Page 85 - Six Sigma for electronics design and manufacturing
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Six Sigma for Electronics Design and Manufacturing
                     54
                     2.3.1 Relationship between z and Cpk
                     Since the formulas for z and Cpk are somewhat similar, the two can
                     be derived from each other, especially if the process is centered (no av-
                     erage shift from nominal):
                                        Cpk = [min of {z 1 , z 2 }]/3
                                 SL
                                                                           (2.13)
                          Cpk = ±
                                             when process is centered (z 1 = z 2 )
                                     = z/3;
                                 3
                     2.3.2 Example calculations of defects and Cpk         (2.12)
                     Example 2.1
                     A check on parts made by a factory indicated that they are made with
                     normal distribution with average = 12.62	 and standard deviation of
                     2.156.
                     a. What is the probability that parts of the following lengths (L) will
                        be made in that factory: L > 18	, L < 8	, and 10	
 L 
 12	?
                     b. If the specifications for the length of the part were 12.62 ± 3	, and
                        the factory made parts with a   = 12.62 and   = 2.156, what are Cp
                        and Cpk and the predicted defect or reject rate (RR)? Repeat the
                        above if the process average is shifted with respect to specification
                        nominal by 1	 to the left and 0.75	 to the right.
                     c. What should the specifications be if the factory decided on the fol-
                        lowing: Cp = 1, Cp = 1.5, and Cp = 2 (six sigma), assuming the av-
                        erage is 12.62	 and the   = 2.156?
                     Solutions to Example 2.1
                     a. From the standard normal distribution (Table 2.3), the area under
                        the curve is used to determine the answers:
                        L > 18	: z 2 = (18 – 12.62)/2.156 = 2.5
                        f(z 2 ) = f(–2.5) = 0.0062 or 0.62% or 6,200 PPM
                        L < 8
                        z 1 = (8 – 12.62)/2.156 = –2.14
                        f(z 1 ) = f(–2.14) = 0.0162 or 1.62% or 16,200 PPM
                        10	
 L 
 12
                        z 2 = (12 – 12.62)/2.156 = –0.29
                        z 1 = (10 – 12.62)/2.156 = –1.22
                        f(z 2 ) – f(z 1 ) = f(–0.29) – f(–1.22) = 0.3859 – 0.1112 = 0.2747

                     or 27.47% or 274,700 PPM
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