ISyE 512 Chapter 6. Control Charts for Variables. Instructor: Prof. Kaibo Liu. Department of Industrial and Systems Engineering UW-Madison

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1 ISyE 512 Chapter 6 Control Charts for Variables Instructor: Prof. Kaibo Liu Department of Industrial and Systems Engineering UW-Madison kliu8@wisc.edu Office: oom 3017 (Mechanical Engineering Building) 1

2 List of Topics in Chapter 6 Control Chart for and Control Chart for and S Operating-Characteristic Function 2

3 eview of the Basic Model of Control Charts Let w be a sample statistic that measures some quality characteristic of interest, and suppose that the mean of w is µ w and the standard deviation of w is σ w. Then the center line, the upper control limit, and the lower control limit become UCL = µ w + k σ w Center line = µ w LCL = µ w - k σ w where k is the "distance" of the control limits from the center line, expressed in standard deviation units

4 Control Chart for and Known µ,σ X Statistical Basis of the Charts suppose {x ij, i=1,,m, j=1,,n} are normally distributed with x ij,~n(µ,σ 2 2 ), thus, ~ N ( µ,( σ / n ) ) X i X bar chart monitors between-sample variability (variability over time) and chart measures within-sample variability (instantaneous variability at a given time) If µ and σ are known, X bar chart is (if k=3) σ µ ± 3σ x µ ± 3 µ ± Aσ n LCL = µ Aσ CL ULC = µ = µ + Aσ A = 3 n 4

5 Control Chart for and Known µ,σ X ange i =max(x ij )-min(x ij ) for j=1,..n If µ and σ are known, the statistical basis of charts is as follows: Define the relative range W=/σ. The parameters of the distribution of W are a function of the sample size n. Denote µ W =E(W)=d 2, σ W =d 3, (d 2 and d 3, are given in Appendix Table VI of Textbook P720) µ =d 2 σ, σ =d 3 σ, which are obtained based on =W σ chart control limits (if k=3) µ ± σ d σ ± 3d σ (d ± 3d ) σ Chart Statistic LCL = D CL = d 2 σ ULC = D 1 2 σ σ D D 1 2 = d = d 2 2 3d + 3d 3 3 5

6 Control Chart for X and Unknown µ and σ Need to estimate µ and σ µ ˆ x = X = m m x m n mn x i ij i = 1 i= 1 j= 1 = ; ; = σ ˆ = d 2 m i= 1 m i X bar chart (if k=3) σˆ / d2 µ ˆ x ± 3ˆ σx x ± 3 x ± 3 x ± A2 n n LCL = x A CL = x ULC = x + A 2 2 A 2 = d 2 3 n A 2 is determined by n, Appendix Table VI 6

7 Control Chart for and Unknown µ and σ (cont s) X m 1 i i d d ˆ d ˆ ; m ˆ = σ = σ = = µ = Need to estimate based on =W σ, σ W =d 3, chart (if k=3) µ, σ d 2 ˆ = σ ) d d 3 (1 d d 3 3 ˆ ˆ ± ± σ ± µ D ULC CL D LCL 4 3 = = = d 3d 1 D d 3d 1 D + = = D 3 and D 4 are determined by n, Appendix Table VI 7

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9 Procedures for Establishment of Control Limits Unknown µ and σ If µ and σ are unknown, we need to estimate µ and σ based on the preliminary in-control data (normally m=20~25). The control limits established using the preliminary data are called trial control limits, which are used to check whether the preliminary data are in control. First check or S chart to ensure all data in-control, and then check X-bar chart Collect Preliminary Data X Estimate or S Establish Trial Control Limits Check Preliminary Data In-control Future Monitoring Update Estimation Eliminate the Outliers due to Assignable Causes Out-of-control 9

10 Example: The data shown here are x-bar and values for 24 samples of size n=5 taken from a process producing bearings. The measurements are made on the inside diameter of the bearing. (a) Set up x-bar and charts on this process. Does the process seem to be in statistical control? If necessary, revise the trial control limits. (b) Estimate the process standard deviation x

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12 X-bar chart and Chart x X Bar Chart UCL=36.7 CL=34 LCL= Chart UCL=9.96 CL=4.7 LCL=0 Sample 12 and 15 out of control 12

13 evised X-bar Chart and Chart emoving the two out of control samples and recalculate control limits for both charts: 40.0 evised X Bar Chart UCL=36.3 CL=33.7 LCL= evised Chart UCL=9.5 CL=4.5 LCL=0 13

14 eview X-bar Chart and Chart X-bar chart and chart (µ, σ known/unknown) - Known µ and σ: A, D 1, D 1, d 2, d 3 - Unknown µ and σ: A 2, D 3, D 4, d 2, d 3 Procedure of control charting Estimation of the process parameters, µ and σ. - Can ONLY be done AFTE the process exhibits in control Other SPC techniques: read text yourself

15 Statistical Basis eview of Control Chart for X-bar Known µ,σ suppose {x j, j=1,,n} are normally distributed with x j ~N(µ,σ 2 ), thus, 2 X ~ N( µ,( σ / If µ and σ are known, X-bar chart is Independently and identically distributed, i.i.d n ) ) α= 2[1-Φ(k)] Let k = 3, A = 3 n LCL CL ULC = = µ µ = µ + A σ A σ 15

16 Interpretation of X-bar and Chart First check the chart and eliminate the assignable causes from chart, and then check the X bar chart Check non-random pattern Cyclic pattern due to temperature, regular rotation of operators or machines, maintenance schedules, tool wear (Fig. 6-8) 16

17 Interpretation of X-bar and Chart Check non-random pattern (cont d) Mixture pattern when the plotted points tend to fall near or slightly outside the control limits. Two overlapping distributions are resulted from too often process adjustment (Fig 6-9). 17

18 Interpretation of X-bar and Chart Check non-random pattern (cont d) Shift in process level due to introduction of new workers, methods, materials, or inspection standard (Fig. 6-10) 18

19 Interpretation of X-bar and Chart Check non-random pattern (cont d) Trend pattern due to gradual tool wear (Fig. 6-11) 19

20 Interpretation of X-bar and Chart Check non-random pattern (cont d) Stratification pattern for the points to cluster around the center line due to incorrect calculation of Control limits or inappropriate reasonable sampling group (Fig. 6-12) 20

21 eview of Type I and Type II Error You Conclude : Statistically in control "In Control" "In Control" Confidence 1 α eality Nature : "Out of Control" Consumer Error, β "Out of Control" Statistically out of control Producer Error, α Pow er 1 β 21

22 Application Conditions of X bar and chart Underlying distribution of the quality characteristics is normal X bar chart is more robust to nonnormality than chart Sample size of 4 or 5 are sufficient to ensure reasonable robustness to the normality assumption for X bar chart Calculation accuracy of Type I error is dependent on the distribution X bar chart (n=4, 5, 6) is not effective to detect a small mean shift (less than 1.5 σ) on the first sample following the shift (OC curve) If n>10, an s chart should be used instead of a chart 22

23 X bar chart OC Curve for x bar and Chart UCL = µ + kσ / n; LCL = µ kσ / n; 0 0 [ Φ k ] µ in control 0 α = 2 1 ( ) µ µ σ 1 = 0 + d, β = P{ LCL x UCL µ = µ + dσ} 1 0 = Pr{ x UCL µ } Pr{ x LCL µ } 1 1 = Φ k d n Φ k d n out of control The expected number of samples taken before the shift is detected (the process is o.o.c) AL out of control = r= 1 rβ r 1 1 (1 β) = 1 β If process is in control: AL is the expected number of samples until a false alarm occurs AL in control = i= 1 k k 1 ( 1 α) ( α) 1 = α

24 OC Curve for X bar chart (Cont d) If the shift is 1.0σ and the sample size is n = 5, then β = Increase sample size reduce β The larger is the mean-shift, the smaller is β 24

25 OC Curve for chart (Cont s) Increase sample size reduce β The larger is the std deviation, the smaller is β 25

26 AL for X bar The larger the mean-shift, the smaller the AL Increasing sample size leads to increased total individual units measured 26

27 Example: A quality characteristic with normal distribution is monitored through use of an x-bar and an chart. These charts have the following parameters (n=4): X-bar chart UCL=626 CL=620 LCL=614 chart UCL= CL=8.236 LCL=0 (a) (b) What is the probability of detecting a shift in the process mean to 610 on the first sample following the shift. Find the average run length for the chart. (Assume sigma does NOT change) What is the probability of detecting the shift in (a) by at least the third sample after the shift? 27

28 S Chart (if k=3) 28

29 S Chart (if k=3) 29

30 Summary of Control Charts (if k=3) Process Parameters X bar chart chart S chart known µ σ LCL = µ Aσ CL = µ ULC = µ + Aσ LCL = D CL = d 2 σ ULC = D 1 2 σ σ LCL = B CL = c 4 σ ULC = B 5 6 σ σ X bar & chart µ ˆ = X σˆ = d 2 LCL = x A CL = x ULC = x + A 2 2 LCL = D CL = ULC = D 3 4 X bar & S chart ˆ = X µ σˆ = S c 4 ; LCL = x A CL = x ULC = x + A 3 3 S S LCL = B CL = S ULC = B 3 4 S S 30

31 Comparison of Chart and S Chart Chart simple for hand calculation (note: no big deal nowadays); good for small sample size; lose information between x min and x max; not used for variable sample size. S Chart when the sample size is large (n>10); Used for variable sample size ; Computation complexity can be simplified by using a computer. 31

32 X bar and S Control Chart with Variable Sample Size Use a weighted average approach in calculating x and S x m i= 1 = m i= 1 n A 3, B 3, and B 4 will use the corresponding sample size of each subgroup (Please see the following Example) i n x i i S = m i= 1 m i= 1 If fixed control limits are preferred, use an average sample size n, or use the modal ( most common) sample size if n i are not very different (n i n 1)S i m 2 i 1/ 2 m: # of samples/subgroups; n i : sample size for each sample/subgroup i 32

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36 4 obs. 5 obs. 36

37 Every sample has its own control chart parameters, UCL and LCL Samples with the same sample size have the same chart parameters 37

38 Example: A normally distributed quality characteristic is monitored through use of an x-bar and an chart. These charts have the following parameters (n=4): X-bar chart UCL=626 CL=620 LCL=614 chart UCL= CL=8.236 LCL=0 Both charts exhibit control. : can be used to estimate process parameters, µ and σ (a) Suppose an S chart were to be substituted for the chart. What would be the appropriate parameters of the S chart? (b) If specifications on the product were 610 ± 15, what would be your estimate of the process fraction nonconforming? (consider each individual product) (c) What could be done to reduce this fraction nonconforming? 38

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40 Estimation of the Process Capability Get process specification limits (USL, LSL) Estimate σ based on ˆ = / ( chart) or σˆ = S / c (S chart) σ d 2 4 Estimate the fraction of nonconforming products p (or p 10 6 PPM nonconforming units) pˆ = Pr{x < LSL} + Pr{x Process-Capability atio > LSL x USL x USL} = Φ( ) + 1 Φ( ) σˆ σˆ ; ; Estimated process stdev C p =1 means the process uses up 100% tolerance band with 0.27% (2700PPM) nonconforming units Percentage of the specification band that the process uses up P=(1/C p )*100% 40

41 Differences among NTL, CL and SL and Impact on Process Capability There is no relationship between control limits and specification limits. C p is an index relating natural tolerance limits to specification limits. Externally determined NTL: natural tolerance limits 3σ n Center line on x bar Distribution of x bar values Distribution of individual process measurement LSL LNTL 3σ LNTL 3σ LSL µ µ UNTL 3σ UNTL 3σ USL USL C p >1, P< 0.27% C p =1, P= 0.27% LNTL LSL 3σ µ USL Externally determined Width defined by NTL s is larger than that defined by CL, why? 3σ C p <1, P> 0.27% UNTL 41

42 Example: Control charts for X and are maintained for an important quality characteristic. The sample size is n = 7; X and are computed for each sample. After 35 samples, we have found that 35 i =1 35 x i = 7805 and i =1200 i=1 (a) Set up X and charts using these data (if k=3). (b) Assuming that both charts exhibit control, estimate the process mean and standard deviation. (c) If the quality characteristic is normally distributed and if the specifications are 220 ± 35, can the process meet the specifications? Estimate the fraction nonconforming. (d) Assuming the variance to remain constant, state where the process mean should be located to minimize the fraction nonconforming. What would be the value of the fraction nonconforming under these conditions? 42

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44 Example 12-3: Samples of n=8 items each are taken from a manufacturing process at regular intervals. A quality characteristic is measured, and x-bar and values are calculated for each sample. After 50 samples, we have x i = 2000 and i= 1 i= 1 i = 250 Assume that the quality characteristic is normally distributed. (a) Compute control limits for the x-bar and control charts. (b) All points on both control charts fall between the control limits computed in (a). What are the natural tolerance limits of the process. (c) If the specification limits are 41±5.0, what is the process capability? 44

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