Designing of an attribute control chart for two-stage process

Abstract

In practice, the products can be manufactured through several stages. In this manuscript, we will propose an attribute control chart plotting the number of defectives for a two-stage process. The in-control average run length is derived and the out-of-control average run lengths are also analyzed according to the process shifts in the first and/or the second stage process. The tables of the average run lengths are given for various specified parameters. An example is given with synthetic data for the illustration of the proposed control chart.

Introduction

Control charts have been considered as a powerful tool in any industry for the improvement of product quality. A control chart constructs lower and upper control limits and plots a suitable statistic at each subgroup. The process is declared as out of control if the plotting statistic is beyond the control limits. The early detection of an out-of-control, if any, is very important to minimize the nonconforming products. It will be also important for a control chart to minimize the type-I error (the probability of declaring the process as out of control when it is actually in control) and type-II error (the chance of declaring the process as in control when it is actually out of control). Basically, control charts are classified as (1) attribute control charts and (2) variable control charts. The former is used when the quality characteristic is obtained from a process of categorizing a product as conforming or nonconforming. The latter is used when the quality of interest is obtained from the measurement process, such as the measurement of diameter of a ball bearing. Both control charts have been widely used in the industry for the improvement of quality. Among variable control charts, the X bar chart is the most popular in the industry, ¹ while among attribute control charts, the $np$ chart is widely used in the industry. More details about the designing and the use of attribute control charts can be read in Epprecht and Costa, ² Laney, ³ Epprecht et al., ⁴ Wu et al.,^5,6 Luo and Wu, ⁷ Wu et al., ⁸ Lucas et al., ⁹ Montgomery, ¹⁰ Wu and Wang, ¹¹ Sim and Lim, ¹² Chen et al., ¹³ Aebtarm and Bouguila, ¹⁵ Aslam et al.^16,17

Usually, attribute control charts are designed to monitor a single process. But, in the industry, there are many situations where manufacturing of the product is based on several production stages or processes. For example, in fire extinguisher manufacturing, cylinders pass through various processes before they are finally produced, as described in Duffuaa et al. ¹⁸ According to Duffuaa et al., ¹⁸ products are processed through several stages and more than one process impacts on product quality characteristics.

According to our best knowledge, the designing of a control chart for a two-stage process has not been studied before. Therefore, there have been not much works for a two-stage or a multi-stage process. In this paper, we will develop an attribute control chart plotting the numbers of defectives for a two-stage process.

Designing of proposed control chart

It is assumed that the two processes considered here are independent of each other so that the statistics obtained from two processes are statistically independent. We propose the following attribute control chart for a two-stage process:

• Step 1: Select a random sample of size ${\mathrm{n}}_1$ from the first process and count the number of defectives (denoted by ${\mathrm{D}}_1$ ). Declare the process as in control if ${\mathrm{D}}_1\le {\mathrm{UCL}}_1$ and go to step 2. Otherwise, declare the process as out of control.

• Step 2: Select a random sample of size ${\mathrm{n}}_2$ from the second process and count the number of defectives (denoted by ${\mathrm{D}}_2$ ). Declare the process as in control if ${\mathrm{D}}_1+{\mathrm{D}}_2\le {\mathrm{UCL}}_2$ . Otherwise, declare the process as out of control.

Suppose that the fractions nonconforming at the first process and at the second process are ${\mathrm{p}}_1$ and ${\mathrm{p}}_2$ , respectively. Then, ${\mathrm{D}}_{\mathrm{i}}$ (i = 1, 2) follows an independent binomial distribution with ${\mathrm{n}}_{\mathrm{i}}$ and ${\mathrm{p}}_{\mathrm{i}}$ . Because we assume that two processes are independent of each other, the number of defectives from the first process, and that from the second process should be independent. Furthermore, it is assumed that the fractions nonconforming at the first and the second processes are ${\mathrm{p}}_{01}$ and ${\mathrm{p}}_{02}$ , respectively, when the processes are in control. We consider the upper control limit (UCL) for the first process in the following form

$${\mathrm{UCL}}_1={\mathrm{n}}_1{\mathrm{p}}_{01}+{\mathrm{k}}_1\sqrt{{\mathrm{n}}_1{\mathrm{p}}_{01}\left(1-{\mathrm{p}}_{01}\right)}$$

(1)

where ${\mathrm{k}}_1$ is the control constant for the first process to be determined. The mean and the variance of ${\mathrm{D}}_1+{\mathrm{D}}_2$ are given as follows

$$\mathrm{E}[{\mathrm{D}}_1+{\mathrm{D}}_2]={\mathrm{n}}_1{\mathrm{p}}_{01}+{\mathrm{n}}_2{\mathrm{p}}_{02}$$

$$\mathrm{Var}[{\mathrm{D}}_1+{\mathrm{D}}_2]={\mathrm{n}}_1{\mathrm{p}}_{01}\left(1-{\mathrm{p}}_{01}\right)+{\mathrm{n}}_2{\mathrm{p}}_{02}\left(1-{\mathrm{p}}_{02}\right)$$

Therefore, the UCL for the combined process, ${\mathrm{UCL}}_2$ , is given as follows

$${\mathrm{UCL}}_2=({\mathrm{n}}_1{\mathrm{p}}_{01}+{\mathrm{n}}_2{\mathrm{p}}_{02})+{\mathrm{k}}_2\sqrt{{\mathrm{n}}_1{\mathrm{p}}_{01}\left(1-{\mathrm{p}}_{01}\right)+{\mathrm{n}}_2{\mathrm{p}}_{02}\left(1-{\mathrm{p}}_{02}\right)}$$

(2)

where ${\mathrm{k}}_2$ is the control constant for the combined process to be determined. We may consider a lower control limit for each stage, but here we only consider an UCL for the simplicity.

The probability of declaring as in-control for first process when it is actually in control is derived as follows

$$\mathrm{P}\left({\mathrm{D}}_1\le {\mathrm{UCL}}_1\right)={\displaystyle \sum }_{\mathrm{i}=0}^{{\mathrm{UCL}}_1}\left(\begin{array}{c}{\mathrm{n}}_1\\ i\end{array}\right){\mathrm{p}}_{01}^{\mathrm{i}}{\left(1-{\mathrm{p}}_{01}\right)}^{{\mathrm{n}}_1-\mathrm{i}}$$

(3)

The probability of declaring as out-of-control for the two-stage process when the processes are actually in control is given by

$${\mathrm{P}}_{\mathrm{out}}^{0,0}=\mathrm{P}\left({\mathrm{D}}_1>{\mathrm{UCL}}_1\right)+\mathrm{P}\left({\mathrm{D}}_1\le {\mathrm{UCL}}_1,{\mathrm{D}}_1+{\mathrm{D}}_2>{\mathrm{UCL}}_2\right)$$

(4)

$$\begin{array}{l}{\mathrm{P}}_{\mathrm{out}}^{0,0}=\left(1-{\displaystyle \sum _{\mathrm{d}=0}^{{\mathrm{UCL}}_1}\left(\begin{array}{c}{\mathrm{n}}_1\\ \mathrm{d}\end{array}\right){\mathrm{p}}_{01}^{\mathrm{d}}{\left(1-{\mathrm{p}}_{01}\right)}^{{\mathrm{n}}_1-\mathrm{d}}}\right)+{\displaystyle \sum _{d=0}^{{\mathrm{UCL}}_1}\left(\begin{array}{c}{\mathrm{n}}_1\\ \mathrm{d}\end{array}\right){\mathrm{p}}_{01}^{\mathrm{d}}{\left(1-{\mathrm{p}}_{01}\right)}^{{\mathrm{n}}_1-\mathrm{d}}}\\ \left(1-{\displaystyle \sum _{\mathrm{i}=0}^{{\mathrm{UCL}}_2-\mathrm{d}}\left(\begin{array}{c}{\mathrm{n}}_2\\ i\end{array}\right){\mathrm{p}}_{02}^{\mathrm{i}}{\left(1-{\mathrm{p}}_{02}\right)}^{{\mathrm{n}}_2-\mathrm{i}}}\right)\end{array}$$

(5)

Hence, the in-control average run length (ARL) which is indication when the process will be out-of-control on average is given as follows

$$\begin{array}{c}{\mathrm{ARL}}_{00}=\frac{1}{{\mathrm{P}}_{\mathrm{out}}^{0,0}}\\ =\frac{1}{\begin{array}{l}1-{\displaystyle \sum _{\mathrm{d}=0}^{{\mathrm{UCL}}_1}\left(\begin{array}{c}{\mathrm{n}}_1\\ \mathrm{d}\end{array}\right){\mathrm{p}}_{01}^{\mathrm{d}}{\left(1-{\mathrm{p}}_{01}\right)}^{{\mathrm{n}}_1-\mathrm{d}}}\\ +{\displaystyle \sum _{\mathrm{d}=0}^{{\mathrm{UCL}}_1}\left(\begin{array}{c}{\mathrm{n}}_1\\ \mathrm{d}\end{array}\right){\mathrm{p}}_{01}^{\mathrm{d}}{\left(1-{\mathrm{p}}_{01}\right)}^{{\mathrm{n}}_1-\mathrm{d}}}\\ \left(1-{\displaystyle \sum _{\mathrm{i}=0}^{{\mathrm{UCL}}_2-\mathrm{d}}\left(\begin{array}{c}{\mathrm{n}}_2\\ \mathrm{i}\end{array}\right){\mathrm{p}}_{02}^{\mathrm{i}}{\left(1-{\mathrm{p}}_{02}\right)}^{{\mathrm{n}}_2-\mathrm{i}}}\right)\end{array}}\end{array}$$

(6)

In the next sections, we will derive the out-of-control ARL when (1) just first process is shifted, (2) just second process is shifted and (3) both processes are shifted.

Just first process is shifted

Suppose that the first process has been shifted to ${\mathrm{p}}_{11}={\mathrm{c}}_1{\mathrm{p}}_{01}$ ( ${\mathrm{c}}_1>1)$ while the second process remains in control.

The probability of declaring as out of control for the shifted process is given as follows

$$\begin{array}{l}{\mathrm{P}}_{\mathrm{out}}^{1,0}=\left(1-{\displaystyle \sum _{\mathrm{d}=0}^{{\mathrm{UCL}}_1}\left(\begin{array}{c}{\mathrm{n}}_1\\ \mathrm{d}\end{array}\right){\mathrm{p}}_{11}^{\mathrm{d}}{\left(1-{\mathrm{p}}_{11}\right)}^{{\mathrm{n}}_1-\mathrm{d}}}\right)+{\displaystyle \sum _{d=0}^{{\mathrm{UCL}}_1}\left(\begin{array}{c}{\mathrm{n}}_1\\ \mathrm{d}\end{array}\right){\mathrm{p}}_{11}^{\mathrm{d}}{\left(1-{\mathrm{p}}_{11}\right)}^{{\mathrm{n}}_1-\mathrm{d}}}\\ \left(1-{\displaystyle \sum _{\mathrm{i}=0}^{{\mathrm{UCL}}_2-\mathrm{d}}\left(\begin{array}{c}{\mathrm{n}}_2\\ i\end{array}\right){\mathrm{p}}_{02}^{\mathrm{i}}{\left(1-{\mathrm{p}}_{02}\right)}^{{\mathrm{n}}_2-\mathrm{i}}}\right)\end{array}$$

(7)

The out-of-control ARL when just the first process has shifted will be

$${\mathrm{ARL}}_{10}=\frac{1}{\begin{array}{l}\left(1-{\displaystyle {\sum }_{\mathrm{d}=0}^{{\mathrm{UCL}}_1}\left(\begin{array}{c}{\mathrm{n}}_1\\ \mathrm{d}\end{array}\right){\mathrm{p}}_{11}^{\mathrm{d}}{\left(1-{\mathrm{p}}_{11}\right)}^{{\mathrm{n}}_1-\mathrm{d}}}\right)\\ +{\displaystyle {\sum }_{\mathrm{d}=0}^{{\mathrm{UCL}}_1}\left(\begin{array}{c}{\mathrm{n}}_1\\ \mathrm{d}\end{array}\right){\mathrm{p}}_{11}^{\mathrm{d}}{\left(1-{\mathrm{p}}_{11}\right)}^{{\mathrm{n}}_1-\mathrm{d}}}\\ \left(1-{\displaystyle {\sum }_{\mathrm{i}=0}^{{\mathrm{UCL}}_2-\mathrm{d}}\left(\begin{array}{c}{\mathrm{n}}_2\\ \mathrm{i}\end{array}\right){\mathrm{p}}_{02}^{\mathrm{i}}{\left(1-{\mathrm{p}}_{02}\right)}^{{\mathrm{n}}_2-\mathrm{i}}}\right)\end{array}}$$

(8)

Just second process is shifted

Now, suppose that the second process has been shifted to ${\mathrm{p}}_{12}={\mathrm{c}}_2{\mathrm{p}}_{02}$ ( ${\mathrm{c}}_2>1)$ while the first process remains in control. Then, the probability of declaring as out of control for the shifted process is given as follows

$$\begin{array}{l}{\mathrm{P}}_{\mathrm{out}}^{0,1}=\left(1-{\displaystyle \sum _{\mathrm{d}=0}^{{\mathrm{UCL}}_1}\left(\begin{array}{c}{\mathrm{n}}_1\\ \mathrm{d}\end{array}\right){\mathrm{p}}_{01}^{\mathrm{d}}{\left(1-{\mathrm{p}}_{01}\right)}^{{\mathrm{n}}_1-\mathrm{d}}}\right)+{\displaystyle \sum _{d=0}^{{\mathrm{UCL}}_1}\left(\begin{array}{c}{\mathrm{n}}_1\\ \mathrm{d}\end{array}\right){\mathrm{p}}_{01}^{\mathrm{d}}{\left(1-{\mathrm{p}}_{01}\right)}^{{\mathrm{n}}_1-\mathrm{d}}}\\ \left(1-{\displaystyle \sum _{\mathrm{i}=0}^{{\mathrm{UCL}}_2-\mathrm{d}}\left(\begin{array}{c}{\mathrm{n}}_2\\ \mathrm{i}\end{array}\right){\mathrm{p}}_{12}^{\mathrm{i}}{\left(1-{\mathrm{p}}_{12}\right)}^{{\mathrm{n}}_2-\mathrm{i}}}\right)\end{array}$$

(9)

Therefore, the out-of-control ARL for this case is given by

$${\mathrm{ARL}}_{01}=\frac{1}{\begin{array}{l}\left(1-{\displaystyle {\sum }_{\mathrm{d}=0}^{{\mathrm{UCL}}_1}\left(\begin{array}{c}{\mathrm{n}}_1\\ \mathrm{d}\end{array}\right){\mathrm{p}}_{01}^{\mathrm{d}}{\left(1-{\mathrm{p}}_{01}\right)}^{{\mathrm{n}}_1-\mathrm{d}}}\right)\\ +{\displaystyle {\sum }_{d=0}^{{\mathrm{UCL}}_1}\left(\begin{array}{c}{\mathrm{n}}_1\\ \mathrm{d}\end{array}\right){\mathrm{p}}_{01}^{\mathrm{d}}{\left(1-{\mathrm{p}}_{01}\right)}^{{\mathrm{n}}_1-\mathrm{d}}}\\ \left(1-{\displaystyle {\sum }_{\mathrm{i}=0}^{{\mathrm{UCL}}_2-\mathrm{d}}\left(\begin{array}{c}{\mathrm{n}}_2\\ \mathrm{i}\end{array}\right){\mathrm{p}}_{22}^{\mathrm{i}}{\left(1-{\mathrm{p}}_{22}\right)}^{{\mathrm{n}}_2-\mathrm{i}}}\right)\end{array}}$$

(10)

Both processes are shifted

Now, suppose that the both processes have been shifted such as

$${\mathrm{p}}_{11}={\mathrm{c}}_1{\mathrm{p}}_{01}\mathrm{and}{\mathrm{p}}_{12}={\mathrm{c}}_2{\mathrm{p}}_{02}$$

(11)

The probability of declaring as out-of-control for the two-stage process when the processes are shifted is given by

$$\begin{array}{l}{\mathrm{P}}_{\mathrm{out}}^{1,1}=\left(1-{\displaystyle \sum _{\mathrm{d}=0}^{{\mathrm{UCL}}_1}\left(\begin{array}{c}{\mathrm{n}}_1\\ \mathrm{d}\end{array}\right){\mathrm{p}}_{11}^{\mathrm{d}}{\left(1-{\mathrm{p}}_{11}\right)}^{{\mathrm{n}}_1-\mathrm{d}}}\right)+{\displaystyle \sum _{d=0}^{{\mathrm{UCL}}_1}\left(\begin{array}{c}{\mathrm{n}}_1\\ d\end{array}\right){\mathrm{p}}_{11}^{\mathrm{d}}{\left(1-{\mathrm{p}}_{11}\right)}^{{\mathrm{n}}_1-\mathrm{d}}}\\ \left(1-{\displaystyle \sum _{\mathrm{i}=0}^{{\mathrm{UCL}}_2-\mathrm{d}}\left(\begin{array}{c}{\mathrm{n}}_2\\ \mathrm{i}\end{array}\right){\mathrm{p}}_{12}^{\mathrm{i}}{\left(1-{\mathrm{p}}_{12}\right)}^{{\mathrm{n}}_2-\mathrm{i}}}\right)\end{array}$$

(12)

Hence, the out-of-control ARL for this case is obtained by

$${\mathrm{ARL}}_{11}=\frac{1}{\begin{array}{l}\left(1-{\displaystyle {\sum }_{\mathrm{d}=0}^{{\mathrm{UCL}}_1}\left(\begin{array}{c}{\mathrm{n}}_1\\ \mathrm{d}\end{array}\right){\mathrm{p}}_{11}^{\mathrm{d}}{\left(1-{\mathrm{p}}_{11}\right)}^{{\mathrm{n}}_1-\mathrm{d}}}\right)\\ +{\displaystyle {\sum }_{d=0}^{{\mathrm{UCL}}_1}\left(\begin{array}{c}{\mathrm{n}}_1\\ \mathrm{d}\end{array}\right){\mathrm{p}}_{11}^{\mathrm{d}}{\left(1-{\mathrm{p}}_{11}\right)}^{{\mathrm{n}}_1-\mathrm{d}}}\\ \left(1-{\displaystyle {\sum }_{\mathrm{i}=0}^{{\mathrm{UCL}}_2-\mathrm{d}}\left(\begin{array}{c}{\mathrm{n}}_2\\ \mathrm{i}\end{array}\right){\mathrm{p}}_{22}^{\mathrm{i}}{\left(1-{\mathrm{p}}_{22}\right)}^{{\mathrm{n}}_2-\mathrm{i}}}\right)\end{array}}$$

(13)

These out-of-control ARLs for three cases are obtained using the following procedure:

1. Specify values of ${\mathrm{p}}_{01}$ , ${\mathrm{p}}_{02}$ , ${\mathrm{n}}_1$ , ${\mathrm{n}}_2$ and the target in-control ARL ( $r_0$ ).

2. Determine the control coefficients ${\mathrm{k}}_1$ and ${\mathrm{k}}_2$ such that ${\mathrm{ARL}}_{00}\ge r_0$ .

3. Obtain the out-of-control ARLs for various values of shift constants ${\mathrm{c}}_1$ and/or ${\mathrm{c}}_2$ .

To construct the tables for ARLs, we consider two cases for fractions nonconforming such as ${\mathrm{p}}_{01}={\mathrm{p}}_{02}=0.01$ and ${\mathrm{p}}_{01}={\mathrm{p}}_{02}=0.05$ . Also, we consider three cases for the target in-control ARL such as $r_0=\mathrm{200}$ , $r_0=300$ and $r_0=400$ . Several combinations for ( ${\mathrm{n}}_1$ , ${\mathrm{n}}_2$ ) are considered. Tables are prepared only for ${\mathrm{c}}_1={\mathrm{c}}_2$ , whose values range from 1.0 to 4.0. Note that the case of ${\mathrm{c}}_1={\mathrm{c}}_2=1.0$ indicates that both processes are in control.

Tables 1 and 2 show the ARLs when ${\mathrm{p}}_{01}={\mathrm{p}}_{02}=0.01$ and Tables 3 and 4 show the ARLs when ${\mathrm{p}}_{01}={\mathrm{p}}_{02}=0.05$ . Tables 1 and 3 list the cases of equal sample sizes for the first and the second process, while Tables 2 and 4 show the unequal cases.

Table 1.

The ARLs of the proposed chart with equal sample size when p₀₁ = p₀₂ = 0.01.

Shift	r0 = 200, n1 = n2 = 76; k1 = 4.6633; k2 = 4.6632			r0 = 300, n1 = n2 = 70; k1 = 4.6150; k2 = 3.57482			r0 = 400, n1 = n2 = 66; k1 = 4.2779; k2 = 3.01652
	First process is shifted	Second process is shifted	Both processes are shifted	First process is shifted	Second process is shifted	Both processes are shifted	First process is shifted	Second process is shifted	Both processes are shifted
1.0	203.60	203.60	203.60	301.55	301.55	301.55	401.34	401.34	401.34
1.2	125.88	135.19	88.65	184.04	199.04	128.96	242.77	263.94	169.59
1.4	82.05	92.62	45.42	118.46	135.36	64.88	154.92	178.63	84.30
1.6	55.97	65.53	26.21	79.84	94.99	36.77	103.54	124.68	47.21
1.8	39.72	47.79	16.56	55.99	68.68	22.83	72.03	89.64	28.95
2.0	29.17	35.82	11.24	40.65	51.03	15.22	51.89	66.22	19.08
2.5	15.24	19.30	5.36	20.68	26.89	6.96	25.91	34.38	8.47
3.0	9.13	11.69	3.21	12.08	15.93	4.01	14.88	20.07	4.75
3.5	6.06	7.75	2.24	7.83	10.33	2.69	9.48	12.83	3.11
4.0	4.35	5.51	1.74	5.49	7.20	2.02	6.55	8.80	2.28

ARL: average run length.

Table 2.

The ARLs of the proposed chart with unequal sample size when p₀₁ = p₀₂ = 0.01.

Shift	r0 = 200, n1 = 54; n2 = 14; k1 = 4.7182; k2 = 3.23792			r0 = 300, n1 = 88; n2 = 55; k1 = 4.8822; k2 = 3.23162			r0 = 400, n1 = 95; n2 = 40; k1 = 4.4831; k2 = 3.38462
	First process is shifted	Second process is shifted	Both processes are shifted	First process is shifted	Second process is shifted	Both processes are shifted	First process is shifted	Second process is shifted	Both processes are shifted
1.0	204.16	204.16	204.16	299.83	299.83	299.83	398.61	398.61	398.61
1.2	122.68	177.44	108.89	172.22	210.10	126.36	210.82	301.09	165.82
1.4	79.53	155.27	64.98	106.36	151.88	62.98	122.24	231.89	81.60
1.6	54.65	136.72	42.10	69.64	112.78	35.48	76.19	181.69	45.40
1.8	39.34	121.07	29.04	47.84	85.73	21.94	50.35	144.56	27.73
2.0	29.40	107.78	21.04	34.22	66.52	14.60	34.91	116.61	18.23
2.5	16.09	82.21	11.02	17.02	38.06	6.67	16.46	71.79	8.08
3.0	10.05	64.29	6.78	9.85	23.76	3.85	9.23	47.03	4.54
3.5	6.88	51.34	4.66	6.37	15.90	2.60	5.87	32.41	2.98
4.0	5.06	41.74	3.46	4.48	11.25	1.96	4.09	23.29	2.20

ARL: average run length.

Table 3.

The ARLs of the proposed chart with equal sample size when p₀₁ = p₀₂ = 0.05.

Shift	r0 = 200, n1 = n2 = 91; k1 = 3.6914; k2 = 2.92662			r0 = 300, n1 = n2 = 61; k1 = 4.5488; k2 = 3.23692			r0 = 400, n1 = n2 = 35; k1 = 5.4723; k2 = 3.46042
	First process is shifted	Second process is shifted	Both processes are shifted	First process is shifted	Second process is shifted	Both processes are shifted	First process is shifted	Second process is shifted	Both processes are shifted
1.0	199.93	199.93	199.93	303.13	303.13	303.13	404.01	404.01	404.01
1.2	75.24	82.29	36.93	130.62	135.33	66.05	205.50	206.33	113.69
1.4	32.76	38.17	11.13	63.02	67.24	21.25	113.76	114.66	42.41
1.6	16.23	19.87	4.77	33.46	36.69	9.07	67.50	68.32	19.43
1.8	9.03	11.44	2.65	19.30	21.70	4.79	42.45	43.17	10.41
2.0	5.56	7.19	1.79	11.98	13.75	2.97	28.06	28.67	6.30
2.5	2.37	3.05	1.13	4.75	5.61	1.48	11.91	12.30	2.65
3.0	1.48	1.79	1.02	2.54	3.00	1.12	6.17	6.43	1.62
3.5	1.17	1.32	1.00	1.69	1.95	1.03	3.73	3.90	1.24
4.0	1.05	1.12	1.00	1.32	1.47	1.00	2.54	2.66	1.09

ARL: average run length.

Table 4.

The ARLs of the proposed chart with unequal sample size when p₀₁ = p₀₂ = 0.05.

Shift	r0 = 200, n1 = 90; n2 = 65; k1 = 4.8142; k2 = 3.95512			r0 = 300, n1 = 75; n2 = 22; k1 = 4.8207; k2 = 3.95002			r0 = 400, n1 = 84; n2 = 22; k1 = 5.0647; k2 = 3.25132
	First process is shifted	Second process is shifted	Both processes are shifted	First process is shifted	Second process is shifted	Both processes are shifted	First process is shifted	Second process is shifted	Both processes are shifted
1.0	201.56	201.56	201.56	300.60	300.60	300.60	402.41	402.41	402.41
1.2	74.98	96.93	40.88	99.14	211.77	74.69	119.23	284.98	90.40
1.4	33.18	51.36	12.94	40.63	152.97	25.99	45.17	206.56	29.26
1.6	16.88	29.53	5.64	19.68	112.97	11.56	20.65	152.83	12.35
1.8	9.64	18.21	3.12	10.88	85.10	6.18	10.95	115.20	6.37
2.0	6.06	11.93	2.06	6.69	65.26	3.82	6.54	88.30	3.84
2.5	2.63	5.16	1.21	2.82	35.95	1.78	2.66	48.41	1.74
3.0	1.61	2.85	1.04	1.70	21.49	1.24	1.60	28.69	1.22
3.5	1.23	1.90	1.00	1.28	13.74	1.07	1.22	18.13	1.06
4.0	1.08	1.45	1.00	1.11	9.29	1.02	1.08	12.10	1.01

ARL: average run length.

From Tables 1 –4, we note the following trends:

1. As the fraction nonconforming increases, the out-of-control ARLs decrease more rapidly.

2. As the target in-control ARL increases, smaller sample sizes are required for the two-stage process.

3. According to the sample size allocation at the first and the second process, the ARL performance can be different.

Figure 1 shows the trend in ARLs according to the shift constants when p₀₁ = p₀₂ = 0.01 and $r_0=300$ . From Figure 1, we see that ARLs are decreasing rapidly when both processes, although they are also reasonably fast decreasing when first process is shifted or second process is shifted. We also note that the values of ARLs are more slowly decreasing when only the second process is shifted than when only the first process is shifted.

Figure 1.

Comparison of ARLs according to different process shifts when p₀₁ = p₀₂ = 0.01 and $r_0=300$ .

Example

For the possible application to an industrial data, we illustrate the use of the proposed chart with synthetic data. Suppose that there is a two-stage manufacturing process, where the number of nonconforming items is counted at each process by subgroups. To generate the synthetic data, it is assumed that p₀₁ = p₀₂ = 0.01 and ${\mathrm{n}}_1={\mathrm{n}}_2=67$ . The data are reported in Table 5, where $D_1$ shows the number of nonconforming items at the first process for each of 30 subgroups and $D_2$ shows the number of nonconforming items at the second process.

Table 5.

Cans data.

	D1	D2	D = D1 + D2
1	0	1	1
2	2	0	2
3	0	1	1
4	0	1	1
5	0	1	1
6	1	2	3
7	1	2	3
8	0	0	0
9	1	1	2
10	0	0	0
11	0	2	2
12	2	1	3
13	3	0	3
14	1	0	1
15	1	1	2
16	3	0	3
17	1	0	1
18	2	0	2
19	0	1	1
20	3	1	4
21	0	1	1
22	0	1	1
23	0	0	0
24	1	3	4
25	1	1	2
26	0	0	0
27	2	0	2
28	0	0	0
29	2	2	4
30	0	0	0

The estimated fraction nonconforming for our data is given as follows

$$\begin{array}{l}{\overline{\mathrm{p}}}_{01}=\frac{{{\displaystyle \sum }}_{\mathrm{i}=1}^{\mathrm{m}}{\mathrm{D}}_{\mathrm{i}}}{\mathrm{mn}}=\frac{27}{2010}=0.013\mathrm{and}\\ {\overline{\mathrm{p}}}_{02}=\frac{{{\displaystyle \sum }}_{\mathrm{i}=1}^{\mathrm{m}}{\mathrm{D}}_{\mathrm{i}}}{\mathrm{mn}}=\frac{23}{2010}=0.011\end{array}$$

where $m$ and $n$ show the observations and subgroup number, respectively.

The target in-control ARL is selected as r₀ = 372. The two control constants for this target and the true fractions nonconforming are obtained by

$${\mathrm{k}}_1=4.642,{\mathrm{k}}_2=3.5435$$

Therefore, the control limit for first process when using the estimated fraction nonconforming is calculated as

$${\mathrm{UCL}}_1=67\times 0.013+4.6421\sqrt{67\times 0.013\left(1-0.013\right)}=4.08$$

The ${\mathrm{UCL}}_2$ is given as

$$\begin{array}{c}{\mathrm{UCL}}_2=\left(67\times 0.013+67\times 0.011\right)\\ +3.5435\sqrt{\begin{array}{l}67\times 0.013\left(1-0.013\right)\\ +67\times 0.011\left(1-0.011\right)\end{array}}=5.33\end{array}$$

We plotted the number of nonconforming cans on control chart for the first process in Figure 2.

Figure 2.

Control chart for the first process in our example.

From Figure 2, we note that all the values of $D_1$ are below ${\mathrm{UCL}}_1=4.08$ , which shows that the first process is in control. To check the state of the second process or the combined one, we plotted the total number of nonconforming cans from both processes on Figure 3.

Figure 3.

Control chart for the combined process in our example.

From Figure 3, we note that all the values of $D=D_1+D_2$ are below ${\mathrm{UCL}}_2$ , which indicate that the two-stage process is in control.

Comparative study

In this section, we will compare the performance of the proposed control chart with the Shewhart np control chart using $D=D_1+D_2$ . Therefore, the UCL of the Shewhart control chart is given by (2). However, it is different from the proposed chart in that the process is declared as in-control for the proposed chart when the first stage is in-control and the combined process is in-control. To save the space, we will consider the case when p = 0.05, r₀ = 400 and n = 70 for the proposed control chart and Shewhart control chart. We only consider the case that both stages are shifted. Figure 4 shows the ARLs according to the shift constants for the proposed chart and Shewhart chart. From Figure 4, it can be noted the proposed control chart has smaller ARLs than the Shewhart control chart when both processes are shifted. Therefore, the proposed control chart is more efficient than the Shewhart type control chart in detecting the shift in the manufacturing process.

Figure 4.

The comparisons of the proposed chart and Shewhart control chart when p₀₁ = p₀₂ = 0.05 and $r_0=400$ .

Concluding remarks

In this manuscript, we designed an attribute control chart for a two-stage process. The performance of the proposed chart is assessed in terms of ARLs. The proposed chart is illustrated with a synthetic data set. It is observed that ARLs are decreasing rapidly when both processes, although they are also reasonably fast decreasing when the first process is shifted or the second process is shifted. We also note that the values of ARLs are more slowly decreasing when only the second process is shifted than when only the first process is shifted. These ARLs can be decreased faster if a larger sample size is allocated either the first or the second process.

An attribute control chart for a more general multi-stage process can be extended as a future study. Also, a variable control chart for a two-stage process should be developed in the future.