % d2 ( $ ' UCL = D4 mr = 3.267mr CL = mr LCL = D4 mr = 0
56 Example The purity of a chemical product is measured hourly. The following table shows purity determinations for the last 24 hours. In addition, it shows the moving range values.
purity 81 83 82 80 84 76 83 85 79 82 75 80 moving range 2 1 2 4 8 7 2 6 3 7 5 purity 83 86 84 85 81 83 77 82 75 83 85 86 81.67 moving range 3 3 2 1 4 2 6 5 7 8 2 1 3.96 57 Example The control limits for a control chart for individual measurements are then:
UCL = x + z! /2 mr 1.128 CL = x = 81.67 mr 1.128 = 81.67 + 3 3.96 1.128 3.96 1.128 = 92.20 LCL = x " z! /2 = 81.67 " 3 = 71.14 and for the moving ranges:
UCL = 3.267mr = 3.267 ! 3.96 = 12.94 CL = mr = 3.96, LCL = 0
58 Example Control Chart for Individual Measurements
95
measurement 90 85 80 75 70 65 1 4 7 10 13 16 19 22
sample number purity UCL CL LCL 59 Example Control Chart for Moving Ranges
14 12 10 8 6 4 2 0 1 4 7 10 13 16 19 22
sample number moving range UCL CL LCL moving range 60 Control Chart for Proportions If the process parameter is given by a classification as defective or nondefective, we use an attribute control chart in particular, an NPchart or a Pchart. As in the case of the control chart for the mean, we assume we observe the process using repeated samples of size n. The number of defective units in a sample, say D, has a binomial distribution with parameters n and p, where p is the fraction defective in the population.
61 Binomial Distribution In general, consider: An experiment that consists of n socalled Bernoulli trials. Trials are independent of each other. Each trial results in one of two outcomes, "success" or "failure". Each trial has the same probability of success: p (the probability of failure is 1p). D = the number of successes in n trials. Then D has a binomial distribution with parameters n and p. See also Section 2.6 in the prework coursepack.
62 Control Chart for Proportions Recall that E(D) = pn, V(D) = np(1 ! p) If we have m samples available, we can estimate p by
m ^ = 1 !D = 1 D P = !P m i =1 i mn i =1 i n m 1 63 Control Chart for Proportions: NPchart We can then determine the parameters for the NPcontrol chart:
! UCL = du /2 , CL = np = d, LCL = d!! /2
! /2 ! /2 where du and d! are the smallest resp. largest values such that ! P(D > du /2; p = p) " 1 ! 2 ! P(D < d! /2; p = p) " 1 ! 2 64 Control Chart for Proportions: Pchart Alternatively, we can construct a control chart based on the fraction nonconforming. This gives the following parameters for the Pcontrol chart: UCL = ! du /2 n , CL = p, LCL = ! d! /2 n 65 Control Chart for Proportions If np > 5 we can use the normal approximation to the binomial distribution. This yields the following parameters for the NPchart: UCL = np + z! /2 np(1 " p)
CL = np = d LCL = np " z! /2 np(1 " p) For the Pchart we simply divide these parameters by n.
66 Control Chart for Number of Defects A different attribute control chart is needed if we are interested in the number of defects (per unit of product). As before, we assume we observe the process using repeated samples of size n. Denote the number of defects in unit j by Cj, and the total number of defects in the sample by C. The average number of defects per unit product in a sample is then U = C/n = C.
67 Poisson Distribution Denote the the average number of flaws per unit of product by . Partition the product in a very large number, say , of very small pieces. Assume that the probability that one such small piece contains a defect is equal to p=/. the probability that one such small piece contains more than 1 flaw is negligible. Then the number of flaws in the unit has a binomial distribution with parameters and p= /. For this distribution becomes the socalled Poisson distribution. 68 Poisson Distribution Let Y be a Poisson distribution with parameter It then has the following probability mass function:
P Y = y = e! " ( ) "y y! for y = 0,1,2,... Moreover,
E(Y) = Var(Y) = ! 69 Control Chart for Number of Defects It is often realistic to assume that the number of defects per unit (Cj) is Poisson distributed, with a mean number of defects of , and thus C~Poisson(n). Then we have E(C) = n!, V(C) = n! E(U) = !, V(U) = ! / n Finally, we can estimate by m m ^ = U = 1 "U = 1 "C = C ! m i =1 i m i =1 i
70 Control Chart for Number of Defects: Cchart We can now determine the parameters for the Ccontrol chart:
! ! ^ UCL = cu /2 , CL = nc = nu = n", LCL = c! /2
! /2 ! /2 where cu and c! are the smallest resp. largest values such that ! ^ P(C > cu /2; n" = n") # 1 ! 2 ! ^ P(C < c! /2; n" = n") # 1 ! 2 71 Control Chart for Defects per Unit: Uchart Alternatively, we can construct a control chart based on the number of defects per unit. This gives the following parameters for the Ucontrol chart: UCL = ! cu /2 n ^ , CL = u = ", LCL = ! c! /2 n 72 Control Chart for Number of Defects ^ If n! > 5 we can use the normal approximation to the Poisson distribution. This yields the following parameters for the Cchart:
UCL = nu + z! /2 nu ^ CL = nu = nc = n" LCL = nu # z! /2 nu
For the Uchart we simply divide these parameters by n.
73 Example We have observed solder defects on 24 samples of five printed circuit boards each:
# defects 7 6 8 # defects/unit 1.4 1.2 1.6 # defects 4 16 11 # defects/unit 0.8 3.2 2.2 10 24 6 5 4 8 2.0 4.8 1.2 1.0 0.8 1.6 12 8 6 5 9 7 2.4 1.6 1.2 1.0 1.8 1.4 11 15 8 2.2 3.0 1.6 14 8 21 9.71 2.8 1.6 4.2 1.94 Using the Poisson distribution, the control limits for a Uchart are: ! UCL = cu /2 / n = 20 / 5 = 4.0 ^ CL = u = " = 1.94
! LCL = c! /2 / n = 2 / 5 = 0.4
74 Example: UChart
6 5 4 3 2 1 0 1 3 5 7 9 11 13 15 17 19 21 23 sample number # defects/unit # defects/unit UCLU CLU LCLU Samples 5 and 24 are out of control.
75 Example Since the control limits should be for a process that is in control, we discard sample points 5 and 24, and recompute the control limits: ! UCL = cu /2 / n = 18 / 5 = 3.6 ^ CL = u = " = 1.71 ! LCL = c! /2 / n = 1 / 5 = 0.2 Note that we also need to look for assignable causes of the out of control points!
76 Example: UChart
6 # defects/unit 5 4 3 2 1 0 1 4 7 10 13 16 19 22 sample number # defects/unit UCLU CLU LCLU 77 Example Using a normal approximation, the initial control limits for a Uchart are: UCL = 3.81, CL = 1.94, LCL = 0.0722
and the revised ones (after eliminating the same outofcontrol samples): UCL = 3.46, CL = 1.71, LCL = 0.045 ! 0 78 Control Chart Performance Effect of choice of z/2 and n: Increasing the value of z/2 increases the width of the control limits. This decreases the risk of a false alarm, i.e., a point falling beyond the control limits while the process is really in control. This is called a Type I error. 79 Control Chart Performance Unfortunately, decreasing the value of z/2 increases the probability that a measurement falls within the control limits while the process is out of control! This is called a Type II error. Note however that we can reduce both errors by increasing the sample size n! 80 Control Chart Performance We can evaluate decisions regarding sample size and sampling frequency using the average run length (ARL) of a chart. The ARL is defined as the average number of points (samples) before a point (falsely) indicates an outofcontrol situation. Denoting the probability that a point in the chart exceeds the control limits by p, 1 we have (why?) ARL = p
81 Control Chart Performance In Shewhart control charts, with 3sigma control limits (z/2=3), this means ARL = 1 p = 1 0.0027 ! 370 In other words, on average one in every 370 samples gives a false alarm (if the process is in control). 82 Control Chart Performance In the vane opening example (slide 29), suppose that we are sampling every hour. This means that we can expect one false alarm per 370 hours of production. Suppose that when the process goes out of control, the mean shifts to 30. We can now determine the probability that a particular sample detects that the process is out of control. We use the control limits based on the reduced sample
83 Control Chart Performance The probability that a measurement will detect that the process is out of control is equal to:
1 ! P LCL " X " UCL when = 30 = = 1 ! P 30.29 " X " 36.14 when = 30 # 30.29 ! 30 36.14 ! 30 & =1! P % "Z" ( $ 2.18 / 5 2.18 / 5 ' = P 0.30 " Z " 6.30 ( ( ) ) ( ) ) 1 ! (1 ! 0.62) = 0.62
84 Control Chart Performance The expected number of samples (and thus hours) until the shift is detected is equal to: 1 0.62 ! 1.62 If this is unacceptable, we can improve the performance in two ways: Increase the sample size Increase the sampling frequency 85 Control Chart Performance Increasing the sample size from 5 to 10 would narrow the control limits, and the probability that a shift in mean to 30 will be detected by a sample becomes: 1 ! P 30.93 " X " 35.67 when = 30 # 31.13 ! 30 35.27 ! 30 & =1! P % "Z" ( $ 2.18 / 10 2.18 / 10 ' = 1 ! P 1.64 " Z " 7.64 ) 0.95
which reduces the expected time to detection to 1/.95=1.05 hours.
86 ( ) ( ) Control Chart Performance Alternatively, we could increase the sampling frequency to once per half hour, while maintaining a sample size of 5. This reduces the expected time to detection to 1.62/2 = 0.81 hours. So, either doubling the sampling frequency or doubling the sample size will decrease the expected time until detection. 87 Other Control Charts A disadvantage of Shewhart control charts is that they are relatively insensitive to small shifts in the process. Basic Shewhart control charts only use the last observation (sample). The Western Electric rules use past information also, but they increase the false alarm rate. An alternative is the cumulative sum (CUSUM) control chart.
88 Process Capability The performance of the process when operating in control is often called the capability of the process. We judge the performance by a comparison with specification limits, which specify the quality requirements of the process. A centered process is one for which the process mean is equal to the midpoint of the upper and lower specification limits.
89 Process Capability Tools: Tolerance (or tier) chart Plot of all measurements, grouped by sample Histogram Frequency of occurrence of certain values for the quality characteristic 90 Process Capability Reconsider the vane opening example. Suppose that the nominal value of the characteristic of interest is 37.5, with lower and upper specification limits equal to 30 and 45, respectively. From the following graphs, it is clear that the process is not centered. 91 Tolerance Chart 92 Histogram 93 Process Capability Ratio The process capability ratio (PCR) is defined as PCR = USL ! LSL 6" Here 6 is called the width of the process, since most of the units should fall within an interval of this width around the process mean. The PCR has a natural interpretation for centered processes. 94 Process Capability Ratio PCR > 1 It is expected that very few products that do not conform to the specification limits are produced. PCR = 1 Assuming a normally distributed process, the expected fraction nonconforming is 0.27%, or 2700 parts per million. PCR < 1 It is expected that a large number of products will not conform to the specification limits. 95 Process Capability Ratio 96 Process Capability Ratio Suppose that in the vane opening example, we have USL = 45 LSL = 30 The estimate of based on the reduced set of observations is 2.18, which gives = 1.15 6" 6 # 2.18 so the process capability seems ok.
97 PCR = USL ! LSL = 45 ! 30 Process Capability Ratio If the process is not centered, the PCR measures potential capability rather than actual capability. We can refine the definition of PCR to account for noncentered processes: # USL ! ! LSL & PCR k = min % , ( 3" ' $ 3"
where is the process mean.
98 Process Capability Ratio Again considering the cigarettelighter example, where we estimated =10.71: # USL ! ! LSL & PCR k = min % , ( 3" ' $ 3" # 45 ! 33.2 33.2 ! 30 & = min % , ( $ 3 ) 2.18 3 ) 2.18 ' = min 1.80,0.49 = 0.49
so the actual process capability is poor!
99 ( ) Process Capability Ratio 100 Motorola's SixSigma Criterion PCR > 2 If the process mean shifts by 1.5, the expected fraction nonconforming is only 3.4 parts per million Even if the process mean shifts by as much as 3, the expected fraction nonconforming is still only 0.27%, or 1,350 parts per million 101 Statistical Quality Control The discussed techniques do not improve the quality of the process or product they simply quantify and monitor problems and/or changes that take place in the process. At least as important are the actions that follow a warning sign of a SQC tool. 102 ...
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This note was uploaded on 05/12/2010 for the course ESI 6321 taught by Professor Josephgeunes during the Spring '07 term at University of Florida.
 Spring '07
 JosephGeunes

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