Page 121 - Six Sigma for electronics design and manufacturing

Six Sigma for Electronics Design and Manufacturing
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fuses and counts the number of defectives. Calculate the Cp, Cpk,
population , and the control limits of the nP chart. Assume that the
fuses were made in a normally distributed manufacturing process and
that the process is centered (process average = specification nominal)
and has a two-sided distribution of defects.
Example 3.8 solution
One-sided RR = 0.01/2 = 0.005 = f(–z); therefore z = 2.575 from the z
table (Table 2.3). Cp = Cpk = z/3 = 0.86 = ± SL/3 for no shift.
Using the formula for the nP charts:
s = ( 1 · 0 .9 9 ) = 0.995
UCL np = n p + 3 s = 1 + 3 · 0.995 = 3.985
LCL np = n p – 3 s = 1 – 3 · 0.995 = 0
Note that in this example, the specification limits were not given,
yet the implied Cp and Cpk could be calculated. If a process average
shift to the specification limits is given (such as ±1.5 ), it is still pos-
sible to calculate Cpk if we assume that the rejects are mostly gener-
ated by one side of the distribution.
Example 3.9
A company’s quality team was sent to audit a supplier plant making
their parts given specifications of 8 ± 3. They read a variable control
chart with n = 4 and X = 8.1, UCL x = 11.1 and LCL x = 5.1. What are
the quality data for the population of parts delivered to the company?
Example 3.9 solution
From the X control chart:
3 s = 3
s = 1
= s · n = 2
average shift = 0.1
Cp = 3/(3 · 2) = 0.5
Cpk = min of [(3 – 0.1)/6 = 0.48 or (3 + 0.1)/6 = 0.517] = 0.48
z l = 1.55
f(–z l ) = 0.0606
z 2 = 1.45
f(–z 2 )= 0.0735
Total rejects = 0.0606 + 0.0735 = 0.1341 or 13.41% or 134,100 PPM