Design for Quality and Product Excellence

1 © 2011 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from

the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 12

Design for Quality and Product Excellence

Teaching Notes

The precise manner in which a person or team approaches product design, solving problems to

achieve product excellence, or developing product reliability is not as critical as doing it in a

systematic fashion. Students have been exposed to process management and improvement in

Chapter 7, but they may still have some difficulty in understanding how measurement

(metrology) and Six Sigma projects can be used at the design stage to make frequent, but gradual

changes as an approach to process improvement.

Key objectives for this chapter should include:

To explore the typical structured product development process consisting of idea

generation, preliminary concept development, product/process development, full-scale

production, product introduction, and market evaluation.

To learn that concurrent, or simultaneous, engineering is an effective approach for

managing the product development process by using multi-functional teams to help

remove organizational barriers between departments and therefore reduce product

development time. Design reviews help to facility product development by stimulating

discussion, raising questions, and generating new ideas

To introduce the concept of Design for Six Sigma (DFSS) consisting of a set of tools and

methodologies used in the product development process to ensure that goods and services

meet customer needs and achieve performance objectives, and that the processes used to

make and deliver them achieve Six Sigma capability. DFSS consists of four principal

activities of: Concept development, Design development, Design optimization, and

Design verification. These activities are often incorporated into a variation of the

Design for Quality and Product Excellence 2

© 2011 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from


DMAIC process, known as DMADV, which stands for Define, Measure, Analyze,

Design, and Verify.

To define concept development as the process of applying scientific, engineering, and

business knowledge to produce a basic functional design that meets both customer needs

and manufacturing or service delivery requirements. This involves developing creative

ideas, evaluating them, and selecting the best concept.

To explore Quality Function Deployment (QFD) -- a planning process to guide the

design, manufacturing, and marketing of goods by integrating the voice of the customer

throughout the organization. A set of matrices, often called the House of Quality, is used

to relate the voice of the customer to a product’s technical requirements, component

requirements, process control plans, and manufacturing operations.

To investigate good product design, which anticipates issues related to cost,

manufacturability, and quality. Improvements in cost and quality often result from

simplifying designs, and employing techniques such as design for manufacturability

(DFM) – the process of designing a product for efficient production at the highest level of

quality.

To study social responsibilities in the design process including product safety and

environmental concerns, which have made Design for Environment (DfE) and design

for disassembly important features of products, because they permit easy removal of

components for recycling or repair, eliminate other environmental hazards, and makes

repair more affordable.

To explore Design for Excellence (DFX), an emerging concept that includes many

design-related initiatives such as concurrent engineering, design for manufacturability

design for assembly, design for environment and other ―design for‖ approaches. DFX

objectives include higher functional performance, physical performance, user

friendliness, reliability and durability, maintainability and serviceability, safety,

compatibility and upgradeability, environmental friendliness, and psychological

characteristics.

To introduce concept engineering (CE) -- a focused process for discovering customer

requirements and using them to select superior product or service concepts that meet those

requirements.

To investigate manufacturing specifications, consisting of nominal dimensions and

tolerances. Nominal refers to the ideal dimension or the target value that manufacturing

seeks to meet; tolerance is the permissible variation, recognizing the difficulty of meeting




a target consistently. Tolerance design involves determining the permissible variation in

a dimension.

Design optimization includes setting proper tolerances to ensure maximum product

performance and making designs robust; that is, insensitive to variations in

manufacturing or the use environment.

A scientific approach to tolerance design uses the Taguchi loss function. Taguchi

assumes that losses can be approximated by a quadratic function so that larger deviations

from target correspond to increasingly larger losses. For the case in which a specific

target value, T, is determined to produce the optimum performance, and in which quality

deteriorates as the actual value moves away from the target on either side (called

―nominal is best‖), the loss function is represented by L(x) = k(x - T)2.

To examine the characteristics of Design Failure Mode And Effects Analysis (DFMEA)

-- a methodology to identify all the ways in which a failure can occur, to estimate the

effect and seriousness of the failure, and to recommend corrective design actions.

To study the dimensions of reliability—the ability of a product to perform as expected

over time. Formally, reliability is defined as the probability that a product, piece of

equipment, or system performs its intended function for a stated period of time under

specified operating conditions. In practice, the number of failures per unit time determines

reliability during the duration under consideration (called the failure rate), look at

functional failure at the start of product life (The early failure period is sometimes called

the infant mortality period), reliability failure after some period of use.

To understand why reliability is often modeled using an exponential probability

distribution and use the reliability function, specifying the probability of survival, which

is: R(T) = 1 – e- T

.

To explore systems composed of individual components with known reliabilities,

configured in series, in parallel, or in some mixed combination, and how it ties into

various aspects of design, including optimization, tolerance design, and design

verification.

To learn that design optimization includes setting proper tolerances to ensure maximum

product performance and making designs robust; a scientific approach to tolerance design

uses the Taguchi loss function. Techniques for design verification include formal

reliability evaluation, using techniques such as accelerated life testing and burn-in.




To appreciate that the purpose of a design review is to stimulate discussion, raise

questions, and generate new ideas and solutions to help designers anticipate problems

before they occur.

To understand techniques for design verification including formal reliability evaluation.

These include accelerated life testing, which involves overstressing components to

reduce the time to failure and find weaknesses; and burn-in, or component stress testing,

which involves exposing integrated circuits to elevated temperatures in order to force

latent defects to occur.

To appreciate that Six Sigma performance depends on reliable measurement systems.

Common types of measuring instruments used in manufacturing today fall into two

categories: ―low-technology‖ and ―high-technology.‖ Low-technology instruments are

primarily manual devices that have been available for many years; high-technology

describes those that depend on modern electronics, microprocessors, lasers, or advanced

optics.

To define metrology--the science of measurement – broadly as the collection of people,

equipment, facilities, methods, and procedures used to assure the correctness or adequacy

of measurements, and is a vital part of global competitiveness, including characteristics

such as: accuracy, precision, repeatability or equipment variation, reproducibility or

operator variation, calibration and traceability.

To appreciate that process capability is the range over which the natural variation of a

process occurs as determined by the system of common causes; that is, what the process

can achieve under stable conditions. The relationship between the natural variation and

specifications is often quantified by a measure known as the process capability index, Cp.

To learn that a process capability study is a carefully planned study designed to yield

specific information about the performance of a process under specified operating

conditions. Three types of studies are a peak performance study, process characterization

study, and component variability study.

ANSWERS TO QUALITY IN PRACTICE KEY ISSUES

Testing Audio Components at Shure, Inc.

1. The general definition of reliability as: the probability that a product, piece of

equipment, or system performs its intended function for a stated period of time under

specified operating conditions, is thoroughly tested by Shure. Tests are tailored to




various market segments, according to the type of use (or abuse) the equipment is likely

to incur. For the consumer market, Shure uses the cartridge drop and scrape test, which

is particularly important to test for, in the light of how ―scratch‖ DJ’s use the equipment.

For presentation and installation audio systems, they use the microphone drop test and

perspiration test. For mobile communications, the two above tests, temperature, and

cable and cable assembly flex tests are applicable. For the performance audio, the

microphone drop test, perspiration test, sequential shipping, cable and cable assembly

flex, and temperature storage would all be appropriate. The purpose of the tests is to

simulate actual operating conditions so that the products can sustain accidents and rough

handling and perform effectively over a useful life. Quality characteristics that are

studied are achieved reliability and performance.

2. For the microphone drop test, the measures are probably variable measures of sound and

response levels, within an acceptable range. Thus, standard variables control charts may

be used. For the perspiration test, it may be that a p-chart or u-chart is used for attribute

measures. The cable and cable assembly flex test might use a p-chart to measure the

percentage of cables tested that failed due to rocking motions or twisting motions. The

sequential shipping tests would probably show varying proportions of failures due to

dropping, vibration, and rough handling. These might be sorted out using a Pareto chart.

Then efforts could be made to improve the most frequently occurring causes. The

cartridge drop and scrape test could also use p- or np-charts (see Chapter 13) to show

results per sample of 100 repetitions of the test. The temperature tests would most likely

use standard variables charts to measure whether test performance was within control

limits, or not.

Applying QFD in a Managed Care Organization

1. Although this example of QFD involved the design of a tangible items, it is more difficult

to implement in a service context, as opposed to a pure manufacturing context, because

both customer requirements and technical requirements are harder to quantify and assess

that with tangible products.

2. The detailed calculations in the Importance of the hows row and Percentage of

importance of the hows row used to arrive at these figures can be shown and verified on a

spreadsheet. Note that some discrepancies involving incorrect multiplication, were found

in part of the QFD ―House of Quality.‖




Design for Qualtiy and Product Excellence 7

© 2011 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied,

duplicated, or posted to a publicly accessible website, in whole or in part.

Direction of Rate of Co.

Rate of Absol. % Font

Use of Gloss. Q&A Tbl. of Lang.

Improvement Import. Now Plan Improv. Wgt. Improve size Update Photos colors Terms Sect. Contnt. Frindly.

Ease-use 4.5 3.2 4.5 1.4 6.3 25.2% 3 1 3 3 9 3 9 3

Accuracy 5.0 3.1 4.6 1.5 7.4 29.5% 9 1 3 1

Timeliness 3.2 3.8 3.8 1.0 3.2 12.7% 9 1

Clarity 3.8 2.6 3.9 1.5 5.7 22.7% 1 1 3 1 9 3 1 3

Conciseness 2.5 4.1 4.1 1.0 2.5 9.9% 1 1

Import. of hows

108.1 427.9 153.4 98.2 460.0 244.7 249.1 173.0

% of Import. of hows

5.65% 22.35% 8.01% 5.13% 24.03% 12.78% 13.01% 9.04%

The numbers in the original table were verified by the calculations shown above (some columns of the original table were rearranged

for convenience of calculation). The rates of improvement, absolute weights, and percent improvements, based on the given values for

―rate of importance‖ and ―company now‖ and ―plan‖ were validated. As in the original table, the ―importance of hows‖ and ―percent

of importance of hows‖ turned out to be accurately calculated. Specific factors shown as the most important were ―glossary terms‖ and

―updates.‖




3. The lessons that can be learned and applied to other service organizations that seek to

design or redesign their products and services include the facts that QFD provides for a

systematic approach to linking the ―voice of the customer‖ to operational requirements.

By doing so, operating efficiencies can be realized and customer satisfaction can be

enhanced. In addition, employee satisfaction often can be improved, as well, as found in

the case. It must be recognized that time and effort is involved in gathering, sorting, and

analyzing the characteristics and factors. Also, there is subjectivity in applying ratings and

weights to variables. Hence, the results are not easy to predict and guarantees are limited.

ANSWERS TO REVIEW QUESTIONS

1. Product design and development consists of six steps:

Idea Generation. New or redesigned product ideas should incorporate customer

needs and expectations.

Preliminary Concept Development. In this phase, new ideas are studied for

feasibility.

Product/Process Development. If an idea survives the concept stage, the actual

design process begins by evaluating design alternatives and determining

engineering specifications for all materials, components, and parts. This phase

usually includes prototype testing, design reviews, and development, testing, and

standardization of the manufacturing processes

Full-Scale Production. If no serious problems are found, the company releases the

product to manufacturing or service delivery teams.

Market Introduction. The product is distributed to customers.

Market Evaluation. An ongoing product development process that relies on

market evaluation and customer feedback to initiate continuous improvements.

2. Competitive pressures are forcing companies to reduce time to market, which means that

the time for product development is also squeezed. The problems incurred in speeding up

the process are well known. If done too hastily, the result will be the need to revise or

scrap the design, cost increases or project over-runs, difficulty in manufacturing the

product, early product failure in the field, customer dissatisfaction, and/or lawsuits due to

product liability. One of them most significant impediments to rapid design is poor intra-

organizational coordination. Reducing time to market can only be accomplished by

process simplification, eliminating design changes, and improving product

manufacturability. This requires involvement and cooperation of many functional groups




to identify and solve design problems in order to reduce product development and

introduction time.

3. Design for Six Sigma (DFSS) uses a set of tools and methodologies in the product

development process to ensure that goods and services will meet customer needs and

achieve performance objectives, and that the processes used to make and deliver them

achieve Six Sigma capability. DFSS consists of four principal activities:

Concept development, in which product functionality is determined based upon

customer requirements, technological capabilities, and economic realities;

Design development, which focuses on product and process performance issues

necessary to fulfill the product and service requirements in manufacturing or delivery;

Design optimization, which seeks to minimize the impact of variation in production

and use, creating a ―robust‖ design; and

Design verification, which ensures that the capability of the production system meets

the appropriate sigma level

4. Concept engineering (CE) emerged from a consortium of companies that included

Polaroid and Bose along with researchers at MIT. CE is a focused process for

discovering customer requirements and using them to select superior product or service

concepts that meet those requirements, and it puts the voice of the customer into a

broader context and employees numerous other techniques to ensure effective processing

of qualitative data. Five major steps comprise the process:

Understanding the customer’s environment. This step involves first project planning

activities such as team selection, identifying fit with business strategy, and gaining

team consensus on the project focus. It also includes collecting the voice of the

customer to understand the customer’s environment – physical, psychological,

competitive, and so on.

Converting understanding into requirements. In this step, teams analyze the customer

transcripts to translate the voice of the customer into more specific requirements using

the KJ method. This step focuses on identifying the technical requirements we

discussed in the context of QFD, selecting the most significant requirements, and

―scrubbing‖ the requirements to refine them into clear and insightful statements.

Operationalizing what has been learned. Involves determining how to measure how

well a customer requirement is met. The principal requirement is to focus on

throughput time, so the concept of ―quickly‖ needs to be operationalized and

measured. Once potential metrics are defined, they are evaluated to reduce the

number of metrics that need to be used while ensuring that they cover all key

requirements. This usually requires some sort of customer questionnaire to identify

the importance of the requirements and prioritized them.

Concept generation. This step generates ideas for solutions that will potentially meet

customers’ needs. The approach requires brainstorming ideas that might resolve each




individual customer requirement, selecting the best ones, and then classifying them

under the traditional functional product characteristics. This helps to develop a

―market in‖ rather than a ―product out‖ orientation. Creative thinking techniques are

applied here to increase the number and diversity of potential ideas.

Concept selection. The potential ideas are evaluated for their capability to meet

requirements, tradeoffs are assessed, and prototyping may begin. The process ends

with reflection on the final concept to test whether the decision ―feels right‖ based on

all the knowledge that has been acquired.

Concept engineering is an important tool for assuring quality because it provides a

systematic process that leaves a strong audit trail back to the voice of the customer. This

makes it difficult to challenge the results of skeptics and convert them. The process also

helps to build consensus and gives design teams confidence in selling their concept to

management. However, it takes a lot of discipline and patience.

5. QFD benefits companies through improved communication and teamwork between all

constituencies in the production process, such as between marketing and design, between

design and manufacturing, and between purchasing and suppliers. Product objectives are

better understood and interpreted during the production process. Use of QFD determines

the causes of customer dissatisfaction, making it a useful tool for competitive analysis of

product quality by top management. Productivity as well as quality improvements

generally follow QFD. QFD reduces the time for new product development. QFD allows

companies to simulate the effects of new design ideas and concepts. Companies can

reduce product development time and bring new products into the market sooner, thus

gaining competitive advantage.

6. In the QFD development process, a set of matrices is used to relate the voice of the

customer to a product’s technical requirements, component requirements, process control

plans, and manufacturing operations. The first matrix, called the House of Quality,

provides the basis for the QFD concept.

Building the House of Quality consists of six basic steps:

* Identify customer requirements.

* Identify technical requirements.

* Relate the customer requirements to the technical requirements.

* Conduct an evaluation of competing products or services




* Evaluate technical requirements and develop targets.

* Determine which technical requirements to deploy in the remainder of the

production/delivery process.

The first House of Quality in the QFD process provides marketing with an important tool

to understand customer needs and gives top management strategic direction. Three other

―houses of quality‖ are used to deploy the voice of the customer to (in a manufacturing

setting) component parts characteristics, process plans, and quality control. The second

house applies to subsystems and components. At this stage, target values representing the

best values for fit, function, and appearance are determined. In manufacturing, most of

the QFD activities represented by the first two houses of quality are performed by product

development and engineering functions.

In the last two stages, the planning activities involve supervisors and production line

operators. In the third house, the process plan relates the component characteristics to key

process operations, the transition from planning to execution. Key process operations are

the basis for a control point. A control point forms the basis for a quality control plan

delivering those critical characteristics that are crucial to achieving customer satisfaction.

This is specified in the last house of quality. These are the things that must be measured

and evaluated on a continuous basis to ensure that processes continue to meet the

important customer requirements defined in the first House of Quality.

7. Product design can have a major impact on manufacturability. If careful thought and

planning is not done by the designer (or design team), the end product can end up being

difficult or impossible to build due to placement of components, methods for

attachments, ―impossible‖ tolerances, difficulties in attaching or fastening components

and/or difficulties in getting the whole assembled ―system‖ to work smoothly, even with

the highest quality components. In addition time, materials, and other resources may be

wasted unnecessarily due to a poor manufacturing design.

The concept of Design for Manufacturability (DFM) is the process of designing a product

so that it can be produced efficiently at the highest level of quality. Its goal is to improve

quality, increase productivity, reduce lead time (time to market, as well as manufacturing

time) and maintain flexibility to adapt to future market conditions.

8. Key design practices for high quality in manufacturing and assembly include: 1) analyze all

design requirements to assess proper dimensions and tolerances, 2) determine process

capability, 3) identify and evaluate possible manufacturing quality problems, 4) select

manufacturing processes that minimize technical risks, and 5) evaluate processes under

actual manufacturing conditions.




9. Social responsibilities in the design process include safety and environmental concerns,

which have made Design for Environment (DFE) and Design for Disassembly important

features of products. Legal and environmental issues are becoming critical in designing

products and services, today. Product safety and its consequences, product liability, should

be of primary concern because of the damage that hazardous designs can do to consumers of

the product. Also, liability lawsuits can do major damage to the financial health of an

organization, as well as its image and reputation in the marketplace. Records and

documentation relating to the design process are the best defense against liability lawsuits.

These would include records on prototype development, testing, and inspection results.

Environmental issues involve questions of whether ―environmentally friendly‖ designs

(those that minimize damage to the environment in manufacture and product use) are being

developed, what impacts will the design of the product have on the environment when it is

scrapped, and how can consumers be given the most value for their money, while balancing

the other two issues? The above questions can often be addressed by considering it as a

―design for environment‖ concept (often combined with and ―design for disassembly‖).

What is the best design for repairability/recylability?

10. Design for Excellence (DFX) is an emerging concept that includes many design-related

initiatives such as concurrent engineering, design for manufacturability design for

assembly, design for environment and other ―design for‖ approaches. DFX objectives

include higher functional performance, physical performance, user friendliness, reliability

and durability, maintainability and serviceability, safety, compatibility and upgradeability,

environmental friendliness, and psychological characteristics. DFX represents a total

approach to product development and design involves the following activities:

Constantly thinking in terms of how one can design or manufacture products better,

not just solving or preventing problems

Focusing on ―things done right‖ rather than ―things gone wrong‖

Defining customer expectations and going beyond them, not just barely meeting them

or just matching the competition

Optimizing desirable features or results, not just incorporating them

Minimizing the overall cost without compromising quality of function

11. Manufacturing specifications consist of nominal dimensions and tolerances. Nominal

refers to the ideal dimension or the target value that manufacturing seeks to meet;

tolerance is the permissible variation, recognizing the difficulty of meeting a target

consistently. Traditionally, tolerances are set by convention rather than scientifically. A

designer might use the tolerances specified on previous designs or base a design decision

on judgment from past experience. Setting inappropriate tolerances can be costly, since

tolerance settings often fail to account for the impact of variation on product

functionality, manufacturability, or economic consequences. The Taguchi loss function is




a scientific approach to tolerance design. Taguchi assumed that losses can be

approximated by a quadratic function so that larger deviations from target cause

increasingly larger losses.

12. The Taguchi loss function is a useful concept for process design. Taguchi suggests that

there is not strict cut-off point that divides good quality from poor quality. Rather, he

assumed that losses can be approximated by a quadratic function so that larger deviations

from target correspond to increasingly larger losses. For the case in which a specific

target value, T, is determined to produce the optimum performance, and in which quality

deteriorates as the actual value moves away from the target on either side (called

―nominal is best‖), the loss function is represented by L(x) = k(x - T)2

where x is any

actual value of the quality characteristic and k is some constant. Thus, (x – T) represents

the deviation from the target, and the loss increases by the square of the deviation.

13. The purpose of Design Failure Mode and Effects Analysis (DFMEA) is to identify all the

ways in which a failure can occur, to estimate the effect and seriousness of the failure,

and to recommend corrective design actions. A DFMEA usually consists of specifying the

following information for each design element or function: Failure modes; effect of the

failure on the customer; severity, likelihood of occurrence, and detection rating; potential

causes of failure, and corrective actions or controls. A simple example of a DFMEA for

an ordinary household light socket is provided in the chapter.

14. Reliability has grown increasingly important among the quality disciplines due to safety

needs of consumers, the search for competitive advantage by companies, growing consumer

awareness, and rising expectations and the difficulty of achieving high reliability in more

sophisticated and complex modern products.

15. Reliability is the probability that a product, piece of equipment, or system performs its

intended function for a stated period of time under specified operating conditions. There are

four key components of this definition, including probability, time, performance, and

operating conditions. All of these have to be considered in a comprehensive definition of

reliability. Probability allows comparison of different products and systems, time allows us

to measure the length of life of the product, performance relates to the ability of the product

to do what it was designed to do, and operating conditions specify to amount of usage and

the environment in which the product is used.

16. A functional failure is one incurred at the start of the product's life due to defective

materials, components, or work on the product. A reliability failure is one that is incurred

after some period of use. For example, if a new TV set suffers a blown picture tube during

the first week, it's a functional failure. There was obviously a defect in the manufacture of

the tube. If the vertical hold feature of the set goes out (perhaps 3 days after the 1 year




warranty is up), that is a reliability failure. It should reasonably be expected to last much

longer than one year, but it didn't.

17. Failure rate is defined as the number of failures per unit of time during a specified time

period being considered. For example, if 15 MP-3 players were tested for 500 hours and

there were two failures of the units, the failure rate would be: 2 / (15 x 500) = 1 / 3750 or

0.000267.

18. The cumulative failure rate curve plots the cumulative percent of failures against time on

the horizontal axis. The failure rate curve is obtained by determining the slope of the failure

rate curve at a number of points to obtain the instantaneous failure rate (failures per unit

time) at that point. A plot of these values yields the failure rate curve.

19. The average failure rate over any interval of time is the slope of the line between the two

endpoints of the interval on the failure rate curve.

20. The product life characteristics curve, is the so-called "bath-tub curve" because of its shape.

It is actually the failure rate curve, described above. Such curves can be used to understand

the distinctive failure rate patterns of various designs and products, over time.

21. The reliability function represents the probability that an item will not fail within a certain

period of time, T. It is directly related to the cumulative distribution function: F(T) =

1 - e- T

, that yields the probability of failures. Since F(T) is the probability of failure, the

reliability function, R(T) can be defined as the complement, e.g. probability of not failing:

R(T) = 1 - (1 - e- T

) = e- T

It can also be expressed using the mean time to failure (MTTF) value as: R(T) = e-T/

22. The reliability of series, parallel, and series parallel is relatively easy to compute, given the

reliability of components in each system. For the series system, RS = R1R2R3. Thus

reliabilities are multiplicative.

For a parallel system, the relationships are a little more complex, since the units are

designed to use redundant components, so that if one unit fails the system can continue to

operate. The system reliability is computed as:

RS = 1 - [(1 - R1)(1 - R2)(1 - Rn)]

For series-parallel systems, the equivalent reliabilities of each parallel sub-system are

calculated, successively, until there are no more parallel sub-systems. The system is then




reduced to a serially equivalent system in which all component reliabilities can be

multiplied to get the final reliability value.

23. The purpose of a design review is to stimulate discussion, raise questions, and generate

new ideas and solutions to help designers anticipate problems before they occur. To

facilitate product development, a design review is generally conducted in three major

stages of the product development process: preliminary, intermediate, and final. The

preliminary design review establishes early communication between marketing,

engineering, manufacturing, and purchasing personnel and provides better coordination of

their activities. It usually involves higher levels of management and concentrates on

strategic issues in design that relate to customer requirements and thus the ultimate

quality of the product. The preliminary design review evaluates such issues as the

function of the product, conformance to customer’s needs, completeness of

specifications, manufacturing costs, and liability issues.

After the design is well established, an intermediate review takes place to study the

design in greater detail to identify potential problems and suggest corrective action.

Personnel at lower levels of the organization are more heavily involved at this stage.

Finally, just before release to production, a final review is held. Materials lists, drawings,

and other detailed design information are studied with the purpose of preventing costly

changes after production setup.

24. Methods of product testing for reliability include: life testing, accelerated life testing,

environmental testing and vibration and shock testing. In life and accelerated life testing the

product is tested until it fails. The latter speeds up the process by overstressing the item to

hasten its eventual failure. Environmental and shock tests are performed to determine the

product's ability to survive and operate under adverse conditions of heat, cold, or shock.

25. Latent defects are frequently found in electronic devices, such as semi-conductors. The term

refers to the fact that a certain small proportion of the units will have defects which show up

during the early life of the product, perhaps the first 1,000 hours of operation. Then the

remaining components, after the "infant mortality" period has passed, the remaining

components may operate for years without many failures.

26. Robust designs are those that are insensitive to variations in manufacturing or in the use

environment.

27. Common types of measuring instruments (see Bonus Materials folder on the Premier

website) used in manufacturing today fall into two categories: ―low-technology‖ and

―high-technology.‖ Low-technology instruments are primarily manual devices that have

been available for many years and include rulers, calipers, mechanical micrometers, go-no




go gauges, etc.; high-technology describes those that depend on modern electronics,

microprocessors, lasers, or advanced optics, such as micrometers with digital readouts,

electronic optical comparators, and computerized coordinate measuring machines.

28. Metrology is the science of measurement. It formerly included only the measurement

processes involved in gauging the physical attributes of objects. Today, metrology is

much more broadly defined as: the collection of people, equipment, facilities, methods,

and procedures used to assure correctness or adequacy of measurements. It is vital to

quality control because of the increasing complexity of modern manufacturing and

service operations. In particular, the increasing emphasis and oversight of government

agencies, the implications of measurement errors on safety and product liability, and the

need for reliance on improved quality control methods, such as SPC, make metrology an

important branch of science.

29. Accuracy is defined as the closeness of agreement between an observed value and an

accepted reference value or standard. Accuracy is measured as the amount of error in a

measurement in proportion to the total size of the measurement. One measurement is

more accurate than another if it has a smaller relative error.

Precision is defined as the closeness of agreement between randomly selected individual

measurements or results. Precision, therefore, relates to the variance of repeated

measurements. A measuring instrument having a low variance is said to be more precise

than another having a higher variance.

Reproducibility is the variation in the same measuring instrument when it is used by

different individuals to measure the same parts. Causes of poor reproducibility include

poor training of the operators in the use of the instrument or unclear calibrations on the

gauge dial.

30. Calibration is the comparison of a measurement device or system having a known

relationship to national standards to another device or system whose relationship to

national standards is unknown. Calibration is necessary to ensure the accuracy of

measurement and hence to have confidence in the ability to distinguish between

conforming and nonconforming production. Measurements made with uncalibrated or

inadequately calibrated equipment can lead to erroneous and costly decisions.

31. Repeatability and reproducibility (R&R) require a study of variation and can be addressed

through statistical analysis. R&R studies must be done systematically, and require quite a

number of steps. A repeatability and reproducibility study is conducted in the following

manner (Note: formulas are omitted for the sake of brevity).




1. Select m operators and n parts. Typically at least 2 operators and 10 parts are

chosen. Number the parts so that the numbers are not visible to the operators.

2. Calibrate the measuring instrument.

3. Let each operator measure each part in a random order and record the results.

Repeat this for a total of r trials. At least two trials must be used. Let Mijk

represent the kth measurement of operator i on part j.

4. Compute the average measurement for each operator and the difference between

the largest and smallest average.

5. Compute the range for each part and each operator (these values show the

variability of repeated measurements of the same part by the same operator);

compute the average range for each operator; compute the overall average range.

6. Calculate ―control limits‖ on the individual ranges Rij , using a constant (D4) that

depends on the sample size (number of trials, r) and can be found in a table for

control charts. Any range value beyond the control limits might result from some

assignable cause, not random error. Possible causes should be investigated and, if

found, corrected. The operator should repeat these measurements using the same

part. If no assignable cause is found, these values should be discarded and all

statistics in step 5 as well as the control limit should be recalculated.

Once these basic calculations are made, an analysis of repeatability and

reproducibility can be performed, equipment variation (EV) is computed as

reproducibility, and operator variation as appraisal variation (AV).

Constants K1 and K2 are chosen and depend on the number of trials and number of

operators, respectively. These constants provide a 99 percent confidence interval

on these statistics. An overall measure of repeatability and reproducibility (R&R)

is given by:

Repeatability and reproducibility are often expressed as a percentage of the

tolerance of the quality characteristic being measured. The American Society for

Quality suggests the following guidelines for evaluating these measures of

repeatability and reproducibility:

Under 10% error: This rate is acceptable.

10 to 30% error: This rate may be acceptable based on the importance of the

application, cost of the instrument, cost of repair, and so on.

Over 30% error: Generally, this rate is not acceptable. Every effort should be

made to identify the problem and correct it.

32. Process capability is the range over which the natural variation of a process occurs as

determined by the system of common causes. It is the ability of the combination of




people, machines, methods, materials, and measurements to produce a product or service

that will consistently meet design specifications. Process capability is measured by the

proportion of output that can be produced within design specifications; in other words, it

is a measurement of the uniformity of the product.

33. A process capability study is a carefully planned study designed to yield specific

information about the performance of a process under specified operating conditions.

Three types of studies are often conducted. A peak performance study is focused on

determining how a process performs under actual operating conditions. A process

characterization study is designed to determine how a process performs under actual

operating conditions. A component variability study has the goal of determining the

relative contribution of different sources of total variation. The six steps involved in

making a process capability study are listed in the chapter.

34. The following are brief definitions of the various process capability indexes:

Cp is the ratio of the specification width to the natural tolerance of the process

Cpl is the lower one-sided index that relates the distance from the process mean to the

lower tolerance limit to its 3 natural spread

Cpu is the upper one-sided index that relates the distance from the process mean to the

upper tolerance limit to its 3 natural spread

These indexes are calculated to determine the ability of a process to meet or exceed

design specifications and are only meaningful when a process is known to be under

control. General a process is considered to be capable if its index is 1.0 or above. These

indexes may be used to establish quality policy in operating areas or with a supplier by

stating an acceptable standard, such as: all capability indexes must be at 2.0 (called 6

quality) or above if the process is to be considered acceptable for elimination of

inspection processes by customers.

SOLUTIONS TO PROBLEMS

Note: Data sets for several problems in this chapter are available in the Excel workbook

C12Data on the Premium website for this chapter accompanying this text. Click on the

appropriate worksheet tab as noted in the problem (e.g., Prob. 12-5) to access the data.

1. Tonia’s Tasty Tacos conducted consumer surveys and focus groups and identified the

most important customer expectations as




• Tasty, moderately healthy food

• Speedy service

• An easy-to-read menu board

• Accurate order filling

• Perceived value

Develop a set of technical requirements to incorporate into the design of a new facility

and a House of Quality relationship matrix to assess how well your requirements address

these expectations. Refine your design as necessary, based upon the initial assessment.

Answer

1. Analysis of customer responses for Tonia’s Tasty Tacos indicates that there are likely to be

several strong relationships between customer requirements and associated technical

requirements of the product that Tonia designs (for example. a burrito), such as value vs.

price; nutrition vs. calories (and other nutritional content values, such as sodium, and

percent trans-fat).

Note the three customer response categories that are unrelated to the design of the burritos --

order accuracy, speedy service, and menu board. These factors would require a separate

analysis as part of a facility and process design.




PARTIAL HOUSE OF QUALITY MATRIX

FOR TONIA’S TASTY TACOS

Price Size Calor

ies

Sodium % t-Fat Imprtnce

12 3 45

Compet.

Eval. 12 3 4 5

Selling

Pts. 1 2 3 4 5

Taste Moistness

Flavor

Visual Visually

Appealing

Health Nutritious

Value Good Value

Competitive

Evaluation:

= Very strong relationship

= Strong relationship

= Weak relationship

2. Newfonia, Inc., is working on a design for a new smartphone. Marketing staff conducted

extensive surveys and focus groups with potential customers to determine the

characteristics that the customers want and expect in a smartphone. Newfonia’s studies

have identified the most important customer expectations as

• Initial cost

• Reliability

• Ease of use

• Features

• Operating cost

• Compactness

Develop a set of technical requirements to incorporate into the design of a House of

Quality relationship matrix to assess how well your requirements address these

expectations. Refine your design as necessary, based upon the initial assessment.




Answer

2. Analysis of customer responses for Newfonia’s proposed smartphone indicates the

likelihood of several strong relationships between customer requirements and associated

technical requirements of the design, such as value vs. price; features vs. compactness; and

ease of use vs. features. Operating costs may possibly be distantly related to initial cost and

features. Technical characteristics required to translate the ―voice of the customer‖ into

operational or engineering terms might be measures of purchase cost, operating programs

(e.g., BranchOS, or other similar systems), number and type of features, weight,

dimensions, battery life, cost of replacement batteries, and peripherals.

3. Tonia’s Tasty Tacos (Problem 1) acquired some additional information. It found that

consumers placed the highest importance on healthy food, followed by value, followed by

order accuracy and service. The menu board was only casually noted as an important

attribute in the surveys. Tonia faces three major competitors in this market: Grabby’s,

Tacoking, and Sandy’s. Studies of their products yielded the information shown in the

table in C12Data file for Prob.12-3 on the Premium website for this chapter. Results of

the consumer panel ratings for each of these competitors can also be found there (a 1–5

scale, with 5 being the best). Using this information, modify and extend your House of

Quality from Problem 1 and develop a deployment plan for a new burrito. On what

attributes should the company focus its marketing efforts?

Answer

3. With the new data given for Tonia's customers, a partial House of Quality for the design

of the burritos can be built, as shown below. Note that the relationships between customer

requirements (flavor, health, value) and associated technical requirements (% fat,

calories, sodium, price) of the burrito design are strong.

The inter-relationships of the roof are not shown (limitations of MSWord software), these

may be sketched in. For example, they would show a strong inter-relationship between fat

and calories.





FOR TONIA’S TASTY TACOS

Price Size Calor

ies

Sodium % t-Fat Imprtnce

12 3 45

Compet.

Eval. 12 3 4 5

Selling

Pts. 1 2 3 4 5

Taste Moistness x G Q S

Flavor x S G Q *

Visual Visually

Appealing x G S Q

Health Nutritious x QS G *

Value Good Value x Q SG *

Competitive

Evaluation: Grabby’s

5

3

5

5

5

Tacoking 3 5 2 2 2

Sandy's 4 4 3 3 4

Targets $0.2

6/

oz.

7.0/

oz.

80/

oz.

85 mg. 13%

Deployment * * *



= Weak relationship

Tonia’s Tasty Tacos technical requirements must be placed on a more equal basis, which

would best be shown as units/ounce, except for the percent fat value. These are shown

below.

Company Price/oz. Calories/oz. Sodium/oz. % Fat

Grabby's $ 0.282 80 13.63 13

Tacoking $ 0.300 85 12.67 23

Sandy's $ 0.292 90 13.33 16

Although Tonia’s is low in price per ounce, as well as calories, and percent fat, this analysis

suggests that Tonia’s should try to increase its size and visual appeal, while continuing to

reduce the cost per ounce. At the same time, it should build on the strength of the nutrition

trend by keeping the sodium and percent fat low, as did Grabby's, and slightly reducing the




number of calories per ounce to be even more competitive. If Tonia’s can design a

flavorful, healthy, 7 oz. taco and sell it at an attractive price (say, $1.85 or less), it should

be a very profitable undertaking.

4. Newfonia, Inc. (Problem 2), faces three major competitors in this market: Oldphonia,

Simphonia, and Colliefonia. It found that potential consumers placed the highest

importance on reliability (measured by such things as freedom from operating system

crashes and battery life), followed by compactness (weight/bulkiness), followed by

flexibility (features, ease of use, and types of program modules available). The operating

cost was only occasionally noted as an important attribute in the surveys. Studies of their

products yielded the information shown in the table in C12Data file for Prob.12-4 on the

Premium website for this chapter. Results of the consumer panel ratings for these

competitors are also shown in that spreadsheet. Using this information, modify and

extend your House of Quality from Problem 2 and develop a deployment plan for the new

smartphone. On what attributes should the company focus its marketing efforts?

Answer

4. With the new data given for Newfonia’s potential customers, a partial House of Quality

for the design of the smartphone can be built, as shown below. Note the strong

relationships between customer requirements and associated technical requirements of the

smartphone design.

The inter-relationships of the roof are not shown (limitations of MSWord software), but

these may be sketched in. For example, they would show a strong inter-relationship between

size and weight.





FOR NEWPHONIA’S SMARTPHONE CASE

Cost Size

(in.)

Wt.

(oz.)

Featr.

(num.)

Opr.P

rog.

Bat.

Life

Opr.

Cost

Importan

ce

12 3 45

Compet

Eval.

12 3 45

Selling

Pts.

1 2 3 4

5

Reliable Keeps

operating x G S H *

Compact Fits

pocket

x GSH

Not heavy x S G Q

Features Calendar,

contact

mgt., etc.

x G S H *

Ease of

use

Intuitive

operations x QS G *

Value Good

value x Q SG *

Competitive

Evaluation:

Oldphonia

3

4

5

4

5

5

5

Simfonia 5 4 3 2 2 2 3 = Very strong relationship

Colliefonia 4 4 3 3 4 3 4 = Strong relationship

Targets $250 5 x

3.2

6 oz. 10 Win.

CE

35 Mod

. = Weak relationship

Deployment * * *

This analysis suggests that Newfonia should try to position itself between Simfonia and

Colliefonia in price and features. It should build on the strength of the customer’s reliability

concern, keeping battery life near 35 hours and use a proven operating program, such as

BranchOS. Enough features (10) should be offered to be competitive. If Newfonia can

design a high-value smartphone and sell it at an attractive price (say, $250 or less), it

should be a very profitable undertaking.

5. A genetic researcher at GenLab, Ltd. is trying to test two laboratory thermometers (that

can be read to 1/100,000th of a degree Celsius) for accuracy and precision. She measured

25 samples with each and obtained the results found in the C12Data file for Prob.12-5 on

the Premium website for this chapter. The true temperature being measured is 0 degrees




C. Which instrument is more accurate? Which is more precise? Which is the better

instrument?

Answer

5. Accuracy of: Thermometer A Thermometer B

Abs [0.00031- 0] Abs [-.00005 - 0]

100 x ------------------------ = 0.031 % 100 x ----------------------- = 0.005%

1 deg. 1 deg.

Thermometer B is more accurate.

The Excel-calculated (see spreadsheets Prob12-5a.xls and Prob12-5b.xls on the Premium

website for details) statistics and frequency distribution, shows that Thermometer B is

also more precise than Thermometer A, as indicated by a smaller standard deviation.

Thermometer B is a better instrument, because it is likely that it can be adjusted to center

on the nominal value of 0.

Frequency Table - Problem 12-5a

Upper Cell

Boundaries Frequencies

Cell 1 -0.00251 1

Cell 2 -0.00169 1

Cell 3 -0.00086 3

Cell 4 -0.00003 5

Cell 5 0.00080 5

Cell 6 0.00163 6

Cell 7 0.00246 4




Standard Statistical Measures

Mean 0.000312

Median 0.000246

Mode #N/A

Standard deviation 0.001343

Variance 0.000002

Max 0.002456

Min -0.002514

Range 0.004970




Frequency Table Problem 12-5b

Upper Cell


Cell 1 -0.00221 1

Cell 2 -0.00070 7

Cell 3 0.00005 7

Cell 4 0.00156 7

Cell 5 0.00232 3


Mean -0.000046

Median -0.000123

Mode #N/A


Variance 0.000001

Max 0.002316

Min -0.002209

Range 0.004525




6. Two scales were at Aussieburgers, Ltd. used to weigh the same 25 samples of hamburger

patties for a fast-food restaurant in Australia. Results are shown in C12Data file for

Prob.12-6 on the Premium website for this chapter. The samples were weighed in grams,

and the supplier has ensured that each patty weighs 114 grams. Which scale is more

accurate? Which is more precise? Which is the better scale?

Answer

6. See spreadsheets Prob12-6a.xls and Prob12-6b.xls for details.

Accuracy of: Scale A Scale B

Abs[113.96 -114] Abs[115.92 - 114]

100 x ------------------------ = 0.035 % 100 x ----------------------- = 1.685%

114 114

Scale A is more accurate.




The frequency distribution, taken from the Excel printout, shows that Scale B is more

precise than Scale A.

Scale B is a better instrument, because it is likely that it can be adjusted to center on the

nominal value of 0.

Scale A

Frequency Table - Problem 12-6a

Upper Cell


Cell 1 112.00 3

Cell 2 112.67 0

Cell 3 113.33 5

Cell 4 114.00 9

Cell 5 114.67 0

Cell 6 115.33 6

Cell 7 116.00 2


Mean 113.96

Median 114.00

Mode 114.00


Variance 1.29

Max 116.00

Min 112.00

Range 4.00




Scale B

Frequency Table Problem 12-6b

Upper Cell


Cell 1 114.00 3

Cell 2 115.33 5

Cell 3 116.00 10

Cell 4 117.33 5

Cell 5 118.00 2


Mean 115.92

Median 116.00

Mode 116.00


Variance 1.24

Max 118

Min 114

Range 4.00




7. A blueprint specification for the thickness of a dishwasher part at PlataLimpia, Inc. is

0.325 ± 0.025 centimeters (cm). It costs $15 to scrap a part that is outside the

specifications. Determine the Taguchi loss function for this situation.

Answer

7. The Taguchi Loss Function for PlataLimpia, Inc. part is: L(x) = k (x - T)2

$15 = k (0.025)2

k = 24000

L(x) = k (x - T)2 = 24000 (x - T)

2

8. A team was formed to study the dishwasher part at PlataLimpia, Inc. described in

Problem 7. While continuing to work to find the root cause of scrap, they found a way to

reduce the scrap cost to $10 per part.

a. Determine the Taguchi loss function for this situation.

b. If the process deviation from target can be held at 0.015 cm, what is the Taguchi loss?




Answer

8. The Taguchi Loss Function is: L(x) = k (x - T)2

a) $10 = k (0.025)2

k = 16000

L(x) = k (x - T)2 = 16000 (x - T)

2

b) L(x) = 16000 (x - T)2

L(0.015) = 16000 (0.015)

2 = $3.60

9. A specification for the length of an auto part at PartsDimensions, Inc. is 5.0 ± 0.10

centimeters (cm). It costs $50 to scrap a part that is outside the specifications. Determine

the Taguchi loss function for this situation.

Answer


$50 = k (0.10)2

k = 5000

L(x) = k (x - T)2 = 5000 (x - T)

2

10. A team was formed to study the auto part at PartsDimensions described in Problem 9.

While continuing to work to find the root cause of scrap, the team found a way to reduce

the scrap cost to $30 per part.

a. Determine the Taguchi loss function for this situation.

b. If the process deviation from target can be held at 0.020 cm, what is the Taguchi loss?




Answer


a) $30 = k (0.10)2

k = 3000

L(x) = k (x - T)2 = 3000 (x - T)

2

b) L(x) = 3000 (x - T)2

L(0.020) = 3000 (0.020)

2 = $ 1.20

11. Ruido Unlimited makes electronic soundboards for car stereos. Output voltage to a

certain component on the board must be 12 ± 0.2 volts. Exceeding the limits results in an

estimated loss of $50. Determine the Taguchi loss function.

Answer


$50 = k (0.2)2

k = 1250

L(x) = k (x - T)2 = 1250 (x - T)

2

12. An electronic component has a specification of 100 ± 3 ohms. Scrapping the component

results in a $81 loss.

a. What is the value of k in the Taguchi loss function?

b. If the process is centered on the target specification with a standard deviation of 1

ohm, what is the expected loss per unit?




Answer

12. For a specification of 100 ± 3 ohms:

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

$81 = k ( 3 )2

k = 9

b) EL(x) = k ( 2 + D

2) = 9 ( 1

2 + 0

2 ) = $9

13. An automatic cookie machine must deposit a specified amount of 25 ± 0.2 grams (g) of

dough for each cookie on a conveyor belt. If the machine either over- or underdeposits the

mixture, it costs $0.02 to scrap the defective cookie.

a. What is the value of k in the Taguchi loss function?

b. If the process is centered on the target specification with a standard deviation of 0.06 g,

what is the expected loss per unit?

Answer

13. For a specification of 25 ± 0.2 grams

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

$0.02 = k ( 0.2 )2

k = 0.5

b) For = 0.06

EL(x) = k ( 2 + D

2) = 0.5 ( 0.06

2 + 0

2 ) = $0.0018

14. A computer chip is designed so that the distance between two adjacent pins has a

specification of 2.000 ± 0.002 millimeters (mm). The loss due to a defective chip is $2. A

sample of 25 chips was drawn from the production process and the results, in mm, can be

found in the C12Data file for Prob.12-14 on the Premium website for this chapter.

a. Compute the value of k in the Taguchi loss function.

b. What is the expected loss from this process based on the sample data?




Answer

14. For a specification of 2.000 ± .002 mm and a $2 scrap cost:

Analysis of the dataset for problem 12-14 provides the following statistics:

x = 2.00008; D = 2.00008 - 2.00 = 0.00008

= 0.00104

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

$2 = k (0.002)2 k = 500,000

b) EL(x) = k ( 2 + D

2) = 500,000 ( 0.00104

2 + 0.00008

2 ) = $0.544

15. In the production of transformers, any output voltage that exceeds 120 ± 15 volts is

unacceptable to the customer. Exceeding these limits results in an estimated loss of $450.

However, the manufacturer can adjust the voltage in the plant by changing a resistor that

costs $2.25.

a. Determine the Taguchi loss function.

b. Suppose the nominal specification is 120 volts. At what tolerance should the

transformer be manufactured, assuming that the amount of loss is represented by the cost

of the resistor?

Answer

15. a) The Taguchi Loss function is: L(x) = k (x - T)2

450 = k (15)2

k = 0.5

So, L(x) = 0.5 (x-T)2




b) $2.25 = 0.5 (x-120)2

4.50 = (x - 120)2

(x - T)Tolerance = 50.4 = 2.12 volts

2.12 = x - 120

x = 122.12

16. At Elektroparts Manufacturers’ integrated circuit business, managers gathered data from a

customer focus group and found that any output voltage that exceeds 120 ± 5 volts was

unacceptable to the customer. Exceeding these limits results in an estimated loss of $200.

However, the manufacturer can still adjust the voltage in the plant by changing a resistor

that costs $2.00.

a. Determine the Taguchi loss function.

b. Suppose the nominal specification remains at 120 volts. At what tolerance should the

integrated circuit be manufactured, assuming that the amount of loss is represented by the

cost of the resistor?

Answer

16. a) The Taguchi Loss function is: L(x) = k (x - T) 2

200 = k (5)2

k = 8

So, L(x) = 8 (x-T)2




b) The Taguchi Loss function is: L(x) = k (x - T) 2

$2.00 = 8 (x-120)2

0.25 = (x - 120)2

(x - T)Tolerance = 25.0 = 0.5 volts

0.5 = x - 120

x = 120.5

17. Two processes, P and Q, are used by a supplier to produce the same component, Z, which

is a critical part in the engine of the Air2Port 778 airplane. The specification for Z calls

for a dimension of 0.24 mm ± 0.03. The probabilities of achieving the dimensions for

each process based on their inherent variability are shown in the table found in the

C12Data file for Prob.12-17 on the Premium website for this chapter. If k = 60,000, what

is the expected loss for each process? Which would be the best process to use, based on

minimizing the expected loss?

Answer

17. For the Air2Port 778 plane parts (see spreadsheets Prob12-17.xls for detailed

calculations):

Specifications are 24 +/- 3 mm

L(x) = 60000 (x - T)2

For a typical calculation:

L(0.21) = 60000 (0.21 - 0.24)2 = $ 54.00

Weighted loss = 0.12 X $54.00 = $ 6.48




Air2Port Airplane Co.

Calculation of Taguchi Loss Values

Value

Loss ($)

Process P

Probability

Weighted

Loss ($)

Process Q

Probability

Weighted

Loss ($)

0.20 96.00 0 0.00 0.02 1.92

0.21 54.00 0.12 6.48 0.03 1.62

0.22 24.00 0.12 2.88 0.15 3.60

0.23 6.00 0.12 0.72 0.15 0.90

0.24 0.00 0.28 0.00 0.30 0.00

0.25 6.00 0.12 0.72 0.15 0.90

0.26 24.00 0.12 2.88 0.15 3.60

0.27 54.00 0.12 6.48 0.03 1.62

0.28 96.00 0 0.00 0.02 1.92

Expected Loss 20.16 16.08

Therefore, Process Q incurs a smaller loss than Process P, even though some output of Q

falls outside specifications.

18. The average time to handle a call in a the Call-Nowait call processing center has a

specification of 6 ±1.25 minutes. The loss due to a mishandled call is $16. A sample of

25 calls was drawn from the process and the results, in minutes, can be found in the

C12Data file for Prob.12-18 on the Premium website for this chapter.

a. Compute the value of k in the Taguchi loss function.

b. What is the expected loss from this process based on the sample data?

Answer

18. For a specification of 6 1.25 minutes and a $16 call mishandling cost:

x = 6.016; D = 6.016 - 6.00 = 0.016

= 0.8957

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

$16 = k (1.25 )2 ; k = 10.24




b) E [L(x) = k ( 2 + D

2)] = 10.24 ( 0.8957

2 + 0.016

2 ) = $8.218

19. Compute the average failure rate during the intervals 0 to 40, 40 to 70, and 70 to 100, and

0 to 100, based on the information in Figure 12.28.

Answer

19. Based on the Cumulative Failure Rate curve:

From 0 - 40, slope = 29.5 / 40 = 0.738

From 40 - 70, slope = (40 – 29.5) / (70 - 40) = 0.350

From 70 - 100, slope = (90 - 40) / (100 - 70) = 1.678

From 0 - 100, slope = 90 / 100 = 0.9

See the spreadsheet Prob12-19 for more details, including a diagram showing the failure

rate curve.

20. The life of a cell phone battery is normally distributed with a mean of 900 days and

standard deviation of 50 days.

a. What fraction of batteries is expected to survive beyond 975 days?

b. What fraction will survive fewer than 800 days?

c. Sketch the reliability function.

d. What length of warranty is needed so that no more than 10 percent of the batteries will

be expected to fail during the warranty period?




Answer

20. a) P(x > 975) = 0.5 - P(900 < x < 975)

975-900

P(900 < x < 975)= P (z < ----------------) = P (0 < z < 1.5) = 0.4332

50

Therefore, P(x > 875) = 0.5 - 0.4332 = 0.0668 or 6.68% should survive beyond 975 days.

800 -900

b) P (x < 800) = P (z < -------------) = P (z < -2.0 )

50

Therefore, P (x < 880) = 0.5 - 0.4772 = 0.0228 or 2.28% should survive less than 800 days.

c) The reliability function looks approximately as follows (see spreadsheet Prb12-20.xls for

details):

d) Let xw be the limit of the warranty period.

P (x < xw) = 0.10; z = -1.28, for z = x-900 = -1.28 , xw = 836 hours for the

50 warranty limit.




21. Lifetred, Inc., makes automobile tires that have a mean life of 75,000 miles with a

standard deviation of 2,500 miles.

a. What fraction of tires is expected to survive beyond 77,250 miles?

b. What fraction will survive fewer than 68,750 miles?

c. Sketch the reliability function.

d. What length of warranty is needed so that no more than 10 percent of the tires

will be expected to fail during the warranty period?

Answer

21. a) P(x > 77250) = 0.5 - P(75000 < x < 77250)

77250 - 75000

P(75000 < x < 77250)= P(z < -------------------) = P (0 < z < 0.9) = 0.3159

2500

Therefore, P(x > 77250) = 0.5 - 0.3159 = 0.1841 or 18.41% should survive beyond 77250

miles.

68750- 75000 - 6250

b) P (x < 68750) = P(z < -------------------- ) = P(z < ---------- ) = P (z < -2.50) =

2500 2500

0.5 - P(68750 < x < 75000) = 0.5 - 0.4938 = 0.0062

Therefore, P (x < 68750) = 0.0062 or 0.62% should survive less than 68750 miles.

c) The reliability function looks approximately as follows:




d) Let xw be the limit of the warranty period.

P (x < xw) = 0.10; z = -1.28, for z = x-75000 = -1.28 , xw = 71,800 miles for the

2500 warranty limit.

22. Massive Corporation’s tested five motors in an 800-hour test. Compute the failure rate if,

three failed after 200, 375, and 450 hours and the other two ran for the full 800 hours

each.

Answer

22. Massive Corporation’s motors have a failure rate of:

= 3 = 3 = 0.001143 failures / hour

[(2 x 800) + 200 +375 + 450] 2625

23. Livelong, Inc.’s computer monitors have a failure rate of 0.00005 units per hour.

Assuming an exponential distribution, what is the probability of failure within 10,000

hours? What is the reliability function?




Answer

23. The reliability function for Livelong’s monitors is R(T) = 1 - F(T) = e- T

= 0.00005; Use F(T) = P(x < 10000)

F(T) = P(x < 10000) = 1 - e-0.00005 (10000)

= 1- 0.607 = 0.393 or 39.3% probability that a

monitor will survive less than 10,000 hours.

24. An electronic component in a satellite radio has failure rate of = .000015. Find the mean

time to failure (MTTF). What is the probability that the component will not have failed after

12,000 hours of operation?

Answer

24. The MTTF is = 1 ; so, = 66666.67

R (T) = e- T/

= e- 12000 / 66666.67

= e -0.18

= 0.835 or 83.5% probability of surviving for at

least 12,000 hours

25. The MTBF of an integrated circuit made by IceeU, Inc. is 18,000 hours. Calculate the

failure rate.

Answer

25. The failure rate ( ) for IceeU, Inc.’s integrated circuits is:

= 1 = 0.000056 failures / hr.

18000

26. A manufacturer of MP3 players purchases major electronic components as modules. The

reliabilities of components differ by supplier (see diagram, below). Suppose that the

configuration of the major components is given by:




The components that can be purchased from three different suppliers. The reliabilities of

the components are as follows:

Component Supplier 1 Supplier 2 Supplier 3

A .97 .92 .95 B .85 .90 .90 C .95 .93 .88

Transportation and purchasing considerations require that only one supplier be chosen.

Which one should be selected if the radio is to have the highest possible reliability?

Answer

26. Supplier 1: RaRbc = (0.97) [1 - (1 - 0.85)(1 - 0.95)] = 0.963

Supplier 2: RaRbc = (0.92) [1 - (1 - 0.90)(1 - 0.93)] = 0.914

Supplier 3: RaRbc = (0.95) [1 - (1 - 0.90)(1 - 0.88)] = 0.939

Therefore, choose Supplier 1.

27. An electronic missile guidance system consists of the following components:

Components A, B, C, and D have reliabilities of 0.98, 0.95, 0.85, and 0.99, respectively

(see the following diagram). What is the reliability of the entire system?

Answer

27. The reliability of the parallel Rcc shown in the diagram above the problem, is calculated as:

Rcc = 1 - (1 - 0.85) 2 = 0.98

RaRbRccRd = (0.98)(0.95) (0.98)(0.99) = 0.903




28. A Bestronics store processes customers through 3 work stations when they wish to buy a

certain popular product. Modular components for the product must be checked

electronically at two work stations before final checkout, where the cashier collects cash or

credit cards for the sale. a) If workstation 1 has reliability of testing equipment of 0.98,

workstation 2 has reliability of testing equipment of 0.92 and the final checkout register has

a reliability of 0.90, what is the overall checkout system reliability? b) If the store manager

wants to ensure at least a 90% system reliability can she do so by dedicating two final

checkout registers to the process, in parallel, each having a 0.90 reliability, with the same

reliability at workstations 1 and 2?

Answer

28. a) RaRbRc = (0.98)(0.92)(0.90) = 0.811

b) RaRbc = (0.98) (0.92) [1 - (1 - 0.90)(1 - 0.90)] = 0.893

No, this will not provide the minimum required system reliability. The manager must find a

way to improve reliability of one or more workstations or checkout registers.

29. Manuplex, Inc. has a complex manufacturing process, with three operations that are

performed in series. Because of the nature of the process, machines frequently fall out of

adjustment and must be repaired. To keep the system going, two identical machines are

used at each stage; thus, if one fails, the other can be used while the first is repaired (see

accompanying figure).




The reliabilities of the machines are as follows: Machine Reliability

A .70 B .80 C .95

a. Analyze the system reliability, assuming only one machine at each stage (all the

backup machines are out of operation).

b. How much is the reliability improved by having two machines at each stage?

Answer

29. a) RaRbRc = (0.70)(0.80)(0.95) = 0.532

b) RaaRbbRcc = [1 - (1 - 0.70)2] [1 - (1 - 0.80)

2] [1 - (1 - 0.95)

2] =

(0.91) (0.96) (0.9975) = 0.871

The improvement is significant, rising 0.339 from 0.532 to 0.871

30. An automated production system at Autoprod, Inc. consists of three operations: turning,

milling, and grinding. Individual parts are transferred from one operation to the next by a

robot. Hence, if one machine or the robot fails, the process stops.

a. If the reliabilities of the robot, turning center, milling machine, and grinder are 0.98,

0.90, 0.93, and 0.85, respectively, what is the reliability of the system?

b. Suppose that two grinders are available and the system does not stop if one fails. What

is the reliability of the system?

Answer

30. a) RtRmRg = (0.98)(0.90)(0.93)(0.85) = 0.697

b) RtRmReg = (0.98)(0.90)(0.93)[1 - (1 - 0.85)2] = 0.802




GAGE R&R PROBLEMS

31. A gauge repeatability and reproducibility study at Frankford Brake Systems collected the

data found in the C12Data file for Prob.12-31 on the Premium website for this chapter.

Analyze these data. The part specification is 1.0 ± 0.06 mm.

Answer

31. Detailed calculations for the first operator are as follows:

x 1 = ( Mijk) /nr = 29.720 / 30 = 0.9907

R 1 = ( Rij) / n = 0.280 / 10 = 0.028

Use this method to calculate values for the second operator:

x 2 = 29.901 / 30 = 0.9967; R 2 = 0.380/ 10 = 0.038

x D = max { x i} - min { x i} = 0.9967 - 0.9907 = 0.006

R = ( R i) / m = (0.028 + 0.038) / 2 = 0.033

D4 = 2.574 ; UCLR = D4 R = (2.574) (0.033) = 0.0849, all ranges below

K1 = 3.05 ; K2 = 3.65 (from Table 12.3)

EV = K1 R = (3.05) (0.033) = 0.10065

AV = (K x ) - (EV / nr) 2 D2 2 = 0.0119

RR = (AV) + (EV) 22 = 0.1014

Equipment variation = 100 (0.10065 / 0.12) = 83.88%

Operator variation = 100 (0.0119 / 0.12) = 9.92%

R & R variation = 100 (0.1014 / 0.12) = 84.50%




For detailed spreadsheet data, see Prob12-31RR.xls. Spreadsheet results confirm prior

calculations, as follows:

Tolerance analysis

Average range 0.033 Repeatability (EV) 0.101 83.88%

X-bar range ( x D) 0.006 Reproducibility (AV) 0.012 9.93%

Repeatability and Reproducibility (R&R) 0.101 84.46%

Control limit for individual ranges 0.085

Note: any ranges beyond this limit may be the result of assignable causes. Identify and correct. Discard values and recompute statistics.

Concentrate on reducing equipment variation

Note that the calculator values, shown in the detailed calculations above, and computer

values do not match precisely, because a greater number of decimal places are used by the

computer to carry out calculations. All formulas are identical, however.

32. A gauge repeatability and reproducibility study was made at Precision Parts, Inc., using

three operators, taking three trials each on identical parts. The data that can be found in

the C12Data file for Prob.12-32 on the Premium website for this chapter were collected.

Do you see any problems after analyzing these data? What should be done? The part

specification for the collar that was measured was 1.6 ± 0.2 inches.

Answer

32. Detailed calculations for the first operator are as follows:

x 1 = ( Mijk) /nr = 48.48 / 30 = 1.616

R 1 = ( Rij) / n = 1.33 / 10 = 0.133

Use this method to calculate values for the second operator:

x 2 = 46.74 / 30 = 1.558; R 2 = 1.58/ 10 = 0.158




Also, use this method to calculate values for the third operator:

x 3 = 47.05 / 30 = 1.568; R 3 = 0.610/ 10 = 0.061

x D = max { x i} - min { x i} = 1.616 – 1.558 = 0.058

R = ( R i) / m = (0.133 + 0.158 + 0.061) / 3 = 0.117

D4 = 2.574 ; UCLR = D4 R = (2.574) (0.117) = 0.3012, all ranges below

K1 = 3.05 ; K2 = 3.65 (from Table 12.3)

EV = K1 R = (3.05) (0.117) = 0.3569

AV = (K x ) - (EV / nr) 2 D2 2 = 0.1424

RR = (AV) + (EV) 22 = 0.3843

Equipment variation = 100 (0.3569 / 0.40) = 89.23%

Operator variation = 100 (0.1424 / 0.40) = 35.60%

R & R variation = 100 (0.3843 / 0.40) = 96.08%

Note that the range in sample 7 exceeded the control limit of 0.301 by for the first

operator. This point could have been due to a misreading of the gauge. If so, this sample

should be thrown out, another one taken, and the values recomputed.

For detailed spreadsheet data, see Prob12-32RR.xls. Spreadsheet results confirm prior

calculations, as follows: Tolerance analysis

Average range 0.117 Repeatability (EV) 0.358 89.47%

X-bar range ( x D) 0.058 Reproducibility (AV) 0.142 35.58%

Repeatability and Reproducibility (R&R) 0.385 96.28%

Control limit for individual ranges 0.302

Note: any ranges beyond this limit may be the result

of assignable causes. Identify and correct. Discard

values and recompute statistics.

Concentrate on reducing equipment variation




Note also that the calculator values, shown in the detailed calculations above, and

computer values do not match precisely, because a greater number of decimal places are

used by the computer to carry out calculations. All formulas are identical, however.

33. A machining process at Mach3 Tool Co. has a required dimension on a part of 0.575

± 0.007 inch. Twenty-five parts each were measured as found in the C12Data file for

Prob.12-33 on the Premium website for this chapter. What is its capability for

producing within acceptable limits?

Answer

33. For sample statistics at Mach3 Tool Co. of: x = 0.5750; = 0.0065 and a tolerance of

0.575 ± 0.007

Cp = UTL - LTL = 0.582 - 0.568 = 0.359; not capable - unsatisfactory

6 6 (0.0065)

See spreadsheet Prob12-33.xls for more descriptive analysis.

Note: There is some rounding error in the above calculations. See spreadsheet for more

descriptive analysis.

Nominal specification 0.5750 Average 0.5750 Cp 0.3567

Upper tolerance limit 0.5820 Std. deviation 0.0065 Cpl 0.3547

Lower tolerance limit 0.5680 Cpu 0.3587

Cpk 0.3547

34. Adjustments were made in the process at Mach3 Tool Co., discussed in Problem 33 and

25 more samples were taken. The results are given in the C12Data file for Prob.12-34 on

the Premium website for this chapter. What can you observe about the process? Is it now

capable of producing within acceptable limits?

Answer

34. For sample statistics of: x = 0.5755; = 0.0017 and a tolerance of 0.575 ± 0.007




The standard deviation is smaller than previously, indicating less ―spread‖ within the

data. See spreadsheet P12-35.xls for more descriptive analysis.

Cp = UTL - LTL = 0.582 - 0.568 = 1.373; The process capability is now inside

6 6 (0.0017) the tolerance limits, at an acceptable

level.

Note, however, that the other process capability indexes, below, show that there are still

some slight problems with process centering that must be addressed.

Nominal specification 0.5750 Average 0.5755 Cp 1.3838

Upper tolerance limit 0.5820 Std. deviation 0.00169 Cpl 1.4866

Lower tolerance limit 0.5680 Cpu 1.2810

Cpk 1.2810

35. From the data for Kermit Theatrical Products, construct a histogram and estimate the

process capability. If the specifications are 24 ± 0.03, estimate the percentage of parts that

will be nonconforming. Finally, compute Cp, Cpu, and Cpl. Samples for three parts were

taken as shown in the C12Data file for Prob12-35 on the student Premium website for

this chapter.

Answer

35. Summary statistics and the histogram from spreadsheet Prob12-35.xls show:

Column 1

Mean 24.0014

Standard Error 0.00097

Median 24.001

Mode 24.000

Standard Deviation 0.00967

Sample Variance 9.4E-05

Kurtosis 0.53132

Skewness 0.05271

Range 0.058

Minimum 23.971

Maximum 24.029

Confidence Level(95.0%)

0.00192




Bin Frequency

23.971 1

23.977 0

23.983 0

23.988 7

23.994 14

24.000 26

24.006 20

24.012 19

24.017 7

24.023 5

More 1

Histogram

0

5

10

15

20

25

30

23.9

71

23.9

77

23.9

83

23.9

88

23.9

94

24.0

00

24.0

06

24.0

12

24.0

17

24.0

23

More

Bin

Fre

qu

en

cy

Frequency

For sample statistics of: x = 24.0014; = 0.0097

Specification limits for the process are: 23.97 < < 24.03

z = 24.0300 - 24.0014 = 2.95 P( z > 2.94) = (0.5 - 0.4984) = 0.0016 that

0.0097 items will exceed upper limit

z = 23.9700 - 24.0014 = -3.24 P( z < -3.24) = 0.00 that items

0.0097 will exceed lower limit

Therefore, the percent outside is: 0.0016, or 0.16 %




Cp = UTL - LTL = 24.030 - 23.970 = 1.031

6 6 (0.0097)

Cpu = UTL - x = 24.030 - 24.0014 = 0.983

3 3 (0.0097)

Cpl = x - LTL = 24.0014 - 23.970 = 1.079

3 3 (0.0097)

The process capability indexes are slightly out of tolerance for the upper index, and within

minimum limits for the lower and overall index. These results indicate that the process may

be minimally adequate if it can be centered on the nominal dimension of 24. However, the

ideal situation would be to launch process improvement studies so that the capability

indexes could be at least doubled.

36. Samples for three parts made at River City Parts Co. were taken as shown in the C12Data

file for Prob.12-36 on the Premium website for this chapter. Data set 1 is for part 1, data

set 2 is for part 2, and data set 3 is for part 3.

a. Calculate the mean and standard deviations for each part and compare them to the

following specification limits:

Part Nominal Tolerance

1 1.750 ± 0.045 2 2.000 ± 0.060 3 1.250 ± 0.030

b. Will the production process permit an acceptable fit of all parts into a slot with a

specification of 5 ± 0.081 at least 99.73 percent of the time?

Answer

36. a) Sample statistics as shown in spreadsheet Prob.12-36.xls are:

Data set 1: x = 1.7446; s = 0.0163; 3s = 0.0489

Data set 2: x = 1.9999; s = 0.0078; 3s = 0.0234




Data set 3: x = 1.2485; s = 0.0052; 3s = 0.0156

Part 1 will not consistently meet the tolerance limit since its ± 3s value is greater than the

tolerance limit. Parts 2 and 3 are well within their tolerance limits, since their ± 3s values

are smaller than the stated tolerances.

b) x T = 4.9930 ; Estimated Process = s +s +s12

22

3 2 =

0.0163 +0.0078 + 0.00522 2 2 = 0.0188

Process limits: 4.9930 ± 3(0.0188) or

4.9366 to 5.0494 vs. specification limits of

4.919 to 5.081 for a confidence level of 0.9973.

The parts will fit within their combined specification limit with a 0.9973 confidence

level.

37. Omega Tecnology Ltd. (OTL) is a small manufacturing company that produces various

parts for tool manufacturers. One of OTL’s production processes involves producing a

Teflon® spacer plate that has a tolerance of 0.05 to 0.100 cm in thickness. On the

recommendation of the quality assurance (QA) department and over objections of the

plant manager, OTL just purchased some new equipment to make these parts. Recently,

the production manager was receiving complaints from customers about high levels of

nonconforming parts. He suspected the new equipment, but neither QA nor plant

management would listen.

The manager discussed the issue with one of his production supervisors who mentioned

that she had just collected some process data for a study that the quality assurance

department was undertaking. The manager decided that he would prove his point by

showing that the new equipment was not capable of meeting the specifications. The data

provided by the supervisor are shown in the C12Data file for Problem 12-37 on the

Premium website for this chapter. Perform a process capability study on these data and

interpret your results.




Answer

37. Omega Technology Ltd.’s process capability results from the Excel spreadsheet software

are shown below. (See spreadsheet Prob12-37.xls for details.)

Average 0.0764 Cp 0.8019

Standard deviation 0.0104 Cpl 0.8468

Cpu 0.7569

Cpk 0.7569

These data show that the process has a rather low overall capability, with Cp = 0.8019

and a total of 1.71% of the values falling outside of the specification limits of 0.05 - 0.10

Process statistics: x = 0.0764, = 0.0104

z = 0.10 - 0.0764 = 2.27 P( z > 2.27) = (.5- 0.4884) = 0.0116 that the part

0.0104 will exceed upper limit

z = 0.05 - 0.0764 = -2.54 P( z < -2.54) = (.5- 0.4945) = 0.0055 that the part

0.0104 will exceed lower limit

Therefore, the percent outside is: 0.0171, or 1.71 %

38. Suppose that a refrigeration process at Coolfoods, Ltd. Has a normally distributed output

with a mean of 25.0 and a variance of 1.44.

a. If the specifications are 25.0 ± 3.25, compute Cp, Cpk, and Cpm. Is the process capable and

centered?

b. Suppose the mean shifts to 23.0 but the variance remains unchanged. Recompute and

interpret these process capability indexes.

c. If the variance can be reduced to 40 percent of its original value, how do the process

capability indices change (using the original mean of 25.0)?

Answer

38. (a) x = 25.0; = 1.2




Cp = UTL - LTL = 28.25 – 21.75 = 0.903

6 6 (1.2)

Cm

= Cp / ] / )targetmean[(1 22 = 0.903 / ]2.1 / )0.250.25[(1 22 = 0.903

Cpu

= UTL - x = 28.25 - 25.0 = 0. 903

3 3 (1.2)

Cpl

= x - LTL = 25.0 - 21.75 = 0.903

3 3 (1.2)

Conclusion: The process is centered on the mean, but it does not have adequate

capability at this time.

Cpk

= min (Cpl

, Cpu

) = 0.903

(b) x = 23; = 1.2

Cp = UTL -LTL = 28.25 – 21.75 = 0.903 This result has not changed.

6 6 (1.2)

Cmodified

= Cp / ] / )targetmean[(1 22 = 0.903 / ]2.1 / )0.250.23[(1 22 = 0.584

Because of the shift away from the target, capability is lower.

Cpu = UTL - x = 28.25 - 23.0 = 1.458

3 3 (1.2)

Cpl

= x - LTL = 23.0 - 21.75 = 0.347 ; Cpk

= min (Cpl

, Cpu

) = 0.347

3 3 (1.2)

Conclusion: The process is skewed and still does not have adequate capability at this

time.




(c) 2

new = 0.4 (1.44) = 0.576 new = 0.759

Cp = UTL -LTL = 28.25 – 21.75 = 1.427

6 6 (0.759)

Cmodified

= Cp / ] / )targetmean[(1 22 = 1.427/ ]0.759 / )0.250.25[(1 22 = 1.427

If there is no shift away from the target, capability is equal to Cp.

Cpu

= 28.25 - 25.0 = 1.427

3 (0.759)

Cpl

= 25.0 - 21.75 = 1.427

3 (0.759)

Cpk

= min (Cpl

, Cpu

) = 1.427

Reducing the variance brings the Cpl

and Cpu

to the point of adequacy, provided the

process can remain centered.

39. A process has upper and lower tolerance limits of 5.80 and 5.00, respectively. If the

customer requires a demonstrated Cp of 2.0, what must the standard deviation be? If both

Cpu and Cpl must also be 2.0, determine the process mean, using that calculated standard

deviation, assuming a normal distribution of output.

Answer

39. Cp = UTL -LTL = 2.0 = 5.80 - 5.00 = 0.8 ; Therefore, = 0.0667

6 6 6

Cpu

= UTL - x = 5.80 - x = 2.0; Therefore, we get: x = 5.4

3 3

Cpl

= x - LTL = x - 5.00 = 2.0; Therefore, we get: x = 5.4

3 3




40. Clearly demonstrate that Six Sigma requires Cp = 2.0 and Cpk = 1.5.

Answer

40.

As explained in Ch. 11:

The easiest way to understand this is to think of the distance from the target to the upper

or lower specification (half the tolerance), measured in terms of standard deviations of the

inherent variation, as the sigma level. A k-sigma quality level satisfies the equation:

k * process standard deviation = tolerance / 2

Note that in Figure 11.1, if the design specification limits were only 4 standard deviations

away from the target, the tails of the shifted distributions begin to exceed the

specification limits by a significant amount.

Table 11.1 shows the number of defects per million for different sigma quality levels and

different amounts of off-centering. Note that a quality level of 3.4 defects per million can

be achieved in several ways, for instance;

with 0.5-sigma off-centering and 5-sigma quality

with 1.0-sigma off-centering and 5.5-sigma quality

with 1.5-sigma off-centering and 6-sigma quality




In many cases, controlling the process to the target is less expensive than reducing the

process variability. This table can help assess these trade-offs.

The sigma level can easily be calculated on an Excel spreadsheet using the formula:

=NORMSINV(1-Number of Defects/Number of Opportunities) + SHIFT

or equivalently,

=NORMSINV(1-dpmo/1,000,000) + SHIFT

SHIFT refers to the off-centering as used in Table 11.1. Using the airline example

discussed earlier, if we had 3 lost bags for 8000(1.6) = 12,800 opportunities, we would

find =NORMSINV(1-3/12800) + 1.5 = 4.99828 or about a 5-sigma level.

Using data from problem 40, above, we can show that

Cp = UTL -LTL = 2.0 = 5.80 - 5.00 = 0.8 ; Therefore, = 0.0667

6 6 6

Cpu

= UTL - x = 5.80 - x = 2.0; Therefore, we get: x = 5.4

3 3

Cpl

= x - LTL = x - 5.00 = 2.0; Therefore, we get: x = 5.4

3 3

This meets the requirement that:

k * process standard deviation = tolerance / 2

6 * .0667 = 5.80 - 5.00 = 0.400

2

With a mean shift of 1.5

Cmodified

= Cp / ] / )targetmean[(1 22 = 2.0 / ]0667. / )4667.54.5[(1 22 = 1.414

Because of the shift away from the target, capability is lower




SUGGESTIONS FOR PROJECTS, ETC.

1. Customer attributes and technical requirements might be:

Attributes Technical Requirements

a. Book purchase:

Hours Schedule of open hours

Organization By dept./course/professor

Pre-processing availability Reservations on Internet

Ease of payment Cash, check, credit card

Time Speed of checkout

Value Lowest available price

Empathy Understanding/willingness of

personnel to solve problems

b. Registration:

Convenience Time, dates, Internet, phone

Speed Process standards

Costs Fees

Accuracy Error prevention

Empathy Understanding/willingness of

personnel to solve problems

c. Hotel room - business:

Convenience Business location , dates, methods

Speed - check in/out Process standards, system knowledge

Technology FAX, Internet connection

Amenities Restaurant, in-room work areas,

Internet connections, exercise facilities

Costs Fees - related to services


d. Hotel room - family:

Convenience Location near recreation, moderate

dining facilities, dates, methods

Speed - check-in Process standards, system knowledge

Amenities Play, family-related facilities

Costs Fees - related to family budget


CCoonnssttrruuccttiioonn ooff tthhee mmaattrriixx iiss lleefftt ttoo tthhee ssttuuddeenntt..




2. Customer requirements would likely include freshness, taste, consistency, appearance of the

product; knowledge, attentiveness, friendliness of customer service personnel; speed and

accuracy of the cooks and order fillers; accuracy and friendliness of the counter personnel.

Technical requirements might be explored to determine what would be required to deliver

the product to in-house versus delivery customers. The former would require wait staff

training in customer service techniques, while the latter would require knowledgeable

drivers substituting (in some ways) for wait staff. Technology for deliver orders would

involve equipment to receive orders via FAX machines or over the Internet. Regarding

cooks and order fillers (common to both in-house and external product delivery), how much

of the assembly of the pizza should be done by hand (versus machine-made or machine

assisted)? This is merely suggestive of the types of questions concerning customer and

technical requirements that students should consider. Construction of the House of Quality

is left to the student.

3. For a glider the following customer attributes and technical requirements might be:

Attributes Technical Requirements

Ease of assembly "Design for assembly"; Simple instructions

Easy to fly "Launch" mechanism

Flight characteristics Wing, tail, body design

Durability Quality of wood

Value Price/durability ratio

4. The best way to prioritize the voice of the customer would be to have a focus group of

typical customers, such as craftspeople, "do-it-yourselfer's", hobbyists to provide input on

how they used the screwdriver and their priorities. Below is a possible configuration of the

matrix, with priorities for a ―serious‖ craftsperson. Such a person would look for quality and

functionality over price or ―extra‖ features, such as ratchets or interchangeable bits.




HOUSE OF QUALITY MATRIX

FOR A SIMPLE SCREWDRIVER

Price

Interchg

Bits

Steel

Shaft

Rubber

Grip

Ratchet

Capabil.

Plastic

Handle

Easy to use

Does not rust

Durable

Comfortable

Versatile

Inexpensive

Priority 3 1 6 5 2 4



= Weak relationship

5. Answers will vary, depending on the service processes chosen by the students. The case on

―applying QFD to a University Support System‖ might be used as a starting point to

determine what types of information to look for to complete this project.

6. This exercise will give students an appreciation of the challenges of designing and

calibrating measurement systems.

7. Answers will vary, depending on the individual websites and topics chosen by the students.

8. This exercise is designed to further students' awareness of the breadth of the "quality

movement" and help them confirm how and whether the theory of quality is being applied

in practical settings in business and industry. Measurement is commonly used in

manufacturing and services, but it varies widely in precision and accuracy concepts,

depending on the size of the firm and the industry.




ANSWERS TO CASE QUESTIONS

Case - Applying Quality Function Deployment to a University Support Service

1. The answer to the question of whether students agree or disagree with the relative

importance rankings obtained from the study of the RRC at Tennessee Tech ultimately

depends on students’ opinions. However, a strong case might be made that the relative

importance score would depend on the situation. For small, rush, duplicating jobs, prompt

service would seem to be of greatest importance. For research jobs where specific

information has to be found, knowledge and courtesy of the employees would be highly

desirable, as well as accuracy, which might be a close second in importance. For the

inexperienced user, such as a freshman student, empathy and willingness to help would

possibly be ranked as the two highest criteria.

2. Concentrating on the top four characteristics, the following weighted scores can be

calculated:

Resources (personnel) 135

Customer handling 68

Information handling 87

Attitudes and morale 68

The three areas on which the analysts focused were document handling, training, and

layout. The above weighted scores would seem to lend little support to the need to deploy

a new document handling process (45 point score), nor to improve the layout (6 point

score), which have very little impact on customer quality criteria. Training may be

required, but the focus on document handling would seem to be unnecessary.

3. Given the high ranking of resources (personnel), it appears that more attention should be

paid to selection and retention issues. Information handling, in second place, also has a

major impact, with customer handling, and attitudes and morale tied for third place.

These categories could be improved by training and by process analysis to determine if

the best processes were being used. As a result, it could be predicted that morale and

customer satisfaction would likely increase.

Case - Black Elk Medical Center

1. The next steps would include gathering data using the checklist form that the committee

designed. The committee might also want to develop process flow charts, while waiting

for the fall data to be gathered and analyzed.




2. The data from the checklist should be put into a format, perhaps in a spreadsheet, where it

could be analyzed. Analysis tools might include Pareto analysis and histograms.

Segmentation should also be used to find out the incidence of falls in likely locations.

3. Improved processes and systems should be developed based on analysis of the checklist

and process charting. The ―significant few‖ causes, based on the Pareto analysis, should

be addressed first, in order to achieve the greatest immediate impact. Analysis of the

process flow chart could reveal places where processes could be simplified, and might

also identify conditions that would contribute to patient falls, so they could be eliminated.

Bonus Materials

Case - Hydraulic Lift Company

1. The key to the calculation of an estimated process capability for this case is to calculate an

estimated standard deviation for each condition. Using the simplifying assumption that the

sample standard deviation is a good approximation of the population standard deviation will

allow us to make a reasonable estimate, even though for the cases of the small sample sizes

of 30 or 35 that assumption would be open to argument by statisticians.

We will concentrate on the calculation of Cp for only case (a) and (e), since it is obvious that

the capability became drastically worse during the experimental stages from (b) to (d).

Reading the data from the histograms, we can use the calculation of the sample standard

deviation with grouped data from the chapter. The frequency histogram for condition (a)

shows:




mp,

Group x Frequency fx fx2

_________________________________________________

1 45 3 135 6075

2 50 6 300 15000

3 55 0 0 0

4 60 16 960 57600

5 65 4 260 16900

6 70 22 1540 107800

7 75 6 450 33750

8 80 23 1840 147200

9 85 5 425 36125

10 90 10 900 81000

11 95 0 0 0

12 100 5 500 50000

7310 551450

fx 7310

x = n = 100 = 73.1

s = [ / ( ) [( ( /fx n fx) n) / (n -1)]2 21 =

[ / ] [( ) / / ]551450 99 7310 100 992 = 13.138

cp = UTL LTL = 100 - 50 = 0.63; not acceptable

6 6(13.138)




The frequency histogram for condition (e) shows:

mp,

Group x Frequency fx fx2

_______________________________________

1 60 2 120 7200

2 65 0 0 0

3 70 12 840 58800

4 75 7 525 39375

5 80 11 880 70400

6 85 3 255 21675

2620 197450

fx 2620

x = n = 35 = 74.857

s = [ / ( ) [( ( /fx n fx) n) / (n -1)]2 21 =

[ / ] [( / / ]197450 34 35 342620)2 = 6.241

cp = UTL LTL = 100 - 50 = 1.34; not outstanding, but much better

6 6 (6.241)

2. The process used here was obviously a systematic process of problem solving similar to the

one suggested in this chapter. The first step was a) to understand the "mess." A Pareto-like

approach found that 50% of the defective items were due to dimensional problems on one

diameter of the valve stem. Next, b) find facts on the process capability; c) specific

problems were identified: over adjustment by the operator, and lack of ability of the

machine to hold tolerances; d) ideas on machine adjustments and improvements were

generated; e) solutions were implemented, with the machine being adjusted and later

overhauled; f) as Deming said, "Do it over, again and again." [Step 7, added to the process,

is continuous improvement].

Case - Bloomfield Tool Co.

1. See spreadsheet 12blomrrcase.xls for details. Note that there are some rounding errors

below that make answers on the spreadsheet appear slightly different.

Calculations for the repeatability and reproducibility (R&R) study are as follows:




x 1 = ( Mijk) /nr = 2.3378 / 30 = 0.0779

R 1 = ( Rij) / n = 0.1030/ 15 = 0.0069

Use this method to calculate values for operator 2

x 2 = 2.2974 / 30 = 0.0766; R 2 = 0.0945/ 15 = 0.0063

x D = max { x i} - min { x i} = 0.0779 - 0.0765 = 0.0014

R = ( R i ) / m = (0.0069 + 0.0063) / 2 = 0.0066

D4 = 3.267; UCLR = D4 R = (3.267) (0.0066) = 0.0215, all ranges below

K1 = 4.56 ; K

2 = 3.65 (from Table 11.2 in the text)

EV = K1 R = (4.56) (0.0066) = 0.0301

OV = (K x ) - (EV / nr) 2 D2 2 = [(3.65)(0.0013)] - [( 0.0301) / 30] 2 2 =

- 0.0000077

NOTE: According to the Measurement Systems Analysis Reference Manual published by

the Automotive Industry Action Group (AIAG), Troy MI, 1990, p. 44:

"... if a negative value is calculated under the square root sign, the [OV]

defaults to zero (0)."

RR = (EV) + (OV) 2 2 = (0.0301) + (0) 2 2 = 0.0301

Therefore, in this case, RR = EV

% Equipment variation (related to tolerance) = 100 (0.0301 / 0.05) = 60.2%

Operator variation (related to tolerance) = 0%

R & R variation (related to tolerance) = 60.2%

Concentrate on reducing equipment variation, not operator variation.




Total Variation: TV= (RR) + (PV) 2 2 , where PV is Part Variation

PV = Rp K

3

Rp = Range of part averages for the entire sample: 0.1013 to 0.0516 = 0.0497

K3 = 1.45 from Table 11.5

PV = (0.0497) (1.45) = 0.0721

Thus TV = (RR) + (PV) 2 2 = (0.0301) + (0.0721) 2 2 = 0.0781

%OV = 0

%EV = % RR, related to TV = 100 (0.0301 / 0.0781) = 38.5

%PV, related to TV = 100 (0.0721 / 0.0781) = 92.3

NOTE: The sum of the above percentages will not add to 100.

Based on the "rules" for process capability given in the text, it can be assumed that the

equipment and the process need to be improved, since none of the percentages fall below

the 30% or 10% minimums. The operators are consistent in their measurements, so their

methods are not in question at this point.

Worn or faulty gauges should be discarded and the rest should be calibrated.

CCaassee -- TThhee PPIIVVOOTT IInniittiiaattiivvee –– PPaarrtt IIII

The MAIC Process

1. It appeared fairly difficult to find the true ―root causes‖ of the problem. This was

illustrated by the fact that manual strapping errors were not initially found to be a

significant problem. Also, it was necessary to prepare over 100 different graphical

analyses before patterns could begin to be found. It is likely that the ―tribal knowledge‖

that everyone believed to be true, had flaws which temporarily sidetracked the

investigation.

2. Data gathering and analysis should have been (and probably was) used to prove the

feasibility of each of the solutions selected for implementation. The most expensive

solution was, no doubt, the purchase of the strapping machine. Cost-benefit and/or return




on investment analysis should be made. The customer charge for incorrect deposits was

also one that required analysis of error patterns and possibly a customer survey to

determine the impact of implementing the charge. Eliminating double keying should be

reasonably easy to test, since a pilot operation could be set up in parallel with the current

operations. The vacation scheduling policy change could be tested using both an

employee survey and a pilot test. The dollar loss corrective action should also be

subjected to cost-benefit analysis.

3. The two most difficult areas in which to ―hold the gains‖ would probably be in the

personnel areas. Thus the dollar loss corrective action would be difficult to administer

and could cost both supervisors and employees a great deal of discomfort and uncertainty.

Also, while not as uncomfortable, the vacation policy schedule change could be irritating

to employees and provide temptations for supervisors to make numerous exceptions.

Documents

Design for Quality and Product Excellence