CHAPTER 7

6Design for Quality and Product Excellence

CHAPTER 12Design for Quality and Product ExcellenceTeaching NotesThe 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 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 products 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 reliabilitythe 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 ISSUESTesting 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 DJs 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.

Direction ofRate ofCo.Rate ofAbsol.%FontUse ofGloss.Q&ATbl. ofLang.

ImprovementImport.NowPlanImprov.Wgt.ImprovesizeUpdatePhotoscolorsTermsSect.Contnt.Frindly.

Ease-use4.53.24.51.46.325.2%31339393

Accuracy5.03.14.61.57.429.5%9131

Timeliness3.23.83.81.03.212.7%91

Clarity3.82.63.91.55.722.7%11319313

Conciseness2.54.14.11.02.59.9%11

Import. of hows108.1427.9153.498.2460.0244.7249.1173.0

% of Import. of hows5.65%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 level4.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 customers 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 customers 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 products 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 customers 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 PROBLEMSNote: 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.Tonias 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 Tonias 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 TONIAS TASTY TACOS

PriceSizeCaloriesSodium% t-FatImprtnce

12 3 45Compet.

Eval.12 3 4 5Selling

Pts.1 2 3 4 5

TasteMoistness((

Flavor( ( ((

VisualVisually Appealing(

HealthNutritious( (( (

ValueGood 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. Newfonias 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 Newfonias 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.Tonias 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: Grabbys, Tacoking, and Sandys. 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 15 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.

PARTIAL HOUSE OF QUALITY MATRIX

FOR TONIAS TASTY TACOS

PriceSizeCaloriesSodium% t-FatImprtnce

12 3 45Compet.

Eval.12 3 4 5Selling

Pts.1 2 3 4 5

TasteMoistness((xG Q S

Flavor( ( (( xS G Q*

VisualVisually Appealing( xG S Q

HealthNutritious( (( (xQS G*

ValueGood Value( ( x Q SG*

Competitive Evaluation: Grabbys53555

Tacoking35222

Sandy's44334

Targets$0.26/ oz.7.0/ oz.80/ oz.85 mg.13%

Deployment ***

( = Very strong relationship

(= Strong relationship

( = Weak relationship

Tonias 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 Tonias is low in price per ounce, as well as calories, and percent fat, this analysis suggests that Tonias 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 Tonias 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 Newfonias 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.

PARTIAL HOUSE OF QUALITY MATRIX

FOR NEWPHONIAS SMARTPHONE CASECostSize

(in.)Wt.

(oz.)Featr.

(num.)Opr.Prog.Bat.

LifeOpr.

CostImportance

12 3 45Compet Eval.

12 3 45Selling Pts.

1 2 3 4 5

ReliableKeeps operating(((( x G S H*

CompactFits pocket( x GSH

Not heavy(( ( ( xS G Q

FeaturesCalendar, contact mgt., etc.( (( x G S H*

Ease of use Intuitive operations( ( ( xQS G*

ValueGood value( (( x Q SG*

Competitive Evaluation: Oldphonia3454555

Simfonia5432223( = Very strong relationship

Colliefonia4433434(= Strong relationship

Targets$2505 x 3.26 oz.10Win.

CE35Mod.( = 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 customers 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

BoundariesFrequencies

Cell 1-0.002511

Cell 2-0.001691

Cell 3-0.000863

Cell 4-0.000035

Cell 50.000805

Cell 60.001636

Cell 70.002464

Standard Statistical Measures

Mean0.000312

Median0.000246

Mode#N/A

Standard deviation0.001343

Variance0.000002

Max0.002456

Min-0.002514

Range0.004970

Frequency Table Problem 12-5b

Upper Cell

BoundariesFrequencies

Cell 1-0.002211

Cell 2-0.000707

Cell 30.000057

Cell 40.001567

Cell 50.002323

Standard Statistical Measures

Mean-0.000046

Median-0.000123

Mode#N/A

Standard deviation0.001204

Variance0.000001

Max0.002316

Min-0.002209

Range0.004525

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

BoundariesFrequencies

Cell 1112.003

Cell 2112.670

Cell 3113.335

Cell 4114.009

Cell 5114.670

Cell 6115.336

Cell 7116.002

Standard Statistical Measures

Mean113.96

Median114.00

Mode114.00

Standard deviation 1.14

Variance 1.29

Max116.00

Min112.00

Range 4.00

Scale B

Frequency Table Problem 12-6b

Upper Cell

BoundariesFrequencies

Cell 1114.003

Cell 2115.335

Cell 3116.0010

Cell 4117.335

Cell 5118.002

Standard Statistical Measures

Mean115.92

Median116.00

Mode116.00

Standard deviation1.12

Variance1.24

Max118

Min114

Range4.00

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

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

$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

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

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 = $ 1.2011.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

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

$50 = k (0.2)2

k = 1250

( L(x) = k (x - T)2 = 1250 (x - T)212.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 + D2) = 9 ( 12 + 02 ) = $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 + D2) = 0.5 ( 0.062 + 02 ) = $0.001814. 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:

= 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 + D2) = 500,000 ( 0.001042 + 0.000082 ) = $0.54415.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 = = 2.12 volts

2.12 = x - 120

( x = 122.1216.

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 = = 0.5 volts

0.5 = x - 120

( x = 120.517.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 = $ 54.00Weighted loss = 0.12 X $54.00 = $ 6.48

Air2Port Airplane Co.

Calculation of Taguchi Loss Values

Value Loss ($)Process P ProbabilityWeighted Loss ($)Process Q ProbabilityWeighted Loss ($)

0.2096.0000.000.021.92

0.2154.000.126.480.031.62

0.2224.000.122.880.153.60

0.236.000.120.720.150.90

0.240.000.280.000.300.00

0.256.000.120.720.150.90

0.2624.000.122.880.153.60

0.2754.000.126.480.031.62

0.2896.0000.000.021.92

Expected Loss20.1616.08

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 = 10.24b)E [L(x) = k (( 2 + D2)] = 10.24 ( 0.89572 + 0.0162 ) = $8.21819.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 Corporations 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 Corporations 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 Livelongs 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:

ComponentSupplier 1Supplier 2Supplier 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?

Answer27. 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:

MachineReliability

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 PROBLEMS31.

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:

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

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

Use this method to calculate values for the second operator:

2 = 29.901 / 30 = 0.9967;

2 = 0.380/ 10 = 0.038

D = max {

i} - min {

i} = 0.9967 - 0.9907 = 0.006

= ((

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

D4 = 2.574 ; UCLR = D4

= (2.574) (0.033) = 0.0849, all ranges below

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

EV = K1

= (3.05) (0.033) = 0.10065

AV =

= 0.0119

RR = = 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 range0.033Repeatability (EV)0.101 83.88%

X-bar range (

D)0.006Reproducibility (AV)0.012 9.93%

Repeatability and Reproducibility (R&R)0.101 84.46%

Control limit for individual ranges0.085

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 variationNote 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:

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

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

Use this method to calculate values for the second operator:

2 = 46.74 / 30 = 1.558;

2 = 1.58/ 10 = 0.158

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

3 = 47.05 / 30 = 1.568;

3 = 0.610/ 10 = 0.061

D = max {

i} - min {

i} = 1.616 1.558 = 0.058

= ((

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

D4 = 2.574 ; UCLR = D4

= (2.574) (0.117) = 0.3012, all ranges below

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

EV = K1

= (3.05) (0.117) = 0.3569

AV =

= 0.1424

RR = = 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 range0.117Repeatability (EV)0.35889.47%

X-bar range (

D)0.058Reproducibility (AV)0.14235.58%

Repeatability and Reproducibility (R&R)0.38596.28%

Control limit for individual ranges0.302

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: = 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 specification0.5750Average0.5750Cp0.3567

Upper tolerance limit0.5820Std. deviation0.0065Cpl0.3547

Lower tolerance limit0.5680Cpu0.3587

Cpk0.3547

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: = 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 specification0.5750Average0.5755Cp1.3838

Upper tolerance limit0.5820Std. deviation0.00169Cpl1.4866

Lower tolerance limit0.5680Cpu1.2810

Cpk1.2810

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

Mean24.0014

Standard Error0.00097

Median24.001

Mode24.000

Standard Deviation0.00967

Sample Variance9.4E-05

Kurtosis0.53132

Skewness0.05271

Range0.058

Minimum23.971

Maximum24.029

Confidence Level(95.0%)0.00192

BinFrequency

23.9711

23.9770

23.9830

23.9887

23.99414

24.00026

24.00620

24.01219

24.0177

24.0235

More1

For sample statistics of: = 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 - = 24.030 - 24.0014 = 0.983

3 3 (0.0097)

Cpl = - 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:

PartNominalTolerance

11.750 0.045

22.000 0.060

31.250 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: = 1.7446; s = 0.0163; 3s = 0.0489

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

Data set 3: = 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)

T = 4.9930 ; Estimated ( Process = =

= 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 OTLs 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.)

Average0.0764Cp0.8019

Standard deviation0.0104Cpl0.8468

Cpu0.7569

Cpk0.7569

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: = 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) = 25.0; ( = 1.2

Cp = UTL - LTL = 28.25 21.75 = 0.903

6 ( 6 (1.2)

Cm = Cp / = 0.903 / = 0.903

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

3 ( 3 (1.2)

Cpl = - 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)

= 23; ( = 1.2

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

6 (

6 (1.2)

Cmodified = Cp / = 0.903 / = 0.584

Because of the shift away from the target, capability is lower.

Cpu = UTL - = 28.25 - 23.0 = 1.458

3 ( 3 (1.2)

Cpl = - 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)( 2new = 0.4 (1.44) = 0.576 ( (new = 0.759

Cp = UTL -LTL = 28.25 21.75 = 1.427

6 (

6 (0.759)

Cmodified = Cp /= 1.427/= 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 - = 5.80 - = 2.0; Therefore, we get: = 5.4

3 ( 3 (

Cpl = - LTL = - 5.00 = 2.0; Therefore, we get: = 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 qualityIn 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 - = 5.80 - = 2.0; Therefore, we get: = 5.4

3 ( 3 (

Cpl = - LTL = - 5.00 = 2.0; Therefore, we get: = 5.