A textbook of Computer Based Numerical and Statiscal Techniques part 50 pptx

11.17 From mean chart we see that 2nd, 3rd, 6th and 7th samples lies outside the control limits.. In a glass factory, the task of quality control was done with the help of mean x — and

Sol Mean Chart

10 10

x

Mean Range of 10 sample ranges 58 5.8

10 10

R

R=∑ = =

As we have, for n=5, A2 =0.58, D3 =0, D4 =2.115

3-σ control limits for x – chart are:

2

2

UCL

44.2 0.58 5.8 47.567 LCL

44.2 0.58 5.8 40.836

x

x

x

X A R

X A R

X

= +

= + × =

= −

= − × =

= = Range Chart: 3-σ Control Limits for R chart are:

4 3

UCL 2.115 5.8 12.267

R R R

D R

D R R

= = × =

= = × =

= =

60 50

40

30 20

10

0

Mean Chart

Sample Number

FIG 11.17

From mean chart we see that 2nd, 3rd, 6th and 7th samples lies outside the control limits Hence the process is out of control This shows that some assignable causes of variation are operating which should be detected and removed

9 8 7 6 5 4 3 2 1 0

Sample Number

FIG 11.18

Since all the points with in the control limits Hence the process is in statistical control

Example 2 The following are the mean lengths and ranges of lengths of a finished product from

10 samples each of size 5 The specification limits for length are 200 5 cm± Construct x – and R-chart and examine whether the process is under control and state your recommendation.

Assume for n = 5, A 2 = 0.577, D 3 = 0, D 4 = 2.115.

Sol In given problem specification limits for length are given 200 ± 5 cm Hence standard deviation is unknown

(1) Control Limits for x –-chart are:

Central limit, CLx= µ = 200

UCLx = µ +A R2 =200 0.577 4.7+ ×

=202.712; 47

10 10

R

R=∑ = LCLx = µ −A R2 =200 0.577 4.7− × = 197.288 R=4.7

(2) Control limits for R-Chart are:

4

3

UCL 9.941 2.115 4.7

CL 4.7

R R R

D R R

D R

= =

= = = ×

from control charts for mean and range, the process is in statistical control in R—-Chart because

all points lies with in the control limits where as in x—-chart, process is out of control because sample 5, 6 and 8 lies outside the control limits The process therefore should be halted to check

whether there are any assignable causes If assignable causes found, the process should be re-adjusted to remove assignable cause

205 204 203 202 201 200 199 198 197 196 195

Sample Number Mean Chart

FIG 11.19

Range Chart

10 8 6 4 2 0

Sample Number

FIG 11.20

Example 3 In a glass factory, the task of quality control was done with the help of mean (x —

) and standard deviation σ charts 18 samples of 10 items each were chosen and then values ∑X and ∑S were found to be 595.8 and 8.28 respectively Determine the 3-σ limits for mean and standard deviation chart Given that n = 10, A 1 = 1.03, B 3 = 0.28, B 4 = 1.72, ∑S = 8.28.

Sol

No of samples 18

S

—

18

S

= 8.28

18 = 0.46 hence, 3-σ control limits for standard deviation chart are:

UCL–

S = B4.S— = 1.72 × 0.48 = 0.7912 LCLS– = B3.S— = 0.28 – 0.46 = 0.1288

CLS– = 0.46 3-σ control limits for mean chart (x—) are:

X— = ∑

18

x

= 595.8

18 = 33.1

UCLX — = x— + A1σ

= 33.1 + 1.03 × 0.46 UCLX — = 33.57

LCLX — = x— – A1σ

= 33.1 – 1.03 × 0.46 LCLX — = 32.63

CLX — = 33.1

Example 4 If the average fraction defective of a large sample of a product is 0.1537, calculate the control limits when subgroup size is 2,000.

Sol Here, Sample size n = 2,000 for each sample

Average fraction defective = 0.1537 i.e., P = 0.1537

Q = 0.8463

Hence, 3–σ control limits for P-Chart are :

±3 PQ

P

n

UCLP = + 0.1537 0.8463×

0.1537 3

2,000 UCLP = 0.1537 + 0.02418 = 0.17788

LCLP = − 0.1537 0.8463×

0.1537 3

2,000 LCLP =0.1537 0.02418− =0.12952

CLP =0.5137

Example 5 The following data gives the number of defectives in 10 independent samples of varying sizes from a production process.

Draw the control chart for fraction defective.

Sol (In problem 4 sample size is fixed whereas in this problem sample size is variable) Since it is a problem of variable sample size so control chart for fraction defective can be drawn in two ways

(1) By first way, we set up two sets of control limits, one based on the maximum sample size, n=3125 and the second based on minimum sample size n=1, 250

(a) For n=3,125; UCL 0.200,= LCL=0.159

(b) For n=1, 250; UCL 0.212,= LCL=0.147

12 10 8 6 4 2 0

Sample Number

FIG 11.21 Control Chart for Fraction Defective

Since there are 4 points lies outside (based on minimum sample size) of control limits,

so process is of out of control

(2) By second way, 3-σ limit for each sample separately obtained by using formula

3 PQ

P n

±

where Total no of defectives

Total sample size

d P

n

∑

andn is corresponding sample size

17790

d

n

∑

∑

∴ ( )P Q = 0.1791 × 0.8209 = 0.1470231

n

P Q

n 3 ×

P Q

500 400 300 200 100 0

Sample Size

Sample points corresponding to sample no 1, 2, 4, 7 and 9 lie outside the control limits Hence, process is out of control.

FIG 11.22

Example 6 A daily sample of 30 items was taken over a period of 14 days in order to establish attributes control limits If 21 defectives were found, what should be upper and lower control limits of the proportion of defectives?

Sol Since a sample of 30 items is taken daily over a period of 14 days

Total No of items inspected = 30 × 14 = 420

No of defective found = 21

n = 30

∴Average fraction defective P — = 21

420 = 0.05

n

+ where Q = − 1 P

3 0.05 3

30

P

n

UCLP = 0.05 + 3 × 0.0398 UCLP = 0.1694

3 P P

P

n

−

−

= 0.05 – 0.1194 < 0 (negative)

Example 7 The past record of a factory using quality control melthods show that on the average

4 articles produced are defective out of a batch of 100 What is the maximum number of defective articles likely to be encountered in the batch of 100, when the production process is in a state of control? Sol n = Sample size = 400

P = Process fraction defective = 4

100 = 0.04

Q = 1 – P = 0.96

Let d be the number of defectives in a sample size of n i.e., np The 3–σ limit for number

of defectives are given by

( ) 3 ( )

E d ± s E d

400 0.04 3 400 0.04 0.96

= × ± × ×

16 3 15.36 16 3 3.9192

= ± = ± ×

16 11.7576

= ± =(4.2424, 27.7576)

Therefore if the production process is in a statistical control, the number of defective items

to be encountered in a batch of 400 should lie within the control limits, viz (4.2424, 27.7576), i.e., (4, 28) Hence the maximum number of defective items in this batch is 28.

Example 8 In a blade manufacturing factory, 1000 blades are examined daily Following information shows number of defective blades obtained there Draw the np-chart and give your comment?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

9 10 12 8 7 15 10 12 10 8 7 13 14 15 16

Date

No of

Defective

Sol Here n=10000, k=15 (sample no.)

If P denotes the fraction defectives produced by the entire process then

15 1000

P P kn

∑

×

∴ np=1000 0.011 11× =

Hence control limits are

( )

11 3 11 1 0.011 UCL 20.894

11 3 11 1 0.011 LCL 1.106

np

= =

= = − −

= + − −

=

= − − −

= − − −

= Since all the 15 points lies within the control limits, the process is under control

20 15 10 5 0

Date

FIG 11.23

Example 9 The number of mistakes made by an accounts clerk is given below:

Establish a suitable control chart and state how it should be used in future in order to control the mistakes of the clerk.

Sol The control chart to be used for the given problem is the number of defects chart i.e., C-chart.

Average no of mistakes

c— = 24 1.2

20 20

C

Thus the control limits for c—-chart are;

(i) UCL = c+ 3 c = 1.2 3 1.2 + = 4.49

(ii) CL = c— = 1.2

(iii) LCL = c− 3 c = 1.2 3 1.2 − = 2.09 ≈ 0

3 The number of mistakes during the 16th week lies outside the UCL the process is not

under control

Now to establish the suitable control chart for future, we homogenize the data for future control by eliminating the data corresponding to the 16th week

17 0.895

19

new

C = = Hence the revised control limits for c chart are:

UCL 3 0.895 3 0.895 3.73 LCL 3 0.895 3 0.895 1.94 0

17

19

C

= + = + =

= − = − = − ≅

= = =

So the revised C-chart for revised control limit is in statistical control, i.e., all the points

lies within the control limits

8

6

4

2

0

Week

FIG 11.24

Example 10 During the examination of equal length of cloth, the following are the number of defects observed.

Draw a control chart for the number of defects and comment whether the process is under control

or not?

Sol Let the no of defects per unit (equal length) be denoted by c.

The average no of defects in 10 samples

36 3.6

20 10

c

c=∑ = = Hence 3–σ limit for c-chart are:

c±3 c =3.6 3 3.6± =3.6 3 1.8974± × =3.6 5.6922± UCLC– = 3.6 + 5.6922 = 9.2922 LCLC– = 3.6 – 5.6922 = – 2.0922 ≈ 0

CLC– = 3.6 (LCLC– = 0 because no of defects per unit cannot be negative)

8 6 4 2 0

Sample Number

FIG 11.25

Since all the points are within the control limits therefore the process is in statistical control

Example 11 An automobile producer wishes to control the number of defects per automobile The data for 16 such automobiles is shown below:

1 Set up the control lmits for c-charts.

2 Do these data come from a controlled process ? If not, calculated the revised control charts limits.

Sol Here k = 16

Average no of defects in 16 units

16

k

= ∑ = =

Thus, the control limits for c-chart are:

UCL = c+3 c = 2.625+3 2.625 = 2.625+4.861 = 7.486

CL = c = 2.625 LCL = c−3 c = 2.625 3 2.625− = 2.625−4.861 = −2.236 ≈ 0

10 8 6 4 2 0

Sample Number

FIG 11.26