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Chapter 6 – Part 4

Chapter 6 – Part 4. Process Capability. Meaning of Process Capability. The capability of a process is the ability of the process to meet the specifications. A process is capability of meeting the specification limits if at least 99.73 % of the product falls within the specification limits.

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Chapter 6 – Part 4

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  1. Chapter 6 – Part 4 Process Capability

  2. Meaning of Process Capability • The capability of a process is the ability of the process to meet the specifications. • A process is capability of meeting the specification limits if at least 99.73% of the product falls within the specification limits. • This means that the fraction of product that falls outside the specification limits is no greater than 0.0027, or that no more that 3 out of 1,000 units is “out of spec.” • Our method of computing process capability assumes that the process is normally distributed.

  3. Control Limits vs. Spec. Limits • Control limits apply to sample means, not individual values. • Mean diameter of sample of 5 parts, X-bar • Spec limits apply to individual values • Diameter of an individual part, X

  4. Mean= Target Control Limits vs. Spec. Limits Samplingdistribution, X-bar Processdistribution, X USL LSL Lowercontrollimit Uppercontrollimit

  5. Requirements for Assessing Process Capability • To assess capability of a process, the process must be in statistical control. • That is, all special causes of variation must be removed prior to assessing capability. • Also, process performance characteristic (e.g., diameter, bake time) must be normally distributed.

  6. Cp Index USL = upper specification limit LSL = Lower specification limit

  7. Cp Index • We want the spread (variability) of the process • to be as ??? • If the spread of the process is very ????, the • capability of the process will be very ????

  8. Cp Index USL LSL Width of spec limits = USL - LSL Processdistribution, X Spread of Process = USL - LSL

  9. Process is Barely Capable if Cp = 1 .9973 .00135 .00135 X LSL USL Spread of process matches the width of specs. 99.73% of output is within the spec. limits.

  10. Process Barely Capable if Cp = 1 If , what does this imply regarding the spec. limits? Cp=1 LSL = USL =

  11. Process Barely Capable if Cp = 1

  12. Process is Capable if Cp > 1 >.9973 < .00135 < .00135 X LSL USL Spread of process is less than the width of specs. More than 99.73% of output is within the spec. limits.

  13. Process is Not Capable if Cp < 1 < .9973 > .00135 > .00135 X LSL USL Spread of process is greater than the width of specs. Less than 99.73% of output is within the spec. limits.

  14. Estimating the Standard Deviation

  15. Estimating the Standard Deviation

  16. X Sugar Example Ch. 6 - 3

  17. Capability of Sugar Process USL = 20 grams LSL = 10 grams

  18. Capability of Sugar Process • Since Cp <1, the process is not capability of meeting the spec limits. • The fraction of defective drinks (drinks with either too much or not enough sugar) will exceed .0027. • That is, more than 3 out of every 1000 drinks produced can be expected to be too sweet or not sweet enough. • We now estimate the process fraction defective, p-bar.

  19. LSL USL F1 F2 Mean Estimated Process Fraction Defective • What is the estimated process fraction defective -- the percentage of product out of spec? p-bar = F1 + F2

  20. Estimated Process Fraction Defective • We can then use Cp to determine the p-barbecause there is a simple relationship between Cp and z: z = 3Cp (See last side for deviation of this result.) • Suppose, Cp =0.627 z = 3(0.627) =1.88

  21. Estimated Process Fraction Defective • The z value tells us how many standard deviations the specification limits are away from the mean. • A z value of 1.88 indicates that the USL is 1.88 standard deviations above the mean. • The negative of z, -1.88, indicates that the LSL is 1.88 standard deviations below the mean. • We let Area(z) be the area under the standard normal curve between 0 and z.

  22. Process Fraction Defective Area(z) = Area(1.88) = 0.4699 LSL USL F2 0 z =1.88 F2 = % above USL = .5000 - 0.4699 = .0301

  23. z Table (Text, p. 652)

  24. LSL USL F1 F2 0 Process Fallout p-bar = 2[.5 – Area(z)] = F1 + F2 0.4699 z =1.88 p-bar = 2(.5 – .4699) = 2(.0301)=.0602

  25. Process Fallout – Two Sided Spec.

  26. Recommended Minimum Cp

  27. Recommended Minimum Cp

  28. Soft Drink Example Cp =0.33 z = 3Cp = 3(0.33) = 0.99 Area(z) = Area(0.99) = 0.3389 p-bar = 2[.5 - Area(0.99)] = 2[.5 - 0.3389] = 0.3222

  29. Capability Index Based on Target • Limitation of Cp is that it assumes that the process is mean is on target. Process Mean = Target Value = (LSL + USL)/2

  30. CT Capability Index • With Cp, capability value is the same whether the process is centered on target or is way off. • Cp is not affected by location of mean relative to target. • We need capability index that accounts for location of the mean relative to the target as well as the variance. • CT is an index that accounts for the location of mean relative to target.

  31. CT Capability Index

  32. CT Capability Index If process is centered on target, If process is off target,

  33. Example of CT LSL = 10, USL = 20, estimated standard deviation = 5.0 and estimate process mean = 15.33. Compute CT.

  34. CT Capability Index If process mean is adjusted to target,

  35. CT Capability Index • Cpis the largest value that CTcan equal. • Since Cp = 2.2 and CT = .44, the difference is the maximum amount by which we can increase CTby adjusting the mean to the target value.

  36. Conclusion?

  37. Derivation of z = 3Cp

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