Abstract. The process capability index Cpk has been widely used in the
manufacturing industry to provide numerical measures on process perfor-mance. Since Cpkis a yield-based index which is independent of the target T,
it fails to account for process centering with symmetric tolerances, and pre-sents an even greater problem with asymmetric tolerances. Pearn and Chen (1998) considered a new generalization C00pkwhich was shown to be superior to other existing generalizations of Cpk for processes with asymmetric
toler-ances. In this paper, we investigate the relation between the fraction non-conforming and the value of C00pk. Furthermore, we derive explicit forms of the cumulative distribution function and the probability density function for the natural estimator ^Cpk00, under the assumption of normality. We also develop a decision making rule based on the natural estimator ^Cpk00 , which can be used to test whether the process is capable or not.
Key words: Asymmetric tolerance; Decision making rule; Normally distrib-uted process; Process yield.
1 Introduction
Process capability indices (PCIs), providing numerical measures of whether or not the ability of a manufacturing process meets a preset level of production tolerance, have recently been a research focus in quality assurance literature. Examples include Kane (1986), Chan et al. (1988), Zhang et al. (1991), Boyles (1991 and 1994), Pearn et al. (1992), Va¨nnman and Kotz (1995), Va¨nn-man(1997), Pearn and Chen (1998), Chen et al. (1999), and Kotz and Johnson (2002).
Among various capability indices that have been introduced, Cpk is
de-fined as
DOI 10.1007/s001840300300
The C
00pkindex for asymmetric tolerances:
Implications and inference
W. L. Pearn1, P. C. Lin2 and K. S. Chen3
1Department of Industrial Engineering and Management, National Chiao Tung University,
Hsinchu, Taiwan, ROC
2Center of General Education, National Chin-Yi Institute of Technology, Taichung, Taiwan,
ROC
3Department of Industrial Engineering and Management, National Chin-Yi Institute of
yield. In fact, we can calculate the process yield as 2Uð3CpkÞ 1 < %Yield < Uð3CpkÞ
if the process is normally distributed, where UðÞ is the cumulative function for the standard normal distribution. To investigate the relationship between the capability indices and the process yield, Boyles (1994) considered the index Spk, a generalization of Cpk, which is defined as:
Spk¼ 1 3U 1 1 2U USL l r þ1 2U l LSL r ; ð3Þ
where U1is the inverse function of U. For normally distributed process, the index Spk is a one-to-one transformation of fraction nonconforming
(per-centage of defective items). We note that given Spk¼ c, we can calculate the
process yield as 2Uð3cÞ 1. Therefore, Spkrepresents the actual process yield
unlike Cpk which is only approximately related to process yield (Boyles
(1994)).
A process is said to have a symmetric tolerance if the target value T is the midpoint of the specification interval (LSL, USL), i.e. T¼ m. Although cases with symmetric tolerances are common in practical situations, cases with asymmetric tolerancesðT 6¼ mÞ often occur in the manufacturing industry.
Since both Cpkand Spkare independent of target value T, then they do not
take into account the asymmetry of the tolerance. Both Cpk and Spkfail to
distinguish between on-target and off-target processes for processes with asymmetric tolerances. Hence, both Cpk and Spk cannot provide consistent
and reasonable measures on process capability for processes with asymmetric tolerances. To overcome the problem, Pearn and Chen (1998) proposed the index C00pk which was shown to be superior to other existing generalizations of Cpk for processes with asymmetric tolerances. Under the assumption of
normality, Pearn and Chen (1998) considered the natural estimator ^C00pk of C00pk, and obtained the exact formula for the r-th moment. In this paper, we investigate the relation between the fraction nonconforming and the value of C00pk. Furthermore, we derive explicit forms of the cumulative distribution function and the probability density function of the natural estimator ^C00pk, under the assumption of normality. We show that the cumulative distribution function and the probability density function of the natural estimator ^C00pkcan be expressed in terms of a mixture of the chi-square distribution and the normal distribution. The explicit forms of the cumulative distribution func-tion and the probability density funcfunc-tion considerably simplify the complexity for analyzing the statistical properties of the natural estimator ^C00pk. We also analyze the bias and the MSE of the natural estimator ^C00pk for normally
distributed processes with an asymmetric tolerance. In addition, we develop a decision making rule based on the natural estimator ^C00pk, which can be used to test whether the process is capable or not.
2 The generalization C00pk
Pearn and Chen (1998) proposed index C00pk, a generalization of Cpk for
processes with asymmetric tolerances. The generalization C00pk is defined as: C00pk¼d
A
3r ; ð4Þ
where Du¼ USL T, D‘¼ T LSL, d¼ minfDu;D‘g,
A¼ maxfdðl TÞ=Du, dðT lÞ=D‘g. Obviously, if T ¼ m (symmetric
tolerance), then d¼ Du¼ D‘ ¼ d, A¼ jl mj and C00pk reduces to the
original index Cpk. We note that the index C00pkobtains the maximal values at
l = T, regardless of whether the preset specification tolerances are symmetric or asymmetric.
If l¼ T, then A ¼ 0 and C00pk¼ d=ð3rÞ. Therefore, if l ¼ T and C00pk¼ c, then r ¼ d=ð3cÞ. Since both C00pkand Spkare functions ofðl; rÞ, we
denote them by C00pkðl; rÞ and Spkðl; rÞ. Figures 1 and 2 display the
con-tours of C00pkðl; rÞ ¼ c (bold) and Spkðl; rÞ ¼ SpkðT; d=ð3cÞÞ (thin) for
c¼ 1=3; 2=3; 1; 4=3; 5=3, and 2 (top to bottom in plots) with asymmetric cases (LSL,T, USL)¼ (10, 40, 50) and (LSL, T, USL) ¼ (10,34, 50), i.e., D‘:d : Du¼ 3 : 2 : 1 and D‘:d : Du¼ 6 : 5 : 4, respectively. Since C00pk
SpkðT; d=ð3C00pkÞÞ for all values of ðl; rÞ, we conclude that given a process
with C00pkðl; rÞ ¼ c the fraction nonconforming would be guaranteed to be less than that of a process with Spkðl; rÞ ¼ SpkðT; d=ð3cÞÞ which is
2f1 U½3SpkðT; d=ð3cÞÞg. For a given threshold value of C00pk, we note that
these contours are used to form boundaries, separating the acceptable values from the unacceptable values of ðl; rÞ. In addition, we have
Fig. 1. Contours of C00pk(l, r) = c (bold) and Spk(l, r) = Spk(T, d*/(3c)) (thin) for c = 1/3, 2/3, 1,
4/3, 5/3, and 2 (top to bottom in plot) with (LSL, T, USL) = (10, 40, 50).
Spk T; d 3c ¼1 3U 1 1 2U 3c minf1; rg þ1 2U 3c maxf1; rgð Þ ;
where r¼ D‘=Du. For example, c¼ 1 with r ¼ 3 gives the fraction
noncon-forming would be guaranteed to be less than 2f1 U½3SpkðT; d=ð3cÞÞg ¼
2 ½Uð3c = minf1; rgÞ þ Uð3c maxf1; rgÞ ¼ 2 ½Uð3Þ þ Uð9Þ ¼ 1350 PPM for the asymmetric case (LSL, T, USL)¼ (10, 40, 50) in Figure 1, c ¼ 1 with r¼ 3=2 gives the fraction nonconforming would be guaranteed to be less than 2 ½Uð3Þ þ Uð9=2Þ ¼ 1353 PPM for the asymmetric case (LSL, T, USL)¼ (10, 34, 50) in Figure 2, where ‘‘PPM’’ denotes the ‘‘Parts-Per-Million’’.
3 Distribution of ^C00pk
To estimate the generalization C00pk, Pearn and Chen (1998) considered the natural estimator ^C00pk defined in the following. The natural estimator ^C00pk is obtained by replacing the process mean l and the process variance r2by their
conventional estimators X and S2, which may be obtained from a stable
process. ^ C00pk¼d
^A
3S ; ð5Þ
where d¼ minfDu;D‘g, A^¼ maxfdðX TÞ=Du, dðT XÞ=D‘g,
X¼Pni¼1Xi=n and S¼ fðn 1Þ1Pni¼1ðXi XÞ2g1=2 . Obviously, if T¼ m
(symmetric tolerance), then d¼ Du¼ D‘¼ d, ^A¼ jX mj and ^C00pk
re-duces to ^Cpk, the natural estimator of Cpkdiscussed by Kotz et al (1993).
We now define B¼ n1=2ðd=rÞ, K ¼ ðn 1ÞS2=r2, Z¼ n1=2ðX TÞ=r,
Y¼ ½maxfðd=DuÞZ; ðd=D‘ÞZg2, d¼ n1=2ðl TÞ=r. Then, the estimator
^
C00pk can be written as: ^ C00pk¼ ffiffiffiffiffiffiffiffiffiffiffi n 1 p ðB pffiffiffiffiYÞ 3pffiffiffiffiffiffiffinK : ð6Þ
Fig. 2. Contours of C00pk(l, r) = c (bold) and Spk(l, r) = Spk(T, d*/(3c)) (thin) for c = 1/3, 2/3, 1,
Under the assumption of normality of X, K is distributed as v2n1, a
chi-square distribution with n 1 degrees of freedom, and Z is distributed as the normal distribution Nðd; 1Þ with mean d and variance 1. Let UðÞ and /ðÞ be the cumulative distribution function and the probability density function of the standard normal distribution Nð0; 1Þ, respectively. Then, the cumulative distribution function and the probability density function of Z can be ex-pressed as: Uðz dÞ and /ðz dÞ, respectively. Hence, the cumulative dis-tribution function of Y can be expressed as:
FYðyÞ ¼ U½ðDu=dÞ ffiffiffiy
p
d U½ðD‘=dÞ ffiffiffiy
p
d: ð7Þ
The probability density function of Y can be expressed as: fYðyÞ ¼ 1 2dpffiffiffiy Du/½ðDu=d Þpffiffiffiy d þ D ‘/½ðD‘=dÞ ffiffiffiy p þ d ð Þ: ð8Þ
FYðyÞ and fYðyÞ can be used to derive the sampling distribution of ^C00pk (see
Appendix A).
Therefore, the cumulative distribution function of ^C00pkcan be obtained as the following. FC^00 pk ðxÞ ¼ R 1 B2 FKðLðx; yÞÞfYðyÞdy; x <0; 1 FYðB2Þ; x¼ 0; 1R B2 0 FKðLðx; yÞÞfYðyÞdy; x > 0; 8 > > > > > < > > > > > : ð9Þ
and the probability density function of ^C00pkcan be derived as:
fC^00 pk x ð Þ ¼ R 1 1 fKLðx; B2tÞfYB2t 2Lðx;B 2 tÞ x dt; x < 0, R1 0 fK Lðx; B2tÞ fY B2t 2Lðx;B2tÞ x dt; x > 0; 8 > > > < > > > : ð10Þ where B¼ n1=2ðd=rÞ, L(x, y) ¼ ðn 1ÞðB y1=2Þ2=ð9nx2Þ, F KðÞ is the
cumulative distribution function of K, fKðÞ is the probability density
func-tion of K, FYðÞ is the cumulative distribution function of Y expressed as Eq.
(7), and fYðÞ is the probability density function of Y expressed as Eq. (8).
As an illustration for some of the results obtained, we plot the probability density functions of ^C00pkfor an asymmetric case (D‘:d : Du¼ 6 : 5 : 4) with
r¼ d=3, n¼ 1:0; 0:5; 0:5; 1:0, and n ¼ 10; 20; 50, where n ¼ ðl TÞ=r and d¼ minfDu;D‘g. Figures 3 and 4 display the plots of the probability
density functions of ^C00pk for n¼ 1:0ðC00pk¼ 0:78Þ and n ¼ 1:0ðC 00
pk¼ 0:67Þ,
respectively. From Figures 3 and 4 we observe that for n¼ 10 the distribu-tions are skew and have large spread. We also observe that as n increases the spread decreases and so does the skewness. We also observe that the esti-mated index ^C00pk is approximately unbiased for sample size n > 50.
Pearn and Chen (1998) derived the r-th moment of ^C00pkwithout using the distribution of ^C00pk. We note that the estimator ^C00pkis biased. The magnitude
of the bias is Bð ^C00pkÞ ¼ Eð ^C00pkÞ C00pk. The mean square error can be expressed
as MSEð ^C00pkÞ ¼ Varð ^C00pkÞ þ B2ð ^C00 pkÞ, where Varð ^C 00 pkÞ ¼ Eð ^C 00 pkÞ 2 E2ð ^C00 pkÞ is The Cpkindex for asymmetric tolerances: Implications and inference 123
the variance of ^C00pk. To investigate the behavior of the estimator ^C00pk, the bias and the mean square error are calculated (using Maple computer software) for various values of n¼ ðl TÞ=r, b ¼ d=r, d‘¼ d/D‘, du¼ d=Du, and
sample size n.
Tables 1, 2, and 3 display the values of C00pk, Bð ^C00pkÞ and MSEð ^C00pkÞ for n¼ 1:0 (0.5)1.0, ðd‘;duÞ ¼ ð5=6; 5=4Þ, and n ¼ 10ð10Þ50, with b ¼ 3, 4, and
5, respectively. We note that the specification with ðd‘;duÞ ¼ ð5=6; 5=4Þ is
asymmetric.
From Tables 1, 2, and 3, we observe that as the sample size n increases, both the bias and the mean square error of ^C00pkdecrease. Figure 5 displays the plot of the bias of ^C00pk(vs. n) with n¼ 0, 1.0, and 1.0 (from bottom to top in the plot) for fixed b¼ 3, d‘¼ 5=6, du¼ 5=4. Figure 6 displays the plot of the
MSE of ^C00pk(vs. n) with n¼ 1:0; 1:0, and 0 (from bottom to top in the plot) for fixed b¼ 3, d‘¼ 5=6, du¼ 5=4.
Fig. 3. The pdf of ^C00pkwith r = d*/3, n =)1.0, and n = 10 (bottom), 20 (middle), and 50 (top) for the asymmetric case Dl: d : Du= 6 : 5 : 4.
Fig. 4. The pdf of ^C00pkwith r = d*/3, n = 1.0, and n = 10 (bottom), 20 (middle), and 50 (top) for
From Tables 1, 2, and 3, we also observe that as the value of b increases, both the bias and the mean square error of ^C00pkincrease for fixed d‘, du;n, and
n. Figure 7 displays the plot of the bias of ^C00pk (vs. n) with b¼ 3, 4, and 5 (from bottom to top in the plot) for fixed n¼ 0:5, d‘ ¼ 5=6, du¼ 5=4.
Figure 8 displays the plot of the MSE of ^C00pk(vs. n) with b¼ 3, 4, and 5 (from bottom to top in the plot) for fixed n¼ 0:5, d‘¼ 5=6, du¼ 5=4.
We note that ^C00pkis a biased estimator. The results in Tables 1–3, Figures 5 and 7 indicate that the bias of ^C00pk is positive when l6¼ T. That is, C00pk is generally overestimated by ^C00pk. On the other hand, when l¼ T, we have A¼ 0 and C00pk= d=ð3rÞ, the bias of ^C00pktends to be negative for some cases as shown in Tables 1–3 and Figure 5. Thus, ^C00pkis smaller than C00pkand the bias is negative when l = T. This is partially contributed by the fact that ^A
Table 2. The values of C00pk, B( ^C 00 pk) and MSE( ^C 00 pk) for b = 4, n = )1.0(0.5)1.0, d‘= 5/6, du= 5/4, and n = 10(10)50 n n =) 1.0 n =) 0.5 n = 0 n = 0.5 n = 1.0
bias MSE bias MSE bias MSE bias MSE bias MSE
10 0.1047 0.1264 0.1105 0.1485 0.0490 0.1474 0.1053 0.1427 0.0942 0.1115 20 0.0464 0.0449 0.0505 0.0535 0.0041 0.0551 0.0482 0.0523 0.0418 0.0403 30 0.0298 0.0270 0.0327 0.0322 )0.0058 0.0341 0.0312 0.0317 0.0268 0.0244 40 0.0220 0.0192 0.0241 0.0230 )0.0094 0.0248 0.0230 0.0227 0.0198 0.0175 50 0.0174 0.0150 0.0191 0.0179 )0.0110 0.0195 0.0182 0.0177 0.0156 0.0136 C00pk 1.1111 1.2222 1.3333 1.1667 1.0000
Table 1. The values of C00pk , Bð ^C00pk) and MSE( ^C00pk) for b = 3, n = )1.0(0.5)1.0, d‘= 5/6,
du= 5/4, and n = 10(10)50
n n =) 1.0 n =) 0.5 n = 0 n = 0.5 n = 1.0
bias MSE bias MSE bias MSE bias MSE bias MSE
10 0.0733 0.0651 0.0791 0.0806 0.0175 0.0807 0.0739 0.0785 0.0628 0.0575 20 0.0325 0.0234 0.0366 0.0295 )0.0099 0.0311 0.0342 0.0296 0.0278 0.0214 30 0.0209 0.0141 0.0237 0.0179 )0.0147 0.0195 0.0223 0.0181 0.0179 0.0131 40 0.0154 0.0101 0.0175 0.0128 )0.0160 0.0142 0.0164 0.0130 0.0132 0.0094 50 0.0122 0.0079 0.0139 0.0099 )0.0162 0.0113 0.0130 0.0101 0.0104 0.0073 C00 pk 0.7778 0.8889 1.0000 0.8333 0.6667
Table 3. The values of C00pk, B( ^C00pk) and MSE( ^C00pk) for b = 5, n =)1.0(0.5)1.0, d‘= 5/6,
du= 5/4, and n = 10(10)50
n n =) 1.0 n =) 0.5 n = 0 n = 0.5 n = 1.0
bias MSE bias MSE bias MSE bias MSE bias MSE
10 0.1361 0.2092 0.1419 0.2380 0.0804 0.2357 0.1367 0.2286 0.1256 0.1871 20 0.0603 0.0739 0.0644 0.0851 0.0180 0.0867 0.0621 0.0826 0.0557 0.0669 30 0.0387 0.0444 0.0416 0.0511 0.0032 0.0532 0.0401 0.0499 0.0358 0.0403 40 0.0285 0.0316 0.0307 0.0365 )0.0028 0.0385 0.0296 0.0356 0.0263 0.0288 50 0.0226 0.0246 0.0243 0.0283 )0.0058 0.0302 0.0235 0.0277 0.0209 0.0224 C00pk 1.4444 1.5556 1.6667 1.5000 1.3333
is calculated to be positive (see Eq.(5)) even when l= T and A¼ 0. Clearly, the presence of ^A in Eq. (5) reduces the value of the calculated ^C00pk. As the sample size n increases, the mean square error of ^C00pk decreases. Proper sample sizes for capability estimation are essential. The smaller the sample size is, the higher the value of ^C00pk is required to justify the true process capability.
4 A decision making rule for testing C00pk
Using the index C00pk, the engineers can access the process performance and monitor the manufacturing processes on routine basis. To obtain a decision making rule we consider a testing hypothesis with the null hypothesis C00pk C (the process is incapable) and the alternative hypothesis C00pk>C (the process is capable). The null hypothesis will be rejected if ^C00pk>ca, where the
con-stant ca, called the critical value, is determined so that the significance level of
the test is a, i.e., Pð ^C00pk>cajC00pk¼ CÞ ¼ a. The decision making rule to be Fig. 6. MSE plot of C00pk(vs. n) for b = 3, dl= 5/6, du= 5/4 with n = 1.0, -1.0, and 0 (from
bottom to top in the plot).
Fig. 5. Bias plot of ^C00pk (vs. n) for b = 3, dl= 5/6, du= 5/4 with n = 0, 1.0, and –1.0 (from
used is then that, for given values of risk a and sample size n, the process will be considered capable if ^C00pk>caand incapable if ^C00pk ca.
We note that by setting n¼ ðl TÞ=r and b ¼ d=r, the index C00pk can be rewritten as C00pk¼ ½b þ n= maxf1; rg=3 for n < 0 and C00pk¼ ½b n minf1; rg=3 for n 0 where r ¼ D‘=Du. Hence, the value of C00pkcan be
calculated given values of n, b, and r. For example, ifðn; b; rÞ ¼ ð1; 3; 3=2Þ then C00pk¼ ½3 þ ð1Þ= maxf1; 3=2g=3 ¼ 7=9 ¼ 0:7778. If C00pk¼ C, we have b¼ 3C n= maxf1; rg for n < 0 and b ¼ 3C þ n minf1; rg for n 0. In addition, since B¼ n1=2ðd=rÞ and b ¼ d=r then B2 ¼ nb2. Therefore, if
C00pk¼ C then B2¼ n 3Cð n= maxf1; rgÞ 2; n < 0 n 3Cð þ n minf1; rgÞ2; n 0: ( ð11Þ
Fig. 7. Bias plot of ^C00pk(vs. n) for n = 0.5, dl= 5/6, du= 5/4 with b = 3, 4, and 5 (from bottom
to top in the plot).
Fig. 8. MSE plot of ^C00pk(vs. n) for n = 0.5, dl= 5/6, du= 5/4 with b = 3, 4, and 5 (from bottom
to top in the plot).
We can use the central chi-square distribution and the normal distribution to find the critical value ca satisfying Pð ^C00pk>cajC00pk¼ CÞ ¼ a, i.e.,
1 FC^00
pkðcaÞ ¼ a given C
00
pk¼ C. We note that ca is larger than zero in
gen-eral, hence we can find ca by Eq. (9)
ZB2
0
FKðLðca;yÞÞfYðyÞdy ¼ a; ð12Þ
where B2 is given in Eq. (11) and Lðca;yÞ ¼ ðn 1ÞðB y1=2Þ2=ð9nc2aÞ.
We point out that if T¼ m (symmetric tolerance) then C00pk reduces to Cpk and ^C00pk reduces to ^Cpk. We note that the critical values ca for n¼ n0
and n¼ n0 are the same when T¼ m (for the proof see Appendix B).
Tables 4a–7b display the critical values ca for C¼ 1:00; 1:33; 1:66, and 2.00
with sample sizes n¼ 10ð10Þ100, jnj ¼ 0:0ð0:1Þ1:0 and a-risk ¼ 0:01 and 0.05 for T¼ m.
To test if the process meets the capability (quality) requirement, we first determine the value of C and the a-risk. Since both the process parameters l and r are unknown, then parameter n¼ ðl TÞ=r is also unknown. But, we
0.80 2.139 1.621 1.461 1.379 1.328 1.292 1.266 1.245 1.228 1.214 0.90 2.141 1.621 1.461 1.379 1.328 1.292 1.266 1.245 1.228 1.214 1.00 2.141 1.621 1.461 1.379 1.328 1.292 1.266 1.245 1.228 1.214
Table 4b. Critical values cafor C = 1.00 with |n| = 0.0(0.1)1.0, and sample sizes n = 10(10)100
for T = m and a-risk = 0.05
|n| n = 10 20 30 40 50 60 70 80 90 100 0.00 1.523 1.300 1.227 1.188 1.163 1.146 1.133 1.123 1.115 1.108 0.10 1.572 1.339 1.262 1.221 1.195 1.177 1.163 1.152 1.143 1.135 0.20 1.610 1.367 1.284 1.240 1.212 1.191 1.176 1.164 1.153 1.145 0.30 1.639 1.384 1.296 1.249 1.218 1.196 1.180 1.166 1.155 1.146 0.40 1.659 1.393 1.301 1.252 1.220 1.197 1.180 1.167 1.156 1.147 0.50 1.671 1.397 1.303 1.252 1.220 1.197 1.180 1.167 1.156 1.147 0.60 1.679 1.399 1.303 1.252 1.220 1.197 1.180 1.167 1.156 1.147 0.70 1.683 1.399 1.303 1.252 1.220 1.197 1.180 1.167 1.156 1.147 0.80 1.685 1.399 1.303 1.252 1.220 1.197 1.180 1.167 1.156 1.147 0.90 1.686 1.399 1.303 1.252 1.220 1.197 1.180 1.167 1.156 1.147 1.00 1.686 1.399 1.303 1.252 1.220 1.197 1.180 1.167 1.156 1.147
can estimate n by calculating the value ^n¼ ðX TÞ=S from the sample. If the estimated value ^C00pk is larger than the critical value cað ^C00pk>caÞ, then we
conclude that the process meets the capability requirement ðC00pk>CÞ. Otherwise, we do not have sufficient information to conclude that the process
Table 5a. Critical values cafor C = 1.33 with |n| = 0.0(0.1)1.0, and sample sizes n = 10(10)100
for T = m and a-risk = 0.01
|n| n = 10 20 30 40 50 60 70 80 90 100 0.00 2.606 2.017 1.837 1.746 1.689 1.650 1.620 1.597 1.579 1.564 0.10 2.667 2.062 1.877 1.782 1.723 1.682 1.652 1.628 1.608 1.592 0.20 2.716 2.094 1.902 1.804 1.742 1.698 1.666 1.641 1.620 1.603 0.30 2.751 2.113 1.915 1.813 1.749 1.704 1.670 1.644 1.622 1.605 0.40 2.776 2.124 1.921 1.817 1.751 1.705 1.671 1.644 1.623 1.605 0.50 2.792 2.129 1.923 1.817 1.751 1.705 1.671 1.644 1.623 1.605 0.60 2.802 2.131 1.924 1.817 1.751 1.705 1.671 1.644 1.623 1.605 0.70 2.807 2.132 1.924 1.817 1.751 1.705 1.671 1.644 1.623 1.605 0.80 2.809 2.132 1.924 1.817 1.751 1.705 1.671 1.644 1.623 1.605 0.90 2.811 2.132 1.924 1.817 1.751 1.705 1.671 1.644 1.623 1.605 1.00 2.811 2.132 1.924 1.817 1.751 1.705 1.671 1.644 1.623 1.605
Table 5b. Critical values cafor C = 1.33 with |n| = 0.0(0.1)1.0, and sample sizes n = 10(10)100
for T = m and a-risk = 0.05
|n| n = 10 20 30 40 50 60 70 80 90 100 0.00 2.062 1.750 1.647 1.593 1.558 1.534 1.516 1.502 1.491 1.481 0.10 2.111 1.789 1.682 1.626 1.590 1.564 1.545 1.530 1.518 1.508 0.20 2.149 1.815 1.703 1.644 1.605 1.578 1.557 1.541 1.528 1.516 0.30 2.176 1.831 1.714 1.651 1.611 1.582 1.560 1.543 1.529 1.518 0.40 2.194 1.840 1.719 1.654 1.612 1.583 1.561 1.544 1.529 1.518 0.50 2.206 1.843 1.720 1.654 1.612 1.583 1.561 1.544 1.529 1.518 0.60 2.213 1.845 1.720 1.654 1.612 1.583 1.561 1.544 1.529 1.518 0.70 2.217 1.845 1.720 1.654 1.612 1.583 1.561 1.544 1.529 1.518 0.80 2.219 1.845 1.720 1.654 1.612 1.583 1.561 1.544 1.529 1.518 0.90 2.219 1.845 1.720 1.654 1.612 1.583 1.561 1.544 1.529 1.518 1.00 2.220 1.845 1.720 1.654 1.612 1.583 1.561 1.544 1.529 1.518
Table 6a. Critical values cafor C = 1.66 with |n| = 0.0(0.1)1.0, and sample sizes n = 10(10)100
for T = m and a-risk = 0.01
|n| n = 10 20 30 40 50 60 70 80 90 100 0.00 3.287 2.535 2.306 2.190 2.117 2.067 2.029 2.000 1.977 1.958 0.10 3.349 2.580 2.346 2.226 2.151 2.099 2.060 2.030 2.006 1.986 0.20 3.396 2.611 2.370 2.246 2.168 2.114 2.074 2.042 2.016 1.995 0.30 3.431 2.630 2.382 2.255 2.175 2.119 2.077 2.044 2.018 1.997 0.40 3.455 2.639 2.387 2.258 2.177 2.120 2.078 2.045 2.019 1.997 0.50 3.469 2.644 2.389 2.258 2.177 2.120 2.078 2.045 2.019 1.997 0.60 3.479 2.646 2.389 2.258 2.177 2.120 2.078 2.045 2.019 1.997 0.70 3.483 2.646 2.389 2.258 2.177 2.120 2.078 2.045 2.019 1.997 0.80 3.486 2.646 2.389 2.258 2.177 2.120 2.078 2.045 2.019 1.997 0.90 3.487 2.646 2.389 2.258 2.177 2.120 2.078 2.045 2.019 1.997 1.00 3.487 2.646 2.389 2.258 2.177 2.120 2.078 2.045 2.019 1.997 The Cpkindex for asymmetric tolerances: Implications and inference 129
meets the present capability requirement. In this case, we would believe that C00pk C (the process is incapable).
We also can calculate the p-value, i.e. the probability that ^C00pkexceed the observed estimated index given the values of C, n¼ ðl TÞ=r, r ¼ D‘=Du, 0.80 2.756 2.293 2.140 2.058 2.006 1.970 1.943 1.922 1.904 1.890 0.90 2.756 2.293 2.140 2.058 2.006 1.970 1.943 1.922 1.904 1.890 1.00 2.757 2.294 2.140 2.058 2.006 1.970 1.943 1.922 1.904 1.890
Table 7a. Critical values cafor C = 2.00 with |n| = 0.0(0.1)1.0, and sample sizes n = 10(10)100
for T = m and a-risk = 0.01.
|n| n = 10 20 30 40 50 60 70 80 90 100 0.00 3.991 3.070 2.790 2.647 2.559 2.498 2.451 2.416 2.387 2.364 0.10 4.052 3.115 2.829 2.683 2.592 2.529 2.482 2.446 2.416 2.391 0.20 4.099 3.145 2.853 2.703 2.609 2.543 2.495 2.456 2.426 2.400 0.30 4.133 3.163 2.864 2.711 2.615 2.548 2.498 2.459 2.427 2.401 0.40 4.156 3.172 2.869 2.713 2.616 2.549 2.498 2.459 2.428 2.402 0.50 4.170 3.176 2.870 2.714 2.616 2.549 2.498 2.459 2.428 2.402 0.60 4.178 3.178 2.871 2.714 2.616 2.549 2.498 2.459 2.428 2.402 0.70 4.183 3.179 2.871 2.714 2.616 2.549 2.498 2.459 2.428 2.402 0.80 4.185 3.179 2.871 2.714 2.616 2.549 2.498 2.459 2.428 2.402 0.90 4.186 3.179 2.871 2.714 2.616 2.549 2.498 2.459 2.428 2.402 1.00 4.186 3.179 2.871 2.714 2.616 2.549 2.498 2.459 2.428 2.402
Table 7b. Critical values cafor C = 2.00 with |n| = 0.0(0.1)1.0, and sample sizes n = 10(10)100
for T = m and a-risk = 0.05.
|n| n = 10 20 30 40 50 60 70 80 90 100 0.00 3.160 2.665 2.502 2.417 2.362 2.324 2.295 2.273 2.255 2.240 0.10 3.209 2.704 2.537 2.449 2.393 2.353 2.324 2.301 2.282 2.266 0.20 3.246 2.729 2.557 2.466 2.407 2.366 2.334 2.310 2.290 2.273 0.30 3.272 2.744 2.567 2.472 2.412 2.369 2.337 2.312 2.291 2.274 0.40 3.289 2.751 2.571 2.474 2.413 2.370 2.337 2.312 2.291 2.274 0.50 3.300 2.755 2.572 2.475 2.413 2.370 2.337 2.312 2.291 2.274 0.60 3.306 2.756 2.572 2.475 2.413 2.370 2.337 2.312 2.291 2.274 0.70 3.309 2.756 2.572 2.475 2.413 2.370 2.337 2.312 2.291 2.274 0.80 3.311 2.756 2.572 2.475 2.413 2.370 2.337 2.312 2.291 2.274 0.90 3.312 2.756 2.572 2.475 2.413 2.370 2.337 2.312 2.291 2.274 1.00 3.312 2.756 2.572 2.475 2.413 2.370 2.337 2.312 2.291 2.274
and sample size n, and then compare this probability with the significance level a. If the estimated index value is c0, given the values of C, n, r, and
sample size n, then the p-value can be calculated as: p-value¼ Pð ^C00pk>c0jC00pk¼ CÞ¼1FC^00 pkðc0Þ ¼ ZB2 0 FKðLðc0;yÞÞ 1 2 ffiffiffipy Du d/½ Du d ffiffiffi y p npffiffiffinþD‘ d/½ D‘ d ffiffiffi y p þnpffiffiffin dy; ð13Þ where Du=d ¼ 1= minf1; rg, D‘=d ¼ maxf1; rg, B2 is given in Eq. (11) and
Lðc0;yÞ ¼ ðn 1ÞðB y1=2Þ2=ð9nc20Þ. The numerical calculations can be easily
carried out using the Maple computer software, to integrate the function based on the central chi-square distribution and the normal distribution. If the p-value is smaller than the a-risk, than we conclude that the process meets the capability requirement (C00pk>C). Otherwise, we do not have sufficient information to conclude that the process meets the present capability requirement. In this case, we would believe that C00pk C (the process is incapable).
As an example, we consider the following normally distributed process with asymmetric specification tolerances LSL¼ 20, T¼ 26:5, and USL¼ 32. We note that d ¼ ðUSL LSLÞ=2 ¼ 6, D‘ ¼ T LSL ¼ 6:5,
Du¼ USL T ¼ 5:5, d¼ minfD‘;Dug ¼ 5:5, r ¼ D‘=Du¼ 1:18. To test if
the process meets the capability (quality) requirement, we first determine C¼ 1:33, i.e., we define a process with C00pk>1:33 is capable. If the sample size n¼ 100, the sample mean X ¼ 27, and the sample standard deviation S¼ 1:10. We can calculate ^A ¼ maxfdðX TÞ=Du, dðT XÞ=D‘g ¼ 0:5,
^
n¼ ðX TÞ=S ¼ 0:45, and ^C00pk¼ 1:515. We find the corresponding p-value is 0.055 using the Maple computer software to calculate Eq. (13). We conclude that the process meets the capability requirement if the a-risk is set larger than 0.055. If the a-risk is set smaller than 0.055, we do not have sufficient information to conclude that the process meets the present capa-bility requirement.
5 An application example
The example presented in the following concerns with the capability of a process, which produces electronic telecommunication amplifiers (see Pearn et al. (2001)). The original data and a complete description of this process are given in Juran Institute (1990). The quality characteristic of interest is the gain (the boosting ability) of an amplifier. The design of the amplifiers had called for a gain of 10 decibels (dB) and allowed the amplifiers to be con-sidered acceptable if the gain fell between 7.75 dB and 12.25 dB, i.e. (LSL, T,USL)¼ ð7:75; 10; 12:25Þ. A sample of the gains of 120 amplifiers was taken by the quality improvement team to estimate the capability of the manufacturing process producing the amplifiers. Chou et al. (1998) noted that the data follow a non-Normal distribution. The data were then fitted by an SB
distribution. They also transformed the data to approximate Normality using the estimated transformation
Z¼ 0:96 þ 0:98 ln X 7:59 4:68þ 7:59 X
: ð14Þ
We note that a significant error may be introduced if someone use the original specification limits, (LSL, T, USL)¼ ð7:75; 10; 12:25Þ, to evaluate the quality through the transformed data. Using the estimated transformation Eq. (14), we have the transformed specificationðLSL0;T0;USL0Þ ¼ ð2:31; 1:00; 5:06Þ as well as the transformed data.
Table 8 displays the sample of the original gains of 120 amplifiers listed in Juran Institute (1990). A histogram of the data, with the specification limits, is given in Figure 9. Table 9 displays the corresponding transformed amplifier gain data, using the estimated transformation in Eq. (14). A histogram of the transformed data, with the transformed specification limits, is given in Figure 10. We may now apply a normal-based SPC procedure to the trans-formed data. We note that the transtrans-formed specification (LSL0, T0, USL0) is asymmetric. Therefore, we apply the proposed generalization C00pk to the transformed data. To test if the quality of the amplifiers meets the quality
8.6 9.2 8.5 9.6 9.0 10.7 8.6 10.0 8.8 8.6
Fig. 9. Histogram of the 120 untransformed amplifier gain data with (LSL, T, USL) = (7.75, 10, 12.25).
requirement, we first determine C¼ 1:00, i.e., we define a process with C00pk>1:00 is capable. We then calculate d¼ ðUSL0 LSL0Þ=2 ¼ 3:685, d¼ minfDu, D‘g ¼ minf4:06; 3:31g ¼ 3:31, n ¼ 120, Z ¼Pni¼1Zi=n¼ 0:000713,
S2¼Pni¼1ðZi ZÞ2=ðn 1Þ ¼ 0:985, S ¼ 0:993, ^A¼ maxfdðZ T0Þ=Du,
dðT0 ZÞ=D‘g ¼ maxf0:815; 0:999g ¼ 0:999, ^n¼ ðZ T0Þ=S ¼ 1:007,
and ^C00pk¼ ðd ^AÞ=ð3SÞ ¼ 0:776. We then find the corresponding p-value be 0.9999 using the Maple computer software to calculate Eq. (13). Obvi-ously, the quality of the amplifiers does not meet the quality requirement: C00pk>1:00.
While all the 120 amplifiers fell within the specification limits, the low value of ^C00pk shows that the average quality of the amplifiers significantly deviates from the target value, which is unsatisfactory causing the commu-nication failed. The quality improvement team could now concentrate their investigations to find problems causing the manufacturing line incapable, and find ways to make the process average closer to the target value. Some quality improvement activities involving Taguchi’s parameter designs should be ini-tiated to identify the significant factors causing the process failing to cluster around the target value.
Table 9. The transformed amplifier gain data
)1.1 1.4 )0.1 0.8 )2.0 0.9 2.9 )1.3 0.4 0.1 )0.9 0.0 1.1 0.5 0.3 )1.6 0.6 1.8 )2.0 )0.7 0.2 )0.6 0.7 2.0 )1.6 )0.4 )0.2 )2.0 1.4 )0.4 2.6 )1.3 )1.6 )0.7 )0.2 1.0 0.5 0.1 0.3 1.6 0.4 0.8 )0.2 )0.9 0.0 )0.3 0.6 0.5 )0.1 )0.7 )0.6 0.2 1.1 )2.0 )1.1 )0.1 )1.3 0.3 )0.6 )2.0 )1.6 )0.4 0.3 )0.2 1.2 )1.6 0.9 )0.7 0.1 0.7 0.9 1.5 )0.3 0.5 )0.1 )0.9 1.4 0.8 0.2 )1.3 )0.2 0.9 )0.4 0.0 0.2 )0.6 )1.1 0.6 )0.2 0.4 )1.1 1.1 0.7 )0.7 )1.3 0.9 0.1 0.0 )1.1 0.8 )0.4 )0.9 0.1 1.2 0.6 )0.7 0.0 0.2 1.3 )0.6 )0.3 0.3 )0.4 0.7 0.1 1.6 )0.3 1.0 )0.1 )0.3
Fig. 10. Histogram of the 120 transformed amplifier gain data with (LSL’, T’, USL’) = (-2.31, 1.00, 5.06).
statistical properties of ^Cpk is greatly simplified. We also analyzed the bias and the MSE of the estimated index ^C00pkfor normally distributed processes. Furthermore, we also developed a decision making rule, based on the natural estimator ^C00pk. The function of p-value was given and the numerical calcu-lations of p-value can be easily carried out using mathematical computer softwares, e.g., Mathematica, Maple, and MatLab. Therefore, the practitio-ners can use the proposed decision making rules to test whether the process with asymmetric tolerance is capable or not.
References
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Boyles RA (1994) Process capability with asymmetric tolerances. Commun Statist-Simul 23:615– 643
Chan LK, Cheng SW, Spiring FA (1988) A new measure of process capability: Cpm. Journ Qual
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Chen KS, Pearn WL, Lin PC (1999) A new generalization of Cpmfor processes with asymmetric
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Pearn WL, Lin PC, Chen KS (2001) Estimating process capability index C00pmk for asymmetric tolerances. Metrika 54:261–279
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Appendix A: Derivation of Eq. (9)
Under the assumption of normality, the cumulative distribution function and the probability density function of ^C00pkcan be derived as follows.
[Case I]: For x > 0, using the technique of conditioning ^C00pkon Y in Eq. (6), we may obtain FC^00 pk ðxÞ ¼ Pð ^C00pk xÞ ¼ P ffiffiffiffiffiffiffiffiffiffiffi n 1 p ðB pffiffiffiffiYÞ 3pffiffiffiffiffiffiffinK x ! ¼ 1 P pffiffiffiffiffiffiffinK< ffiffiffiffiffiffiffiffiffiffiffi n 1 p ðB pffiffiffiffiYÞ 3x ! ¼ 1 Z 1 0 P pffiffiffiffiffiffiffinK< ffiffiffiffiffiffiffiffiffiffiffi n 1 p ðB pffiffiffiyÞ 3x ! fYðyÞ dy:
Since K is distributed as v2n1, then P pffiffiffiffiffiffiffinK< ffiffiffiffiffiffiffi n1 p ðBpffiffiffiyÞ 3x ¼ 0 for y > B2 and x > 0. Hence, FC^00 pk ðxÞ ¼ 1 Z B2 0 P pffiffiffiffiffiffiffinK< ffiffiffiffiffiffiffiffiffiffiffi n 1 p ðB pffiffiffiyÞ 3x ! fYðyÞdy ¼ 1 Z B2 0 P K < Lðx,yÞð Þ fYðyÞdy ¼ 1 Z B2 0
FKðLðx,yÞÞ fYðyÞdy; for x > 0; ðA1Þ
where B¼ n1=2ðd
=rÞ, Lðx; yÞ ¼ ðn 1ÞðB y1=2Þ2=
ð9nx2Þ, F
KðÞ is the
cumulative distribution function of K, and fYðÞ is the probability density
function of Y expressed as Eq. (8).
[Case II]: Since K is distributed as v2
n1, then P ffiffiffiffiffiffiffi nK p ffiffiffiffiffiffiffin1 p ðBpffiffiffiyÞ 3x ¼ 0 for x < 0 and y < B2. Hence,
F^ C00pk ðxÞ ¼ Z 1 B2 P ffiffiffiffiffiffiffi nK p ffiffiffiffiffiffiffiffiffiffiffi n 1 p ðB pffiffiffiyÞ 3x ! fYðyÞ dy ¼ Z 1 B2 P Kð Lðx; yÞÞ fYðyÞ dy ¼ Z 1
B2 FKðLðx; yÞÞ fYðyÞ dy; for x < 0: ðA2Þ
where FYðÞ is the cumulative distribution function of Y expressed as Eq. (7).
Combining Eqs. (A1), (A2) and (A3), we obtain Eq. (9) for the cumulative distribution function of ^C00pk. Taking the derivative of the cumulative distribution function of ^C00pk with Leibniz’s rule and changing the variable with t¼ y/B2, we obtain the probability density function of ^
C00pk in Eq. (10).
Appendix B: Symmetry property of n
Following we will show that given the same values of C, n, and a the critical values ca for n¼ n0 and n¼ n0 are the same when T¼ m.
Since if T¼ m then r ¼ D‘=Du¼ 1 and by Eq. (11) B2¼ D2¼
nðd=rÞ2 ¼ nð3C þ jnjÞ2 given C00pk¼ C. Furthermore, d ¼ n1=2ðl TÞ=r ¼
n1=2n and f
Y(y) expressed as Eq. (8) reduces to
fYðyÞ ¼ 1 2 ffiffiffipy /½ ffiffiffi y p pffiffiffinn þ /½pffiffiffiyþpffiffiffinn ¼ 1 2 ffiffiffipy /½ ffiffiffi y p pffiffiffinjnj þ /½pffiffiffiyþpffiffiffinjnj ; for T¼ m: ðA4Þ Therefore, the Eq. (12) reduces to
ZD2
0
FKðLðca;yÞÞfYðyÞ dy ¼ a; ðA5Þ
where Lðca;yÞ ¼ ðn 1ÞðD y1=2Þ2=ð9nc2aÞ with D ¼ n1=2ð3C þ jnjÞ given
C00pk¼ C and fYðyÞ expressed as Eq. (A4). We get the same equation if we
substitute n by n0and (n0) into Eq. (A5) given the same values of C, n, and
a. Therefore, Eq. (A5) is an even function of n for case T¼ m. Hence, given the same values of C, n, and a the critical values cafor n¼ n0and n¼ n0are
