Chapter 8: Cusum EWMA Charts

1. Chapter 8: Cusum & EWMA Charts

2. Cusum charts Shewhart charts are not always sensitive to shifts in parameter values The Cusum technique is sensitive We will look at Cusum charts for changes in the mean

3. Cusum charts Deviation from a reference value, k, is maintained Xi – k, where k is a constant C1 = X1 – k C2 = (X2 – k) + (X1 – k) = (X2 – k) + C1 C3 = (X3 – k) + C2 : Cm = (Xm – k) + Cm-1 The Ci values are plotted to form a rudimentary Cusum chart

4. Cusum charts Consider a desired level of a process at its mean m0 If the mean output of a process, Xbar stays around m0, the cusum will be roughly horizontal With approximately the same number of values above and below m0

5. Example Given 20 values from a N(0, 1) followed by 20 values from a N(1, 1) These observations are sample means Assume the reference value is zero First, these values are plotted on a Shewhart chart and an R chart Then, they are plotted on a rudimentary Cusum chart

8. Note on Shewhart chart Although there are two observations that seem somewhat high, the shift is not detected

9. Rudimentary Cusum chart

10. Notes on rudimentary Cusum chart There is no slope on the first 20 samples But, after the first 20 samples, the slope is definitely steep If the alarm value (h) is 5, a call to action would have been signaled on the 24th observation What is the appropriate value of h?

11. One-sided Cusum We have been talking about two-sided Cusum charts Initially, Cusum charts were one-sided A variation of Cm = S(Xbari – k) is plotted where k is the reference value Suppose that there is a quality level, m0, that is considered acceptable and another level, m1, that is considered rejectable

12. One-sided Cusum Reference value, k k = (m0 + m1)/2 If Cm falls below zero, reset to zero This is the variation Cm > h is a signal that the process mean has shifted to some value greater than k

13. One-sided Cusum The proper value of h Base it on the ARL The ARL should be large if the process mean is stable at m0 The ARL should be small if the process mean has shifted to m1 ARL at m0, L0 ARL at m1, L1

14. ARLs for several Cusum schemes

15. Example Assume m0 = 10 and m1 = 10.4 Given s = .6 Find a Cusum scheme that comes close to L0 = 500 and L1 =3 From the Table B = 1.04 A = 2.26

16. Example, cont. K = (10 + 10.4)/2 = 10.2 B = [.2 SQRT(n)/.6 n = 9.7 10 A = [h SQRT(10]/.6 = 2.26 h = .43 Summary: Take samples of n = 10, and when Qm > .43 that is the signal that the process is out of control. When Qm < 0, reset it to zero.

17. Using a nomogram Procedure Connect the desired L0 and L1 Results in a point on the B scale Determine n from n = [Bs/ABS(k – m0)]2 Usually round n up unless it is slightly above an integer Recompute B using the rounded n

18. Using a nomogram Procedure Connect the new value of B to the desired value on the L0 scale Note the value on the L1 scale Determine h from the value of A The Cusum scheme is now specified

19. Using a nomogram Procedure Alternate Cusum scheme is obtained by connecting a point on the B scale to the desired value on the L1 scale and noting the value on the L0 scale The final value on the A scale is read resulting in another Cusum scheme

20. Using a nomogram Procedure Two additional Cusum schemes may be obtained by rounding n in the other direction There will be four Cusum schemes Choose on the basis of how close the schemes come to the desired ARL values (If n happens to be an integer, there will only be one Cusum scheme)

21. Example One-sided Cusum scheme with ARL0 = 400 when the mean is 80 (acceptable quality) and ARL1 = 5 when the mean is 100 (rejectable quality) Process output is normally distributed Standard deviation is 20

22. Example, cont. k = (100 + 80)/2 = 90 Connect L1 = 5 and L0 = 400 Read B = .722 [10 SQRT(n)]/20 = .722 n = 2.08 Round n to 2 [10 SQRT(2)]/20 = .707

23. Example, cont. Connect B = .707 and L0 =400 reading A = 3.16 [h SQRT(2)/20] = 3.16 h = 44.69 Summary: Compute S(Xbari – 90). If this value becomes negative, start anew If the summation exceeds 44.69, the process is out of control

24. Example, cont. The line connecting L0 = 400 and A = 3.16 also intersects L1 = 5.2 The scheme k = 90, h = 44.69, n = 2 yields the desired ARL at m0 = 80, but a slightly worse ARL at m1 = 100

25. Example, cont. Alternate Connecting B = .707 and L1 = 5, we could have found a scheme that holds the ARL at m1 = 100, but has L0 =300 at m0 = 80

26. Example, cont. More alternates Since n = 2.08 and was rounded down to 2, a conservative approach would be to round n up to 3 [10 SQRT(3)]/20 = .866 Connect B = .866 to L0 = 400 [h SQRT(n)]/s = 2.6 h = [2.6 (20)]/SQRT(3) = 30.02

27. Example, cont. More alternates The line intersects L1 = 3.8 which is better than the called for ARL at m0 = 80 Connecting the points B = .866 and L1 = 5 yields an extremely large ARL at m1 = 100 Which scheme is the best? Probably the very first scheme

28. First example Consider the 40 values, 20 from N(0, 1) followed by 20 from N(1,1) We are concerned about increases from m0 = 0 to m0 = 1 with L0 = 500 Here n = 1 and s = 1 B = [.5 SQRT(1)]/1 = .5 A = h = 4.42 and L1 = 9.5 (compared to 44 on a Shewhart chart)

29. First example, 2-sided Suppose that we are concerned with decreases to m2 = -1 as well as increases to m1 = 1 We just determined that h = 4.42 The ARLs will be 1/L0 = 1/500 + 1/500 giving L0 =250 1/L1 = 1/9.5 + 1/9.5 giving L1 = 4.75

31. 8-1.2 The Tabular or Algorithmic Cusum for Monitoring the Process Mean (two sided) Accumulate derivations from the target ?0 above the target with one statistic, C+ Accumulate derivations from the target ?0 below the target with another statistic, C— C+ and C- are one-sided upper and lower cusums, respectively.

32. 8-1.2 The Tabular or Algorithmic Cusum for Monitoring the Process Mean The statistics are computed as follows: The Tabular Cusum starting values are K is the reference value (or allowance or slack value) If either statistic exceeds a decision interval h, the process is considered to be out of control. Often taken as h = 5?

33. 8-1.2 The Tabular or Algorithmic Cusum for Monitoring the Process Mean Example 8-1 ?0 = 10, n = 1, ? = 1 Interested in detecting a shift of 1.0? = 1.0(1.0) = 1.0 Out-of-control value of the process mean: ?1= 10 + 1 = 11 k = ½ and h = 5? = 5 The equations for the statistics are then:

34. 8-1.2 The Tabular or Algorithmic Cusum for Monitoring the Process Mean Example 8-1 If an adjustment has to be made to the process, may be helpful to estimate the process mean following the shift. The estimate can be computed from N+, N- are counters, indicating the number of consecutive periods that the cusums C+ or C- have been nonzero.

35. Example 8-1 Pgs. 411-414 Note on page 414, the new mean is estimated as m0 + k + C29+/N+ = 10 + .5 + 5.28/7 = 11.25

36. 8-1.2 The Tabular or Algorithmic Cusum for Monitoring the Process Mean Example 8-1 The cusum control chart indicates the process is out of control. The next step is to search for an assignable cause, take corrective action required, and reinitialize the cusum at zero. If an adjustment has to be made to the process, may be helpful to estimate the process mean following the shift.

37. 8-2. The Exponentially Weighted  Moving Average Control Chart The Exponentially Weighted Moving Average Control Chart Monitoring the Process Mean The exponentially weighted moving average (EWMA) is defined as where 0 < ? ? 1 is a constant. z0 = ?0 (sometimes z0 = )

38. 8-2.1 The Exponentially Weighted Moving Average Control Chart Monitoring the Process Mean The control limits for the EWMA control chart are where L is the width of the control limits.

39. 8-2.1 The Exponentially Weighted Moving Average Control Chart Monitoring the Process Mean As i gets larger, the term [1- (1 - ?)2i] approaches zero. This indicates that after the EWMA control chart has been running for several time periods, the control limits will approach steady-state values given by

40. 8-2.2 Design of an EWMA Control Chart The design parameters of the chart are L and ?. The parameters can be chosen to give desired ARL performance. In general, 0.05 ? ? ? 0.25 works well in practice. L = 3 works reasonably well (especially with the larger value of ?. L between 2.6 and 2.8 is useful when ? ? 0.1 Similar to the cusum, the EWMA performs well against small shifts but does not react to large shifts as quickly as the Shewhart chart. EWMA is often superior to the cusum for larger shifts particularly if ? > 0.1

41. Example 8-2 Pgs. 428-431

42. First example in the notes for this chapter N(0,1) to N(1,1), 2-sided, l = .2, L =3 See next slide

44. 8-2.4 Robustness of the EWMA to Non-normality As discussed in Chapter 5, the individuals control chart is sensitive to non-normality. A properly designed EWMA is less sensitive to the normality assumption.

45. Assignment Suggestion: 8-1, 8-7, 8-15, 8-19

46. End