TQM &amp; SPC Problem Solutions

# 10 a x 100 r 4 x chart ucl 100 05774 102308 lcl 100

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10. a. x = 100, R = 4 X Chart: UCL = 100 + 0.577(4) = 102.308 LCL = 100 - 0.577(4) = 97.692 R Chart: UCL = 2.115(4) = 8.46 LCL = 0(4) = 0 The average and range for the process for the last 5 days are 99.8 and 12. The average is within the control limits, but the range falls above the UCL for the range. The process is out of control. b. To provide nearly 100 pounds of fish every day, the supplier needs to analyze existing processes (cause-effect diagram, for example) to see whether there are opportunities for improvement. After implementing some changes in the process he needs to observe the outcomes for the average and range again and to check if they are within control limits. 11. a. X-Bar Chart: R-Chart: Center Line: 10.00 Center Line: 0.22 Upper Control Limit: 10.22 Upper Control Limit: 0.56 Lower Control Limit: 9.78 Lower Control Limit: 0.00 NAME: **************** CHAPTER 9, PROBLEM 11 SECTION: ******* 16-Dec-02 Observation Sample Sample Sample 1 2 3 Average Range ======= ==================== ========= ========= ======= 1 10.01 9.90 10.03 9.98 0.13 2 9.87 10.20 10.15 10.07 0.33 3 10.04 9.89 9.76 9.90 0.28 4 10.17 9.94 9.83 9.98 0.34 5 10.21 10.13 10.04 10.13 0.17 6 10.16 10.02 9.85 10.01 0.31 7 10.14 9.89 9.80 9.94 0.34 8 9.86 9.91 9.99 9.92 0.13 9 10.18 10.04 9.93 10.05 0.25 9 - 6

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10 9.91 9.87 10.06 9.95 0.19 Observation Sample Sample Sample 1 2 3 Average Range ======= ========== ========= ========= ========== ======= 11 10.08 10.14 10.03 10.08 0.11 12 9.82 9.87 9.92 9.87 0.10 13 10.14 10.06 9.84 10.01 0.30 14 10.16 10.17 10.19 10.17 0.03 15 10.13 9.84 9.92 9.96 0.29 16 10.16 9.81 9.83 9.93 0.35 17 10.20 10.10 10.03 10.11 0.17 18 9.87 9.93 10.06 9.95 0.19 19 9.84 9.91 9.99 9.91 0.15 20 10.06 10.19 10.01 10.09 0.18 ========== ======= Grand Averages 10.00 0.22 Enter Values of A2 1.023 D3 0.000 D4 2.575 RESULTS: Average Range ======== ======== Center Line 10.00 0.22 Upper Control Limit (UCL) 10.22 0.56 Lower Control Limit (LCL) 9.78 0.00 9 - 7
11. b. None of the samples are out of control since all points are within the control limits. AVERAGE CHART for Chapter 9 Problem 11 9.6000 9.8000 10.0000 10.2000 10.4000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 sample average Center Line UCL LCL RANGE CHART for Chapter 9 Problem 11 0.0000 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Sample Range Center Line UCL LCL c. Yes, the process appears to be sufficiently stable to begin using it as a basis for calculating X-Bar and R charts. None of the points are out of control in average or range. 12. a. c p = (200 - 160) / ( 6 * 6) = 40/36 9 - 8

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= 1.1111 c pk = MIN ( (200-170)/(3 * 6), (170-160)/(3 * 6) ) = MIN (30/18, 10/18) = 0.5555 b. To improve c pk to 1.0, you could either change the mean, the specifications or the standard deviation. If the mean is changed to 180, for example, the c pk = 1.18. Likewise, if the standard deviation is reduced to 3.333, the c pk = 1.0 without changing the mean. Finally, the specifications could be changed, for example, by reducing the lower specification to 152 yielding a c pk = 1.
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