1. 1 Process Capability Assessment

2. 2 Process Capability vs. Process Control u Evaluating Process Performance – Ability of process to produce parts that conform to engineering specifications(CONFORMANCE) – Ability of process to maintain a state of statistical control; i.e., be within control limits(CONTROL)

3. 3 – Process Capability…dealing with individual measurements, e.g., X (LSL, USL) Linkages Between Process Control & Process Capability u Process must be in statistical control before assessing process capability. Why? u Statistical aspects – Process Control…use summary statistics from a sample (subgroup); dealing with sampling distributions, e.g., and R (LCL, UCL)

4. 4 u Process may be in statistical control, but not capable (of meeting specifications) – Process is off-center from nominal (bias) – Process variability is too large relative to specifications (variation) – Process is both off-center and has large variation.

5. 5 Relationship Between Process Variability and Product Specification (a) Process variation is small relative to the specifications so that the process mean can shift about without causing the process to degrade its capability. This will reduce the defects per million (DPM), reduce the cost of quality (COQ), and hence increase profitability. Upper Specification Lower Specification

6. 6 Relationship Between Process Variability and Product Specification (b) Process variation is large relative to the specifications such that the process must remain well centered for the process capability to be maintained at a tolerable level. Variation, however, must be reduced. This will increase process capability, reduce DPMs, reduce COQ, and increase profitability. Upper Specification Lower Specification

7. 7 Relationship Between Process Variability and Product Specification (c) Process variation is large relative to the specifications so that the process cannot be considered capable regardless of the process centering. Hence we have a severe and urgent problem. Process variation must be reduced drastically. Upper Specification Lower Specification

8. 8 Statistical Assessment of Process Capability u Get Process in Statistical Control u Statistical Assessment (Minitab or Excel) – Construct histogram of individual measurements – Compute probability of exceeding specifications P() Empirically (observed) Mathematically: … assume N( ,  ) and compute (expected) Convert to defects per million (DPM) and sigma capability Compute process capability indices … C p, C pk

9. 9 Alternatives for Improving Process Capability u If bias – Recenter and recompute P(), dpm, and sigma capability u If too much variation – Sort by 100% inspection  – Widen tolerance  – Use a more precise process (e.g., better or new technology) to reduce variation  – Use statistical methods to identify variation reduction opportunities for existing process

10. 10 Summary: Comparison of Specification Limits and Control Limits u Spec limits or tolerances for product quality characteristics are: – Characteristic of the part/item (product) in question – Based on functional design considerations – Related to/compared with an individual part measurement – Used to establish a part’s conformance to design intent

11. 11 Control limits on a control chart are: u Characteristic of the process in question u Based on the process mean and variation  Dependent on sampling parameters, viz., sample size and  -risk (Type I error) u Used to identify presence/absence of special-cause variation in the process

12. 12 Some Common Indices of Process Capability C p Formula Specification Range Variation of Distribution of Individual Product USL ~ LSL N(  X,  X ) rejectreject -3  X (1) +3  X (1) (1) TX -3  X (2) +3  X (2) (2) C p (1) < C p (2)

13. 13 Relationship Between C p, DPM, and Sigma of Process (Assumes No Bias) *DPM = (Probability of Exceeding Specs) * 10 6

14. 14 C pk u Purpose: To promote adherence of process mean to target (nominal) value of spec. u Formulas:

15. 15 u Example USLLSL100 190 x N(130, 10) T = 145 (  X -T) = bias

16. 16 Questions?

17. 17 PCB Exercise (USL = +8, LSL = -8)

18. 18 Microeconomics of Quality: Loss Due to Variation (Taguchi) u Linking Cost of Quality Due to Bias and Variation to DPM and Process Capability in PCB Manufacture u Variation is Related to Functional Form (Distribution) of Process Output

19. 19 Loss Function Representation of Quality for PCBs LSL (-8 microns) Target (T) (Normal)USL (+8 microns) Quality Loss Reject (Scrap) Failure Cost @ USL = $2.40/unitFailure Cost @ USL = $2.40/unit Failure Rate (Probability) = 100%Failure Rate (Probability) = 100% Probability Distribution of Quality Characteristic Produced by Process Quadratic Loss Function Poor Fair Good Best Good Fair Poor X (Microns)

20. 20 Loss Function Approach u Measured loss Function: L M (X) L M (X) = k (x - T) 2 where k is an unknown constant x is a value of the quality characteristic T is the target u Determining the Constant k L M (x) @ USL = k (USL - T) 2 where L M (x) @ USL is a known measured cost of scrap (=Cost of a failure @ USL * failure rate (probability)) USL & T are known k = (2.40)(1.0)/(8) 2 =0.0375

21. 21 u Total Loss Function (Measured + Hidden): L T (x) LT(x) = ak (x - T) 2 where a is the hidden “cost of quality” multiplier (6 < a < 50) If we assume a = 28, then ak = 28 * 0.0375 = 1.05 u Evaluation of Expected Total Process loss: E T {L(x)} E T {L(x)} = ak {  x 2 + (  x - T 2 )} Where  x 2 is process variance and (  x - T) is process bias (mean from target)

22. 22 Effect of Bias & Variation (Variance) of Process Variable on Cost of Quality in Millions of Dollars (ak = 1.05; produce 10 million units/year)  

23. 23 Questions?

24. 24 (  x -T) = 0  x = 6 E T {L(x)} = 1.05 {6 2 + 0 2 } * 10 7 = $378 million Figure 2. Evaluation of Quality Loss Function (N(0,6 2 )) T Loss ($) LSLUSL x (microns) Probability Distribution Loss Function

25. 25 Analysis (Figure 2.) Process Capability No. Standard Deviation from Mean (z) z = 3Cp = 3 (0.444) = 1.332 This is a 1.332  Process.

26. 26 Defects Per Million (dpm): Excel dpm = (1-NORMSDIST(1.332)) * 2 * 10 ^ 6 = 182,423 dpm M LSL-1.332RRUSL+1.332A

27. 27 (  x -T) = 0  x = 2 E T {L(x)} = 1.05 {2 2 + 0 2 } * 10 7 = $42 million Figure 3. Evaluation of Quality Loss Function (N(0,2 2 )) T Loss ($) LSLUSL x (microns) Probability Distribution Loss Function

28. 28 Analysis (Figure 3) Process Capability No. Standard Deviation from Mean (z) z = 3Cp = 3 (1.333) = 4.000 This is a 4  Process.

29. 29 Defects Per Million (dmp): Excel dpm = (1-NORMSDIST(4.000)) * 2 * 10 6 = 63 dpm Expected Cost Change (ECC)

30. 30 E T {L(x)} = 1.05 {2 2 + 2 2 } * 10 7 = $84 million Figure 4. Quality Loss Consequences of Shifting the Process Mean Toward the Upper Specification TLSLUSL x (microns) (  x -T) = 2  x = 2 (  x -T) = 2

31. 31 Analysis (Figure 4) Process Capability

32. 32 Defects Per Million (dpm): Excel dpm = (1-NORMSDIST(3))+ NORMSDIST (-5.01)) * 10^6 = 1349.97 +.27 = 1350.24 Expected Cost Change (ECC) No. Standard Deviation from Mean (z) Z USL = 3 (1) = 3 Z LSL = 3(-1.67) = -5.01

33. 33 Questions?

34. 34 Normal Distribution: N(0, 2.67 2 ) x (microns) Loss ($) LSLUSL (  x -T) = 0  x = 2.67 (Note: 3  x = 8, thus  x = 8/3 = 2.67) E T {L(x)} = 1.05 {2.67 2 + 0 2 } * 10 7 = $74.85 million

35. 35 Uniform Distribution: U(0, 4.62 2 ) x (microns) Loss ($) LSLUSL (  x -T) = 0  x = 4.62 (Note:  x 2 = (b-a) 2 /12 = 21.33;  x = 4.62) E T {L(x)} = 1.05 {4.62 2 + 0 2 } * 10 7 = $224.12 million

36. 36 Analysis (Figure 5) Process Capability Normal Distribution Uniform Distribution (Inspection; Adjustment)

37. 37 Defects Per Million (dmp): Excel Normal Distribution dpm = (1-NORMSDIST(3.000)* 2 * 10 6 = 2700 Uniform Distribution Theoretically, there are no units exceeding the specification limits; however, there are many more units further away from the target value (T) than with a normal distribution. This accounts for the higher variance (4.62 2 vs 2.67 2 ) and, as we shall see, higher cost of quality. No. Standard Deviation from Mean (z) Normal Distribution Z = 3 Cp= 3 (1.000) = 3.000 This is a 3  Process. Uniform Distribution Z = 3 Cp= 3 (0.577) = 1.731  Process

38. 38 Cost of Quality Normal (N) Distribution E{L(x)} = ak {  2 + (  - T) 2 } = ak * 10 7 {2.67 2 + 0 2 } = ak * 10 7 * 7.13 Uniform (U) Distribution E{L(x)} = ak * 10 7 { 4.62 2 + 0 2 } = ak * 10 7 * 21.34 Expected Cost Change (ECC): Normal vs Uniform

39. 39 Summary

40. 40 Comparison of Process (1 vs 2) (.,.)  (  - T),   bias, variation A (0, 6) B (0, 2) Process T = 0 Max Min Max Min

41. 41 Comparison of Process (1 vs 3) A (0, 6) C (2, 2) Process T = 0 Max Min Max Min (.,.)  (  - T),   bias, variation

42. 42 Questions?

43. 43 Understanding the Differences 3  CapabilityHistorical Standard 4  CapabilityCurrent Standard 6  CapabilityNew Standard

44. 44 Understanding the Differences 3  Floor space of a small hardware store 1.5 misspelled words per page in a book $2.7 million indebtedness per $1 billion in assets 3 1/2 months per century Coast-to- coast trip 4  Floor space of a typical living room 1 misspelled word per 30 pages in a book $63,000 indebtedness per $1 billion in assets 2 1/2 days per century 45 minutes of freeway driving (in any direction) 5  Size of the bottom of your telephone 1 misspelled word in a set of encyclopedias $570 indebtedness per $1 billion in assets 30 minutes per century A trip to the local gas station SigmaAreaSpellingMoneyTimeDistance 6  Size of a typical diamond 1 misspelled word in all of the books contained in a small library $2 indebtedness per $1 billion assets 6 seconds per century 4 steps in any direction

45. 45 Understanding the Difference Suppose a process produced 294,118 units of product. If the process capability was 4  then the defects produced could be represented by the matrix of dots given below. If the capability was 6 , only one dot would appear in the entire matrix.

46. 46 Understanding the Difference 4  Capability: Defect Dots = 1849 6  Capability: Defect Dots = 1

47. 47 The Inspection Exercise The Inspection Exercise Task: Count the number of times the 6th letter of the alphabet appears in the following text. The Necessity of Training Farm Hands for First Class Farms in the Fatherly Handling of Farm Live Stock is Foremost in the Eyes of Farm Owners. Since the Forefathers of the Farm Owners Trained the Farm Hands for First Class Farms in the Fatherly Handling of Farms Live Stock, the Farm Owners Feel they should Carry on with the Family Tradition of Training Farm Hands of First Class Farmers in the Fatherly Handling of Farm Live Stock Because They Believe it is the basis of Good Fundamental Farm Management.

48. Questions?