Institute of Technology of Cambodia

Quality Control

1. Set up X-Bar and R charts on this process Sample Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 34.5 34.2 31.6 31.5 35.0 34.1 32.6 33.8 34.8 33.6 31.9 38.6 35.4 34.0 37.1 34.9 33.5 31.7 34.0 35.1 33.7 32.8 33.5 34.2 34.0042 R 3 4 4 4 5 6 4 3 7 8 3 9 8 6 5 7 4 3 8 4 2 1 3 2 4.7083

Control Limits for X-Bar Chart n=5 A2 = 0.5770 UCL = + A2 = 34.0042 + (0.5770*4.7083) = 36.7209 Center Line = = 34.0042 LCL = - A2 = 34.0042 - (0.5770*4.7083) = 31.2875

Lecturer: Mr. HANG Samnang

1

Institute of Technology of Cambodia

Quality Control

45 40 35 30 25 20 15 10 5 0 0 5 10 15 20 25 30 X-Bar UCL LCL Center Line

We note that two points, sample 12 and sample 15, plot above the upper control limit, so the process is not in control. These two points must be investigated to see whether an assignable cause can be determined. By eliminating the sample 12 and 15, the new center line and revised control limits are calculated as: Revised UCL = + A2 = 33.6546 + (0.5770*4.5000) = 36.2543 Revised Center Line = = 33.6546 Revised LCL = - A2 = 33.6546 - (0.5770*4.5000) = 31.0550

After revising, the sample will be in control limit in the next sample.

Lecturer: Mr. HANG Samnang

2

Institute of Technology of Cambodia

Quality Control

45 40 35 30 25 20 15 10 5 0 0 10 20 30 Revised X-chart Revised UCL Revised LCL Center Line Revised

Control Limits for R-Chart n=5 D3 = 0, D4 = 2.1140 UCL = D4 = 2.1140*4.7083 = Center Line = = 4.7083 LCL = D3 = 012 10 8 R-Bar 6 4 2 0 0 5 10 15 20 25 30 UCL Center Line LCL

It is in the statistical control.

Lecturer: Mr. HANG Samnang

3

Institute of Technology of Cambodia

Quality Control

2. a/ Set up X-Bar and S-control-charts on this process Sample X1 X2 Number 1 2.50 0.50 2 0.00 0.00 3 1.50 1.00 4 0.00 0.50 5 0.00 0.00 6 1.00 -0.50 7 1.00 -1.00 8 0.00 -1.50 9 -2.00 -1.50 10 -0.50 3.50 11 0.00 1.50 12 0.00 -2.00 13 -1.00 -0.50 14 0.50 1.00 15 1.00 0.00 X3 2.00 0.50 1.00 -2.00 0.00 0.00 -1.00 -0.50 1.50 0.00 0.00 -0.50 -0.50 -1.00 1.50 X4 -1.00 1.00 -1.00 0.00 -0.50 0.00 -1.00 1.50 1.50 -1.00 0.00 0.00 -1.00 -0.50 1.50 X5 1.00 1.50 0.00 -1.00 0.50 0.00 0.00 0.00 0.00 -1.50 2.00 -0.50 0.00 -2.00 1.00 X6 -1.00 1.00 -1.50 1.50 1.00 0.50 1.50 0.00 0.00 -1.50 -1.50 2.00 0.50 -1.00 -1.00 X7 0.50 -1.00 -1.00 -1.50 -0.50 -1.00 0.00 0.00 0.50 -1.00 0.50 1.50 0.50 -1.50 0.00 X8 1.50 1.00 -1.00 0.00 -0.50 1.00 1.00 -1.00 1.00 -1.00 -0.50 0.00 -1.50 0.00 1.00 X9 0.50 1.50 1.00 -2.00 0.00 -2.00 0.00 0.50 0.00 1.00 2.00 0.50 -1.00 1.50 -2.00 X10 -1.50 -1.00 -1.00 -1.50 0.00 1.00 0.00 -0.50 1.00 0.50 -1.00 -1.00 -1.00 1.50 -1.50 0.50 0.45 -0.10 -0.60 0.00 0.00 0.05 -0.15 0.20 -0.15 0.30 0.00 -0.55 -0.15 0.15 -0.0033 R 4.00 2.50 3.00 2.50 1.50 3.00 2.50 3.00 3.50 5.00 3.50 4.00 2.00 3.50 3.50 3.1333

R=max(X)-min(X) Control Limits for X-Bar Chart n = 10 A2 = 0.3080 UCL = + A2 = -0.0033+ (0.3080*(3.1333)) = 0.9618 Center Line = = -0.0033 LCL = - A2 = -0.0033 - (0.3080**(3.1333)) = -0.9684

It is in statistical control limit.

Lecturer: Mr. HANG Samnang

4

Institute of Technology of Cambodia1.50

Quality Control

1.00

0.50 X-Bar 0.00 0 -0.50 5 10 15 20 UCL LCL Center Line

-1.00

-1.50

Control Limits for S-Chart Sample Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 s 1.3333 0.9265 1.1255 1.1738 0.4714 0.9718 0.8960 0.8182 1.1832 1.5284 1.2065 1.1547 0.6852 1.2483 1.2704 1.0662

Lecturer: Mr. HANG Samnang

5

Institute of Technology of Cambodia

Quality Control

S=

n = 10

B3 = 0.2840, B4 = 1.7160

UCL = B4 = 1.7160*1.0662 = 1.8296 Center Line = = 1.0662 LCL = B3 = 0.2840*1.0662 = 0.3028

2.0000 1.8000 1.6000 1.4000 1.2000 1.0000 0.8000 0.6000 0.4000 0.2000 0.0000 0 5 10 15 20 S-Chart UCL LCL Center Line

It is in statistical control limit. b/ Set up an R-chart Control Limits for R-Chart n = 10 D3 = 0.2230, D4 = 1.7770 UCL = D4 = 1.7770*3.1333 = 5.5679 Center Line = = 3.1333 LCL = D3 = 0.2230*3.1333 = 0.6987

Lecturer: Mr. HANG Samnang

6

Institute of Technology of Cambodia

Quality Control

6.00 5.00 4.00 3.00 2.00 1.00 0.00 0 5 10 15 20

R-Chart ULC LCL Center Line

It is in statistical control limit. Compare S-Chart and R-Chart: S-Chart is more difficult to calculate than R-Chart. However, S-Chart is more accurated than RChart because ULC, LCL and Center Line of S-Chart has a good interval (ULC = 1.8296, Center Line = 1.0662, LCL = 0.3028) so we can get the high quality of the product in manufacturing. ULC, LCL and Center Line of R-Chart has not a good interval as S-Chart (ULC = 5.5679, Center Line = 3.1333, LCL = 0.6987) so we can get the high quality of the product in manufacturing.

Lecturer: Mr. HANG Samnang

7

Institute of Technology of Cambodia

Quality Control

3. Construction a fraction nonconforming control chart for these data: Sample Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Number of Nonconforming Assemblies 7 4 1 3 6 8 10 5 2 7 6 15 0 9 5 1 4 5 7 12 117 Fraction NonConforming 0.07 0.04 0.01 0.03 0.06 0.08 0.10 0.05 0.02 0.07 0.06 0.15 0.00 0.09 0.05 0.01 0.04 0.05 0.07 0.12 0.0585

UCL =

= 0.0585 +

= 0.1290

Center Line = = 0.0585 LCL =

= 0.0585 -

= -0.0119

Lecturer: Mr. HANG Samnang

8

Institute of Technology of Cambodia

Quality Control

0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0.00 -0.02 0 5 10 15 20 25 P-Chart UCL LCL Center Line

We note that two points, sample 12, plot above the upper control limit, so the process is not in control. These two points must be investigated to see whether an assignable cause can be determined. By eliminating the sample 12, the new center line and revised control limits are calculated as:

= 0.0537

UCL =

= 0.0537 +

= 0.1213

Center Line = = 0.0537 LCL =

= 0.0537 -

= -0.0139

Lecturer: Mr. HANG Samnang

9

Institute of Technology of Cambodia

Quality Control

0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0.00 -0.02 -0.04 0 5 10 15 20 25 P revise UCL Revise LCL Revise Center Line revise

After revising, the sample will be in control limit in the next sample.

4. a. Estimate the potential quality (Cp) =

, but =C4 = 0.9400 1.1170

n=5=

So:=

= 2.9842

Lecturer: Mr. HANG Samnang

10

Institute of Technology of Cambodia

Quality Control

b. Estimate the actual capability (Cpk) = = 100 = min (Cpu, Cpl) = min ( ) = min ( )

= min (1.4921, 4.4763) = 1.4921 So Cpk = 1.4921 5. Construction single-sampling plans We have = 0.05, = 0.10 a. PQL (P1) = 0.04, CQL (P2) = 0.21 Using the Poisson unity values: Determine the operating ratio R =

=

Table T5.1, we get n = 3 and np1 = 1.366 n=

=

Thus: C = 3, n = 34

Pa 0.95 0.75 0.50 0.25 0.10

np 1.366 2.535 3.672 5.109 6.681

n 34 34 34 34 34

p 0.0402 0.0746 0.1080 0.1503 0.1965

Lecturer: Mr. HANG Samnang

11

Institute of Technology of Cambodia

Quality Control

Sampling Plan C=3 , n=341.00 0.90 0.80 0.70 0.60 0.50 0.40 0.30 0.20 0.10 0.00 0.0000 0.0500 0.1000 0.1500 0.2000 0.2500

Sampling Plan C=3 , n=34

b. PQL (P1) = 0.03, CQL (P2) = 0.13 Using Binomial nomograph We have 1- = 0.95, = 0.10 From graph binomial, link 1- at the right scale to P1 and at the right scale to P2 We get n = 50 and c = 3

Pa 0.95 0.75 0.50 0.25 0.10

np 1.366 2.535 3.672 5.109 6.681

n 50 50 50 50 50

p 0.0273 0.0507 0.0734 0.1022 0.1336

Lecturer: Mr. HANG Samnang

12

Institute of Technology of Cambodia

Quality Control

Sampling n=50, c=31.00 0.90 0.80 0.70 0.60 0.50 0.40 0.30 0.20 0.10 0.00 0.0000

Sampling n=50, c=3

0.0500

0.1000

0.1500

c. PQL (P1) = 0.02, CQL (P2) = 0.06 Using Thorndyke Chart Operating ratio R =

=

3

According to the Thorndyke, we get C = 9 , np of (1-) = 5.2 and np of = 14 And n1 = n2 = = 260 = 233

So C=9, n = 260

Lecturer: Mr. HANG Samnang

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