mont4e_sm_ch16_sec10 - Section 16-10 16-47 a) Yes, this...

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Unformatted text preview: Section 16-10 16-47 a) Yes, this process is in-control. CUSUM chart with h = 4 and k = 0.5 is shown. CUSUM Chart of viscosity 4 3 2 Cumulative Sum 1 0 -1 -2 -3 -4 1 2 3 4 5 6 7 8 9 Sample 10 11 12 13 14 15 LCL=-3.875 0 UCL=3.875 b) Yes, this process has shifted out-of-control. For the CUSUM estimated from all the data observation 20 exceeds the upper limit. 16-29 CUSUM Chart of viscosity 7.5 5.0 Cumulative Sum UCL=3.88 2.5 0.0 0 -2.5 LCL=-3.88 -5.0 2 4 6 8 10 12 Sample 14 16 18 20 For a CUSUM with standard deviation estimated from the moving range of the first 15 observation, the moving range is 1.026 and the standard deviation estimate is 0.9096. If this standard deviation is used with a target of 14.1, the following CUSUM is obtained. CUSUM Chart of C1 7.5 5.0 Cumulative Sum UCL=3.64 2.5 0.0 0 -2.5 LCL=-3.64 -5.0 1 3 5 7 9 11 Sample 13 15 17 19 16-48 a) CUSUM Control chart with k = 0.5 and h = 4 16-30 CUSUM Chart for Purity Upper CUSUM 3 3.2 Cumulative Sum 2 1 0 -1 -2 -3 Lower CUSUM -3.2 10 20 0 Subgroup Num ber The CUSUM control chart for purity does not indicate an out-of-control situation. The SH values do not plot beyond the values of -H and H. b) CUSUM Control chart with k = 0.5 and h = 4 CUSUM Chart for New Purity Data Upper CUSUM 3 3.2 Cumulative Sum 2 1 0 -1 -2 -3 Lower CUSUM -3.2 10 15 20 25 0 5 Subgroup Number The process appears to be moving out of statistical control. 16-49a) ^ = 0.1736 b) CUSUM Control chart with k = 0.5 and h = 4 16-31 CUSUM Chart for Diameter Upper CUSUM 0.678191 Cumulative Sum 0.5 0.0 -0.5 Lower CUSUM -6.8E-01 10 15 20 25 0 5 Subgroup Number The process appears to be out of control at the specified target level. 16-50 a) CUSUM Control chart with k = 0.5 and h = 4 CUSUM Chart for Concentration Upper CUSUM 30 32 Cumulative Sum 20 10 0 -10 -20 -30 Lower CUSUM -32 10 20 0 Subgroup Number The process appears to be in statistical control. b) With the target = 100 a shift to 104 is a shift of 104 100 = 4 = 0.5. From Table 16-9 with h = 4 and a shift of 0.5, ARL = 26.6 16-51 - 0 51 - 50 = 0.5 standard deviations. From Table 16-9, ARL = 38.0 = 2 - 0 51 - 50 = 1 standard deviation. From Table 16-9, ARL = b) If n = 4, the shift to 51 is a shift of = / n 2/ 4 10.4 a) A shift to 51 is a shift of 16-32 16-52 a) The process appears to be in control. EWMA Chart of C1 91.0 UCL=90.800 90.5 EWMA 90.0 _ _ X=90 89.5 LCL=89.200 89.0 2 4 6 8 10 12 Sample 14 16 18 20 b) The process appears to be in control. EWMA Chart of C1 91.5 91.0 90.5 EWMA 90.0 89.5 89.0 88.5 2 4 6 8 10 12 Sample 14 16 18 20 LCL=88.614 _ _ X=90 UCL=91.386 c) For part (a), there is no evidence that the process has shifted out of control. 16-33 EWMA Chart of C1 91.0 UCL=90.800 90.5 EWMA 90.0 _ _ X=90 89.5 LCL=89.200 89.0 2 4 6 8 10 12 14 Sample 16 18 20 22 24 For part b), there is no evidence that the process has shifted out of control. EWMA Chart of C1 91.5 91.0 90.5 EWMA 90.0 89.5 89.0 88.5 2 4 6 8 10 12 14 Sample 16 18 20 22 24 LCL=88.614 _ _ X=90 UCL=91.386 16-53 a) The estimated standard deviation is 0.169548. b) The process appears to be in control. 16-34 EWMA Chart of C1 10.2 UCL=10.1695 10.1 EWMA 10.0 _ _ X=10 9.9 LCL=9.8305 9.8 2 4 6 8 10 12 14 Sample 16 18 20 22 24 c) The process appears to be out of control at the observation 13. EWMA Chart of C1 10.3 10.2 10.1 EWMA 10.0 9.9 9.8 9.7 2 4 6 8 10 12 14 Sample 16 18 20 22 24 LCL=9.7063 _ _ X=10 UCL=10.2937 16-54 a) The process appears to be in control. 16-35 EWMA Chart of C1 110 UCL=108.00 105 EWMA 100 _ _ X=100 95 LCL=92.00 90 2 4 6 8 10 12 Sample 14 16 18 20 b) The process appears to be in control. EWMA Chart of C1 115 110 105 EWMA 100 95 90 85 2 4 6 8 10 12 Sample 14 16 18 20 LCL=86.14 _ _ X=100 UCL=113.86 c) Since the shift is is preferred. 16-55 0.5 , smaller is preferred to detect the shift fast. so the chart in part (a) a) The shift of the mean is 1 . So we prefer smaller ARL 10.3. = 0.1 and L=2.81 since this setting has the = 0.5 and L=3.07 since this setting has the b) The shift of the mean is 2 X . So we prefer smaller ARL 3.63 c) The shift of the mean is 3 X . Solving of 9. 2 2/ = 3 for n gives us the required sample size n 16-36 16-56 a) With a target = 100 and a shift to 102 results in a shift of 102 - 100 = 0.5 standard deviations. 4 From Table 16-9, ARL = 38. The hours of production are 2(38) = 76. b) The ARL = 38. However, the time to obtain 38 samples is now 0.5(38) = 19. c) From Table 16-9, the ARL when there is no shift is 465. Consequently, the time between false alarms is 0.5(465) = 232.5 hours. Under the old interval, false alarms occurred every 930 hours. d) If the process shifts to 102, the shift is - 0 = 102 - 100 = 1 standard deviation. From Table 16-9, the / n 4/ 4 ARL for this shift is 10.4. Therefore, the time to detect the shift is 2(10.4) = 20.8 hours. Although this time is slightly longer than the result in part (b), the time between false alarms is 2(465) = 930 hours, which is better than the result in part (c). ...
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This note was uploaded on 10/12/2009 for the course IND E 315 taught by Professor Kailashkapur during the Spring '09 term at University of Washington.

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