Page 220 - The Handbook for Quality Management a Complete Guide to Operational Excellence

206 P r o c e s s C o n t r o l Q u a n t i f y i n g P r o c e s s Va r i a t i o n 207

6 ˆ σ 6 × 0 009448
.
C = 100 × engineering tolerance = 100 × 0 04 = 142 2%
.
R
.
engineering tolerance 0.04
.
C = = = . 0 527
M 8 ˆ σ 8 × . 0 00948
Z 1 9 .
Z = Minimum { 1 9 } . 1 9 = MIN = = .
. , . 2 3 = 9 C
0 63
MIN PK 3 3
Assuming that the target is precisely 1.000, we compute:
C 0 703
.
0 692
C PM = p = = .
( X T ) 2 (0 99832 − 1 000 ) 2
.
.
−
1 + 1 +
ˆ σ 2 0.000948 2
Since the minimum acceptable value for C is 1, the 0.703 result indi-
P
cates that this process cannot meet the requirements. Furthermore, since
the C index doesn’t consider the centering process, we know that the
P
process can’t be made acceptable by merely adjusting the process closer
to the center of the requirements. Thus, we would expect the Z , Z , and
L
U
Z values to be unacceptable as well.
MIN
The C value always provides the same conclusions as the C index. The
R
P
number implies that the “natural tolerance” of the process uses 142.2 percent
of the engineering requirement, which is, of course, unacceptable.
The C index should be 1.33 or greater. Obviously it is not. If this were
M
a machine capability study, the value of the C index would indicate that
M
the machine was incapable of meeting the requirement.
The value of C is only slightly smaller than that of C . This indicates
PK
P
that we will not gain much by centering the process. The actual amount
we would gain can be calculated by assuming the process is exactly cen-
tered at 1.000 and recalculating Z MIN . This gives a predicted total reject rate
of 3.6 percent instead of 4.0 percent.

09_Pyzdek_Ch09_p151-208.indd 207 11/21/12 1:42 AM