Select better suppliers based on manufacturing precision for processes with multivariate data

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Select better suppliers based on

manufacturing precision for processes

with multivariate data

C.H. Yen a & W.L. Pearn b

a

Department of Industrial Engineering and Management , Diwan College of Management , Taiwan

b

Department of Industrial Engineering and Management , National Chiao Tung University , Taiwan

Published online: 09 Apr 2009.

To cite this article: C.H. Yen & W.L. Pearn (2009) Select better suppliers based on manufacturing

precision for processes with multivariate data, International Journal of Production Research, 47:11, 2961-2974, DOI: 10.1080/00207540701796985

To link to this article: http://dx.doi.org/10.1080/00207540701796985

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International Journal of Production Research Vol. 47, No. 11, 1 June 2009, 2961–2974

Select better suppliers based on manufacturing precision

for processes with multivariate data

C.H. Yena* and W.L. Pearnb

a_{Department of Industrial Engineering and Management, Diwan College of Management, Taiwan;} b

Department of Industrial Engineering and Management, National Chiao Tung University, Taiwan

(Received 27 April 2007; final version received 1 October 2007)

Process capability indices have been widely used in the manufacturing industry for measuring process reproduction capability according to manufacturing specifications. Processes with univariate data have been investigated extensively, but are comparatively neglected for processes with multivariate data. Chou (Chou, Y.M., 1994. Selecting a better supplier by testing process capability indices. Quality Engineering, 6, 427–438) developed a procedure using univariate Cpto determine whether or not two processes are equally capable, which allows

one to select the supplier with better quality. However, for processes with multiple characteristics, no methods are available for comparing two processes with multivariate data. In this paper, we consider the supplier selection problem based on manufacturing precision in which the processes involve multiple quality characteristics. We derive the distribution of the corresponding test statistic, and calculate critical values required for the comparison purpose. A real-world application is presented for justification.

Keywords: process capability; multiple quality characteristics; critical value

1. Introduction

Process capability indices have been widely used in the manufacturing industry for measuring process reproduction capability conforming to the manufacturing specifica-tions. In current industry practice customers often require their suppliers to provide process capability for certain product characteristics in the supply chain partnership. Process capability indices also can be used as a benchmark for quality improvement activities. Cp, Cpkand Cpm, are some well-known indices used in the industry for evaluating

process performance, but limited to cases with single engineering specification. Most research work has dealt with statistical properties estimating/testing the univariate indices. Kotz and Lovelace (1998), Kotz and Johnson (2002) and Spiring et al. (2003) provided a compact survey and comprehensive discussions on process capability indices over recent years.

Process capability indices, which establish the relationships between the actual process performance and the manufacturing specifications, have been the focus of recent research

*Corresponding author. Email: [email protected]

ISSN 0020–7543 print/ISSN 1366–588X online

ß 2009 Taylor & Francis

DOI: 10.1080/00207540701796985 http://www.informaworld.com

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in quality assurance and process capability analysis. The precision index Cp is the first

process capability index which appeared in the literature, which is defined in Kane (1986) as:

Cp¼

USL LSL

6 ,

where USL and LSL are the upper and lower specification limits, and is the process standard deviation. The index Cpwas designed to measure the magnitude of the overall

process variation relative to the manufacturing tolerance, which is used for controlled normal processes. Clearly, the index measures manufacturing precision, which reflects product quality consistency (uniformity), an important criterion for judging product quality. A small value of Cp implies that the product quality is not consistent causing

complaints from the customers not only damaging marketing potentials but also incurring more repair cost.

The use of capability indices was first explored within the automotive industry. Ford Motor Company (1984) has used Cp to keep track of the process performance and to

reduce process variation. Recently, the manufacturing industry has been making extensive efforts to implement statistical process control in their plants and supply bases. Capability indices have received increasing usage not only in capability assessments, but also in the evaluation of purchasing decisions, which are becoming the standard tools for quality reporting. Proper understanding and use are essential for the company to maintain capable product supplies. Process precision measures using Cp for normal, truncated

normal, contaminated normal processes based on one single, multiple, (X, R), or (X, S) control chart samples, have been investigated extensively. Examples include Kane (1986), Cheng and Spiring (1989), Chou and Owen (1989), Chan et al. (1991), Kirmani et al. (1991), Kocherlakota (1992), Pearn et al. (1992, 1998, 2004, 2006), Pearn and Wu (2004), Pearn and Chang (2005). Clearly, in using the Cpindex for measuring process precision,

product uniformity would be the primary concern rather than the process yield.

Although some multivariate capability indices have been proposed, no statistical properties of those indices are discussed. In this paper, we develop a statistical test procedure using the estimator of MCp(multivariate extension of Cp) to judge whether the

capability of one process is superior to another process. A real-world application is presented to justify the proposed methodology. Practitioners can use the proposed procedure in making reliable decisions for their in-plant applications.

2. Process with multiple characteristics

Most research has been devoted to capability measures with a single process/quality characteristic. However, it is quite common that the manufactured product involves more than one quality characteristic. That is, it requires several different characteristics for adequate product description. Each of those characteristics must satisfy certain specifications. The assessed quality of a product depends on the combined effects of those characteristics rather than on their individual values. For example, automobile paint usually has a range of light reflective abilities and a range of adhesion abilities (see Taam et al. 1993). A paint that satisfies one criterion but not the other is considered undesirable. Those characteristics are related to each other through the composition of the paint. It is therefore natural to consider a bivariate characterisation of this paint.

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For tolerance region of processes with multiple characteristics, most researchers take an ellipsoidal or a rectangular region. For more complex engineering specifications, the tolerance region would be rather complicated. For instance, a drawing of a connecting rod in a combustion engine consists of crank-bore inner diameter, pin-bore inner diameter, rod length, bore true-location and so on (see Taam et al. 1993). In multivariate processes, we usually assume that the process characteristics X follows the multivariate normal distribution Nv(,

P

), where v is the characteristic dimensions, is the mean vector, and P

is the variance-covariance matrix of X. Also T is the target vector, X is the sample mean vector and S is the sample covariance matrix. For processes with multivariate data, Taam et al. (1993) defined the multivariate capability index MCp

MCp¼

vol:ðmodified tolerance regionÞ vol:½ðX Þ01_{ðX Þ k}

vðqÞ

¼ vol:ðmodified tolerance regionÞ ð2

v,0:9973Þ v=2

j j1=2½ðv=2 þ 1Þ1, ð1Þ where kv(q) is the 99.73th percentile of the 2distribution with v degrees of freedom, j

P jis the determinant ofP, and () is the gamma function. Note that, if MCpis less than 1,

then the process variation is greater than the specified range of variation. It indicates that the process precision is not adequate with respect to the specifications (product quality is not uniform/consistent).

3. Comparing two multivariate processes using MCp

An estimator of MCpcan be expressed as

MC

^ p¼

vol:ðmodified tolerance regionÞ vol:ðestimated 99:73% process regionÞ¼

vol:ðmodified tolerance regionÞ ð2

v,0:9973Þ v=2

S

j j1=2½ðv=2 þ 1Þ1 ð2Þ where S is the sample variance-covariance matrix, and jSj is the determinant of S. From Equation (1), MC^ p can be rewritten as MCpðj j=S j jÞ1=2. Let ~X ¼ ðX1, X2, . . . ,XnÞ0be an

n-dimensional vector of measurements taken from a multivariate normal distribution with mean vector ¼ (1, 2, . . . , v)0, target vector T, process variance-covariance matrixP.

Using the following theorem (Theorem 1), we may obtain the distribution of MC^ p.

Chou (1994) developed a procedure using univariate Cpto determine whether or not two

processes are equally capable, which allows one to select the supplier with better quality. However, for processes with multiple characteristics (multivariate data), no methods are available for comparing two processes with multivariate data. For this purpose, we consider the problem of comparing two multivariate processes using MCp. The hypothesis

testing would be as follows: H0: MCp1MCp2(process I is not better than process II)

versus H1: MCp14MCp2(process I is better than process II). The critical value c can be

determined as: P MC ^ p1 MC ^ p2 > cjMCp1¼MCp2 8 < : 9 = ;¼ ) P MCp1ðjS1j=j1jÞ1=2 MCp2ðjS2j=j2jÞ1=2 > cjMCp1¼MCp2 ( ) ¼ )P ðjS1j=j1jÞ 1=2 S2 j j=j2j ð Þ1=2> c ¼ ) P ðjS2j=j2jÞ S1 j j=j1j ð Þ> c 2 ¼ ð3Þ

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Theorem 1: The distribution of the generalised variance jSj of a sample X1, X2, . . . , Xnfrom

Nv(,

P

) is the same as the distribution of jPj/(n 1)vtimes the product of v independent factors, the distribution of the ith factor being the 2 distribution with n i degrees of freedom.

Proof: See Anderson (2003), p. 268.

Lemma 1: Let x 2_n and y 2_n1 be independently distributed. Let z2¼4xy. Then z 2_2n2.

Proof: See Srivastava and Khatri (1979), p. 82.

From the above theorem, |S|/|P| is distributed as 2

n12n2 2nv=ðn 1Þv.

For v ¼ 2, Equation (3) becomes

P 2 n21 2 n22 ðn21Þ2 ðn11Þ2 2 n11 2 n12 > c2 ( ) ¼ ) P ð 2 2n24Þ 2 =4 ðn21Þ2 ðn11Þ2 ð2 2n14Þ 2₌₄> c 2 ( ) ¼ (Referring to Lemma 1, we can derive 2_n12_n2 ð2_2n4Þ2=4, see Corollary 1 in the Appendix.) )P ð 2 2n24Þ 2 ðn21Þ2 ðn11Þ2 ð2 2n14Þ 2 > c 2 ( ) ¼ ) P 2_2n₂₄=2n24 2 2_2n₁₄=2n14 2 ðn11Þ2 ðn21Þ2 ð2n24Þ2 ð2n14Þ2 > c2 8 > < > : 9 > = > ;¼ )P F2n24,2n14 2 > c2ðn21Þ 2 ðn11Þ2 ð2n14Þ2 ð2n24Þ2 ¼ ) F2n24,2n14,1¼c ðn21Þ ðn11Þ ð2n14Þ ð2n24Þ : Thus, the critical value can be expressed as

c ¼ F2n24,2n14,1 ðn11Þ ðn21Þ ð2n24Þ ð2n14Þ : ð4Þ

For v ¼ 3, Equation (3) can be expressed as

P 2 n21 2 n22 2 n23 ðn21Þ3 ðn11Þ3 2 n11 2 n12 2 n13 > c2 ( ) ¼ ) P 2 2n24 2 2n14 !2 2 n23 2 n13 ðn11Þ3 ðn21Þ3 > c2 8 < : 9 = ;¼ )P 2 2n24 2n24 2 2n14 2n14 0 @ 1 A 22 n23 n23 2 n13 n13 ð2n24Þ2 ð2n14Þ2 ðn23Þ ðn13Þ ðn11Þ3 ðn21Þ3 > c2 8 > < > : 9 > = > ;¼ )P F2n24,2n14 2 Fn23,n13> c 2ð2n14Þ2 ð2n24Þ2 ðn13Þ ðn23Þ ðn21Þ3 ðn11Þ3 ¼: ð5Þ

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Let z ¼ xy, where x ðF2n24, 2n14Þ

2_{, y F}

n23, n13, then Equation (5) can be expres-sed asRc2ð2n14Þ2ðn13Þðn21Þ3=ð2n24Þ2ðn23Þðn11Þ3

0 fzðzÞdz ¼1 . Thus, the critical value can be

expressed as c ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F1 z ð1 Þ ð2n24Þ2 ð2n14Þ2 ðn23Þ ðn13Þ ðn11Þ3 ðn21Þ3 s , ð6Þ where fZðzÞ ¼ Z1 0 kðn1, n2Þ xðð2n24Þ=42Þ z x ðn23Þ=21 1 þð2n24Þ ð2n14Þ ffiffiffi x p ð2n2þ2n18Þ=2 1 þ ðn23Þ ðn13Þðz=xÞ ðn2þn16Þ=2dx, for x, z 0 kðn1, n2Þ ¼ ½ð2n2þ2n18Þ=2ððn2þn16Þ=2Þ 2n24 2n14 ð2n24Þ=2 _n 23 n13 ðn23Þ=2 ½ð2n24Þ=2½ð2n14Þ=2½ðn23Þ=2½ð2n13Þ=2 , FzðtÞ ¼ Rt

0fzðzÞdz ¼1 , and F1z ðÞis the inverse function of FzðÞ; see Corollary 2 in

the Appendix.

Tables 1 to 3 display the critical values for various sample sizes in the case with v ¼ 2 under test levels ¼ 0.05, ¼ 0.025 and ¼ 0.01. Tables 4 to 6 display the critical values for various sample sizes in the case of v ¼ 3 under ¼ 0.05, ¼ 0.025 and ¼ 0.01. For practical and convenient purpose, a step-by-step procedure is provided below: Step 1: Determine the sample size ni for each supplier and the -risk (the chance of

incorrectly rejecting a better supplier).

Step 2: Take a random sample from each process and calculate the sample covariance matrix.

Step 3: Calculate the test statistic MC^ p1=MC ^

p2and the critical value c.

Table 1. Critical value for testing H0: MCp1MCp2under ¼ 0.05 (v ¼ 2).

n2 nl 10 20 30 40 50 60 70 80 90 100 10 2.33 2.31 2.29 2.29 2.28 2.28 2.28 2.27 2.27 2.27 20 1.81 1.74 1.71 1.69 1.68 1.68 1.67 1.67 1.65 1.66 30 1.68 1.60 1.56 1.54 1.52 1.51 1.51 1.50 1.50 1.49 40 1.62 1.53 1.49 1.46 1.45 1.44 1.43 1.42 1.42 1.41 50 1.59 1.49 1.44 1.42 1.40 1.39 1.38 1.37 1.37 1.36 60 1.57 1.46 1.42 1.39 1.37 1.36 1.35 1.34 1.34 1.33 70 1.55 1.45 1.40 1.37 1.35 1.34 1.33 1.32 1.31 1.31 80 1.54 1.43 1.38 1.35 1.33 1.32 1.31 1.30 1.30 1.29 90 1.53 1.42 1.37 1.34 1.32 1.31 1.30 1.29 1.28 1.28 100 1.52 1.41 1.36 1.33 1.31 1.30 1.29 1.28 1.27 1.27

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Table 2. Critical value for testing H0: MCp1MCp2under ¼ 0.025 (v ¼ 2). n2 nl 10 20 30 40 50 60 70 80 90 100 10 2.76 2.69 2.67 2.65 2.64 2.64 2.63 2.63 2.63 2.63 20 2.06 1.94 1.89 1.87 1.85 1.84 1.83 1.83 1.82 1.82 30 1.89 1.75 1.70 1.67 1.65 1.63 1.62 1.62 1.61 1.61 40 1.81 1.67 1.61 1.57 1.55 1.54 1.53 1.52 1.51 1.50 50 1.76 1.62 1.55 1.52 1.50 1.48 1.47 1.46 1.45 1.44 60 1.74 1.58 1.52 1.48 1.46 1.44 1.43 1.42 1.41 1.40 70 1.72 1.56 1.50 1.46 1.43 1.41 1.40 1.39 1.38 1.38 80 1.70 1.55 1.48 1.44 1.41 1.39 1.38 1.37 1.36 1.35 90 1.69 1.55 1.46 1.42 1.40 1.38 1.36 1.35 1.35 1.34 100 1.68 1.52 1.45 1.41 1.39 1.37 1.35 1.34 1.33 1.32

n2 nl 10 20 30 40 50 60 70 80 90 100 10 3.37 3.25 3.20 3.17 3.16 3.15 3.14 3.14 3.13 3.13 20 2.39 2.21 2.14 2.10 2.08 2.06 2.05 2.04 2.03 2.03 30 2.15 1.96 1.88 1.83 1.81 1.79 1.78 1.77 1.76 1.75 40 2.05 1.84 1.76 1.71 1.68 1.66 1.65 1.64 1.63 1.62 50 1.99 1.78 1.69 1.64 1.61 1.59 1.58 1.56 1.55 1.55 60 1.95 1.74 1.65 1.60 1.57 1.54 1.53 1.51 1.50 1.50 70 1.93 1.71 1.62 1.57 1.53 1.51 1.49 1.48 1.47 1.46 80 1.91 1.69 1.59 1.54 1.51 1.49 1.47 1.45 1.44 1.43 90 1.89 1.67 1.58 1.52 1.49 1.47 1.45 1.43 1.42 1.41 100 1.88 1.66 1.56 1.51 1.47 1.45 1.43 1.42 1.41 1.40

n2 n1 10 20 30 40 50 60 70 80 90 100 10 2.94 3.14 3.19 3.21 3.22 3.23 3.24 3.24 3.24 3.25 20 1.95 2.00 1.99 1.99 1.99 1.98 1.98 1.98 1.98 1.97 30 1.73 1.75 1.73 1.72 1.71 1.70 1.70 1.69 1.69 1.69 40 1.64 1.64 1.61 1.60 1.59 1.58 1.57 1.56 1.56 1.56 50 1.59 1.58 1.55 1.53 1.52 1.50 1.50 1.49 1.49 1.48 60 1.56 1.54 1.51 1.48 1.47 1.46 1.45 1.44 1.44 1.43 70 1.53 1.51 1.48 1.45 1.44 1.43 1.42 1.41 1.40 1.40 80 1.51 1.49 1.45 1.43 1.41 1.40 1.39 1.38 1.38 1.37 90 1.50 1.47 1.44 1.41 1.40 1.38 1.37 1.36 1.36 1.35 100 1.49 1.46 1.42 1.40 1.38 1.37 1.36 1.35 1.34 1.34

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Step 4: If MC^ p1=MC ^

p24c, then we reject H0and conclude that MCp1> MCp2. From the

definition of c, it is clear that the value of MC

^

p1=MC ^

p2 must be higher than the original

target value for the true MCp1=MCp2. The power of the test can be also computed below:

The power of the test, , is given by MCp1 MCp2 ¼P MC ^ p1 MC ^ p2 > cjMCp1 MCp2 ¼k 8 < : 9 = ;: ð7Þ

Take v ¼ 2, and 3, Equation (7) can be expressed as P F2n24,2n14 > c ðn21Þ ðn11Þ ð2n14Þ ð2n24Þ ðMCp2Þ ðMCp1Þ , P F2n24,2n14 2 Fn23,n13> c 2ð2n14Þ2 ð2n24Þ2 ðn13Þ ðn23Þ ðn21Þ3 ðn11Þ3 ðMCp2Þ2 ðMCp1Þ2 ( ) :

n2 nl 10 20 30 40 50 60 70 80 90 100 10 3.64 3.81 3.84 3.86 3.87 3.87 3.88 3.88 3.88 3.88 20 2.29 2.28 2.26 2.24 2.23 2.22 2.22 2.21 2.21 2.21 30 2.01 1.96 1.92 1.90 1.88 1.87 1.86 1.85 1.85 1.84 40 1.89 1.82 1.78 1.75 1.73 1.71 1.70 1.70 1.69 1.68 50 1.82 1.75 1.70 1.66 1.64 1.63 1.61 1.60 1.60 1.59 60 1.78 1.70 1.64 1.61 1.59 1.57 1.56 1.55 1.54 1.53 70 1.75 1.66 1.61 1.57 1.55 1.53 1.51 1.50 1.49 1.49 80 1.73 1.64 1.58 1.54 1.52 1.50 1.48 1.47 1.46 1.46 90 1.71 1.62 1.56 1.52 1.49 1.48 1.46 1.45 1.44 1.43 100 1.70 1.60 1.54 1.50 1.48 1.46 1.44 1.43 1.42 1.41

n2 nl 10 20 30 40 50 60 70 80 90 100 10 4.68 4.79 4.81 4.81 4.82 4.82 4.82 4.82 4.82 4.82 20 2.76 2.67 2.62 2.58 2.56 2.55 2.54 2.53 2.52 2.52 30 2.38 2.25 2.18 2.14 2.11 2.09 2.08 2.06 2.06 2.05 40 2.22 2.07 1.99 1.94 1.91 1.89 1.87 1.86 1.85 1.84 50 2.13 1.97 1.88 1.83 1.80 1.78 1.76 1.75 1.74 1.73 60 2.07 1.90 1.82 1.77 1.73 1.71 1.69 1.67 1.66 1.65 70 2.03 1.86 1.77 1.72 1.68 1.66 1.64 1.62 1.61 1.60 80 2.00 1.83 1.74 1.68 1.65 1.62 1.60 1.58 1.57 1.56 90 1.98 1.80 1.71 1.65 1.62 1.59 1.57 1.55 1.54 1.53 100 1.96 1.78 1.69 1.63 1.60 1.57 1.55 1.53 1.52 1.51

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4. A real-world application

To illustrate how the proposed method can be applied to the actual data collected from the factories, we present a real-world example of an electronic component manufacturer making ceramic multilayer capacitors applicable to consumer electronics, telecommunica-tions, automotive parts, and data processing devices. The capacitor consists of a rectangular ceramic block in which a number of interleaved electrodes are contained. A cross section of the ceramic multi-layer capacitor structure is depicted in Figure 1. For a detailed model of the ceramic multilayer capacitor investigated, all the electrical characteristics are displayed in Table 7.

Consider a production process with multiple characteristics following a multivariate normal distribution, which is taken from a multilayer capacitor factory in Taiwan adapting the six-sigma quality improvement program. To compare product quality between a supplier versus another, 50 random samples are taken from the two processes. The quality control of the process involves the simultaneous control of the layer thickness, the layer length, and the layer width. The lower and upper specification limits for layer thickness, layer length, and layer width have been set to [1.45, 1.75], [3.0, 3.4] and [1.45, 1.75], respectively. The sample covariance matrices are summarised below, where S1represents the data from supplier I, and S2represents the data from supplier II.

S1¼ 0:00193 0:00046 0:00086 0:00046 0:00097 0:00075 0:00086 0:00075 0:00167 2 4 3 5, S2 ¼ 0:00236 0:00029 0:00003 0:00029 0:00176 0:00097 0:00003 0:00097 0:00161 2 4 3 5:

Figure 1. Structure of a ceramic multi-layer capacitor.

Table 7. Specification of Y5V/BME/1206/22uF/6.3V.

Capacitance range 22uF, Size 1206 Tolerance on capacitance after 1000 hours 20% to þ 80% Rated voltage UR(DC) 6.3 V

Test voltage (DC) for 1 minute 2.5 UR

Tan D (Note 1) 12.50%

Insulation resistance after 1 minute at UR(DC) Rins. C 500 s

Maximum capacitance change as a function of temperature þ30% to 80%

Ageing Typically, 12.5% per time decade

Terminations NiSn plated

Resistance to soldering heat 260 _{C, 10 sec}

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To compare the two processes, we consider a test with null hypothesis H0: MCp1MCp2against the alternative hypothesis H1: MCp14MCp2, where MCp1and

MCp2represent the process capability indices of the two suppliers respectively. The test

procedure can be described as:

Step 1: For the two suppliers with sample sizes n1¼n2¼50, set the confidence level as

0.05.

Step 2: Calculate the sample covariance. From the above result, we obtain S1¼ 0:00193 0:00046 0:00086 0:00046 0:00097 0:00075 0:00086 0:00075 0:00167 2 4 3 5, S2 ¼ 0:00236 0:00029 0:00003 0:00029 0:00176 0:00097 0:00003 0:00097 0:00161 2 4 3 5: Step 3: Calculate the test statistic MC

^

p1=MC ^

p2and critical value c:

MC^ p1 ¼ 4 3 ð0:3=2Þ ð0:4=2Þ ð0:3=2Þ S1 j j1=2ð 2 3,0:9973Þ 3=2_½ð2:5Þ1 ¼2:13239, MC^ p2 ¼ 4 3 ð0:3=2Þ ð0:4=2Þ ð0:3=2Þ S2 j j1=2ð 2 3,0:9973Þ 3=2_½ð2:5Þ1 ¼1:28415: Therefore, MC ^ p1=MC ^

p2¼2.13239/1.28415 ¼ 1.6605, and c ¼ 1.52 (refer to Table 4).

Step 4: As 1.660541.52, we conclude that with 95% confidence process (supplier) I is superior to process (supplier) II. In order to show the sensitivity of the test procedure, the power curve of the test is depicted in Figure 2 under the ratio value of MCp1/MCp2¼0.8 to 2.3.

5. Conclusions

Processes with multiple quality characteristics are quite common in the manufacturing industry. Processes with univariate data have been investigated extensively, but are

Figure 2. Power curve of testing.

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comparatively neglected for processes with multivariate data. Chou (1994) developed a procedure using univariate Cpto determine whether or not two processes are equally

capable, which allows one to select the supplier with better quality. However, for processes with multiple characteristics, no methods are available for comparing two processes with multivariate data. In this paper, we considered the supplier selection problem based on manufacturing precision in which the processes involve multiple quality characteristics. We developed an effective test procedure for practitioners to make reliable decisions in their in-plant applications involving supplier selections.

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Appendix

Corollary 1: If 2

n1and 2n2are mutually independently distributed, then 2n12n2is distributed

as ð2 2n4Þ2=4.

Proof: Let x12n1and x22n2. The joint pdf of x1and x2is given by

fx1,x2ðx1, x2Þ ¼ ð1=2Þðn1Þ=2xn=23=2₁ ex1=2 ½ðn 1Þ=2 ð1=2Þðn2Þ=2xn=22₂ ex2=2 ½ðn 2Þ=2 : Let z1¼x1 and z2¼2 ffiffiffiffiffiffiffiffiffiffix1x2 p

. Using the transformation method, the solution is x1¼z1 and

x2¼z22=4z1, and the Jacobian of the transformation is

J ¼ 1z2 0 2 4 ffiffiffiz1 p z2 2z1 ¼ z2 2z1 :

So, we find that the joint p.d.f. of z1z2is

fz1,z2ðz1, z2Þ ¼ ð1=2Þðn1Þ=2zn=23=2₁ ez1=2 ½ðn 1Þ=2 ð1=2Þðn2Þ=2ðz2₂=4z1Þn=22eðz 2 2=4z1Þ=2 ½ðn 2Þ=2 z2 2z1 ,0 z1,z2 1:

Then, the marginal density function of z2is obtained as follows:

fz2ðz2Þ ¼ Z1 0 ð1=2Þðn1Þ=2zn=23=2₁ ez1=2 ½ðn 1Þ=2 ð1=2Þðn2Þ=2ðz2 2=4z1Þn=22eðz 2 2=4z1Þ=2 ½ðn 2Þ=2 z2 2z1 dz1 ¼C1zn32 Z1 0 z1=2₁ ez1=2ðz22=4z1Þ=2_dz 1,0 z2 1 where C1¼ ð1=2Þ2n9=2 ½ðn 1Þ=2 ½ðn 2Þ=2: Let hðz2Þ ¼ Z1 0 z1=2₁ ez1=2ðz22=4z1Þ=2_dz 1: Hence h0_ðz 2Þ ¼ z2 4z1 Z1 0 z1=2₁ ez1=2ðz22=4z1Þ=2_dz 1: Now, let z2 2 4z1 ¼w:

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Using variable transformation technique, we obtain h0ðz2Þ ¼ 1 2 Z1 0 w1=2_ew=2ðz2 2=4wÞ=2dw ¼ 1 2 hðz2Þ:

The above equation gives hðz2Þ ¼eðz2=2þC2Þ, where C2 is a constant. Thus, the p.d.f. of z2can be

obtained as below, where C3¼C1eC2. Therefore, we have z222n4,

fz2ðz2Þ ¼C1e

C2_zn3

2 ez2=2¼C3zð2n4Þ=212 ez2=2, 0 z2 1

Corollary 2: x ðF2n24, 2n14Þ 2_{, y F}

n23, n13, if z ¼ xy, then the p.d.f. of z is

fZðzÞ ¼ Z1 0 kðn1, n2Þ 1 2 xð2n24Þ=42 z x ðn23Þ=21 1 þð2n24Þ ð2n14Þ ffiffiffi x p ð2n2þ2n18Þ=2 1 þðn23Þ ðn13Þ z x ðn2þn16Þ=2dx, for x, z 0 where kðn1, n2Þ ¼ ½ð2n2þ2n18Þ=2½ðn2þn16Þ=2 2n24 2n24 ð2n24Þ=2 n23 n13 ðn23Þ=2 ½ð2n24Þ=2½ð2n14Þ=2½ðn23Þ=2½ð2n13Þ=2 : Proof: Let t ðF2n24,2n14Þ, x ¼ t 2 , where ftðtÞ ¼ ½ð2n2þ2n18Þ=2 ½ð2n24Þ=2½ð2n14Þ=2 t ð2n24Þ=21 1 þ 2n24 2n14 t ð2n2þ2n18Þ=2, for t 0:

Using the transformation method, t ¼pffiffiffix, and J ¼ 1=2pffiffiffix(Jacobian), so the p.d.f. of x is

fxðxÞ ¼ ½ð2n2þ2n18Þ=2 ½ð2n24Þ=2½ð2n14Þ=2 1 2x ð2n24Þ=41 1 þ 2n24 2n14 ffiffiffi x p ð2n2þ2n18Þ=2, for x 0.

Now, let z ¼ xy given x ðF2n24,2n14Þ 2_{, y F}

n23,n13. Using the transformation method, x ¼ x,

y ¼ z/x, and J ¼ 1/x (Jacobian), so the joint pdf of x and z is

fx,zðx, zÞ ¼ kðn1, n2Þ 1 2x ð2n24Þ=42 z x ðn23Þ=21 1 þð2n24Þ ð2n14Þ ffiffiffi x p ð2n2þ2n18Þ=2 1 þðn23Þ ðn13Þ z x ðn2þn16Þ=2, for x, z 0, kðn1, n2Þ ¼ ½ð2n2þ2n18Þ=2½ðn2þn16Þ=2 2n24 2n24 ð2n24Þ=2 _n 23 n13 ðn23Þ=2 ½ð2n24Þ=2½ð2n14Þ=2½ðn23Þ=2½ð2n13Þ=2 :

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Then the marginal density function of z is obtained as follows: fZðZÞ ¼ Z1 0 kðn1, n2Þ 1 2x ð2n24Þ=42 z x ðn23Þ=21 1 þð2n24Þ ð2n14Þ ffiffiffi x p ð2n2þ2n18Þ=2 1 þðn23Þ ðn13Þ z x ðn2þn16Þ=2dx, for x, z 0, where kðn1, n2Þ ¼ ½ð2n2þ2n18Þ=2½ðn2þn16Þ=2 2n24 2n24 ð2n24Þ=2 _n 23 n13 ðn23Þ=2 ½ð2n24Þ=2½ð2n14Þ=2½ðn23Þ=2½ð2n13Þ=2 :

The programs for Table 4, 5 and 6. Statistics

Graphics ‘Graphics’

Statistics ‘Continuous Distributions’

n1¼Table [k, {k, 10, 100, 10}]

n2¼Table [p, {p, 10, 100, 10}]

K1¼Table ½k1½i,j,fi,10g,fj,10g;

For½i ¼ 1,i 10,i ¼ i þ 1, For½ j ¼ 1,j 10,j ¼ j þ 1,

A½i,j ¼Gamma ðð2 n½ 2½½j 2 n1½½i 8Þ=2Þ Gamma ððn½ 2½½j n1½½i 6Þ=2Þ ðð2 nð 2½½j 4Þ=2 n1½½i 4ÞÞ

ð2n2½½j4Þ=2

ððnð 2½½j 3Þ=n1½½i 3ÞÞðn2½½j3Þ=2

Gamma ðð2 n½ 2½½j 4Þ=2Þ Gamma ðð2 n½ 1½½i 4Þ=2Þ Gamma ððn½ 2½½j 3ÞÞ=2 Gamma ððn½ 1½½i 3Þ=2Þ

;

P½i,j ¼ Find Root½N Integrate ½A½i,j 0:5 x

ð2n2½½j4Þ=4Þ2_ðz=xÞðn2½½j3Þ=21 1 þ ðð2 n2½½j 4Þ=2 n1½½i 4Þ ffiffiffix p ð2n2½½j2n1½½i8Þ=2 1 þ ððn2½½j 3Þ=n1½½i 3ÞÞ ðz=xÞ ð Þððn2 ½½jn1 ½½i6Þ=2Þ2 0 @ 1 A, fz,0,tg,fx,0,1g ¼¼ 0:95,ft,0,100g; PP1½i,j ¼ t=:P½i,j; k1½i,j ¼ Sqrt PP1½i,j 2 n2½½j 4 2 n1½½i 4 2 n2½½j 3 n1½½i 3 n1½½i 1 n2½½j 1 3 $ % ; K1 (Table 4 result) K2¼Table½k2½i,j,fi,10g,fj,10g; For ½i ¼ 1,i 10,i ¼ i þ 1, For ½j ¼ 1,j 10,j ¼ j þ 1, A½i,j ¼

Gamma ðð2 n½ 2½½j 2 n1½½i 8Þ=2Þ Gamma ððn½ 2½½j n1½½i 6Þ=2Þ ðð2 nð 2½½j 4Þ=2 n1½½i 4ÞÞð2n2½½j4Þ=2nn21½½j3½½i3

ðn2½½j3Þ=2

Gamma ðð2 n½ 2½½j 4Þ=2Þ Gamma ðð2 n½ 1½½i 4Þ=2Þ Gamma ððn½ 2½½j 3Þ=2Þ Gamma ððn½ 1½½i 3Þ=2Þ

; P2½i,j ¼ FindRoot½NIntegrate½A½i,j 0:5 xðð2n2½½j4Þ=4Þ2_ðz=xÞððn2½½j3Þ=2Þ1 1 þ ðð2 n2½½j 4Þ=2 n1½½i 4Þ ffiffiffix p ð2n2½½j2n1½½i8Þ=2 1 þ ððn2½½j 3Þ=n1½½i 3Þ ðz=xÞ ð Þn2 ½½jn1 ½½i62 0 @ 1 A, fz,0,tg,fx,0,1g ¼¼ 0:975,ft,0,100g; PP2½i,j ¼ t=:P2½i,j;

k2½i,j ¼ Sqrt PP 2½i,j ðð2 nð 2½½j 4Þ=2 n1½½i 4ÞÞ2 ððnð 2½½j 3Þ=n1½½i 3ÞÞ ððnð 1½½i 1Þ=n2½½j 1ÞÞ3;

K2

(Table 5 result)

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K3¼Table½k3½i,j,fi,10g,fj,10g;

For ½i ¼ 1,i 10,i ¼ i þ 1, For ½j ¼ 1,j 10,j ¼ j þ 1,

A½i,j ¼Gamma ðð2 n½ 2½½j 2 n1½½i 8Þ=2Þ Gamma ððn½ 2½½j n1½½i 6Þ=2Þ ðð2 nð 2½½j 4Þ=2 n1½½i 4ÞÞ

ð2n2½½j4Þ=2

ððnð 2½½j 3Þ=n1½½i 3ÞÞðn2½½j3Þ=2

Gamma ðð2 n½ 2½½j 4Þ=2Þ Gamma ðð2 n½ 1½½i 4Þ=2Þ Gamma ððn½ 2½½j 3Þ=2Þ Gamma ððn½ 1½½i 3Þ=2Þ ;

P3½i,j ¼ FindRoot½NIntegrate½A½i,j 0:5 xðð2n2½½j4Þ=4Þ2_ðz=xÞððn2½½j3Þ=2Þ1 1 þ ðð2 n2½½j 4Þ=2 n1½½i 4Þ ffiffiffix p ð2n2½½j2n1½½i8Þ=2 1 þ ððn2½½j 3Þ=n1½½i 3Þ ðz=xÞ ð Þn2½½jn1½½i6=2 ! , fz,0,tg,fx,0,1g ¼¼ 0:99,ft,0,100g; PP3½i,j ¼ t=:P3½i,j;

k3½i,j ¼ Sqrt PP3½i,j ðð2 nð 2½½j 4Þ=2 n1½½i 4ÞÞ2 ððnð 2½½j 3Þ=n1½½i 3ÞÞ ððnð 1½½i 1Þ=n2½½j 1ÞÞ3

;

K3

(Table 6 result)