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300 Part IV: Quality Assurance
EXAMPLE 19.14
Using the information given in example 19.12, compute Ĉ pl and Ĉ pu .
Solution:
Cˆ
Cˆ
pl
pu
X LSL
272
264
= = = 103 .
3sˆ 3258 .
( )
USL
X 280
272
= = = 103 .
3sˆ 3258 .
( )
Note that both Ĉ pl and Ĉ pu are equal to Ĉ p , which will always be the case when the process
mean is centered between the specification limits. Moreover, when both Ĉ pl and
Ĉ pu are equal, the percentage of nonconforming units below the lower specification
limit and above the upper specification limit are the same.
EXAMPLE 19.15
Uusing the information given in Example 19.13, compute Ĉ pl and Ĉ pu .
Solution:
Part IV.B.5
Cˆ
Cˆ
pl
pu
X LSL
276
264
= = = 155 .
3sˆ 3258 .
( )
USL
X 280
276
= = = 052 .
3sˆ 3258 .
( )
In this case, the value of Ĉ pl is much larger than one whereas the value of Ĉ pu is much
smaller than one, which indicates that most of the nonconforming units produced by
the process are falling above the upper specification limit. Finally, note that both Ĉ pl and
Ĉ pu are sensitive to where the process mean is located.
Process Capability Index: C pk
In order to overcome the centering problem in C p discussed in the preceding
section, process capability index C pk was introduced. The PCI C pk , which is again
one of the first five used in Japan, is defined as
C min C , C . (19.68)
= ( )
pk pl pu
The index C pk is related to the index C p as
C
pk
( )
= 1
k C
(19.69)
p