Quality Management

“

It costs a lot to produce a bad product.

” Norman Augustine

Cost of quality 1. Prevention costs 2. Appraisal costs 3. Internal failure costs 4. External failure costs 5. Opportunity costs

History: how did we get here… • Deming and Juran outlined the principles of Quality Management.

• Tai-ichi Ohno applies them in Toyota Motors Corp.

• Japan has its National Quality Award (1951).

• U.S. and European firms begin to implement Quality Management programs (1980’s).

• U.S. establishes the Malcolm Baldridge National Quality Award (1987).

• Today, quality is an imperative for any business.

What is quality management all about?

Try to manage all aspects of the organization in order to excel in all dimensions that are important to “customers” Two aspects of quality: features: more features that meet customer needs = higher quality freedom from trouble: fewer defects = higher quality

What does Total Quality Management encompass?

TQM is a management philosophy: • continuous improvement • leadership development • partnership development Cultural Alignment Customer Technical Tools (Process Analysis, SPC, QFD)

Developing quality specifications Input Design Process Design quality Output Dimensions of quality Conformance quality

Continuous improvement philosophy 1. Kaizen: Japanese term for continuous improvement. A step-by-step improvement of business processes.

2. PDCA:

Plan

-

do

-

check

-

act

as defined by Deming.

Plan Do Act Check 3. Benchmarking : what do top performers do?

Tools used for continuous improvement 1. Process flowchart

Tools used for continuous improvement 2. Run Chart Performance Time

Tools used for continuous improvement 3. Control Charts Performance Metric Time

Tools used for continuous improvement 4. Cause and effect diagram (fishbone)

Machine Man Environment Method Material

Tools used for continuous improvement 5. Check sheet Item ------ ------ ------ A √ √ √ B √ √ C √ √ √ √ D √ √ E √ √ √ √ √ √ F G √ √ √

Tools used for continuous improvement 6. Histogram Frequency

Tools used for continuous improvement 60 50 40 30 20 10 7. Pareto Analysis A B C D E F 100% 75% 50% 25% 0%

Summary of Tools 1. Process flow chart 2. Run diagram 3. Control charts 4. Fishbone 5. Check sheet 6. Histogram 7. Pareto analysis

Case: shortening telephone waiting time… • A bank is employing a call answering service • The main goal in terms of quality is “zero waiting time” - customers get a bad impression - company vision to be friendly and easy access • The question is how to analyze the situation and improve quality

The current process

Custome r A Custome r B Operator Receiving Party

How can we reduce waiting time?

Fishbone diagram analysis

Absent receiving party Out of office Not giving receiving party’s coordinates Complaining Customer Working system of operators Absent Not at desk Lunchtime Too many phone calls Lengthy talk Does not know organization well Leaving a message Operator Absent Does not understand customer Makes custome r wait Takes too much time to explain

Reasons why customers have to wait (12-day analysis with check sheet) A B C D E F One operator (partner out of office) Receiving party not present No one present in the section receiving call Section and name of the party not given Inquiry about branch office locations Other reasons Daily average 14.3

6.1

5.1

1.6

1.3

0.8

29.2

Total number 172 73 61 19 16 10 351

Pareto Analysis: reasons why customers have to wait Frequency 300 250 200 150 100 Percentage 87.1% 71.2% 49% 0% A B C D E F

Ideas for improvement 1. Taking lunches on three different shifts 2. Ask all employees to leave messages when leaving desks 3. Compiling a directory where next to personnel’s name appears her/his title

Results of implementing the recommendations 300 200 100 Frequency Before… Percentage 100% 87.1% 71.2% 49% 300 200 Frequency 100 …After Percentage

Improvement

0% 100% 0% A B C D E F B C A D E F

In general, how can we monitor quality…?

By observing variation in output measures!

1. Assignable variation: we can assess the cause 2. Common variation: variation that may not be possible to correct (

random variation

,

random noise

)

Statistical Process Control (SPC) Every output measure has a target value and a level of “acceptable” variation (upper and lower tolerance limits) SPC uses samples from output measures to estimate the mean and the variation (standard deviation) Example We want beer bottles to be filled with

12 FL OZ ± 0.05 FL OZ

Question: How do we define the output measures?

In order to measure variation we need… The average (mean) of the observations:

X

 1

N i N

  1

x i

The standard deviation of the observations:  

i N

  1 (

x i N



X

) 2

What is the key assumption behind SPC?

LESS VARIABILITY implies BETTER PERFORMANCE ! High Cost Low Lower spec Target Upper spec Performance Measure

Capability Index (C pk ) It shows how well the performance measure fits the design specification based on a given tolerance level A process is

k

 capable if

X



k

 

UTL

and 1 

UTL



k



X

and

X



k

 

LTL X



LTL k

  1

Capability Index (C pk ) Another way of writing this is to calculate the capability index:

C pk

 min

X



LTL k

 ,

UTL k

 

X

C pk < 1 means process is not capable at the

k

 level C pk >= 1 means process is capable at the

k

 level

Accuracy and Consistency the target T.

We say that a process is consistent if its standard deviation is low.

Example: Capability Index (C pk )

X

= 10 and

σ = 0.5

LTL

= 9

UTL

= 11

LTL X UTL C pk

 min 10 3   0 9 .

5 or 11 3   0 10 .

5  0 .

667

Example

Consider the capability of a process that puts pressurized grease in an aerosol can. The design specs call for an average of 60 pounds per square inch (psi) of pressure in each can with an upper tolerance limit of 65psi and a lower tolerance limit of 55psi. A sample is taken from production and it is found that the cans average 61psi with a standard deviation of 2psi. 1. Is the process capable at the 3  level?

2. What is the probability of producing a defect?

Solution

LTL = 55 UTL = 65  = 2

C pk C pk

 min(  min(

X

61  3   6

LTL

55 , ,

UTL

3   65  6 61 )

X

 ) min( 1 , 0 .

6667 )  0 .

6667 No, the process is not capable at the 3  level.

Example (contd)

Suppose another process has a sample mean of 60.5 and a standard deviation of 3. Which process is more accurate? This one.

Which process is more consistent? The other one.

Solution

P(defect) = P(X<55) + P(X>65) =P(X<55) + 1 – P(X<65) =P(Z<(55-61)/2) + 1 – P(Z<(65-61)/2) =P(Z<-3) + 1 – P(Z<2) =G(-3)+1-G(2) =0.00135 + 1 – 0.97725 (from standard normal table) = 0.0241

2.4% of the cans are defective.

Control Charts Upper Control Limit Central Line Lower Control Limit Control charts tell you when a process measure is exhibiting abnormal behavior.

Two Types of Control Charts • p Chart This is a plot of

proportions

over time (used for performance measures that are

yes/no attributes

) • X/R Chart This is a plot of

averages

and

ranges

over time (used for performance measures that are

variables

)

Statistical Process Control with

p

Charts UCL = 0.117

p

= 0.066

LCL = 0.015

Statistical Process Control with

p

Charts When should we use

p

charts?

1. When decisions are simple “yes” or “no” by inspection 2. When the sample sizes are large enough (>50) Sample (day) 1 2 3 4 5 6 7 Items 200 200 200 200 200 200 200 Defective 10 8 9 13 15 25 16 Percentage 0.050

0.040

0.045

0.065

0.075

0.125

0.080

Statistical Process Control with

p

Charts Let’s assume that we take

t

samples of size

n

…

p

 ( total number of number of samples) " defects"  (sample size)

s p



p

( 1 

p

)

n UCL



LCL



p



zs p p



zs p

Statistical Process Control with

p

Charts

p

 6  80 200  1 15  0 .

066

s p

 0 .

066 ( 1  0 .

066 )  0 .

017 200

UCL

 0 .

066  3  0 .

017  0 .

117

LCL

 0 .

066  3  0 .

017  0 .

015

Statistical Process Control with

p

Charts UCL = 0.117

p

= 0.066

LCL = 0.015

Statistical Process Control with

X/R

Charts When should we use

X/R

charts?

1. It is not possible to label “good” or “bad” 2. If we have relatively smaller sample sizes (<20)

Statistical Process Control with

X/R

Charts Take

t

samples of size

n

(sample size should be 5 or more)

X

 1

n i n

  1

x i X

is the mean for each sample

R

 max {

x i

}  min {

x i

}

R

is the range between the highest and the lowest for each sample

Statistical Process Control with

X/R

Charts

X



t

1

j t

  1

X j X

is the average of the averages.

R



t

1

j t

  1

R j R

is the average of the ranges

Statistical Process Control with

X/R

Charts define the upper and lower control limits…

UCL X



X



A

2

R LCL X



X



A

2

R UCL R LCL R



D

4

R



D

3

R

Read A 2 , D 3 , D 4 Table TN 8.7

from

Example: SPC for bottle filling…

Sample

1 2 3 4 5 6 7 8 9 10 11.90

12.03

11.92

11.96

11.95

11.99

12.00

12.02

12.01

11.92

Observation (x

i

)

11.92

12.09

11.91

12.03

11.92

11.97

12.02

12.06

12.10

11.98

11.93

12.00

12.03

11.94

12.01

11.91

12.07

12.06

12.04

12.06

12.06

12.05

11.92

11.94

11.94

11.92

12.00

12.07

11.91

12.09

12.01

12.07

12.07

11.98

12.00

12.06

12.07

12.00

11.94

12.07

Average Range (R)

Example: SPC for bottle filling… Calculate the average and the range for each sample…

Sample

1 2 3 4 5 6 7 8 9 10 11.90

12.03

11.92

11.96

11.95

11.99

12.00

12.02

12.01

11.92

Observation (x

i

)

11.92

12.09

11.91

12.03

11.92

11.97

12.02

11.93

12.01

12.06

12.00

11.91

12.10

11.98

12.04

12.06

12.03

11.94

11.92

11.94

12.07

12.06

12.00

12.07

12.06

12.05

11.94

11.92

11.91

12.09

12.01

12.07

12.07

11.98

12.00

12.06

12.07

12.00

11.94

12.07

Average

11.97

12.00

11.99

11.98

12.03

12.01

12.01

12.02

11.97

12.01

Range (R)

0.19

0.15

0.15

0.15

0.15

0.12

0.15

0.13

0.15

0.17

Then…

X

 12 .

00 is the

average

of the

averages R

 0 .

15 is the

average

of the

ranges

Finally… Calculate the upper and lower control limits

UCL X

 12 .

00  0 .

58  0 .

15  12 .

09

LCL X

 12 .

00  0 .

58  0 .

15  11 .

91

UCL R

 2 .

11  0 .

15

LCL R

 0  0 .

15  0  1 .

22

The X Chart UCL = 12.10

X = 12.00

LCL = 11.90

The R Chart UCL = 0.32

R = 0.15

LCL = 0.00

The X/R Chart What can you conclude?

The process is in control UCL X LCL UCL R LCL

Sample n

Example

Defects Sample n Defects 1 2 3 4 5 15 15 15 15 15 3 1 0 0 0 6 7 8 9 10 15 15 15 15 15 2 0 3 1 0 a. Develop a p chart for 95 percent confidence (z = 1.96) b. Based on the plotted data points, what comments can you make?

Solution

Ten defectives were found in 10 samples of size 15.

P

 10 10 ( 15 )  0 .

067

S p



P

( 1 

P

) 

n

.

067 ( 1  .

067 )  0 .

0645 15 UCL = .067 + 1.96(.0645) = .194

LCL = .067 - 1.96(.0645) = -.060 zero Defect proportion on Days 1 and 8 is 0.2, so process out of control.