ISyE 512 Chapter 7. Control Charts for Attributes. Instructor: Prof. Kaibo Liu. Department of Industrial and Systems Engineering UW-Madison

(1)

ISyE 512 Instructor: Kaibo Liu

ISyE 512 Chapter 7

Instructor: Prof. Kaibo Liu

Department of Industrial and Systems Engineering

UW-Madison

Email: [email protected]

Office: Room 3017 (Mechanical Engineering Building)

Control Charts for Attributes

(2)

List of Topics in Chapter 7

• P Control Chart for nonconforming unit

• NP Control Chart for nonconforming unit

• C Control Chart for nonconformities

(3)

How to describe a product

not meet “Quality” requirements?

• Nonconformity

—A departure of a quality characteristic from its

intended level or state that occurs with a severity sufficient to

cause an associated product or service not to meet a

specification requirement.

• Nonconforming unit

—A unit of product or service containing at

least one nonconformity.

• Defect

—A departure of a quality characteristic from its intended

level or state that occurs with a severity sufficient to cause an

associated product or service not to satisfy intended normal, or

reasonably foreseeable usage requirements.

• Defective

(Defective Unit)—A unit of product or service

containing at least one defect, or having several imperfections

that in combination cause the unit not to satisfy intended normal,

or reasonably foreseeable, usage requirements.

3

(4)

Review of Binomial Distribution

Let x = # defective unit in a sample of size n where defective

unit follow a Bernoulli Process (two outcomes, p-constant, x -

independent)

f (x)



n

x





 



p

x

(1



p)

n



x

E(x)



np

V(x)



np(1



p)

(5)

Sample Estimation of p

n

p

x

V

n

p

V

p

n

np

n

x

E

p

E

n

x

p

)

1 (

)

(

1 )

ˆ

(

)

(

)

ˆ

(

ˆ

Let

2 



Proportion of nonconforming parts,

_{Probability of nonconforming for}

each individual part/observation, r.v .

Sample statistic

(6)

Review of the Basic Model of a Control Chart

Let w be a sample statistic that measures some quality

characteristic of interest, and suppose that the mean of w is



w

and

the standard deviation of w is



w

. Then the center line, the upper

control limit, and the lower control limit become

UCL =



w

+ L



w

Center line =



w

LCL =



w

– L



w

where L is the "distance" of the control limits from the center line,

expressed in standard deviation units

(7)

Fraction Nonconforming Control Chart (p-Chart)

n

p

LCL

p

n

p

UCL

n

p

N

p

)

1 (

3 Centerline

)

1 (

3 )

1 (

,

~

ˆ

large

np

For

ˆ



























Binomial (

np

>10,

0.1<=p<=0.9

)



normal

If LCL

_p

<0, set LCL

_p

=0

Remarks: When data points are plotted below LCL, they generally do not represent

a real improvement. Actually, they are often caused by errors in the

inspection rather than a process improvement

Assume p is known

(8)

How to Establish a p-Chart?

• Is there an assignable cause for out-of-control points or a nonrandom

pattern? If so, find the root causes and delete these points, and then update

control limits.

n

p

LCL

p

n

p

UCL

)

1 (

3 Centerline

)

1 (

3 











If p

unknown

, conduct a test and trial control limits with

p

)

p

(

E

m

pˆ

mn

D

p

m 1 i i m 1 i i











p

ˆ



m=20-25 samples for constructing trial control limits

Trial Control

Limits

When a point is ON a control limit, it is considered as either

out-of-control or in-control depending on how the problem asks

(9)

Example: The following data give the number of nonconforming ROM

chips in samples of size 200. Construct a p chart for these data. Assume

that any values beyond the control limits have an assignable cause and

revise the control limits as appropriate.

Sample Nonconforming Sample Nonconforming

1

19

12

18

2

7

13

17

3

11

14

21

4

29

15

16

5

24

16

6

24

17

23

7

15

18

14

8

25

19

4

9

11

20

21

10

21

24

11

37

22

10

9

(10)

n= 200

sample Nonconforming pi UCL LCL

1 19 0.095 0.1507 0.0293 2 7 0.035 0.1507 0.0293 3 11 0.055 0.1507 0.0293 4 29 0.145 0.1507 0.0293 5 24 0.12 0.1507 0.0293 6 24 0.12 0.1507 0.0293 7 15 0.075 0.1507 0.0293 8 25 0.125 0.1507 0.0293 9 11 0.055 0.1507 0.0293 10 10 0.05 0.1507 0.0293 11 37 0.185 0.1507 0.0293 12 18 0.09 0.1507 0.0293 13 17 0.085 0.1507 0.0293 14 21 0.105 0.1507 0.0293 15 16 0.08 0.1507 0.0293 16 16 0.08 0.1507 0.0293 17 23 0.115 0.1507 0.0293 18 14 0.07 0.1507 0.0293 19 4 0.02 0.1507 0.0293 20 21 0.105 0.1507 0.0293 21 24 0.12 0.1507 0.0293 22 10 0.05 0.1507 0.0293 p-bar 0.09 0 0.05 0.1 0.15 0.2 1 3 5 7 9 11 13 15 17 19 21 pi Samples p-char (Example)

Good or NOT?

(11)

Example 13-1: A fraction nonconforming control chart with center line 0.10, UCL

_p

= 0.19,

and LCL

_p

= 0.01 is used to control a process. (Samples on the control limits are considered

as out-of-control)

a. If 3-sigma limits are used, find the sample size for the control chart.

b. Use the Poisson approximation to the binomial to find the probability of type I error.

c. Use the Poisson approximation to the binomial to find the probability of type II error if the

process fraction defective is actually p = 0.20.

Useful Approximation in Control

Charting

(12)

(13)

Variable Sample Size

• Variable width of control limits

corresponding to each sample size

– not appropriate for nonrandom pattern check

• Constant width of control limits using average sample size

– future sample size should not differ greatly

• Standardized Control Chart

– can be used to check a nonrandom pattern

– no reference to the actual process fraction defective

i p i p

n

)

p

1 (

p

3 p

LCL

;

n

)

p

1 (

p

3 p

UCL

i i











ns

observatio

of

#

total

defects

of

#

total

n

D

p

_m 1 i i m 1 i i





 

m

n

m 1 i i





0 CL

;

3 LCL

;

3 UCL

;

p

;

n

)

p

1 (

p

pˆ

Z

i i i











MEAN 0. 22 3 LW L UWL LC L UCL 0 0. 1 0. 2 0. 3 0. 4 0. 5 D ef ect s 0 ₁₀ ₂₀ ₃₀ R ow Numbers

P Chart U sin g Crick etg rap h III

Different

samples,

the same

center line

Different samples, different CL

Standardization

(14)

Table 7.5, P314

I n(i) D(i) pi=D(i)/n(i) sigma=sqrt(pbar*(1-pbar)/n(i)) LCL (ni) UCL(ni) LCL (n bar) UCL(n bar) Zi LCL(stand)UCL(stand) 1 100 12 0.120 0.029 0.007 0.184 0.007 0.185 0.81 -3 3 2 80 8 0.100 0.033 0.000 0.194 0.007 0.185 0.12 -3 3 3 80 6 0.075 0.033 0.000 0.194 0.007 0.185 -0.64 -3 3 4 100 9 0.090 0.029 0.007 0.184 0.007 0.185 -0.20 -3 3 5 110 10 0.091 0.028 0.011 0.180 0.007 0.185 -0.18 -3 3 6 110 12 0.109 0.028 0.011 0.180 0.007 0.185 0.47 -3 3 7 100 11 0.110 0.029 0.007 0.184 0.007 0.185 0.48 -3 3 8 100 16 0.160 0.029 0.007 0.184 0.007 0.185 2.17 -3 3 9 90 10 0.111 0.031 0.003 0.188 0.007 0.185 0.49 -3 3 10 90 6 0.067 0.031 0.003 0.188 0.007 0.185 -0.94 -3 3 11 110 20 0.182 0.028 0.011 0.180 0.007 0.185 3.06 -3 3 12 120 15 0.125 0.027 0.015 0.176 0.007 0.185 1.08 -3 3 13 120 9 0.075 0.027 0.015 0.176 0.007 0.185 -0.78 -3 3 14 120 8 0.067 0.027 0.015 0.176 0.007 0.185 -1.09 -3 3 15 110 6 0.055 0.028 0.011 0.180 0.007 0.185 -1.48 -3 3 16 80 8 0.100 0.033 0.000 0.194 0.007 0.185 0.12 -3 3 17 80 10 0.125 0.033 0.000 0.194 0.007 0.185 0.88 -3 3 18 80 7 0.088 0.033 0.000 0.194 0.007 0.185 -0.26 -3 3 19 90 5 0.056 0.031 0.003 0.188 0.007 0.185 -1.30 -3 3 20 100 8 0.080 0.029 0.007 0.184 0.007 0.185 -0.54 -3 3 21 100 5 0.050 0.029 0.007 0.184 0.007 0.185 -1.56 -3 3 22 100 8 0.080 0.029 0.007 0.184 0.007 0.185 -0.54 -3 3 23 100 10 0.100 0.029 0.007 0.184 0.007 0.185 0.14 -3 3 24 90 6 0.067 0.031 0.003 0.188 0.007 0.185 -0.94 -3 3 25 90 9 0.100 0.031 0.003 0.188 0.007 0.185 0.13 -3 3 sum 2450 234 average 98 pbar= 0.096

p bar=total defective/total samples

0.0 0.1 0.1 0.2 0.2 0.3 1 3 5 7 9 11 13 15 17 19 21 23 25 p sample index

Fig 6-6 control chart with variable sample size 0.0 0.0 0.1 0.1 0.2 0.2 1 3 5 7 9 11 13 15 17 19 21 23 25 p -4.0 -3.0 -2.0 -1.00.0 1.0 2.0 3.0 4.0 1 3 5 7 9 11 13 15 17 19 21 23 25 27 zi index

(15)

np Control Chart

(The number of nonconforming items)

Rather than plotting the fraction nonconforming, we plot the number of

nonconforming items with an “np Chart”:

UCL

_X

= np + 3 np(1–p)

Center line = np

LCL

_X

= np – 3 np(1–p)

• Np and p control charts can be transferred to each other. From p to np, multiple n to

the UCL, CL, and LCL; From np to p, divide n to the UCL, CL, and LCL.

If LCL

_X

<0, set LCL

_X

=0

(16)

Example: The number of transmission cases that required deburring in a

16-day sample of 100 each was as follows:

Day

Number Day

Number

1

5

9

4

2

4

10

6

3

11

15

4

2

12

4

5

6

13

5

6

3

14

7

9

15

3

8

6

16

6 Prepare an np chart with trial control limits. Assume that any points plotting

out of control have assignable causes, and continue to refine the control

limits until no points plot out of control.

(17)

I Di UCL(trial) LCL(trial) UCL LCL

1 5 12.34 0 11.32 0.00 2 4 12.34 0 11.32 0.00 3 3 12.34 0 11.32 0.00 4 2 12.34 0 11.32 0.00 5 6 12.34 0 11.32 0.00 6 3 12.34 0 11.32 0.00 7 9 12.34 0 11.32 0.00 8 6 12.34 0 11.32 0.00 9 4 12.34 0 11.32 0.00 10 6 12.34 0 11.32 0.00 11 15 12.34 0 11.32 0.00 12 4 12.34 0 11.32 0.00 13 5 12.34 0 11.32 0.00 14 7 12.34 0 11.32 0.00 15 3 12.34 0 11.32 0.00 16 6 12.34 0 11.32 0.00

sum 88 UCL(trial) LCL(trial)

p bar=88/(100*16)= 0.055 12.34 -1.34 set to zero

np 5.500

eliminate point 11

pbar=(88-15)/(100*15) 0.048667 UCL LCL

np 4.866667 11.32 -1.59 set to zero

np chart with trial control limits

0.0 5.0 10.0 15.0 20.0 1 3 5 7 9 11 13 15

np chart after eliminate outliers

0.0 5.0 10.0 15.0 20.0 1 3 5 7 9 11 13 15 Di UCL(trial) LCL(trial) UCL LCL

17

(18)

np Chart Properties

• Advantage

– np chart is a scaling of the vertical axis by the

constant n, provide the same information as p

chart

– np chart needs less calculation ( no need to

calculate D

_i

/n

_i

)

– often used when n is constant and p is small

• Limitation

– not easy for interpretation when n is varied (UCL

LCL and Ctr line all vary)

(19)

OC Curve and ARL

• Type II error for the p chart

• ARL

₀

=ARL

_in-control

=1/

• ARL

₁

=ARL

_{out-of-control}

=1/(1-)

19 𝜷 = 𝑷 𝒑 < 𝑼𝑪𝑳

_𝒑

𝒑

_𝟏

− 𝑷 𝒑 ≤ 𝑳𝑪𝑳

_𝒑

𝒑

_𝟏

(20)

Example: A control chart is used to control the fraction nonconforming for a plastic part manufactured in an

injection molding process. Ten subgroups yield the following data:

Sample Number

Sample Size

No. Nonconforming

1

100

10

2

100

15

3

100

31

4

100

18

5

100

26

6

100

12

7

100

25

8

100

15

9

100

8

10

100

8 a. Set up a control chart for the number nonconforming in samples of n = 100. (Samples on the control limits are

assumed to be in-control.)

b. For the chart established in part (a), what is the probability of detecting a shift in the process fraction

nonconforming to 0.30 on the first sample after the shift has occurred?

(21)

(22)

Example: Consider the 3 sigma control chart (CL = 0.10, UCL = 0.19, LCL

= 0.01). Find the average run length if the process fraction nonconforming

shifts to 0.20.

(23)

Control Charts for Nonconformities

(Defects) - C and U Charts

• Why need it:

– Control the total number of nonconformities in a sample or

the average number of nonconformities per unit

• nonconformity/defect: Each specific point at which a specification is not

satisfied, e.g.,

– weld spots on a car

– paint dent on a car body

• A unit may not be “nonconforming”, even though it has several

nonconformities. So, nonconforming



defects or nonconformities

• Assumption: The occurrence of nonconformities in

an

inspection unit

(or a sample)

of constant size is well

modeled by the Poisson distribution.

– The number of potential location for nonconformities can be very large, but

the probability of occurrence of a nonconformity at any location is small and

constant

(24)

What is an Inspection Unit

• An inspection unit could be a single unit of product

• Or it can be a group of several units, e.g., 144

microprocessors, 5 cars

• It is NOT necessarily integer

(25)

Statistical Basis of C Chart

Let X = # of nonconformities in an inspection unit

Assume X ~ Poisson ( E(X) = C )

For large C



X ~ N ( E(X) = C , Var(X) = C )

x=0,1,2,...

!

)

(

x

c

e

x

p

x

c





• c chart: total number of defects in an inspection

unit or sample

Nonconformities

(26)

Control Charts for

Nonconformities - c Chart

• Control limits for the c chart with a known c (know mean and variance)

• If unknown c, c is estimated from preliminary samples of inspection units for

constructing trial control limits

• The preliminary samples are examined by the control chart using the trial

control limits for checking out-of-control points

3

3 UCL

c

CL

c

LCL

c

 



 

If LCL<0, set LCL=0

samples

of

number

samples

all

in

defects

of

#

total

m

c

cˆ

m 1 i i







Nonconformities

₃

3 UCL

c

CL

c

LCL

c

 



 

(27)

Example (Textbook Ex 7.3): 26 successive samples of 100 PCB’s

What is the inspection unit?

100 boards

(28)

26

516 

c



3 c

c



3

67 .

19

24

472 .

L



c



C

97 .

32

3 





c

UCL

36 .

6

3 





c

LCL

(29)

(30)

Statistical Basis of u Chart

• X: # nonconformities in a sample of n inspection units (n is not necessarily

integer), X ~ Poisson (c), E(X)=V(X)=c

Y = average # of nonconformities

per inspection

unit

in a sample = X / n

E ( Y ) = c / n = u

Var ( Y ) = c / n

2 _{= u / n}

•u chart: average number of defects per inspection unit, in a sample

size of n inspection units

𝑼𝑪𝑳/𝑳𝑪𝑳=c±𝟑 c

(31)

Control Charts for Nonconformities

Per Unit - u Chart

• c: total nonconformities in a sample of n inspection units (n is not

necessary be integer)

• u: average # of nonconformities per inspection unit in a sample

• If unknown u, is estimated from preliminary samples of

inspection units for constructing trial control limits

n

u

3 u

UCL

u

CL

n

u

3 u

LCL









m

u

;

n

c

u

m 1 i i i i i





u

31

(32)

(33)

(34)

Example Find the 3-sigma control limits for:

a. A c chart with process average equal to four nonconformities.

b. A u chart with c = 4 and n = 4.

(35)

Variable Sample Size of Control Charts for

Nonconformities

• If sample size varies, one should use a u chart rather than a c

chart

• Approaches

– Control limits varies with each sample size, but the center

line is constant

– Use a control limits based on an average sample size

– Use a standardized control chart (this is preferred option),

with UCL=3, LCL=-3, Center line=0.

• This chart can be used for pattern recognition

u CL ; n u 3 u UCL ; n u 3 u LCL i i     

m

n

m 1 i i





i i i

n

u

Z





35

(36)

(37)

(38)

Example: Following are the number of nonconformities in 20 samples of 50 letter-quality

printer cases. Develop the trial control limits for a c chart. If any values are out of control,

assume that the cause is assignable. Modify the control limits accordingly.

Sample

Nonconf.

Sample

Nonconf.

1

19

11

37

2

21

12

16

3

14

13

4

23

14

28

5

13

15

17

6

21

16

29

7

15

17

25

8

24

18

15

9

20

19

11

10

19

20

19

0 5 10 15 20 25 30 35 40 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 b e r o f N o n -c o n fo rm it ie s

(39)

Day

Rolls

Produced

Imperfections

Number of

Day

Produced

Rolls

Imperfections

Number of

1

18

12

11

18

2

18

14

12

18

14

3

24

20

13

18

9

4

22

18

14

20

10

5

22

15

20

14

6

22

12

16

20

13

7

20

11

17

24

16

8

20

15

18

24

18

9

20

12

19

22

20

10

20

10

20

21

17 Example A paper mill uses a control chart to monitor the imperfection in finished rolls of paper.

Production output is inspected for 20 days, and the resulting data are shown below. Use these data to set up

a control chart for nonconformities per roll of paper. What kind of control chart you should use and why?

What are the CL and UCL/LCL?

(40)

Example (Cont’d)

-2 -1 0 1 2 3 4 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 z -s c o re Standardized u chart

18 [0.1088, 1.2926]

20 [0.1392, 1.2622]

21 [0.1527, 1.2487]

22 [0.1653, 1.2361]

24 [0.1881, 1.2133]

n

_i

[LCL

_i

, UCL

_i

]

0 4 8 12 16 20 Day 0 0.3 0.6 0.9 1.2 1.5 U

Control Chart for Pape r Impe rfe ctions

(41)

OC Curve and ARL for c and u Charts

• Type II error for the c chart (OC curve see Fig 7-19, P332) (if assume

points of the control limits are considered as out-of-control)

• Type II error for the u chart (if assume points of the control limits are

considered as out-of-control)

• ARL

₀

=ARL

_in-control

=1/



• ARL

₁

=ARL

_{out-of-control}

=1/(1-



)

}

|

{

}

|

{

}

|

{

}

|

{

1

1 u

nLCL

x

P

u

nUCL

x

P

u

LCL

y

P

u

UCL

y

P

u



















}

|

{

}

|

{

x

UCL

c

₁

P

x

LCL

c

₁

P



_c





_c





41

(42)

Example (Textbook Problem 7-56) A control chart is to be established on a process producing

refrigerators. The inspection unit is one refrigerator, and a control chart for nonconformities is to

be used. As preliminary data, 16 nonconformities were counted in inspecting 30 refrigerators.