The specification for a plastic handle calls for a length of 6.0 inches ± 0.5 in

Question

The specification for a plastic handle calls for a length of 6.0 inches ± 0.5 inches (5.5 to 6.5 inches). The process is known to operate at a mean thickness of 6.1 inches. The minimum acceptable process capability is 2.0. The standard deviation () of the process is currently 0.08 inches.

a) Can the company meet the customer’s specification requirements at this time? If it cannot, explain if it is due to a drifting of the mean or too much variability.

b) Suppose that the mean of the process has now shifted to 5.95 inches. What is the maximum standard deviation () of this process if the company wants to ensure that it can maintain a Cpk of 2.0?

c) The specification limits have not changed. Suppose that the mean of the process has now shifted to 5.9 inches with a standard deviation () of 0.09. What is the range (upper and lower limits) on the mean of the process to maintain a Cpk of 2.0 or greater?

d) Suppose that the mean of the process is operating at a mean of 6.1 inches and a standard deviation of 0.1 inches and follows a normal probability distribution. The lower spec (specification) limit is 5.95 and the upper spec limit is 6.2. What percent of the values are outside of the spec limits?  

Explanation / Answer

Ans-

The ¯x and R charts for the process are shown in Figure 1 on the following page. The estimated population mean and standard deviation are µˆ = x¯¯ = 11.0 (1a) ˆ = R¯ d2 = 27.3 (1b) with both quantities expressed in ten-thousandths on an inch. 1.1.1 Specification limits If the specification limits are set at 0±0.01, then, the appropriate process capability ratios are Cp = USLLSL 6ˆ = 1.220 (2a) Cpk = USL µˆ 3ˆ = 1.087 (2b) Cpkm = Cpk s 1+ µ T 2 = 1.008 (2c) These values indicate a fairly capable process (capability greater than 1) with a slight off-center bias as noted by Cpkm. 1.1.2 Improvements To quantify performance improvement when the above process is centered on target, we look at the ratio of Cp/Cpk = 1.073. The process would be improved approximately 7% if centered. 1.2 Process in statistical control A process in statistical control has x¯¯ = 39.7 and R¯ = 2.5, with specification limits at 40±5. To determine the potential process capability for this process, we first estimate the standard deviation using ˆ = R¯ d2 = 2.5 1.128 = 2.216 (3) where d2 = 1.128 for n = 2. Using this standard deviation estimate, the maximum process capability is Cp = USLLSL 6ˆ = 0.752 (4) 1 -20 -10 0 10 20 30 40 50 2 4 6 8 10 12 14 16 18 20 (a) 0 20 40 60 80 100 120 140 2 4 6 8 10 12 14 16 18 20 (b) Figure 1: ¯x and R charts for the hole diameter problem 2 By contrast, the actual process capability is Cpk = 39.735 3 · 2.216 = 0.701 (5) 1.2.1 Cpk vs. Cpkm We compare Cpk with its relative Cpkm 1 . The value for the latter [see (2c)] Cpkm = 0.695 (6) This value is quite close to that of Cpk implying a slight deviation from center. 2 Thermoforming Production Process Given are x¯¯ = 34.75 and S¯ = 0.17. Control charts for sample size of n = 3 are shown in Figure 2 on the next page, and for n = 5 are shown in Figure 3 on page 5. 2.1 Western Electric rules Looking at the control chart with n = 3 shown in Figure 2 on the next page, sample 12 is the first to plot outside the control limits thus issuing an alarm. On the other hand, the upward trend in the data is noticeable from sample 2 on, but sample 7 plots below sample 6, thereby failing to issue an alarm due to Western Electric rule #5 (six consecutive points increase). In the case of n = 5, as shown in Figure 2 on page 5, the first alarm is signaled as early as sample #7, and definitely by sample #9. It is important to note that this represents at best a total of 35 samples produced before detection, while the case with n = 3 signals the alarm after 36 of the same samples. 2.2 Number of bad parts produced Given that historically Cp = Cpk = 1, then it can be derived that the upper and lower specification limits are LSL = x¯¯3S¯ = 34.24 (7a) USL = x¯¯+3S¯ = 35.26 (7b) As discussed above in Section 2.1, the number samples before detection would be roughly about 35 for either n = 3 or n = 5. Inspection of the data reveals that in the first 35 data points, only one bad one was produced x31 = 35.470 > 35.26 = USL. 2.3 Average Run Length for µ shift The risk associated with detecting a mean shift of 0.25 is = Lk n Lk n = 30.25 5 32 5 0.9925 (8) where we have used n = 5 as the number of points per sample. The average run length, ARL, for this process is then ARL = 1 1 = 133.2 (9) 1 In the 5th edition, the question asks redundantly to compare Cpkm to itself. We correct the typo and tackle the slightly more difficult problem presented. 3 34.5 35 35.5 36 5 10 15 20 25 ¯x Sample number (a) -0.1 0 0.1 0.2 0.3 0.4 0.5 5 10 15 20 25 s Sample number (b) Figure 2: Control chart for thermoforming process with n = 3 4 34.6 34.8 35 35.2 35.4 35.6 35.8 36 36.2 36.4 2 4 6 8 10 12 14 16 ¯x Sample number (a) -0.1 0 0.1 0.2 0.3 0.4 0.5 2 4 6 8 10 12 14 16 s Sample number (b) Figure 3: Control chart for thermoforming process with n = 5, units in ten-thousandth of an inch 5 160 158 150 151 153 154 158 162 186 195 179 184 175 192 186 197 190 189 185 182 181 180 183 186 206 210 216 212 211 202 205 197

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