Analyzing Control Charts: Stability and Variation Insights

Interpret the following charts to determine whether the processes are stable. A) Process is no are above the C B)Process is no (5/6) are in zon deviations from C) Process is sta does not violate D) Process is no points (1-7) goi E) Process is no (4/6) are in zon deviations from

F) Process is sta does not violate

ot stable - 5 consecutive points (1 - 5) CL ot stable - 2 out of 3 consecutive points ne A and farther than 2 standard m center line able - process is flowing correctly, and e any of special cause variation rules ot stable - process shows a trend of 7 ing downward ot stable - 2 out of 3 consecutive points ne A and farther than 2 standard m center line, large jumps in data points Control Chart Special Cause Var 1) Any point above UCL or bel 2) 2 out of 3 consecutive poin beyond 3) 4 out of 5 consecutive poin beyond 4) 7 or more consecutive poin centreline 5) A trend of 7 or more points 6) 8 consecutive points witho 7) 15 consecutive points in zo 8) 14 consecutive points alter

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able - process is flowing correctly, and e any of special cause variation rules

riation Rules: low LCL. nts in zone A or nts in zone B or nt on one side of s moving up or down. out any point in zone C one C rnating up and down.

Sample 1 2 3 4 5 6 7 8 68.51 68.94 68.66 68.49 68.64 68.34 68.99 68.92 68.46 68.20 68.44 68.94 68.63 68.42 68.94 68.91 68.54 68.54 68.55 68.56 68.62 68.99 68.95 68.97 68.34 68.56 68.77 68.62 68.32 68.02 68.95 68.93 68.46 68.70 68.70 68.69 68.34 68.03 68.94 68.96 68.46 68.70 68.64 68.56 68.24 68.47 68.97 68.95 Range 0.20 0.74 0.33 0.45 0.40 0.97 0.05 0.06 UCL R CHART 0.8 0.8 0.8 0.8 0.8 0.8 0.8 0.8 LCL R CHART 0 0 0 0 0 0 0 0 R Bar 0.40 0.40 0.40 0.40 0.40 0.40 0.40 0.40 X Bar 68.46 68.61 68.63 68.64 68.47 68.38 68.96 68.94 UCL X Bar 68.8268 68.827 68.827 68.827 68.827 68.827 68.827 68.827 LCL X Bar 68.4428 68.443 68.443 68.443 68.443 68.443 68.443 68.443 X Double Bar 68.63 68.63 68.63 68.63 68.63 68.63 68.63 68.63 A finishing process packages assemblies into boxes. You have noticed variability problem because some products fit too tightly into the boxes and others fit too Using xbar and R charts, plot and interpret the process.

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n = 6 a 2 = 0.48 d3 = 0 x dbl bar = 68.63 d4 = 2.00 r bar = 0.40 a 2 = 0.48 y in the boxes and desire to improve the process to fix the o loosely. Following are width measurements for the boxes. 1 2 3 0.00 0.20 0.40 0.60 0.80 1.00 1.20 Range U 1 2 3 68.00 68.10 68.20 68.30 68.40 68.50 68.60 68.70 68.80 68.90 69.00 69.10 X Bar U After reviewing both the x-bar and r charts, both contain points outside of the control limits which implies that there are special causes of variation present and the process is not in control. The cause of this variation within the box width will need to be further investigated and corrective actions taken.

X Bar U

4 5 6 7 8 R Chart CL R CHART LCL R CHART R Bar 4 5 6 7 8 X Bar Chart UCL X Bar LCL X Bar X Double Bar

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UCL X Bar LCL X Bar X Double Bar

Upper Limit USL 33 Lower Limit LSL 32 STD DEV ( ) 𝜎 0.22 (mean) 𝜇 32.6 A) Is the process capable? Cpu 0.606061 Cpl 0.909091 Cpk 0.606061 B) If not, what proportion will meet tolerances? Cumulative Z USL 1.818182 Z LSL -2.727273 Proportion Meeting Tolerance 96.2% Tolerances for a new assembly call for weights between 32 and 33 pounds. T pounds with a population standard deviation of .22 pounds. The process pop a. Is the process capable? b. If not, what proportion will meet tolerances? Since the Cpk value is less than 1.25 (0.606061 < 1.25) the process is n

Cpu = (USL - )/(3 ) 𝜇 𝜎 Cpk = Capability index Capability Index Mean Cpl = ( - LSL)/(3 ) 𝜇 𝜎 USL = Upper specification limit 1.25 Capable Cpk = min(Cpu,Cpl) LSL = Lower specification limit 1.33 Highly capable Cpu = Upper capability index 2.00 World-class capa Cpl = Lower capability index = Computed population process mean 𝜇 = Estimated process standard deviation 𝜎 e Normal Probability from Z table 0.96562 Z Score USL = (USL - )/( ) 𝜇 𝜎 0.00326 Z Score LSL = ( - LSL )/( ) 𝜇 𝜎 Proportion Meeting Tolerance = (Z USL - Z LSL) The assembly is made using a process that has a mean of 32.6 pulation is normally distributed. not capable

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ning able (Six Sigma)

Sample Defectives ProportionCL UCL LCL 1 57 0.57 0.4128 0.5605 0.2651 2 28 0.28 0.4128 0.5605 0.2651 3 45 0.45 0.4128 0.5605 0.2651 4 32 0.32 0.4128 0.5605 0.2651 5 30 0.3 0.4128 0.5605 0.2651 6 48 0.48 0.4128 0.5605 0.2651 7 32 0.32 0.4128 0.5605 0.2651 8 24 0.24 0.4128 0.5605 0.2651 9 25 0.25 0.4128 0.5605 0.2651 10 27 0.27 0.4128 0.5605 0.2651 11 28 0.28 0.4128 0.5605 0.2651 12 29 0.29 0.4128 0.5605 0.2651 13 65 0.65 0.4128 0.5605 0.2651 14 66 0.66 0.4128 0.5605 0.2651 15 69 0.69 0.4128 0.5605 0.2651 16 70 0.7 0.4128 0.5605 0.2651 17 26 0.26 0.4128 0.5605 0.2651 18 13 0.13 0.4128 0.5605 0.2651 19 45 0.45 0.4128 0.5605 0.2651 20 46 0.46 0.4128 0.5605 0.2651 21 47 0.47 0.4128 0.5605 0.2651 22 48 0.48 0.4128 0.5605 0.2651 23 28 0.28 0.4128 0.5605 0.2651 24 29 0.29 0.4128 0.5605 0.2651 25 75 0.75 0.4128 0.5605 0.2651 Suppose that a company makes the following product with the following numbe a p chart to see whether the process is in control (n = 100). Interpret the results. 1 2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Based o samples The roo action t under c

Sample Size (n) 100 P Bar 0.4128 CL = P Bar Valu 0.4128 1- P Bar 0.5872 ers of defects. Construct 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 P Chart Proportion CL UCL LCL on the p chart, the process is not in control with multiple s going above and below the UCL and LCL respectively. ot cause of the problem will need to be identifed and taken aggainst these causes to get the process back control.

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Sample Defects CL UCL LCL C Bar 5.4333 1 6 5.4333 12.426 0 2 5 5.4333 12.426 0 3 7 5.4333 12.426 0 4 6 5.4333 12.426 0 5 8 5.4333 12.426 0 6 5 5.4333 12.426 0 7 6 5.4333 12.426 0 8 7 5.4333 12.426 0 9 6 5.4333 12.426 0 10 8 5.4333 12.426 0 11 7 5.4333 12.426 0 12 6 5.4333 12.426 0 13 7 5.4333 12.426 0 14 8 5.4333 12.426 0 15 7 5.4333 12.426 0 16 6 5.4333 12.426 0 17 5 5.4333 12.426 0 18 2 5.4333 12.426 0 19 1 5.4333 12.426 0 20 0 5.4333 12.426 0 21 12 5.4333 12.426 0 22 4 5.4333 12.426 0 23 6 5.4333 12.426 0 24 7 5.4333 12.426 0 25 8 5.4333 12.426 0 26 3 5.4333 12.426 0 27 2 5.4333 12.426 0 28 3 5.4333 12.426 0 29 2 5.4333 12.426 0 Construct and interpret a c chart using the following data: Interpret the res 1 2 3 4 5 6 0 2 4 6 8 10 12 14 Based on the c c controlled from 7 - 16) above th points going dow spike almost up from sample 22 points belowe th change needs to causes needs to

30 3 5.4333 12.426 0

sults. 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 C Chart Defects CL UCL LCL chart, the process is not in control. The process starts out m sample 1 to 6, then shows a trend of 10 points (Samples he control line, from Samples 14 - 20 there is a trend of 7 wnward to the minimum LCL, followed by an immediate to the UCL. The process then regained some control 2 to 25, then looses control again with 4 consequtive he CL (Samples 26 - 30). The root cause of this process o be determined and actions against the assignable o be taken.

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Brennan Ross Problem Set 2

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