20/05/1436 University of Hail College of Engineering ISE 320 - Quality Control and Industrial Statistics CHAPTER 04 CONTROL CHARTS FOR ATTRIBUTES Professor Mohamed Aichouni http://faculty.uoh.edu.sa/m.aichouni/ise230-quality/ Introduction • Many quality characteristics cannot be conveniently represented numerically. • In such cases, each item inspected is classified as either conforming or nonconforming to the specifications on that quality characteristic. • Quality characteristics of this type are called attributes. attributes • Examples are nonfunctional semiconductor chips, warped connecting rods, etc,. 1 20/05/1436 Types of Control Charts Control Charts for Variables Data X and R charts: for sample averages and ranges. X and s charts: for sample means and standard deviations. deviations Md and R charts: for sample medians and ranges. X charts: for individual measures; uses moving ranges. Control Charts for Attributes Data p charts: proportion of units nonconforming. np charts: number of units nonconforming. c charts: count of nonconformities. u charts: count of nonconformities per unit. Control Chart Selection Quality Characteristic Variable Attribute Defective n>1? no x and MR yes no n>=10? Defect x and R constant sample size? yes no yes x and s 2 p-chart with variable sample size constant sampling unit? p or np yes no c u 20/05/1436 Type of Attribute Charts p charts • This chart shows the fraction of nonconforming or defective product produced by a manufacturing process. • It is also called the control chart for fraction nonconforming. np charts • This chart shows the number of nonconforming. Almost the same as the p chart. c charts • This shows the number of defects or nonconformities produced by a manufacturing process. u charts • This chart shows the nonconformities per unit produced by a manufacturing process. p charts 3 20/05/1436 p charts • In this chart, we plot the percent of d f ti defectives ((per b batch, t h per d day, per machine, hi etc.). • However, the control limits in this chart are not based on the distribution of rate events but rather on the binomial distribution (of proportions). Formula • Fraction nonconforming: p = (np)/n • where p = proportion or fraction nc in the sample or subgroup, • n = number in the sample or subgroup, • np = number nc in the sample or subgroup. 4 20/05/1436 Example • During the first shift, 450 inspection are made of bookof the month shipments and 5 nc units are found. Production during the shift was 15,000 units. What is the fraction nc? p = (np)/n = 5/450 = 0.011 • The p, is usually small, say 0.10 or less. • If p > 0.10, indicate that the organization is in serious difficulty. p-Chart contruction for constant subgroup size • • • • Select S l t the th quality lit characteristics. h t i ti Determine the subgroup size and method Collect the data. Calculate the trial central line and control limits. • Establish E tabli h the revised e i ed ce central t al li line ea and d co control t ol limits. • Achieve the objective. 5 20/05/1436 Select the quality characteristics The quality characteristic? – A single quality characteristic – A group of quality characteristics – A part – An entire product, or – A number of products. products Determine the subgroup size and method • The size of subgroup is a function of the proportion nonconforming. g • If p = 0.001, and n = 1000, then the average number nc, np = 1. Not good, since a large number of values would be zero. • If p = 0.15, and n = 50, then np = 7.5, would make a good chart. • Therefore, the selection subgroup size requires some preliminary observations to obtain a rough idea of the proportion nonconforming. 6 20/05/1436 Collect the data • The quality technician will need to collect sufficient ffi i t data d t for f att least l t 25 subgroups. b • The data can be plotted as a run chart. • Since the run chart does not have limits, its is not a control chart. Calculate the trial central line and control limits • The formula: p= ∑ np ∑n UCL = p + 3 p (1 − p ) n LCL = p − 3 p (1 − p ) n • = average of p for many subgroups • n = number inspected in a subgroup 7 20/05/1436 ∑ np = 138 = 0.018 p= ∑ n 7500 Subgroup Number Number Inspected n np p 1 300 12 0.040 0.018(1 − 0.018) UCL = 0.018 + 3 300 = 0.041 0.018(1 − 0.018) LCL = 0.018 − 3 300 = −0.005 = 0.0 2 300 3 0.010 3 300 9 0.030 4 300 4 0.013 5 300 0 0.0 6 300 6 0.020 7 300 6 0.020 8 300 1 0.003 19 300 16 0.053 300 2 0.007 7500 138 25 Total Negative value of LCL is possible in a theoritical result, but not in practical (proportion of nc never negative). p Chart 0.053 p UCL 0.04 0.03 0.02 p-bar 0.01 LCL 0 5 10 15 20 Subgroup 8 25 20/05/1436 Establish the revised central line and control limits • Determine the standard or reference value f the for th proportion ti nc, po. pnew np − np ∑ = ∑n − n d d • where npd = number nc in the discarded subgroups • nd = number inspected in the discarded subgroups Revised control limits po = pnew pnew = p (1 − po ) UCL = po + 3 o n LCL = po − 3 po (1 − po ) n • where po is central line 9 138 − 16 = 0.017 7500 − 300 UCL = 0.017 + 3 0.017(1 − 0.017) 300 = 0.039 0.017(1 − 0.017) 300 = −0.005 = 0.0 LCL C = 0.017 01 − 3 20/05/1436 Control Charts for Fraction Nonconforming (p chart) Example 2 Example p • A process that produces bearing housings is investigated. Ten samples of size 100 are selected. Sample # # Nonconf. 1 5 2 2 3 3 4 8 5 4 6 1 7 2 8 6 9 3 10 4 • Is this process operating in statistical control? Example p n = 100, m = 10 Sample # # Nonconf. Fraction Nonconf. 1 5 2 2 3 3 4 8 5 4 6 1 7 2 8 6 9 3 10 4 0.05 0.02 0.03 0.08 0.04 0.01 0.02 0.06 0.03 0.04 m p = 10 ∑ pˆ i i=1 m = 0 . 038 20/05/1436 Example p Control Limits are: 0.038(1 − 0.038) = 0.095 100 UCL = 0.038 + 3 CL = 0.038 0.038(1 − 0.038) = −0.02 → 0 100 LCL = 0.038 + 3 C chart – Example 2 P Chart for C1 Proportion 0.10 3.0SL=0.09536 0.05 P=0.03800 0.00 - 3.0SL=0.000 0 1 2 3 4 5 6 7 Sampl e Number 11 8 9 10 20/05/1436 np Chart • The np chart is almost the same as the p chart. C t l liline = npo Central UCL = npo + 3 npo (1 − po ) LCL = npo − 3 npo (1 − po ) • If po is unknown, it must be determined by collecting data, calculating UCL, LCL. Example 12 Subgroup n np UCL np -bar LCL 1 2 3 4 5 300 300 300 300 300 3 6 4 6 20 12.0 12 0 12.0 12.0 12.0 12.0 5.24 5 24 5.24 5.24 5.24 5.24 0.0 00 0.0 0.0 0.0 0.0 21 22 23 24 25 300 300 300 300 300 2 3 6 1 8 12.0 12.0 12.0 12.0 12.0 5.24 5.24 5.24 5.24 5.24 0.0 0.0 0.0 0.0 0.0 20/05/1436 c Chart • The procedures for c chart are the same a s those for the p chart. chart • If count of nonconformities, co, is unknown, it must be found by collecting data, calculating UCL & LCL. UCL = c + 3 c c= 13 LCL = c − 3 c c = average count of nonconformities g 20/05/1436 Example ID Number Subgroup MY102 1 MY113 MY121 MY125 MY132 MY143 MY150 MY152 MY164 MY166 MY172 MY267 MY278 MY281 MY288 c UCL 2 3 4 5 6 7 8 9 10 11 7 6 6 3 20 8 6 1 0 5 14 12.76 12.76 12.76 12.76 12.76 12.76 12.76 12.76 12.76 12.76 12.76 c -bar c 0141 5.64 c= = = 5.64 5.64 g 0 25 5.64 0 5.64 0 5.64 0 + 3 5.64 UCL = 5.64 5.64 0 5.64 0 LCL = 5.640− 3 5.64 5.64 5.64 = −1.48 0 =0 5.64 0 5.64 0 22 23 24 25 4 14 4 5 12.76 12.76 12.76 12.76 5.64 5.64 5.64 5.64 LCL = 12 .76 0 0 0 0 c-Chart 25 20 Count of Nonconformities c UCL c-bar 15 LCL 10 5 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Subgroup Num ber 14 16 17 18 19 20 21 22 23 24 25 20/05/1436 Revised • Out-of-control: sample no. 5, 11, 23. c new = c − c d 141 − 20 − 14 − 14 = = 4.23 g − gd 25 − 3 UCL = co + 3 co = 4.23 + 3 4.23 = 10.40 LCL = co − 3 co = 4.23 − 3 4.23 = −1.94 = 0 u Chart • The u chart is mathematically equivalent to the c chart. chart u= c n u= ∑c ∑n u UCL = u + 3 n 15 LCL = u − 3 u n 20/05/1436 Example ID Number Subgroup 30-Jan 1 31-Jan 1-Feb 2-Feb 3-Feb 4-Feb 28 F b 28-Feb 1-Mar 2-Mar 3-Mar 4-Mar u= n c u UCL u -Bar LCL 2 3 4 5 6 110 82 96 115 108 56 120 94 89 162 150 82 1.091 1.146 0.927 1.409 1.389 1.464 1.51 1.56 1.54 1.51 1.52 1.64 1.20 1.20 1.20 1.20 1.20 1.20 0.89 0.84 0.87 0.89 0.88 0.76 26 27 28 29 30 101 122 105 98 48 105 143 132 100 60 1.040 1 040 1.172 1.257 1.020 1.250 1.53 1 53 1.50 1.52 1.53 1.67 1.20 1 20 1.20 1.20 1.20 1.20 0.87 0 87 0.90 0.88 0.87 0.73 • For January 30: u Jan 30 = 16 ∑ c = 3389 = 1.20 ∑ n 2823 c 120 = = 1.09 n 110 UCLJan 30 =1.20 + 3 1.20 = 1.51 110 LCL Jan 30 =1.20 − 3 1.20 = 0.89 110 20/05/1436 Additional Example from Manufacturing Surface defects have b been counted t d on 25 rectangular steel plates, and the data are shown below. The control chart for nonconformities is set up using this data. Is the Process under Control? Comment. 17 20/05/1436 Additional Example from Manufacturing Nonconformity Classification • Critical nonconformities – Indicate hazardous or unsafe conditions. • Major nonconformities – Failure • Minor Mi nonconformities f iti 18 20/05/1436 Control Charts for Variables vs. Charts for Attributes • Sometimes, the quality control engineer has a choice between variable control charts and attribute control charts. Advantages of attribute control charts • Allowing for quick summaries, that is, the engineer may simply classify products as acceptable or unacceptable, based on various quality criteria. • Thus, attribute charts sometimes bypass the need for expensive, precise devices and time-consuming measurement procedures. y managers g unfamiliar with • More easilyy understood by quality control procedures. 19 20/05/1436 Advantages of variable control charts • More sensitive than attribute control charts charts. • Therefore, variable control charts may alert us to quality problems before any actual "unacceptables" (as detected by the attribute chart) will occur. • Montgomery (1985) calls the variable control charts leading indicators of trouble that will sound an alarm before the number of rejects (scrap) increases in the production process. Home Work 4.1 • Study the process stability using the data shown on the table on non conforming units in the g manufacturing process. 20 20/05/1436 Conclusion "Quality control truly begins and ends g with education", K. Ishikawa (1990). Lecture Finished Any Question? No Yes Ask questions Teachers answers Train your self (Google, YouTube, course webpage End (See you next lecture) 21

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