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Quality and Performance

Chapter 3

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Quality and Performance

Quality

A term used by customers to describe their general satisfaction with a service or product

Defect

Any instance when a process fails to satisfy its customer

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Costs of Quality

Prevention costs

Costs associated with preventing defects before they happen

Appraisal costs

Costs incurred when the firm assesses the performance level of its processes

Internal Failure costs

Costs resulting from defects that are discovered during the production of a service or product

External Failure costs

Costs that arise when a defect is discovered after the customer receives the service or product

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Total Quality Management and Six Sigma

Total Quality Management

A philosophy that stresses three principles (customer satisfaction, employee involvement and continuous improvement) for achieving high level of process performance and quality

Six Sigma

A comprehensive and flexible system for achieving, sustaining, and maximizing business success by minimizing defects and variability in processes

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Total Quality Management

Figure 3.1

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Total Quality Management

Customer Satisfaction

Conformance to Specifications

Value

Fitness for Use

Support

Psychological Impressions

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Total Quality Management

Employee Involvement

Cultural Change

Quality at the Source

Teams

Employee Empowerment

Problem-solving teams

Special-purpose teams

Self-managed teams

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Total Quality Management

Continuous Improvement

Kaizen

Problem-solving tools

Plan-Do-Study-Act Cycle

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Six Sigma

Process average OK; too much variation

Process variability OK; process off target

Process on target with low variability

Reduce spread

Center process

Figure 3.3

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Six Sigma

Goal of achieving low rates of defective output by developing processes whose mean output for a performance measure is +/- six standard deviations (sigma) from the limits of the design specifications for the service or product.

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Acceptance Sampling

The application of statistical techniques to determine if a quantity of material from a supplier should be accepted or rejected based on the inspection or test of one or more samples.

Acceptable Quality Level

The quality level desired by the consumer.

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Acceptance Sampling

Firm A uses TQM or Six Sigma to achieve internal process performance

Supplier uses TQM or Six Sigma to achieve internal process performance

Yes

Fan motors

Fan blades

Accept blades?

Supplier

Manufactures fan blades

TARGET: Firm A’s specs

Accept motors?

Motor sampling

Blade sampling

Firm A

Manufacturers furnace fan motors

TARGET: Buyer’s specs

Buyer

Manufactures furnaces

Figure 3.4

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Statistical Process Control (SPC)

SPC

The application of statistical techniques to determine whether a process is delivering what the customer wants.

Variation of Outputs

No two services of products are exactly alike because the processes used to produce them contain many sources of variation, even if the processes are working as intended.

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Statistical Process Control (SPC)

Performance Measurements

Variables - Service or product characteristics that can be measured

Attributes - Service or product characteristics that can be quickly counted for acceptable performance

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Statistical Process Control (SPC)

Complete Inspection

Inspect each service or product at each stage of the process for quality

Sampling

Sample Size

Time between successive samples

Decision rules that determine when action should be taken

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Statistical Process Control (SPC)

Figure 3.5

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Statistical Process Control (SPC)

The sample mean is the sum of the observations divided by the total number of observations.

where

xi = observation of a quality characteristic (such as time)

n = total number of observations

= mean

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Statistical Process Control (SPC)

The range is the difference between the largest observation in a sample and the smallest. The standard deviation is the square root of the variance of a distribution.

An estimate of the process standard deviation based on a sample is given by:

where

σ = standard deviation of a sample

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Statistical Process Control (SPC)

Categories of Variation

Common cause - The purely random, unidentifiable sources of variation that are unavoidable with the current process

Assignable cause - Any variation-causing factors that can be identified and eliminated

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Statistical Process Control (SPC)

Control Chart

Time-ordered diagram that is used to determine whether observed variations are abnormal

Controls chart have a nominal value or center line, Upper Control Limit (UCL), and Lower Control Limit (LCL)

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Statistical Process Control (SPC)

Steps for using a control chart

Take a random sample from the process and calculate a variable or attribute performance measure.

If a statistic falls outside the chart’s control limits or exhibits unusual behavior, look for an assignable cause.

Eliminate the cause if it degrades performance; incorporate the cause if it improves performance. Reconstruct the control chart with new data.

Repeat the procedure periodically.

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Control Charts

Samples

Assignable

causes likely

UCL

Nominal

LCL

Figure 3.7

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Nominal

UCL

LCL

Variations

Sample number

Control Charts

(a) Normal – No action

Figure 3.8

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Nominal

UCL

LCL

Variations

Sample number

Control Charts

(b) Run – Take action

Figure 3.8

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Nominal

UCL

LCL

Variations

Sample number

Control Charts

Figure 3.8

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Nominal

UCL

LCL

Variations

Sample number

Control Charts

(d) Exceeds control limits – Take action

Figure 3.8

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Control Charts

Type I error

An error that occurs when the employee concludes that the process is out of control based on a sample result that fails outside the control limits, when it fact it was due to pure randomness

Type II error

An error that occurs when the employee concludes that the process is in control and only randomness is present, when actually the process is out of statistical control

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Control Charts

Variable Control Charts

R-Chart – Measures the variability of the process

-Chart – Measures whether the process is generating output, on average, consistent with a target value

Attribute Control Charts

p-chart – Measures the proportion of defective services or products generated by the process

c-chart – Measures the number of defects when more than one defect can be present in a service or product

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Control Charts for Variables

R-Chart

UCLR = D4R and LCLR = D3R

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Control Charts for Variables

UCL = + A2 and LCL = - A2

where

= central line of the chart, which can be either the average of past sample means or a target value set for the process

A2 = constant to provide three-sigma limits for the sample mean

x-Chart

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Control Charts for Variables

Table 3.1

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Control Charts for Variables

Collect data.

Compute the range.

Use Table 3.1 to determine R-chart control limits.

Plot the sample ranges. If all are in control, proceed to step 5. Otherwise, find the assignable causes, correct them, and return to step 1.

Calculate for each sample and determine the central line of the chart,

Steps to Compute Control Charts:

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Control Charts for Variables

Use Table 3.1 to determine control limits

Plot the sample means. If all are in control, the process is in statistical control. Continue to take samples and monitor the process. If any are out of control, find the assignable causes, correct them, and return to step 1.

Steps to Compute Control Charts:

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Example 3.1

The management of West Allis Industries is concerned about the production of a special metal screw used by several of the company’s largest customers. The diameter of the screw is critical to the customers. Data from five samples appear in the accompanying table. The sample size is 4. Is the process in statistical control?

Sample Number	1	2	3	4	R
1	.5014	.5022	.5009	.5027	.0018	.5018
2	.5021	.5041	.5024	.5020	.0021	.5027
3	.5018	.5026	.5035	.5023	.0017	.5026
4	.5008	.5034	.5024	.5015	.0026	.5020
5	.5041	.5056	.5034	.5047	.0022	.5045

OBSERVATIONS

Average

.0021 .5027

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Example 3.1

Compute the range for each sample and the control limits

UCLR = D4 R =

2.282(0.0021) = 0.00479 in.

0(0.0021) = 0 in.

LCLR = D3 R =

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Example 3.1

Process variability is in statistical control.

Figure 3.9

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Example 3.1

Compute the mean for each sample and the control limits.

= 0.5027 + 0.729(0.0021) = 0.5042 in.

= 0.5027 – 0.729(0.0021) = 0.5012 in.

LCLx = X – A2 R

UCLx = X + A2 R

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Example 3.1

Process average is NOT in statistical control.

Figure 3.10

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Control Charts for Variables

where

σx = σ/ n

σ = standard deviation of the process distribution

n = sample size

x = central line of the chart

z = normal deviate number

If the standard deviation of the process distribution is known, another form of the x -chart may be used:

UCLx = x + zσx and LCLx = x – zσx

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Example 3.2

For Sunny Dale Bank the time required to serve customers at the drive-by window is an important quality factor in competing with other banks in the city.

Mean time to process a customer at the peak demand period is 5 minutes

Standard deviation of 1.5 minutes

Sample size of six customers

Design an -chart that has a type I error of 5 percent

After several weeks of sampling, two successive samples came in at 3.70 and 3.68 minutes, respectively. Is the customer service process in statistical control?

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Example 3.2

x = 5 minutes

σ = 1.5 minutes

n = 6 customers

z = 1.96

The new process is an improvement.

LCLx = x – zσ/n =

5.0 – 1.96(1.5)/6 = 3.80 minutes

The process variability is in statistical control, so we proceed directly to the -chart. The control limits are

UCLx = x + zσ/n =

5.0 + 1.96(1.5)/6 = 6.20 minutes

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Application 3.1

Webster Chemical Company produces mastics and caulking for the construction industry. The product is blended in large mixers and then pumped into tubes and capped.

Webster is concerned whether the filling process for tubes of caulking is in statistical control. The process should be centered on 8 ounces per tube. Several samples of eight tubes are taken and each tube is weighed in ounces.

	Tube Number
Sample	1	2	3	4	5	6	7	8	Avg	Range
1	7.98	8.34	8.02	7.94	8.44	7.68	7.81	8.11	8.040	0.76
2	8.23	8.12	7.98	8.41	8.31	8.18	7.99	8.06	8.160	0.43
3	7.89	7.77	7.91	8.04	8.00	7.89	7.93	8.09	7.940	0.32
4	8.24	8.18	7.83	8.05	7.90	8.16	7.97	8.07	8.050	0.41
5	7.87	8.13	7.92	7.99	8.10	7.81	8.14	7.88	7.980	0.33
6	8.13	8.14	8.11	8.13	8.14	8.12	8.13	8.14	8.130	0.03
								Avgs	8.050	0.38

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Application 3.1

Assuming that taking only 6 samples is sufficient, is the process in statistical control?

UCLR = D4 R =

LCLR = D3 R =

1.864(0.38) = 0.708

0.136(0.38) = 0.052

The range chart is out of control since sample 1 falls outside the UCL and sample 6 falls outside the LCL.

Conclusion on process variability given = 0.38 and n = 8:

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Application 3.1

Consider dropping sample 6 because of an inoperative scale, causing inaccurate measures.

	Tube Number
Sample	1	2	3	4	5	6	7	8	Avg	Range
1	7.98	8.34	8.02	7.94	8.44	7.68	7.81	8.11	8.040	0.76
2	8.23	8.12	7.98	8.41	8.31	8.18	7.99	8.06	8.160	0.43
3	7.89	7.77	7.91	8.04	8.00	7.89	7.93	8.09	7.940	0.32
4	8.24	8.18	7.83	8.05	7.90	8.16	7.97	8.07	8.050	0.41
5	7.87	8.13	7.92	7.99	8.10	7.81	8.14	7.88	7.980	0.33
								Avgs	8.034	0.45

What is the conclusion on process variability and process average?

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Application 3.1

The resulting control charts indicate that the process is actually in control.

Now = 0.45, = 8.034, and n = 8

UCLR = D4 R =

1.864(0.45) = 0.839

LCLR = D3 R =

0.136(0.45) = 0.061

UCL x = x + A2 R =

8.034 + 0.373(0.45) = 8.202

8.034 – 0.373(0.45) = 7.866

LCL x = x – A2 R =

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Control Charts for Attributes

p-charts are used for controlling the proportion of defective services or products generated by the process.

The standard deviation is

p = the center line on the chart

UCLp = + zσp and LCLp = – zσp

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Example 3.3

Hometown Bank is concerned about the number of wrong customer account numbers recorded. Each week a random sample of 2,500 deposits is taken and the number of incorrect account numbers is recorded

Using three-sigma control limits, which will provide a Type I error of 0.26 percent, is the booking process out of statistical control?

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Example 3.3

Sample Number	Wrong Account Numbers	Sample Number	Wrong Account Numbers
1	15	7		24
2	12	8		7
3	19	9		10
4	2	10		17
5	19	11		15
6	4	12		3
		Total		147

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Example 3.3

147

12(2,500)

= = 0.0049

p =

Total defectives

Total number of observations

σp = (1 – )/n

= 0.0049(1 – 0.0049)/2,500 = 0.0014

Calculate the sample proportion defective and plot each sample proportion defective on the chart.

UCLp = p + zσp

= 0.0049 + 3(0.0014) = 0.0091

= 0.0049 – 3(0.0014) = 0.0007

LCLp = p – zσp

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Example 3.3

Fraction Defective

Sample

Mean

UCL

LCL

.0091

.0049

.0007

| | | | | | | | | | | |

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

The process is NOT in statistical control.

Figure 3.11

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Application 3.2

A sticky scale brings Webster’s attention to whether caulking tubes are being properly capped. If a significant proportion of the tubes aren’t being sealed, Webster is placing their customers in a messy situation. Tubes are packaged in large boxes of 144. Several boxes are inspected and the following numbers of leaking tubes are found:

Sample	Tubes	Sample	Tubes	Sample	Tubes
1	3	8	6	15	5
2	5	9	4	16	0
3	3	10	9	17	2
4	4	11	2	18	6
5	2	12	6	19	2
6	4	13	5	20	1
7	2	14	1	Total =	72

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Application 3.2

Calculate the p-chart three-sigma control limits to assess whether the capping process is in statistical control.

The process is in control as the p values for the samples all fall within the control limits.

20(144)

= = 0.025

p =

Total number of leaky tubes

Total number of tubes

p (1 – p )

0.025(1 – 0.025)

144

= 0.01301

σp =

UCLp = p + zσp

= 0.025 + 3(0.01301)= 0.06403

LCLp = p – zσp

= 0.025 – 3(0.01301)= –0.01403 = 0

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Control Charts for Attributes

c-charts – A chart used for controlling the number of defects when more than one defect can be present in a service or product.

The mean of the distribution is and the standard deviation is

UCLc = c + z  c

LCLc = c – z  c

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Example 3.4

The Woodland Paper Company produces paper for the newspaper industry. As a final step in the process, the paper passes through a machine that measures various product quality characteristics. When the paper production process is in control, it averages 20 defects per roll.

a. Set up a control chart for the number of defects per roll. For this example, use two-sigma control limits.

b. Five rolls had the following number of defects: 16, 21, 17, 22, and 24, respectively. The sixth roll, using pulp from a different supplier, had 5 defects. Is the paper production process in control?

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Example 3.4

a. The average number of defects per roll is 20. Therefore

= 20 + 2(20) = 28.94

= 20 – 2(20) = 11.06

UCLc = + z

LCLc = - z

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Example 3.4

The process is technically out of control due to Sample 6. However, Sample 6 shows that the new supplier is a good one.

Figure 3.12

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Application 3.3

At Webster Chemical, lumps in the caulking compound could cause difficulties in dispensing a smooth bead from the tube. Even when the process is in control, there will still be an average of 4 lumps per tube of caulk. Testing for the presence of lumps destroys the product, so Webster takes random samples. The following are results of the study:

Tube #	Lumps	Tube #	Lumps	Tube #	Lumps
1	6	5	6	9	5
2	5	6	4	10	0
3	0	7	1	11	9
4	4	8	6	12	2

Determine the c-chart two-sigma upper and lower control limits for this process.

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Application Problem 3.3

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Process Capability

Process Capability – The ability of the process to meet the design specification for a service or product

Nominal Value

A target for design specifications

Tolerance

An allowance above or below the nominal value

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Process Capability

Minutes

Upper

specification

Lower

specification

Nominal

value

(a) Process is capable

Process distribution

Figure 3.13

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Process Capability

Minutes

Upper

specification

Lower

specification

Nominal

value

(b) Process is not capable

Process distribution

Figure 3.13

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Process Capability

Lower

specification

Mean

Upper

specification

Nominal value

Six sigma

Four sigma

Two sigma

Figure 3.14

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Process Capability

Process Capability Index (Cpk)

An index that measures the potential for a process to generate defective outputs relative to either upper or lower specifications.

where

σ = standard deviation of the process distribution

Cpk = Minimum of ,

– Lower specification

3σ

Upper specification –

3σ

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Process Capability

Process Capability (Cp)

The tolerance width divided by six standard deviations.

Cp =

Upper specification – Lower specification

6σ

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Example 3.5

The intensive care unit lab process has an average turnaround time of 26.2 minutes and a standard deviation of 1.35 minutes.

The nominal value for this service is 25 minutes + 5 minutes.

Is the lab process capable of four sigma-level performance?

Upper specification =

30 minutes

Lower specification

20 minutes

Average service

26.2 minutes

 =

1.35 minutes

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Example 3.5

Cpk = Minimum of 26.2 – 20 , 30 – 26.2

3 ( 1.53) 3( 1.53)

Cpk = Minimum of ,

– Lower specification

3σ

Upper specification –

3σ

Process does not meets 4-sigma level of 1.33

Cpk = Minimum of 1.53, 0.94

Cpk = 0.94

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Process did not meet 4-sigma level of 1.33

30 – 20

6 (1.35)

Example 3.5

Cp =

Upper specification – Lower specification

6σ

= 1.23

Cp =

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Example 3.5

Cp = Upper – Lower

6

Cp = 30 – 20 = 1.39

6 (1.20)

Process meets 4-sigma level of 1.33 for variability

New Data is collected:

Upper specification =

30 minutes

Lower specification

20 minutes

Average service

26.1 minutes

 = 1.20 minutes

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Example 3.5

Cpk = Minimum 26.1 – 20 , 30 – 26.1

3 ( 1.20) 3 ( 1.20)

Cpk = 1.08

Process does not meets 4-sigma level of 1.33

Cpk = Minimum ,

– Lower specification

3σ

Upper specification –

3σ

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Application 3.4

Webster Chemical’s nominal weight for filling tubes of caulk is 8.00 ounces ± 0.60 ounces. The target process capability ratio is 1.33, signifying that management wants 4-sigma performance. The current distribution of the filling process is centered on 8.054 ounces with a standard deviation of 0.192 ounces. Compute the process capability index and process capability ratio to assess whether the filling process is capable and set properly.

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Application 3.4

The value of 0.948 is far below the target of 1.33. Therefore, we can conclude that the process is not capable.

= Minimum = 1.135 = 0.948

8.054 – 7.400

3(0.192)

8.600 – 8.054

3(0.192)

a. Process capability index:

Cpk = Minimum ,

– Lower specification

3σ

Upper specification –

3σ

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Application 3.4

b. Process capability ratio:

Cp =

Upper specification – Lower specification

6σ

= = 1.0417

8.60 – 7.40

6(0.192)

The value of Cp is less than the target for four-sigma quality.

Therefore we conclude that the process variability must be addressed first, and then the process should be retested.

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International Quality Documentation Standards

ISO 9001:2008 – Documentation Standards

ISO 14000:2004 – Environmental Management System

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Benefits of Baldrige Performance Excellence Program

Baldrige Performance Excellence Program

Application process is rigorous and helps organizations define what quality means to them

Investment in quality principles and performance excellence pays off in increased productivity

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Benefits of Baldrige Performance Excellence Program

Seven Major Criteria

Leadership

Strategic Planning

Customer Focus

Measurement, Analysis, and Knowledge Management

Workforce Focus

Operations Focus

Results

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Solved Problem 1

The Watson Electric Company produces incandescent light bulbs. The following data on the number of lumens for 40-watt light bulbs were collected when the process was in control.

	Observation
Sample	1	2	3	4
1	604	612	588	600
2	597	601	607	603
3	581	570	585	592
4	620	605	595	588
5	590	614	608	604

a. Calculate control limits for an R-chart and an -chart.

b. Since these data were collected, some new employees were hired. A new sample obtained the following readings: 625, 592, 612, and 635. Is the process still in control?

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Solved Problem 1

Sample		R
1	601	24
2	602	10
3	582	22
4	602	32
5	604	24
Total	2,991	112
Average

x = 598.2

R = 22.4

604 + 612 + 588 + 600

= 601

x =

R =

612 – 588 = 24

a. To calculate , compute the mean for each sample. To calculate R, subtract the lowest value in the sample from the highest value in the sample. For example, for sample 1,

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Solved Problem 1

The R-chart control limits are

2.282(22.4) = 51.12

UCLR = D4 =

LCLR = D3=

The x-chart control limits are

0(22.4) = 0

UCLx = x + A2 R=

598.2 + 0.729(22.4) = 614.53

598.2 – 0.729(22.4) = 581.87

LCLx = x – A2R=

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Solved Problem 1

b. First check to see whether the variability is still in control based on the new data. The range is 43 (or 635 – 592), which is inside the UCL and LCL for the R-chart. Since the process variability is in control, we test for the process average using the current estimate for . The average is 616 which is above the UCL for the chart. Since the process average is out of control, a service for assignable causes inducing excessive average lumens must be conducted.

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Solved Problem 2

The data processing department of the Arizona Bank has five data entry clerks. Each working day their supervisor verifies the accuracy of a random sample of 250 records. A record containing one or more errors is considered defective and must be redone. The results of the last 30 samples are shown in the table. All were checked to make sure that none was out of control.

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Solved Problem 2

Sample	Number of Defective Records	Sample	Number of Defective Records
1	7	16		8
2	5	17		12
3	19	18		4
4	10	19		6
5	11	20		11
6	8	21		17
7	12	22		12
8	9	23		6
9	6	24		7
10	13	25		13
11	18	26		10
12	5	27		14
13	16	28		6
14	4	29		11
15	11	30		9
				Total 300

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Solved Problem 2

a. Based on these historical data, set up a p-chart using z = 3.

b. Samples for the next four days showed the following:

Sample	Number of Defective Records
Tues	17
Wed	15
Thurs	22
Fri	21

What is the supervisor’s assessment of the data-entry process likely to be?

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Solved Problem 2

a. From the table, the supervisor knows that the total number of defective records is 300 out of a total sample of 7,500 [or 30(250)]. Therefore, the central line of the chart is

= 0.04

300

7,500

p =

The control limits are

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Solved Problem 2

Sample	Number of Defective Records	Proportion
Tues	17	0.068
Wed	15	0.060
Thurs	22	0.088
Fri	21	0.084

b. Samples for the next four days showed the following:

Samples for Thursday and Friday are out of control. The supervisor should look for the problem and, upon identifying it, take corrective action.

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Solved Problem 3

The Minnow County Highway Safety Department monitors accidents at the intersection of Routes 123 and 14. Accidents at the intersection have averaged three per month.

a. Which type of control chart should be used? Construct a control chart with three sigma control limits.

b. Last month, seven accidents occurred at the intersection. Is this sufficient evidence to justify a claim that something has changed at the intersection?

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Solved Problem 3

a. The safety department cannot determine the number of accidents that did not occur, so it has no way to compute a proportion defective at the intersection. Therefore, the administrators must use a c-chart for which

There cannot be a negative number of accidents, so the LCL in this case is adjusted to zero.

b. The number of accidents last month falls within the UCL and LCL of the chart. We conclude that no assignable causes are present and that the increase in accidents was due to chance.

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Solved Problem 4

Pioneer Chicken advertises “lite” chicken with 30 percent fewer calories. (The pieces are 33 percent smaller.) The process average distribution for “lite” chicken breasts is 420 calories, with a standard deviation of the population of 25 calories. Pioneer randomly takes samples of six chicken breasts to measure calorie content.

a. Design an -chart using the process standard deviation. Use three sigma limits.

b. The product design calls for the average chicken breast to contain 400 ± 100 calories. Calculate the process capability index (target = 1.33) and the process capability ratio. Interpret the results.

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Solved Problem 4

a. For the process standard deviation of 25 calories, the standard deviation of the sample mean is

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Solved Problem 4

Because the process capability ratio is 1.33, the process should be able to produce the product reliably within specifications. However, the process capability index is 1.07, so the current process is not centered properly for four-sigma performance. The mean of the process distribution is too close to the upper specification.

The process capability ratio is

b. The process capability index is

= Minimum of = 1.60, = 1.07

420 – 300

3(25)

500 – 420

3(25)

Cp =

Upper specification – Lower specification

6σ

= = 1.33

500 – 300

6(25)

Cpk = Minimum of ,

– Lower specification

3σ

Upper specification –

3σ

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This Week’s Discussions

SouthWest Airlines – what do you know about SW Airlines’ operations? How have they streamlined their operations?

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x = x i

i=1

∑ n

σ= x i −x( )

i=1

∑ n−1

or σ= x 2 −

x i i=1

∑ ⎛

⎝ ⎜⎜⎜

⎞

⎠ ⎟⎟⎟⎟

ni=1

∑ n−1

(

)

ppn

c =

UCLc =c +zσc =

(

)

LCLc =c−zσc =

(

)

(

)

UCL

(

)

LCL

0.077

250

0.04(0.96)

0.04

0.003

250

0.04(0.96)

0.04