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BMM 3633 Industrial Engineering Learning Objectives:  Define the concept and application of SPC chart.  Construct a control chart for variable and.

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Presentation on theme: "BMM 3633 Industrial Engineering Learning Objectives:  Define the concept and application of SPC chart.  Construct a control chart for variable and."— Presentation transcript:

1

2 BMM 3633 Industrial Engineering

3 Learning Objectives:  Define the concept and application of SPC chart.  Construct a control chart for variable and attribute data.  Calculate process capability ratio (C p ) and process capability index (C pk ).

4 Contents:  Statistical Process Control  Control Charts  Development of Control Chart  Control Chart Patterns  Process Capability

5 Statistical Process Control Take periodic samples from process Plot sample points on control chart Determine if process is within limits Prevent quality problems UCL LCL

6 Variation Common Causes Variation inherent in a process Can be eliminated only through improvements in the system Special Causes Variation due to identifiable factors Can be modified through operator or management action Statistical Process Control (cont..)

7 Types of Data Attribute data Product characteristic evaluated with a discrete choice Good/bad, yes/no Variable data Product characteristic that can be measured Length, size, weight, height, time, velocity Statistical Process Control (cont..)

8 Control Charts Graph establishing process control limits Charts for variables Mean (x-bar), Range (R) Charts for attributes p and c

9 Process Control Chart 12345678910 Sample number Uppercontrollimit Processaverage Lowercontrollimit Out of control Control Charts (cont..)

10 A Process is In Control if 1.No sample points outside limits 2.Most points near process average 3.About equal number of points above & below centerline 4.Points appear randomly distributed Control Charts (cont..)

11 Development of Control Chart Based on in-control data If non-random causes present discard data Correct control chart limits

12 Development of Control Chart (cont..) Control Charts for Attributes p Charts Calculate percent defectives in sample c Charts Count number of defects in item

13 Control Charts for Variables Mean chart ( x-Chart ) Uses average of a sample (control the central tendency of the process) Range chart ( R-Chart ) Uses amount of dispersion in a sample (control the dispersion of the sample) Development of Control Chart (cont..)

14 Setting Chart Limits For x-Charts when we know s Upper control limit (UCL) = x + z  x Lower control limit (LCL) = x - z  x wherex=mean of the sample means or a target value set for the process z=number of normal standard deviations  x =standard deviation of the sample means =s/ n s=population standard deviation n=sample size

15 Hour 1 SampleWeight of NumberOat Flakes 117 213 316 418 517 616 715 817 916 Mean16.1  =1 HourMeanHourMean 116.1715.2 216.8816.4 315.5916.3 416.51014.8 516.51114.2 616.41217.3 n = 9 LCL x = x - z  x = 16 - 3(1/3) = 15 ozs For 99.73% control limits, z = 3 UCL x = x + z  x = 16 + 3(1/3) = 17 ozs Example

16 17 = UCL 15 = LCL 16 = Mean Control Chart for sample of 9 boxes Sample number |||||||||||| 123456789101112 Variation due to assignable causes Variation due to natural causes Out of control Example (cont..)

17 For x-Charts when we don’t know s Lower control limit (LCL) = x - A 2 R Upper control limit (UCL) = x + A 2 R whereR=average range of the samples A 2 =control chart factor found in the table x=mean of the sample means Setting Chart Limits (cont..)

18 Control Chart Factors Sample Size Mean Factor Upper Range Lower Range n A 2 D 4 D 3 21.8803.2680 31.0232.5740 4.7292.2820 5.5772.1150 6.4832.0040 7.4191.9240.076 8.3731.8640.136 9.3371.8160.184 10.3081.7770.223 12.2661.7160.284 Setting Chart Limits (cont..)

19 Process average x = 12 ounces Average range R =.25 Sample size n = 5 Example

20 UCL x = x + A 2 R = 12 + (.577)(.25) = 12 +.144 = 12.144 ounces Process average x = 12 ounces Average range R =.25 Sample size n = 5 From Table Example (cont..)

21 UCL x = x + A 2 R = 12 + (.577)(.25) = 12 +.144 = 12.144 ounces LCL x = x - A 2 R = 12 -.144 = 11.857 ounces Process average x = 12 ounces Average range R =.25 Sample size n = 5 UCL = 12.144 Mean = 12 LCL = 11.857 Example (cont..)

22 Exercise OBSERVATIONS (SLIP-RING DIAMETER, CM) SAMPLE k 12345xR 15.025.014.944.994.964.980.08 25.015.035.074.954.965.00 0.12 34.995.004.934.924.994.97 0.08 45.034.915.014.984.894.96 0.14 54.954.925.035.055.014.99 0.13 64.975.065.064.965.035.01 0.10 75.055.015.104.964.995.02 0.14 85.095.105.004.995.085.05 0.11 95.145.104.995.085.095.08 0.15 105.014.985.085.074.995.03 0.10 50.091.15

23 Restaurant Control Limits For salmon filets at Darden Restaurants Sample Mean x Bar Chart UCL = 11.524 x – 10.959 LCL – 10.394 ||||||||| 1357911131517 11.5 – 11.0 – 10.5 – Sample Range Range Chart UCL = 0.6943 R = 0.2125 LCL = 0 ||||||||| 1357911131517 0.8 – 0.4 – 0.0 –

24 Specifications LSL 10 USL 12 Capability Mean = 10.959 Std.dev = 1.88 C p = 1.77 C pk = 1.7 Capability Histogram LSLUSL 10.210.510.811.111.411.712.0 Restaurant Control Limits (cont..)

25 R – Chart Type of variables control chart Shows sample ranges over time Difference between smallest and largest values in sample Monitors process variability Independent from process mean

26 Setting Chart Limits For R-Charts Lower control limit (LCL R ) = D 3 R Upper control limit (UCL R ) = D 4 R where R=average range of the samples D 3 and D 4 =control chart factors from Table

27 Example UCL R = D 4 R = (2.115)(5.3) = 11.2 pounds LCL R = D 3 R = (0)(5.3) = 0 pounds Average range R = 5.3 pounds Sample size n = 5 From Table, D 4 = 2.115, D 3 = 0 UCL = 11.2 Mean = 5.3 LCL = 0

28 Exercise OBSERVATIONS (SLIP-RING DIAMETER, CM) SAMPLE k 12345xR 15.025.014.944.994.964.980.08 25.015.035.074.954.965.00 0.12 34.995.004.934.924.994.97 0.08 45.034.915.014.984.894.96 0.14 54.954.925.035.055.014.99 0.13 64.975.065.064.965.035.01 0.10 75.055.015.104.964.995.02 0.14 85.095.105.004.995.085.05 0.11 95.145.104.995.085.095.08 0.15 105.014.985.085.074.995.03 0.10 50.091.15

29 Mean and Range Charts (a) These sampling distributions result in the charts below (Sampling mean is shifting upward but range is consistent) R-chart (R-chart does not detect change in mean) UCL LCL x-chart (x-chart detects shift in central tendency) UCL LCL

30 R-chart (R-chart detects increase in dispersion) UCL LCL (b) These sampling distributions result in the charts below (Sampling mean is constant but dispersion is increasing) x-chart (x-chart does not detect the increase in dispersion) UCL LCL Mean and Range Charts (cont..)

31 CHAPTER 9 Using x- and R-Charts Together Each measures the process differently Both process average and variability must be in control Mean and Range Charts (cont..)

32 The Normal Distribution  =0 1111 2222 3333 -1  -2  -3  95% 99.74%

33 Steps In Creating Control Charts 1.Take samples from the population and compute the appropriate sample statistic 2.Use the sample statistic to calculate control limits and draw the control chart 3.Plot sample results on the control chart and determine the state of the process (in or out of control) 4.Investigate possible assignable causes and take any indicated actions 5.Continue sampling from the process and reset the control limits when necessary

34 Control Charts for Attributes For variables that are categorical Good/bad, yes/no, acceptable/unacceptable Measurement is typically counting defectives Charts may measure Percent defective (p-chart) Number of defects (c-chart)

35 Control Limits for p-Charts Population will be a binomial distribution, but applying the Central Limit Theorem allows us to assume a normal distribution for the sample statistics UCL p = p + z  p ^ LCL p = p - z  p ^ wherep=mean fraction defective in the sample z=number of standard deviations ð p =standard deviation of the sampling distribution n=sample size ^ p(1 - p) n ð p = ^

36 CHAPTER 9 Control Chart Z Values Smaller Z values make more sensitive charts Z = 3.00 is standard Compromise between sensitivity and errors Control Limits for p-Charts (cont..)

37 Example SampleNumberFractionSampleNumberFraction Numberof ErrorsDefectiveNumberof ErrorsDefective 16.06116.06 25.05121.01 30.00138.08 41.01147.07 54.04155.05 62.02164.04 75.051711.11 83.03183.03 93.03190.00 102.02204.04 Total = 80 (0.04)(1 - 0.04) 100 s p = = 0.02 ^ p = = 0.04 80 (100)(20) Samples of work of 20 clerks; each clerk entered 100 records

38 .11 –.10 –.09 –.08 –.07 –.06 –.05 –.04 –.03 –.02 –.01 –.00 – Sample number Fraction defective |||||||||| 2468101214161820 UCL p = p + z  p = 0.04 + 3(0.02) = 0.10 ^ LCL p = p - z  p = 0.04 - 3(0.02) = 0 ^ UCL p = 0.10 LCL p = 0.00 p = 0.04 Example (cont..)

39 .11 –.10 –.09 –.08 –.07 –.06 –.05 –.04 –.03 –.02 –.01 –.00 – Sample number Fraction defective |||||||||| 2468101214161820 UCL p = p + z  p = 0.04 + 3(0.02) = 0.10 ^ LCL p = p - z  p = 0.04 - 3(0.02) = 0 ^ UCL p = 0.10 LCL p = 0.00 p = 0.04 Possible assignable causes present Example (cont..)

40 Exercise 20 samples of 100 pairs of jeans SampleNumberFractionSampleNumberFraction Numberof ErrorsDefectiveNumberof ErrorsDefective 16.06119.09 212.121214.14 30.001310.10 49.091412.12 510.10158.08 612.121614.14 78.081711.11 816.16186.06 912.12195.05 108.082018.18 Total = 200

41 Control Limits for c-Charts Population will be a Poisson distribution, but applying the Central Limit Theorem allows us to assume a normal distribution for the sample statistics wherec=mean number defective in the sample UCL c = c + 3 cLCL c = c - 3 c

42 c = 54 complaints/9 days = 6 complaints/day |1|1 |2|2 |3|3 |4|4 |5|5 |6|6 |7|7 |8|8 |9|9 Day Number defective 14 – 12 – 10 – 8 – 6 – 4 – 2 – 0 – UCL c = c + 3 c = 6 + 3 6 = 13.35 LCL c = c - 3 c = 6 - 3 6 = 0 UCL c = 13.35 LCL c = 0 c = 6 Example Number of complaints: 3, 0, 8, 9, 6, 7, 4, 9, 8 over 9-day period

43 The number of defects in 15 sample rooms Exercise SampleNumber SampleNumberSampleNumber Numberof Defects Numberof DefectsNumberof Defects 112612118 287131215 3 16891310 4139151417 51810141510 Total = 190

44 Which Control Chart to Use Variables Data Using an x-Chart and R-Chart 1.Observations are variables 2.Collect 20 - 25 samples of n = 4, or n = 5, or more, each from a stable process and compute the mean for the x-chart and range for the R-chart 3.Track samples of n observations each.

45 Attribute Data Using the p-Chart 1.Observations are attributes that can be categorized as good or bad (or pass–fail, or functional–broken), that is, in two states. 2.We deal with fraction, proportion, or percent defectives. 3.There are several samples, with many observations in each. For example, 20 samples of n = 100 observations in each. Which Control Chart to Use (cont..)

46 Attribute Data Using a c-Chart 1.Observations are attributes whose defects per unit of output can be counted. 2.We deal with the number counted, which is a small part of the possible occurrences. 3.Defects may be: number of blemishes on a desk; complaints in a day; crimes in a year; broken seats in a stadium; typos in a chapter of this text; or flaws in a bolt of cloth. Which Control Chart to Use (cont..)

47 Patterns in Control Charts Normal behavior. Process is “in control.” Upper control limit Target Lower control limit

48 Upper control limit Target Lower control limit One plot out above (or below). Investigate for cause. Process is “out of control.” Patterns in Control Charts (cont..)

49 Upper control limit Target Lower control limit Trends in either direction, 5 plots. Investigate for cause of progressive change. Patterns in Control Charts (cont..)

50 Upper control limit Target Lower control limit Two plots very near lower (or upper) control. Investigate for cause. Patterns in Control Charts (cont..)

51 Upper control limit Target Lower control limit Run of 5 above (or below) central line. Investigate for cause. Patterns in Control Charts (cont..)

52 Upper control limit Target Lower control limit Erratic behavior. Investigate. Patterns in Control Charts (cont..)

53 CHAPTER 9 Attribute control charts  50 to 100 parts in a sample Variable control charts  2 to 10 parts in a sample Sample Size Determination

54 Process Capability The natural variation of a process should be small enough to produce products that meet the standards required A process in statistical control does not necessarily meet the design specifications Process capability is a measure of the relationship between the natural variation of the process and the design specifications

55 Process Capability Ratio C p = Upper Specification - Lower Specification 6σ A capable process must have a C p of at least 1.0 Does not look at how well the process is centered in the specification range Often a target value of C p = 1.33 is used to allow for off-center processes Six Sigma quality requires a C p = 2.0

56 C p = Upper Specification - Lower Specification 6σ Insurance claims process Process mean x = 210.0 minutes Process standard deviation s = 0.516 minutes Design specification = 210 ± 3 minutes Example

57 = = 1.938 213 - 207 6(0.516) Example (cont..) C p = Upper Specification - Lower Specification 6σ Insurance claims process Process mean x = 210.0 minutes Process standard deviation s = 0.516 minutes Design specification = 210 ± 3 minutes

58 Process is capable = = 1.938 213 - 207 6(0.516) Example (cont..) C p = Upper Specification - Lower Specification 6σ Insurance claims process Process mean x = 210.0 minutes Process standard deviation s = 0.516 minutes Design specification = 210 ± 3 minutes

59 CHAPTER 9 Exercise Net weight specification = 9.0 oz  0.5 oz Process mean = 8.80 oz Process standard deviation = 0.12 oz

60 CHAPTER 9 In a GE insurance claims process, x = 210.0 minutes, and  = 0.516 minutes. The design specification to meet customer expectations is 210 ± 3 minutes. Determine the process capability ratio. Is the process capable? Exercise

61 Process Capability Index A capable process must have a C pk of at least 1.0 A capable process is not necessarily in the center of the specification, but it falls within the specification limit at both extremes C pk = minimum of, Upper Specification - x Limit 3σ Lower x -Specification Limit 3σ

62 Example New Cutting Machine New process mean x = 0.250 inches Process standard deviation s = 0.0005 inches Upper Specification Limit = 0.251 inches Lower Specification Limit = 0.249 inches

63 C pk = minimum of, (0.251) - 0.250 (3)0.0005 New Cutting Machine New process mean x = 0.250 inches Process standard deviation s = 0.0005 inches Upper Specification Limit = 0.251 inches Lower Specification Limit = 0.249 inches Example (cont..)

64 C pk = = 0.67 0.001 0.0015 New machine is NOT capable C pk = minimum of, (0.251) - 0.250 (3)0.0005 0.250 - (0.249) (3)0.0005 Both calculations result in New Cutting Machine New process mean x = 0.250 inches Process standard deviation s = 0.0005 inches Upper Specification Limit = 0.251 inches Lower Specification Limit = 0.249 inches Example (cont..)

65 CHAPTER 9 Net weight specification = 9.0 oz  0.5 oz Process mean = 8.80 oz Process standard deviation = 0.12 oz Exercise

66 Interpreting C pk C pk = negative number C pk = zero C pk = between 0 and 1 C pk = 1 C pk > 1

67 Any Questions???

68 Thank You


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