tm 620: quality management session seven – 9 november 2010 control charts, part i –variables

TM 620: Quality Management

Session Seven – 9 November 2010

• Control Charts, Part I– Variables

Recall: What is Quality?

• Juran – Quality is fitness for use

=> we should be able to determine a set of measurable characteristics which define quality

Recall: What is Quality?

• Taguchi – Loss from quality is proportional to the amount of variability in the system– Why?

• => if we reduce variation, we reduce loss from quality

Quality Improvement

• The reduction of variability in processes and products

Equivalent definition:

• The reduction of waste

• Waste is any activity for which the customer will not pay

Recall: Cost of Quality

Cost of Failure

Cost of Control

Total Cost

Traditional View

Quality Level

Costs

Traditional Loss Function

x

LSL USL

T

TLSL USL

Example (Sony, 1979)

Comparing cost of two Sony television plants in Japan and San Diego. All units in San Diego fell within specifications. Japanese plant had units outside of specifications.

Loss per unit (Japan) = $0.44Loss per unit (San Diego) = $1.33

How can this be?

Sullivan, “Reducing Variability: A New Approach to Quality,” Quality Progress, 17, no.7, 15-21, 1984.

Example

x

T

U.S. Plant (2 = 8.33)

Japanese Plant (2 = 2.78)

LSL USL

Taguchi Loss Function

x

T

x

T

Taguchi Loss Function

T

L(x)

x

k(x - T)2

L(x) = k(x - T)2

Estimating Loss Function

Suppose we desire to make pistons with diameter D = 10 cm. Too big and they create too much friction. Too little and the engine will have lower gas mileage. Suppose tolerances are set at D = 10 + .05 cm. Studies show that if D > 10.05, the engine will likely fail during the warranty period. Average cost of a warranty repair is $400.

Estimating Loss Function

10

L(x)

10.05

400

400 = k(10.05 - 10.00)2

= k(.0025)

Estimating Loss Function

10

L(x)

10.05

400

400 = k(10.05 - 10.00)2

= k(.0025)

k = 160,000

Example 2

Suppose we have a 1 year warranty to a watch. Suppose also that the life of the watch is exponentially distributed with a mean of 1.5 years. The warranty costs to replace the watch if it fails within one year is $25. Estimate the loss function.

Example 2

1.5

L(x)

25

1

f(x)

25 = k(1 - 1.5)2

k = 100

Example 2

1.5

L(x)

25

1

f(x)

25 = k(1 - 1.5)2

k = 100

Single Sided Loss Functions

Smaller is better

L(x) = kx2

Larger is better

L(x) = k(1/x2)

Example 2

L(x)

25

1

f(x)

Example 2

L(x)

25

1

f(x)

25 = k(1)2

k = 25

Expected Loss

Expected Loss: Piston DiameterProbability Probability

Diameter Process A Process B9.925 0.000 0.0259.950 0.200 0.0759.975 0.200 0.20010.000 0.200 0.40010.025 0.200 0.20010.050 0.200 0.07510.075 0.000 0.025

Expected Loss: Piston DiameterProbability Probability

Diameter Loss Process A Process B9.925 900 0.000 0.0259.950 400 0.200 0.0759.975 100 0.200 0.20010.000 0 0.200 0.40010.025 100 0.200 0.20010.050 400 0.200 0.07510.075 900 0.000 0.025

Expected Loss

Expected Loss: Piston DiameterProbability Weighted Probability

Diameter Loss Process A Loss Process B9.925 900 0.000 0.0 0.0259.950 400 0.200 80.0 0.0759.975 100 0.200 20.0 0.20010.000 0 0.200 0.0 0.40010.025 100 0.200 20.0 0.20010.050 400 0.200 80.0 0.07510.075 900 0.000 0.0 0.025

Expected Loss

Expected Loss: Piston DiameterProbability Weighted Probability Weighted

Diameter Loss Process A Loss Process B Loss9.925 900 0.000 0.0 0.025 22.59.950 400 0.200 80.0 0.075 30.09.975 100 0.200 20.0 0.200 20.010.000 0 0.200 0.0 0.400 0.010.025 100 0.200 20.0 0.200 20.010.050 400 0.200 80.0 0.075 30.010.075 900 0.000 0.0 0.025 22.5

Expected Loss

Expected Loss

Expected Loss: Piston DiameterProbability Weighted Probability Weighted

Diameter Loss Process A Loss Process B Loss9.925 900 0.000 0.0 0.025 22.59.950 400 0.200 80.0 0.075 30.09.975 100 0.200 20.0 0.200 20.010.000 0 0.200 0.0 0.400 0.010.025 100 0.200 20.0 0.200 20.010.050 400 0.200 80.0 0.075 30.010.075 900 0.000 0.0 0.025 22.5

Exp. Loss = 200.0 145.0

Expected Loss

Recall, X f(x) with finite mean and variance 2.

E[L(x)] = E[ k(x - T)2 ]

= k E[ x2 - 2xT + T2 ]

= k E[ x2 - 2xT + T2 - 2x + 2 + 2x - 2 ]

= k E[ (x2 - 2x+ 2) - 2 + 2x - 2xT + T2 ]

= k{ E[ (x - )2 ] + E[ - 2 + 2x - 2xT + T2 ] }

Expected Loss

E[L(x)] = k{ E[ (x - )2 ] + E[ - 2 + 2x - 2xT + T2 ] }

Recall, Expectation is a linear operator and E[ (x - )2 ] = 2

E[L(x)] = k{2 - E[ 2 ] + E[ 2x - E[ 2xT ] + E[ T2 ] }

Expected Loss

Recall,

E[ax +b] = aE[x] + b = a + b

E[L(x)] = k{2 - 2 + 2 E[ x - 2T E[ x ] + T2 }

=k {2 - 2 + 22 - 2T + T2 }

Expected Loss

Recall, E[ax +b] = aE[x] + b = a + b

E[L(x)] = k{2 - 2 + 2 E[ x - 2T E[ x ] + T2 }

=k {2 - 2 + 22 - 2T + T2 }

=k {2 + ( - T)2 }

Expected Loss

Recall, E[ax +b] = aE[x] + b = a + b

E[L(x)] = k{2 - 2 + 2 E[ x - 2T E[ x ] + T2 }

=k {2 - 2 + 22 - 2T + T2 }

=k {2 + ( - T)2 }

= k { 2 + ( x - T)2 } = k (2 +D2 )

Since for our piston example, x = T,

D2 = (x - T)2 = 0

L(x) = k2

Example

Example (Piston Diam.)

Expected Loss: Piston DiameterProbability Weighted Probability Weighted

Diameter (x-)2

Process A (x-)2

Process B (x-)2

9.925 0.0056 0.000 0.0000 0.025 0.00019.950 0.0025 0.200 0.0005 0.075 0.00029.975 0.0006 0.200 0.0001 0.200 0.000110.000 0.0000 0.200 0.0000 0.400 0.000010.025 0.0006 0.200 0.0001 0.200 0.000110.050 0.0025 0.200 0.0005 0.075 0.000210.075 0.0056 0.000 0.0000 0.025 0.0001

Var = 0.0013 0.0009E[LA(x)] = .0013*k E[LB(x)] = .0009*k

= 200 = 145

Example (Sony)

x

T

U.S. Plant (2 = 8.33)

Japanese Plant (2 = 2.78)

LSL USL

E[LUS(x)] = 0.16 * 8.33 = $1.33

E[LJ(x)] = 0.16 * 2.78 = $0.44

Tolerance (Pistons)

10

L(x)

10.05

400

400 = k(10.05 - 10.00)2

= k(.0025)

k = 160,000

Recall,

Tolerance

Suppose repair for an engine which will fail during warranty can be made for only $200 10

L(x)

10.05

400

LSL USL

200

Tolerance

Suppose repair for an engine which will fail during warranty can be made for only $200

200 =160,000(tolerance)2 10

L(x)

10.05

400

LSL USL

200

Tolerance

Suppose repair for an engine which will fail during warranty can be made for only $200

200 = 160,000(tolerance)2

tolerance = (200/160,000)1/2

= .0354

10

L(x)

10.05

400

LSL USL

200

Statistical Thinking

• All work occurs in a system of interconnected processes

• All process have variation

• Understanding variation and reducing variation are important keys to success

Variability

• A certain amount of variability is inescapable

• Therefore, no two products are identical

• The larger the variability, the greater the probability that the customer will perceive its existence

Sources of Variability

Include:

• Differences in materials

• Differences in the performance and operation of the manufacturing equipment

• Differences in the way the operators perform their tasks

Variability and Statistics

• Variability is difference from the target• Characteristics of quality must be measurable

Therefore,• Variability is described in statistical terms• We will use statistical methods in our quality

improvement activities

Breaking down workplace barriers

• Train people in the basics of the tools and methods

• Deal with those who look down on or fear numerical methods

• Encourage a data-driven culture of decision making

• Lose the jargon• No “lying with statistics”

Recall: Types of Errors

• Type I error – Producers risk– Probability that a good product will be rejected

• Type II error– Consumers risk– Probability that a nonconforming product will be

available for sale

• Type III error– Asking the wrong question

Data on Quality Characteristics

• Attribute data– Discrete– Often a count of some type

• Variable data– Continuous– Often a measurement, such as length,

voltage, or viscosity

Terms

• Specifications

• Target (or Nominal) Value

• Upper Specification Limit

• Lower Specification Limit

• Random Variation

• Non-random Variation

• Process stability

Terms

• Nonconforming: failure to meet one or more of the specifications

• Nonconformity: a specific type of failure

• Defect: a nonconformity serious enough to significantly affect the safe or effective use of the produce or completion of the service

Nonconforming vs. Defective

• A nonconforming product is not necessarily unfit for use

• A nonconforming product is considered defective if if it has one or more defects

Classroom Exercise

• For a product or service in your job:– Name a quality characteristic– Give an example of a nonconformity that is

not a defect– Give an example of a defect

Types of Inspection

• Receiving

• In Process

• Final

• None

• One Hundred Percent

• Acceptance Sampling

Quality Design & Process Variation

60 80 100 120 140

60 140

14060

Lower Spec Limit

Upper Spec Limit

AcceptanceSampling

Statistical ProcessControl

ExperimentalDesign

Variation and Control

• A process that is operating with only common causes of variation is said to be in statistical control.

• A process operating in the presence of special or assignable cause is said to be out of control.

Finding Trends and Special Causes

• Inspection does not tell you about a problem until it becomes a problem

• We need a mechanism to help us spot special causes when they occur

• We need mechanism to help us determine when we have a trend in the data

Statistical Process Control

• Originally developed by Walter Shewhart in 1924 at the Bell Telephone Laboratories

• Late 1920s, Harold Dodge and Harry Romig developed statistically based acceptance sampling

• Not recognized by industry until after World War II

Definition

• Statistical Process Control (SPC):– “a methodology for monitoring a process to

identify special causes of variation and signal the need to take corrective action when it is appropriate”

(Evans and Lindsay)

Statistical Process Control Tools

• The magnificent seven

• The tool most often associated with Statistical Process Control is Control Charts

Common Causes

Special Causes

Histograms do not take into account changes over time.

Control charts can tell us when a process changes

Control Chart Applications• Establish state of statistical control• Monitor a process and signal when it goes out of

control• Determine process capability

• Note: Control charts will only detect the presence of assignable causes. Management, operator, and engineering action is necessary to eliminate the assignable cause.

Capability Versus Control

Control

Capability

Capable

Not Capable

In Control Out of Control

IDEAL

Developing Control Charts

1. Prepare– Choose measurement– Determine how to collect data, sample

size, and frequency of sampling– Set up an initial control chart

2. Collect Data– Record data– Calculate appropriate statistics– Plot statistics on chart

Types of Sampling

• Random Samples – Each piece has an equal chance of being selected for

inspection

• Systematic Samples– According to time or sequence

• Rational subgroups– A group of data that is logically homogeneous– Computing variation between subgroups

Approaches to Rational Subgrouping

• Each sample consists of units that were produced at the same time (or as closely together as possible)– Used when the primary purpose of the control chart is

to detect process shifts

• Each sample consists of units of product that are representative of all units that have been produced since the last sample was taken– Used when the purpose of the control chart is to

make lot sentencing decisions

Next Steps

3. Determine trial control limits– Center line (process average)

– Compute UCL, LCL

4. Analyze and interpret results– Determine if in control

– Eliminate out-of-control points

– Re-compute control limits as necessary

Warning Limits on Control Charts

Some suggest using two sets of limits on control charts:

• Action limits– Set at 3-sigma– When a point plots outside of this limit, a search for an

assignable cause is made and any necessary corrective action is taken

• Warning limits– Set at 2-sigma– When one or more points fall in between the warning

and action limits or very close to the warning limit, be suspicious

Typical Out-of-Control Patterns• Point outside control limits

• Hugging the center line

• Hugging the control limits

• Instability

• Sudden shift in process average

• Cycles

• Trends

Shift in Process Average

Identifying Potential Shifts

Cycles

Trend

Western Electric Sensitizing Rules:

• One point plots outside the 3-sigma control limits

• Two of three consecutive points plot outside the 2-sigma warning limits

• Four of five consecutive points plot beyond the 1-sigma limits

• A run of eight consecutive points plot on one side of the center line

Additional sensitizing rules:

• Six points in a row are steadily increasing or decreasing

• Fifteen points in a row with 1-sigma limits (both above and below the center line)

• Fourteen points in a row alternating up and down• Eight points in a row in both sides of the center

line with none within the 1-sigma limits• An unusual or nonrandom pattern in the data• One of more points near a warning or control limit

Classroom Exercise

• In small groups, choose two of the sensitizing rules. For each of your two rules, make up a (reasonable) situation where that rule would catch a problem and a (reasonable) situation where that rule might falsely identify a problem.

Final Steps

5. Use as a problem-solving tool– Continue to collect and plot data

– Take corrective action when necessary

6. Compute process capability

Capability vs. Stability

A process is capable if individual products consistently meet specification

A process is stable only if common variation is present in the process

Process Capability Calculations

Process Capability

• The ability of the process to perform at the required level for the given quality characteristic

• Two of the methods for determining:– Find the probability that the process will produce a

part below the LSL plus the probability that the process will produce a part above the USL

– Process Capability RatioPCR = CP = (USL – LSL) / 6σ

Process Capability Ratio Note

• There are many ways we can estimate the capability of our process

• If σ is unknown, we can replace it with one of the following estimates:– The sample standard deviation S

– R-bar / d2

PCR and One Sided Specifications

• Upper specification only– CPU = (USL – μ) / 3σ

• Lower specification only– CPL = (μ – LSL) / 3σ

• We can use the same estimate for σ

PCR and an Off-Center Process

• CPK = min (CPU, CPL)

• Generally, if CP = CPK, then the process is centered at the midpoint of the specifications

• If CP ≠ CPK, then the process is off-center

Commonly Used Control Charts

• Variables data– x-bar and R-charts

– x-bar and s-charts

– Charts for individuals (x-charts)

• Attribute data– For “defectives” (p-chart, np-chart)

– For “defects” (c-chart, u-chart)

Control Charts

We assume that the underlying distribution is normal with some mean and some constant but unknown standard deviation .

Letx

x

ni

i

n

1

Distribution of x

Recall that x is a function of random variables, so it also is a random variable with its own distribution. By the central limit theorem, we know that

where,

x N x ( , )

xnx

Control Charts

xx

x

x

Control Charts

x

x

UCL

LCL

UCL & LCL Set atProblem: How do we estimate & ?

3 x

Control Charts

xx

m

m

ii

1

)(1 fm

RR

m

i

Control Charts

xx RALCL 2

xx RAUCL 2

RDLCLR 3

RDUCLR 4

xx

m

m

ii

1

)(1 fm

RR

m

i

Example

• Suppose specialized o-rings are to be manufactured at .5 inches. Too big and they won’t provide the necessary seal. Too little and they won’t fit on the shaft. Twenty samples of 2 rings each are taken. Results follow.

Part Measurements R Chart X ChartNo. 1 2 x R UCL R UCL LCL Xbar

1 0.502 0.504 0.503 0.002 0.0077 0.002 0.5052 0.4964 0.5032 0.495 0.497 0.496 0.002 0.0077 0.002 0.5052 0.4964 0.4963 0.492 0.496 0.494 0.004 0.0077 0.004 0.5052 0.4964 0.4944 0.501 0.498 0.500 0.003 0.0077 0.003 0.5052 0.4964 0.5005 0.507 0.508 0.508 0.001 0.0077 0.001 0.5052 0.4964 0.5086 0.504 0.504 0.504 0.000 0.0077 0.000 0.5052 0.4964 0.5047 0.497 0.496 0.497 0.001 0.0077 0.001 0.5052 0.4964 0.4978 0.493 0.496 0.495 0.003 0.0077 0.003 0.5052 0.4964 0.4959 0.502 0.501 0.502 0.001 0.0077 0.001 0.5052 0.4964 0.502

10 0.498 0.500 0.499 0.002 0.0077 0.002 0.5052 0.4964 0.49911 0.505 0.507 0.506 0.002 0.0077 0.002 0.5052 0.4964 0.50612 0.502 0.499 0.501 0.003 0.0077 0.003 0.5052 0.4964 0.50113 0.495 0.497 0.496 0.002 0.0077 0.002 0.5052 0.4964 0.49614 0.499 0.496 0.498 0.003 0.0077 0.003 0.5052 0.4964 0.49815 0.503 0.507 0.505 0.004 0.0077 0.004 0.5052 0.4964 0.50516 0.507 0.509 0.508 0.002 0.0077 0.002 0.5052 0.4964 0.50817 0.503 0.501 0.502 0.002 0.0077 0.002 0.5052 0.4964 0.50218 0.497 0.493 0.495 0.004 0.0077 0.004 0.5052 0.4964 0.49519 0.504 0.508 0.506 0.004 0.0077 0.004 0.5052 0.4964 0.50620 0.505 0.503 0.504 0.002 0.0077 0.002 0.5052 0.4964 0.504

Avg = 0.501 0.002x R

Std. = 0.0047

X-Bar Control Charts

X-Bar Chart

0.490

0.495

0.500

0.505

0.510

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

x

X-bar charts can identify special causes of variation, but they are only useful if the processis stable (common cause variation).

Control Limits for Range

UCL = D4R = 3.268*.002 = .0065

LCL = D3 R = 0

R Chart

0.000

0.002

0.004

0.006

0.008

0.010

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Observation

Ran

ge

Special Variables Control Charts

• x-bar and s charts

• x-chart for individuals

X-bar and S charts

• Allows us to estimate the process standard deviation directly instead of indirectly through the use of the range R

• S chart limits:– UCL = B6σ = B4*S-bar– Center Line = c4σ = S-bar– LCL = B5σ = B3*S-bar

• X-bar chart limits– UCL = X-doublebar +A3S-bar– Center line = X-doublebar– LCL = X-doublebar -A3S-bar

X-chart for individuals

• UCL = x-bar + 3*(MR-bar/d2)

• Center line = x-bar

• LCL = x-bar - 3*(MR-bar/d2)

Next Class

• Homework– Ch. 11 (12) Disc. Questions 5, 6, 7– Ch. 11 (12) Problems 5

• Topic– Control Charts, Part II

• Preparation– Chapter 12 (13)