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# Seminar Report On Taguchi Methods2

## by pulkit bajaj on Nov 11, 2009

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CONTENTS...

CONTENTS

Chapter 1
1.1 Introduction 4
1.2 Definitions of quality 5
1.2.1 Traditional and Taguchi definition of Quality 6
1.3 Taguchi’s quality philosophy 7
1.4 Objective of Taguchi Methods 9
1.5 8-Steps in Taguchi Methodology 9

Chapter 2 (Loss Function) 10
2.1 Taguchi Loss Function 10
2.2 Variation of Quadratic Loss function 16 Chapter 3 (Analysis of Variation) 18
3.1 Understanding Variation 18
3.2 What is ANOVA 18
3.2.1 No Way ANOVA 18
3.2. 1.1 Degree of Freedom 19
3.2.2 One Way ANOVA 23
3.2.3 Two Way ANOVA 29
3.3 Example of ANOVA 35
Chapter 4 (Orthogonal Array) 45
4.1 What is Array 45
4.2 History of Array 45
4.3 Introduction of Orthogonal Array 46
4.3.1 Intersecting many factor- A case study 48
4.3.1.1 Example of Orthogonal Array 49

4.3.2 A Full factorial Experiment 57
4.4 Steps in developing Orthogonal Array 59
4.4.1 Selection of factors and/or interactions to be evaluated 59
4.4.2 Selection of number of levels for the factors 59
4.4.3 Selection of the appropriate OA 60
4.4.4 Assignment of factors and/or interactions to columns 61
4.4.5 Conduct tests 63
4.4.6 Analyze results 64
4.4.7 Confirmation experiment 67
4.5 Example Experime

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## Seminar Report On Taguchi Methods2Document Transcript

• 1 Seminar Report On Taguchi’s Methods Submitted by: Submitted to: Ishwar Chander (800982007) Mr. Tarun Nanda Pulkit Bajaj (800982019) (Sr. Lecturer) Hitesh Bansal (800982006) Department of Mechanical Engineering Thapar University, Patiala (Deemed University)
• 2 CONTENTS Chapter 1 1.1Introduction 4 1.2 Definitions of quality 5 1.2.1 Traditional and Taguchi definition of Quality 6 1.3Taguchi’s quality philosophy 7 1.4 Objective of Taguchi Methods 9 1.5 8-Steps in Taguchi Methodology 9 Chapter 2 (Loss Function) 10 2.1 Taguchi Loss Function 10 2.2 Variation of Quadratic Loss function 16 Chapter 3 (Analysis of Variation) 18 3.1 Understanding Variation 18 3.2 What is ANOVA 18 3.2.1 No Way ANOVA 18 3.2. 1.1 Degree of Freedom 19 3.2.2 One Way ANOVA 23 3.2.3 Two Way ANOVA 29 3.3 Example of ANOVA 35 Chapter 4 (Orthogonal Array) 45 4.1 What is Array 45 4.2 History of Array 45 4.3 Introduction of Orthogonal Array 46 4.3.1 Intersecting many factor- A case study 48 4.3.1.1 Example of Orthogonal Array 49
• 3 4.3.2 A Full factorial Experiment 57 4.4 Steps in developing Orthogonal Array 59 4.4.1 Selection of factors and/or interactions to be evaluated 59 4.4.2 Selection of number of levels for the factors 59 4.4.3 Selection of the appropriate OA 60 4.4.4 Assignment of factors and/or interactions to columns 61 4.4.5 Conduct tests 63 4.4.6 Analyze results 64 4.4.7 Confirmation experiment 67 4.5 Example Experimental Procedure 67 4.6 Standard Orthogonal Array 71 Chapter 5 (Robust Designing) 5.1 What is robustness 72 5.2 The Robustness Strategy uses five primary tools 72 5.2.1P-Diagram 73 5.2.2 Quality Measurement 74 5.2.3 Signal To Noise Ratio 75 5.3 Steps in Robust Para design 76 5.4 Noise Factor 77 5.5 OFF-LINE and ON-LINE Quality Control 78 5.5.1 OFF-LINE Quality Control 78 5.5.2 ON-LINE Quality Control 78 5.5.1.1 Product Design 79 5.5.1.2 Process Design 79 5.5.2.1 Product Quality Control method (On Line Quality Control Stage 1) 80 5.5.2.2 Customer Relations (On Line Quality Control Stage 2) 80 References 84
• 4 Preface When Japan began its reconstruction efforts after World War II, Japan faced an acute shortage of good quality raw material, high quality manufacturing equipment and skilled engineers. The challenge was to produce high quality products and continue to improve quality product under these circumstances. The task of developing a methodology to meet the challenge was assigned to Dr. Genichi Taguchi, who at that time was manager in charge of developing certain telecommunication products at the electrical communication laboratories (ECL) of Nippon Telecom and Telegraph Company(NTT).Through his research in the 1950’s and the early 1960’s. Dr Taguchi developed the foundation of robust design and validated his basics philosophies by applying them in the development of many products. In recognisation of this contribution, Dr Taguchi received the individual DEMING AWARD in 1962, which is one of the highest recognition in the quality field. CHAPTER 1
• 6 1.2 Definitions of Quality  “Fitness for use” Dr. Juran (1964)  The leading promoter of the “zero defects” concept and author of Quality is Free (1979) defines quality as” Conformance to requirements. Philips Crosby  Quality should be aimed at needs of consumer, present and future. Dr. Deming  The totality of features and characteristics of a product or service that bear on its ability to satisfy given needs. The American Society for Quality Control (1983)  The aggregate of properties of a product determining its ability to satisfy the needs it was built to satisfy. (Russian Encyclopaedia)  The totality of features and characteristics of a product and service that bear on its ability to satisfy a given need. (European Organization for Quality Control Glossary 1981) Although these definitions are all different, some common threads run through them: • Quality is a measure of the extent to which customer requirements and expectations are satisfied. • Quality is not static, since customer expectation can change. • Quality involves developing product or service specifications and standards to meet customer needs (quality of design) and then manufacturing products or providing services which satisfy those specifications and standards (quality of conformance). It is important to note that the above quality definitions which are before 1950’s does not refer to grade or features. For example Honda Car has more features and is generally considered to be a higher grade car than Maruti. But it does not mean that it is of better quality. A couple with two children may find that a Maruti does a much better job of meeting their requirements in terms of case of loading and unloading ,comfort when the entire family is in the car, gas mileage, maintenance, and of course ,basic cost of vehicle. 1.2.1 Traditional and Taguchi definition of Quality
• 7 Traditional Definition: The more traditional "Goalpost" mentality of what is considered good quality says that a product is either good or it isn't, depending or whether or not it is within the specification range (between the lower and upper specification limits -- the goalposts). With this approach, the specification range is more important than the nominal (target) value. But, is the product as good as it can Traditional Quality Definition be, or should be, just because it is within specification. Taguchi Definition: Taguchi says no to above definition. He define the ‘quality’ as deviation from on-target performance. According to him, quality of a manufactured product is total loss generated by that product to society from the time it is shipped. Financial loss or Quality loss Taguchi Quality Definition L(y) = k(y-m) ² y objective characteristic m target value k constant k = Cost of defective product / (Tolerance) ² = A/Δ² 1.3 Taguchi’s Quality Philosophy
• 8 Genichi Taguchi’s impact on the word quality scene has been far- reaching. His quality engineering system has been used successfully by many companies in Japan and elsewhere. He stresses the importance of designing quality into product into processes, rather than depending on the more traditional tools of on-line quality control. Taguchi’s approach differs from that of other leading quality gurus in that he focuses more on the engineering aspects of quality rather than on management philosophy of statistics. Also, Dr. Taguchi uses experimental design primarily as a tool to make products more robust- to make them less sensitive to noise factors. That is, he views experimental design as a tool for reducing the effects of variation on product and process quality characteristics. Earlier applications of experimental design focused more on optimizing average product performance characteristics without considering effects on variations. Taguchi’s quality philosophy seven basic elements: 1. An important dimension of the quality of a manufactured product is the total loss generated by product to society. At a time when the BOTTOM LINE appears to be the driving force for so many organizations, it seems strange to see “loss to society” as part of product quality. 2. In a competitive economy, continuous quality improvement and cost reduction are necessary for staying in business. This is hard lesson to learn. Masaaki Imai (1986) argues very persuasively that the principal difference between Japanese and American management is that American companies look to new technologies and innovation as the major route to improvement, while Japanese companies focus more on “Kaizen” means gradual improvement in everything they do. Taguchi stresses use of experimental designs in parameter design as a way of reducing quality costs. He identifies three types quality costs: R & D costs, manufacturing costs, and operating costs. All three costs can be reduced through use of suitable experimental designs. 3. A continues quality improvement program includes continuous reduction in the variation of product performance characteristics about their target values. Again Kaizen. But with the focus on product and process variability. This does not fit the mold of quality being of conformance to specification. 4. The customer’s loss due to a product’s performance variation is often approximately proportional to the square of the deviation of the performance characteristic from its target value. This concept of a quadratic loss function, says that any deviation from target results in some loss of the customer, but that large deviations from target result in severe losses. 5. The final quality and cost of manufactured product are determined to a large extent by the engineering designs of the product and its manufacturing process. This is so
• 9 simple, and so true. The future belongs to companies, which, once they understand the variability’s of their manufacturing processes using statistical process control, move their quality improvement efforts upstream to product and process design. 6. A product (or process) performance variation can be reduced by exploiting the nonlinear effects of the product (or process) parameters on the performance characteristics. This is an important statement because its gets to the heart of off- line QC. Instead of trying to tighten speciation beyond a process capability, perhaps a change in design can allow specifications to be loosened. As an example, suppose that in a heating process the tolerance on temperature is a function of the heating time in the oven. The tolerance relationship is represents by the band in given figure. For example : If a process specification says the heating process is to last 4.5 minutes, then the temperature must be held between 354.0 degrees and 355.0 degrees, a tolerance interval 1.0 degrees wide. Perhaps the oven cannot hold this tight a tolerance. One solution would be spending a lot of money on a new oven and new controls. Other possibility would be to change the time for the heating process to, say, 3.5 minutes. Then the temperature would need to be held to between 358.0 and 360.6 degrees an interval of width 2.6 degrees. If the oven could hold this tolerance, the most economical decision might be to adjust the specifications. This would make the process less sensitive to variation in oven temperature. Time Temperature Relationship 7. Statistically designed experiments can be used to identify the settings of product parameters that reduce performance variations. And hence improve quality,
• 10 productivity, performance, reliability, and profits, statistically designed experiments will be the strategic quality weapon of the 1990’s. 1.4 Objective of Taguchi Methods The objective of Taguchi’s efforts is process and product-design improvement through the identifications of easily controllable factors and their settings, which minimize the variation in product response while keeping the mean response on target. By setting those factors at their optimal levels, the product can be made robust to changes in operating and environmental conditions. Thus, more stable and higher-quality products can be obtained, and this is achieved during Taguchi’ parameter-design stage by removing the bad effect of the cause rather than the cause of the bad effect. Furthermore, since the method is applied in a systematic way at a pre-production stage (off-line), it can greatly reduce the number of time-consuming tests needed to determine cost-effective process conditions, thus saving in costs and wasted products 1.5 8-Steps in Taguchi Methodology 1. Identify the main function, side effects and failure mode. 2. Identify the noise factor, testing condition and quality characteristics. 3. Identify the objective function to be optimized. 4. Identify the control factor and their levels. 5. Select the Orthogonal Array, Matrix experiments. 6. Conduct the Matrix equipment. 7. Analyze the data; predict the optimum levels and performance. 8. Perform the verification experiment and plan the failure action.
• 11 CHAPTER 2 2.1 Taguchi Loss Function Genichi Taguchi has an unusual definition for product quality: “Quality is the loss a product causes to society after being shipped, other then any losses caused by its intrinsic functions.” By “loss” Taguchi refers to the following two categories. • Loss caused by variability of function. • Loss caused by harmful side effects. An example of loss caused by variability if function would be an automobile that does not start in cold weather. The car’s owner would suffer a loss if he or she had to pay some to start a car. The car owners employer losses the services of the employee who is now late for work. An example of a loss caused by a harmful side effect would be frost by it suffered by the owner of the car which would not start. An unacceptable product which is scrapped or rework prior to shipment is viewed by Taguchi as a cost to the company but not a quality loss.
• 12 2.1.1 Comparing The Quality Levels of SONY TV Sets Made in JAPAN and in SAN DIEGO The front page of the Ashi News on April 17,1979 compared the quality levels of Sony color TV sets made in Japanese plants and those made in San Diego, California, plant. The quality characteristic used to compare these sets was the color density distribution, which affect color balance. Although all the color TV sets had the same design, most American customers thought that the color TV sets made in San Diego plant were of lower quality than those made in Japan. The distribution of the quality characteristic of these color TV sets was given in the Ashi News (shown in Figure).The quality characteristics of the TV sets from Japanese Sony plants are normally disturbed around the target value m. If a value of 10 is assigned to range of the tolerance specifications for this objective characteristic, then the standard deviation of this normally distributed curve can be calculated and is about 10/6. In quality control, the process capability index(Cp) is usually defined as the tolerance specification divided by 6 times the standard deviation of the objective characteristic: Cp=Tolerance/6*Standard deviation
• 13 Therefore, the process capability index of the objective characteristic of Japanese Sony TV sets is about 1. In addition, the mean value of the distribution of these objective characteristics is very close to the target value of m. On the other hand, a higher percentage of TV sets from San Diego Sony are within the tolerance limits than those from Japanese Sony. However, the color density of San Diego TV sets is uniformly distributed rather than normally distributed. Therefore, the standard deviation of these uniformly distributed objective characteristics is about 1/√12 of the tolerance specification. Consequently, the process capability index of the San Diego Sony plant is calculated as follows: Cp=Tolerance/6(Tolerance/√12) = 0.577 It is obvious that the process capability index of San Diego Sony is much lower than that of Japanese Sony. All products that are outside of the tolerance specifications are supposed to be consider defective and not shipped out of the plant. Thus products that are within tolerance specifications are assumed to be pass and are shipped. As a matter of fact, tolerance specifications are very similar to tests in schools, where 60% is usually the dividing line between passing and failing, and 100% is ideal score. In our example of ideal TV sets, the ideal consideration is that the objective characteristics, color density, meet the target value m. The more the color density deviates from the target value, the lower the quality level of TV set. If the deviation of color density is over the tolerance specifications, m ±5, a TV set is considered defective. In the case of school test, 59% or below is failing, while 60% or above is passing. Similarly, the grades between 60% and 100% in evaluating quality can be classified as follows: 60%-69% Passing (D) 70%-79% Fair (C) 80%-89% Good (B)
• 15 is result is obtained because the constant term L(m) and the first derivative term L’(m) are both zero. In addition, the third order and higher order term are assumed to be negligible. Thus, one can express the loss function as a squared term multiplied by constant k: L(y) =k(y-m)² …………… Equation 2.3 When the deviation of the objective characteristic from the target value increases, the corresponding quality loss will increase. When magnitude of deviation is outside of the tolerance specifications, this product should be considered defective. Let the cost due to defective product be A and the corresponding magnitude of the deviation from the target value be Δ. Taking Δ into right hand side of Equation (2.3), one can determine the value for constant k by following Equation: k=cost of defective product/(tolerance)² In the case of the SONY colour TV sets, let the adjustment cost be A= 300 Rs, when the colour density is out of the tolerance specifications. Therefore, the value of k can be calculated by the following equation: k = 600/5² = 12 Rs Therefore, the loss function is L(y) = 12(y – m)² This equation is still valid even when only one unit of product is made. Consider the visitor to the BHEL Heavy Electric Equipment Company in India who was told, “In our company, only one unit of product needs to be made for our nuclear power plant. In fact, it is not really necessary for us to make another unit of product. Since the sample size is only one, the variance is zero. Consequently, the quality loss is zero and it is not really necessary for us to apply statistical approaches to reduce the variance of our product.” However, the quality loss function [L = k(y – m)²] is defined as the mean square deviation of objective characteristics from their target value, not the variance of products. Therefore, even when only one product is made, the corresponding quality loss can still be calculated by Equation (2.3). Generally, the mean square deviation of objective characteristics from their target values can be applied to estimate the mean value of quality loss in Equation (2.3). One can calculate the mean square deviation from target σ² (σ² in this equation is not variance) by the following equation (the term σ² is also called the mean square error or the variance of products): Quality of Sony TV sets where the tolerance specification is 10 and the objective function data corresponds to figure
• 16 Plant Mean Value Loss L Defective Standard of Objective Variation Location Function Deviation (in Rs) Ratio Japan M 10/6 10²/36 33 0.27% San Diego M 10/√12 10²/12 100 0.00 Substituting this equation into Equation (2.3), one gets the following equation: L = kσ² From Equation (2.4), one can evaluate the differences of average quality levels between the TV sets from Japanese Sony and those from San Diego Sony as shown in Table 2.1. From table 2.1 it is clear that although the defective ratio of the Japanese Sony is higher than that of the san Deego Sony, the quality level of the former is 3 times higher than the latter. Assume that one can tighten the tolerance specifications of the TV sets of San Diego Sony to be m +- 10/3. Also assume that these TV sets remain uniformly distributed after the tolerance specifications are tightened. The average quality level of San Diego Sony TV sets would be improved to the following quality level:
• 17 L = 12[(1/ √12) (10) (2/3)] ² = 44Rs where the value of the loss function is considered the relative quality level of the product. This average quality level of the TV sets of San Diego Sony is 56Rs higher than the original quality level but still 11Rs lower than that of Japanese Sony TV sets. In addition, in this type of quality improvement, one must adjust the products that are between the two tolerance limits,m +- 10/3 and m+- 5, to be within m +- 10/3. In the uniform distribution shown in Figure 2.1, 33.3% would need adjustment, which would cost 300Rs per unit. Therefore each TV set from San Diego Sony would cost an additional 100Rs on average for the adjustment: (300)(0.333) = 100Rs Consequently, it is not really a good idea to spend 100Rs more to adjust each product, which is worth only 56Rs.A better way is to apply quality management methods to improve the quality level of products. 2.2 Variation of the Quadratic Loss Function 1. Nominal the best type: Whenever the quality characteristic y has a finite target value, usually nonzero, and the quality loss is symmetric on the either side of the target, such quality characteristic called nominal-the-best type. This is given by equation L(y) =k(y-m)² ………………………Equation 1 Color density of a television set and the out put voltage of a power supply circuit are examples of the nominal-the-best type quality characteristic. 2. Smaller-the-better type: Some characteristic, such as radiation leakage from a microwave oven, can never take negative values. Also, their ideal value is equal to zero, and as their value increases, the performance becomes progressively worse. Such characteristic are called smaller-the-better type quality characteristics. The response time of a computer, leakage current in electronic circuits, and pollution from an automobile are additional examples of this type of characteristic. The quality loss in such situation can be approximated by the following function, which is obtain from equation by substituting m = 0 L(y) =ky² This is one side loss function because y cannot take negative values.
• 18 3. Larger-the-better type: Some characteristics, such as the bond strength of adhesives, also do not take negative values. But, zero is there worst value, and as their value becomes larger, the performance becomes progressively better-that is, the quality loss becomes progressively smaller. Their ideal value is infinity and at that point the loss is zero. Such characteristics are called larger-the-better type characteristics. It is clear that the reciprocal of such a characteristics has the same qualitative behavior as a smaller-the-better type characteristic. Thus we approximate the loss function for a larger-the-better type characteristic by substituting 1/y for y in equation1: L(y) = k [1/y²] 4. Asymmetric loss function: In certain situations, deviation of the quality characteristic in one direction is much more harmful than in the other direction. In such cases, one can use a different coefficient k for the two directions. Thus, the quality loss would be approximated by the following asymmetric loss function: k(y-m) ²,y>m L(y) = k(y-m) ², y≤m
• 19 CHAPTER 3 Introduction to Analysis of variation 3.1 Understanding variation The purpose of product or process development is to improve the performance characteristics of the product or process relative to customer’s needs and expectations. The purpose of experimentation should be to reduce and control variation of a product or a process; subsequently decisions must be made which parameters affect the performance of a product/process. Since variation is a large part of the discussion relative to quality, analysis of variation (ANOVA) will be the statistical method used to interpret experimental data and make necessary decisions. 3.2 What is ANOVA ANOVA is a statistically based decision tool for detecting any differences in average performance of groups of items tested. ANOVA is a mathematical technique which breaks total variation down into accountable sources; total variation is decomposed into its appropriate components We will start with a very simple case and then build up more comprehensive situations Thereafter, we will apply ANOVA to some very specialized experimental situations 3.2.1 No way ANOVA Imagine an engineer is sent to the production line to sample a set of windshield pumps for the purpose of measuring flow rate. The data collected is as under: Pump No. 1 2 3 4 5 6 7 8 Flow rate 5 6 8 2 5 4 4 6 oz/min. 1 oz/min. = 0.473 ml/s
• 20 No-way ANOVA breaks total variation down into only two components 1. The variation of the average (or mean) of all the data points relative to zero 2. The variation of individual data points around the average (traditionally called experimental error) The notations used in the calculation method are as under: y = observation or response or simply data y i = ith response for example y3 = 8 oz/min N = Total number of observations T = Sum of all observations T = Average of all observations = T/N = y In this case N = 8, T = 40 oz/min , and T = 5.0 oz/min What is the reason for variation from pump to pump? The true flow rate is actually unknown; it is only estimated through the use of some flow meter. There will be some unknown measurement error present, but flow rate will nonetheless be observed and accepted as the pump’s performance under the conditions of the test. Also, the pumps were randomly selected. Although the manufacturer produces identical pumps; however there will be slight differences from pump to pump, causing a pump to pump variation in performance. (This is natural variation of the process) It is for this reason the flow rates of pumps are not identical. No-way ANOVA can be illustrated graphically. (Notes of this section must be taken by hand) The magnitude of each observation can be represented by a line segment extending from zero to the observation. These line segments can be divided into two portions: - One portion attributed to the mean; - Other portion attributed to the error The error includes the natural process variation and the measurement error. The magnitude of the line segment due to the mean is indicated by extending a line from the average value to zero. - The magnitude of line segment due to error is indicated by the difference of the average value from each observation.
• 21 To calculate the total variation present we will do a mathematical operation which will allow a clearer picture to develop. The magnitudes of each of the line segments can be squared and then summed to provide a measure of the total variation present SS T = total sum of squares = 5 2 + 6 2 + 8 2 + − − − − − + 6 2 SS T = 222.0 The magnitude of line segment due to mean can also be squared and summed SS m = sum of squares due to mean = N (T ) 2 T  2 T2 40 2 But T = T / N SS m = N   = = = 200.0 N N 8 The portion of the magnitude of the line segment due to error can be squared and summed to provide measure of the variation around the average value n SSe = error sum of squares = ∑i =1 ( yi − y ) 2 SS e = 0 + 1 + 3 + (−3) + 0 2 + (−1) 2 + (−1) 2 + 12 2 2 3 2 SSe = 22.0 Note that 222.0 = 200.0 + 22.0 This demonstrates a basic property of ANOVA that the total sum of squares is equal to the sum of squares due to known components SST = SSm + SSE The formulas for the sum of squares can be written generally N SS T = ∑ y i2 i =1 SS m = T2 N SS = ∑ ( y − T ) N 2 e i I =1 In the above example the error value was calculated, but it is not necessary as SSe = SST - SSm
• 22 3.2.1.1 Degrees of Freedom (dof) To complete ANOVA calculations, one other element must be considered i.e. Degrees of Freedom. The concept of dof is to allow 1 dof for each independent comparison that can be made in the data. Only 1 independent comparison can be made between the mean of all the data (There is only 1 mean). Therefore, only 1 dof is associated with the mean. The concept of dof also applies to the dof associated with the error estimate. With reference to 8 observations, there are 7 independent comparisons that can be made to estimate the variation in data. Data point 1 can be compared to 2, 2 to 3, 3 to 4 etc. which are 7 independent comparisons. One of the questions an instructor dreads most from an audience is, "What exactly is degrees of freedom?" It's not that there's no answer. The mathematical answer is a single phrase, "The rank of a quadratic form." It is one thing to say that degrees of freedom is an index and to describe how to calculate it for certain situations, but none of these pieces of information tells what degrees of freedom means. At the moment, I'm inclined to define degrees of freedom as a way of keeping score. A data set contains a number of observations, say, n. They constitute n individual pieces of information. These pieces of information can be used either to estimate parameters (mean) or variability. In general, each item being estimated costs one degree of freedom. The remaining degrees of freedom are used to estimate variability. All we have to do is count properly. A single sample: There are n observations. There's one parameter (the mean) that needs to be estimated. That leaves (n-1) degrees of freedom for estimating variability.
• 23 Two samples: There are n1 + n2 observations. There are two means to be estimated. That leaves (n1 + n2 − 2) degrees of freedom for estimating variability. Let v = dof vt = Total dof vm = dof associated with mean (always 1 for each sample) ve = dof associated with error Then vt = vm + ve 8=1+7 The total dof equals total number of observations in the data set for this method of ANOVA Summary of No-way ANOVA table: Source SS Dof Mean 200 1 Error 22 7 Total 222 8 One other statistic calculated from ANOVA is variance V. Error variance or just variance is SS e 22 Ve = = = 3.14 ve 7 Also, sample standard deviation s = V Where N ∑( y − y) 2 i s= i =1 N −1 ∑ ( y − y ) = variance i 2 s2 =V = N −1 Which is essentially SS Ve = e ve
• 24 Although the formula above is faster than ANOVA for calculating error variance in this case, but when the experimental situations become more complex ANOVA will become faster method. Error variance is a measure of the variation due to all the uncontrolled parameters, including measurement error involved in a particular experiment (set of data collected). Which is essentially the natural variation of a process 3.2.2 One – Way ANOVA This is next most complex ANOVA to conduct. This situation considers the effect of one controlled parameter upon the performance of product or process, in contrast to no-way ANOVA, where no parameters were controlled. Again let us try to solve this problem through an imaginary, yet potentially real situation. Imagine the same engineer who took samples for flow rate of windshield pumps is charged with the task of establishing the fluid velocity generated by the windshield washer pumps. This means when the fluid velocity is too low, the fluid will merely dribble out, and if too high the air movement past the windshield will not be able to distribute the cleaning fluid adequately to satisfy the car driver. The engineer proposes a test of three different orifice areas to determine which give a proper fluid velocity. Before the test data is collected some notation in order to simplify the mathematical discussion is: A = Factor under investigation (outlet orifice area) A1 = Ist level of orifice area = 0.0015 sq. in A2 = 2nd level of orifice area = 0.0030 sq. in A3 = 3rd level of orifice area = 0.0045 sq. in The same symbol for the level designations will be used to denote the sum of responses Ai = sum of observations under Ai level Ai = Average of observations under Ai level = Ai / n Ai T = sum of all observations T = Average of all observations = T/N n Ai = number of observations under Ai level k A = number of levels of factor A
• 25 With notation in mind, the engineer constructs four pumps with each given orifice area (making 12 to test for three levels) The test data is as under: Level Area sqin Velocity Ft/s Total A1 0.0015 2.2 1.9 2.7 2.0 8.8 A2 0.0030 1.5 1.9 1.7 -* 5.1 A3 0.0045 0.6 0.7 1.1 0.8 3.2 Grand Total 17.1 * Dropped pump and destroyed it, no data A1 = 8.8 ft/s n A1 = 4 A1 = 2.2 ft/s A2 = 5.1 ft/s n A2 = 3 A2 = 1.7 ft/s A3 = 3.2 ft/s n A3 = 4 A3 = 0.8 ft/s kA = 3 T = 17.1 ft/s N = 11 T = 1.6 ft/s Sum of squares (One-way ANOVA) Two methods can be used to complete the calculations - Including the mean - Excluding the mean Method 1 (Including the mean) As before the total variation can be decomposed into its appropriate components: - The variation of the mean of all observations relative to zero – VARIATION DUE TO MEAN - The variation of the mean of observations under each factor level around the average (mean) of all observations –VARIATION DUE TO FACTOR A - The variation of individual observations around the average of observations under each level – VARIATION DUE TO ERROR The calculations are identical to No-way ANOVA example, except for the component of variation due to factor A, outlet orifice area. N SS T = ∑ y i2 = 2.2²+1.9²+ -----------+0.8² = 31.90 i =1 T 2 = 17.1²/11 = 26.583 SS m = N (T ) 2 = N Graphically, also this can be shown (Pl make hand notes here for graphical representation)
• 26 The magnitude of segments for each level of factor A is squared and summed. For ( instance, the length of the line segment due to level A1 is A1 − T . ) There are four observations under A1 condition. The same type of information is collected for other levels of factor A. SS A = n A 1 ( A1 − T ) + n A 2 ( A2 − T ) + n A3 ( A3 − T ) 2 2 2 SS A = 4(0.64545)² + 3(0.14545) ² + 4(-0.75454) ² = 4.007 The above calculation is tedious and is mathematically equivalent to:  k A  Ai2  T 2 SS A = ∑    −   i =1  n Ai   N  8.8 2 5.12 3.2 2 17.12 = 4.007 which is same as above SS A = + + − 4 3 4 11 The variation due to error is given by 2 SS e = ∑ ∑ ( y i − A j ) k A n Ai j =1 i =1 SSe = 02 + ( − 0.3) + 0.52 + ( −0.2) 2 + − − − − 0.32 + 02 2 SS e = 0.600 Error variation is again based on method of least squares but in one way ANOVA the least squares are evaluated around the average of each level of the controlled factor. Error variation is the uncontrolled variation within the controlled group. Again the total variation is SST = SSm + SSA + SSe 31.190 = 26.583 + 4.007 + 0.600 Dof (Including the mean) v t = vm + v A + ve v t = 11, vA = k A − 1 = 2
• 27 ve = 11 – 1 – 2 = 8 One way ANOVA summary (Method 1) Source SS Dof v Variance V Mean m 26.583 1 26.583 Factor A 4.007 2 2.004 Uncontrolled Error e 0.600 8 0.075 Total 31.190 11 In the above table we have been able to estimate variance for both factor A and uncontrolled error. This is what will be of interest to us when we design experiments. Also, if look at the calculations done for Method 1, you will observe that mean does not affect the calculations for the variation due to factor A and error in any way. Thus in most experimental situations, mean has no practical value with the exception of ‘lower is better’ situations where the variation due to mean is a measure of how far the average is from zero and how successful the factor might be in reducing the average to zero. Method 2: When we exclude the mean from ANOVA calculations, the total variation is then calculated as: - The variation of average of observations under each factor level around the average of all observations - The variation of the individual observations around the average of observations under each factor level Again graphically this can be demonstrated: The same concept of summing the squares of the magnitudes of various line segments is applied in method 2 as well. N SST = ∑ ( yi − T ) 2 = 4.607 i =1 Mathematically this is equivalent to N T2 SST = ∑ ( yi ) −2 i =1 N
• 28 This expression will be used to define the total variation by this method. See this equivalent to ( SS T − SS m ) from previous calculations The variation due to factor A and uncontrolled error is calculated identically as in Method 1  k A  Ai2  T 2 SS A = ∑    −   i =1  n Ai   N  k A n Ai 2 SS e = ∑ ∑ ( y i − A j ) j =1 i =1 Dof (Excluding the mean) In method 1, dof was vt = vm + v A + ve Where vm = 1 (always) and vt =N In Method 2(excluded mean), the dof for mean is subtracted from both sides of the above equation So v t = N – 1 = 11 – 1 = 10 v t = vA + ve 10 = ( k A - 1) + ve so ve = 8 One way ANOVA summary (Method 2) Source SS Dof v Variance V Factor A 4.007 2 2.004 Uncontrolled Error e 0.600 8 0.075 Total 4.607 10 The values of variance for factor A and Error in both methods are identical. The value of mean is disregarded in method 2 and is the most popular method.
• 29 Only when the performance parameter is ‘lower is better’ characteristic would the variance due to mean be relevant; this provides a measure of how effective some factor might be in reducing the average to zero. Let us sum up the above discussion Define three "Sums of Squares" Total Corrected Sum of Squares (SST) • Squared deviations of observations from overall average Error Sum of Squares (SSE) • Squared deviations of observations from treatment averages Treatment Sum of Squares (SStrt) • Squared deviations of treatment averages from overall average (times n) Dot Notation a n y.. = ∑ ∑ i =1 j =1 yij y.. = y../N (the overall average) n yi. = ∑ yij j =1 yi. = yi./n (the average within Treatment i) a n Raw SS = ∑ ∑ i =1 j =1 yij2 Total Corrected SS a n SST = ∑ ∑ ( yij - yi. )2 i =1 j =1 This measures the overall variability in the data. SST/(N-1) is just the sample variance of the whole dataset DECOMPOSITION OF SS I will do (hopefully) derivation of the following equation:
• 30 SST = SStrt + SSE a n 2 a n 2 SS = ∑ ∑  y − y  = ∑ ∑  y − y + y − y      T  ij   ij i. i.  i =1 j =1 i =1 j = 1 a n   2 = ∑ ∑   y − y  +  y − y    ij i.  i.     i = 1 j =1 a n 2 a n 2 a n   = ∑ ∑  y − y  + ∑ ∑  y − y  + ∑ ∑ 2  y − y  y − y     ij i.  i.   ij i.  i.     i = 1 j =1 i =1 j =1 i = 1 j =1 a 2 a n   = SS + ∑ n y − y  + ∑ ∑ 2  y − y  y − y    E  i.   ij i.  i.  i =1 i =1 j =1 a n   = SS + SS + 2  y − y  y − y    E TRT ∑ ∑   ij i.  i.   i =1 j =1 = SS + SS +0 E TRT Must show last term is zero a n   a n ∑ ∑ 2  yij − yi.  yi. − y   = 2 ∑  yi. − y  ∑  yij − yi.              i =1 j =1 i =1 j =1       a   n   = 2 ∑ y − y ∑ y − ny            i.   ij i.         i =1   j =1     a a = 2 ∑  y − y ny − ny  = 2 ∑  y − y 0 = 0  i.  i. i.   i.   i =1 i =1 Two –Way ANOVA Two-way ANOVA is the next highest order of ANOVA to review There are two controlled parameters in this experimental situation Let us consider an experimental situation. A student worked at an aluminum casting foundry which manufactured pistons for reciprocating engines. The problem with the process was how to attain the proper hardness (Rockwell B) of the casting for a particular product. Engineers were interested in the effect of copper and magnesium content on casting hardness.
• 31 According to specs the copper content could be 3.5 to 4.5% and the magnesium content could be 1.2 to 1.8%. The student runs an experiment to evaluate these factors and conditions simultaneously. If A = % Copper Content A1 = 3.5 A2 = 4.5 If B = % Magnesium Content B1 = 1.2 B2 = 1.8 The experimental conditions for a two level factors is given by 2 f = 4 which are A1 B1 A1 B2 A2 B1 A2 B2 Imagine, four different mixes of metal constituents are prepared, casting poured and hardness measured. Two samples are measured from each mix for hardness. The result will look like: A1 A2 B1 76, 78 73, 74 B2 77, 78 79, 80 To simplify discussion 70 points from each value are subtracted in the above observations from each of the four mixes. Transformed results can be shown as: A1 A2 B1 6, 8 3, 4 B2 7, 8 9, 10 Two way ANOVA calculations: The variation may be decomposed into more components: 1. Variation due to factor A 2. Variation due to factor B 3. Variation due to interaction of factors A and B 4. Variation due to error An equation for total variation can be written as SS T = SS A + SS B + SS A×B + SS e A x B represents the interaction of factor A and B. The interaction is the mutual effect of Cu and Mg in affecting hardness. Some preliminary calculations will speed up ANOVA
• 32 A1 A2 Total B1 6, 8 3, 4 21 B2 7, 8 9, 10 34 Total 29 26 55 Grand Total A1 = 29 A2 = 26 B1 =21 B2 =34 T = 55, N=8 n A1 = 4 nB1 = 4 n A2 = 4 nB 2 = 4 Total Variation N T2 SST = ∑ ( yi ) − 2 i =1 N 55 2 = 6² + 8² + 3² + ----------- + 10² - = 40.875 8 Variation due to factor A  k A  Ai2  T 2 SS A = ∑    n  − N  i =1  Ai    2 2 2 A1 A2 Ak T2 + + −− − − − − − + − n A1 n A2 n Ak N 29 2 26 2 552 SS A = + − = 1.125 4 4 8 Please carry out a mathematical check here which is Numerator 29 + 26 = 55 and Denominator 4 + 4 = 8 If these conditions are not met then the SS A calculation will be wrong. For a two level experiment, when the sample sizes are equal, the equation above can be simplified to this special formula:
• 33 ( A1 − A2 ) 2 ( 29 −26) 2 SS A = = = 1.125 N 8 Similarly the variation due to factor B SS B = (B1 −B2 )2 = ( 21 − ) 2 34 =21.125 N 8 To calculate the variation due to interaction of factors A and B Let ( A × B) i represent the sum of data under the ith condition of the combination of factor A and B. Also let c represent the number of possible combinations of the interacting factors and n( A×B )i the number of data under this condition.  c  ( A ×B ) 2  T 2 SS A×B ∑ n =   i  −  N −SS A −SS B i =  ( A×B ) i  1  Note that when the various combinations are summed, squared, and divided by the number of data points for that combination, the subsequent value also includes the factor main effects which must be subtracted. While using this formula, all lower order interactions and factor effects are to be subtracted. For the example problem: A1 B1 = 14, A1 B2 = 15, A2 B1 = 7 A2 B2 = 19 And no. of possible combinations c = 4 And since there are two observations under each combination n( A×B )i = 2 14 2 7 2 15 2 19 2 55 2 SS A× B = + + + − − SS A − SS B = 15.125 2 2 2 2 8 Since SS T = SS A +SS B +SS A×B +SS e SS e = 40.875 −1.125 −21.125 −15.125 = 3.500
• 34 Degrees of Freedom (Dof) – Two way ANOVA vt = v A + vB + v A×B + ve vt =N–1 =8–1=7 vA = k A -1=1 vB = k B -1=1 v A×B = (v A )(vB ) = 1 ve = 7 − 1 − 1 − 1 = 4 ANOVA summary Table (Two-way) Source SS Dof v Variance V F A 1.125 1 1.125 1.29 B 21.125 1 21.125 24.14* AxB 15.125 1 15.125 17.29** E 3.500 4 0.875 Total 40.875 7 * at 95% confidence ** at 90% confidence The ANOVA results indicate that Cu by itself has no effect on the resultant hardness, magnesium has a large effect (largest SS) on hardness and the interaction of Cu and Mg plays a substantial part in determining hardness. A plot of these data can be seen. (Take hand notes here) In this plot there exist non parallel lines which indicate the presence of an interaction. The factor A effect depends on the level of factor B and vice versa. If the lines are parallel, there would be no interaction which means the factor A effect would be same regardless of the level of factor B.
• 35 Hardn ess B2 10 8 6 4 B1 2 A1 A2 Geometrically, there is some information available from the graph that may be useful. The relative magnitudes of the various effects can be seen graphically. The B effect is the largest, A x B effect next largest and the A effect is very small. See 29 26 A1 = = 7.1 A2 = = 6.5 4 4 21 34 B1 = = 5.2 B2 = = 8.5 4 4
• 36 Hardn ess 10 B2 8 B effect A effect 6 Mid pt. A2B1 & A2B2 Mid pt. on line B1 A x B effect B1 4 2 A1 A2 So by plotting the data for each factor we could observe the following: (Make hand notes here) In the first case, there is no interaction because the lines are parallel In 2nd case, some interaction In 3rd case, there is a strong interaction 3.3 EXAMPLE OF ANOVA During the late 1980s, Modi Xerox had a large base of customers (50 thousand) for this model spread over the entire country. Many buyers of these machines earn their livelihood by running copying services. Each of these buyers ultimately serves a very large number of customers (end user). When copy quality is either poor or inconsistent, customers earn a bad name and their image and business gets affected. In the late 1980s, the company integrated the total quality management philosophy into its operation and placed the highest focus on customer satisfaction—any problem of field failure was given the highest priority for investigation. The problem of skips was subjected to
• 37 detailed investigation by a cross-functional team from the Production, Marketing, and Quality Assurance Departments. Problem Description The pattern of blurred images (skips) observed in the copy is shown in Figure above. It usually occurs between 10-60 mm from the lead edge of the paper. Sometimes, on a photocopy taken on a company letterhead paper, the company logo gets blurred, which is not appreciated by the customer. This problem was noticed in only one-third of the machines produced by the company, with the remaining two-thirds of machines in the field working well without this problem. The in-house test evaluation record also confirmed the problem in only about 15% of the machines produced. Data analysis indicated that not all the machines produced were faulty. Therefore, the focus of further investigation was to find out what went wrong in the faulty machines or whether there are basic differences between the components used in good and faulty machines. Preliminary Investigation A copier machine consists of more than 1000 components and assemblies. A brainstorming session by the team helped in the identification of 16 components suspected to be responsible for the problem of blurred images. Each Suspected component had at least two possible dimensional characters which could have resulted in the skip symptom. This led to more than 40 probable causes (40 dimensions arising out of 16 components) for the problem. An attempt was made by the team to identify the real causes among these 40 probable causes. Ten bad machines were stripped open and various dimensions of these 16 component were measured. It was observed that all the dimensions were well within specifications Hence, this investigation did not give any clues
• 38 to the problem. Moreover, the time and effort spent in dismantling the faulty machines and checking various dimensions in 16 components was in vain. This gave rise to the thought that conforming to specification does nor always lead to perfect quality. The team needed to think beyond the specification in order to find a solution to the problem Taguchi Experiment An earlier brainstorming session had identified 16 components that were likely to be the cause of this problem. A study of travel documents of 300 problem machines revealed that on 88% of occasions, the problem was solved by replacing one or more of only four parts of the machine. These four parts were from the list of 16 parts identified earlier. They were considered to be Critical and it was decided to conduct an experiment on these four parts. These parts were the following: (a) Drum shaft (b) Drum gear (c) Drum flange (d) Feed shaft Two sets of these parts Were taken for Experiment I, one from an identified problem machine and one from a problem free machine. The two levels considered in the experiment were good and bad; ‘good’ signifying parts from the problem-free machine and ‘bad’ signifying parts from the problem machine. The factors and levels thus identified are given in Table below. A full factorial experiment would have required 16 trials while the experiment was designed in L8(27) fractional factorial using a linear graph and orthogonal array (OA) table developed by Taguchi.
• 39 The linear graph is presented in Fig. 9.24 and the layout in Table 9.14. A master plan for conducting eight experiments was prepared. The response considered was the number of defective copies (copies exhibiting the skip problem) in a 50-copy run. The master plan along with the response is presented in Table 9.15.
• 40 Analysis and Results The response considered was fraction defective (p5 d/n). Data were normalized by the transformation sin ‾ ¹ (p½). Analysis of variance (ANOVA) was performed on normalized data and the results are presented in Table . F(1,.5) at 0.05 = 6.61, F(1,.5) at 0.01=16.26. pA=(3528-32.4)*100/3788 =923 As can be seen from Table factor A is highly significant (the only significant factor), explaining 92.3% of the total variation. In other words, of the four components studied, the drum shaft alone is the source of trouble for skip. The problem was now narrowed down to one component from the earlier list of 16 components, giving a ray of hope for moving towards a solution. Further investigations were carried out on drum shaft design. Drum Shaft Design
• 41 The configuration of the drum shaft is defined by 15 dimensions. A brainstorming session by the team members identified wobbling and increased play in the drum shaft as major causes for this problem. Four dimensions of the drum shaft were suspected to be causing wobbling and excessive play. These dimensions in all 20 machines (10 good and 10 bad) were checked and found to be well within specification Now the question as to where the problem lay arose-definitely not within the specification, perhaps outside the specification? This led u to think beyond the specification in order to find a solution. As a first step, the dimension patterns of good and bad machines were compared. The dimension patterns for four critical dimensions suspected to be the cause of the problem are show ft in Figure below. There is not much difference in pattern between good and bad machines with respect to dimensions B, C, and D. Dimension A, that is, diameter over pin (DOP) dimensions of the drum shaft splines revealed a difference in pattern between the good and bad machines. The DOP of shafts from 10 problem machines were found to lie in the lower half of the specification range, whereas in case of problem-free machines, the DOP was found to be always on the
• 42 upper half of the specification range(shown in fig. above). The DOP dimension in the drum shaft is shown in Figure below. DOP (diameter over pin) is a measure of the tooth thickness, t. Higher DOP means greater tooth thickness of the splines and vice versa. If the DOP in the splines of the drum shaft is on the lower side, then it will increase the clearance, resulting in more play between the drum shaft and the drum gear assembly .
• 43 Here, the image of the original document is transferred on the photoreceptor drum through a series of lenses and mirrors. The photoreceptor drum is coated with a photo-conductor material and it is electrically charged with positive charge. During the transfer of the image from the document, the whole of the drum area is exposed to light except the area where the image is formed, Due to the exposure to light, the photo-conductor material becomes a conductor and the charge is neutralized, except in the image area. This image is called ‘latent image’. Subsequently, this image is transferred to paper through toner and developer. During the transfer of image, the drum should rotate at a uniform speed. Any jerk to the photoreceptor drum during rotation will cause distortion or blur on the latent image. The photoreceptor drum is driven by the drum shaft anti drum gear assembly. An excessive play between drum shaft and drum gear gets magnified and produces jerks in the photoreceptor unit. Bad machine dimension pattern clearly indicates the possibility of an excessive play between drum shaft and drum gear assembly. A sketch of the Photoreceptor assembly is shown below here.
• 44 A lower DOP results in Producing a large gap between the drum shaft and drum gear which causes excessive play in the drum shaft. Technically, excessive play between the drum and drive gear can cause the skip problem This theory was further confirmed when this model (X) was compared with model Y and model Z where no skip problem was observed. In models Y and Z, the drum shaft and drive gear were integrated into a single unit. This Probably explains the zero play and no skip defect. The drum and drive gear assembly of the three models X, Y, and Z are shown in figure. for comparison For further validation of this point, play between the drum shaft and drive gear was eliminated by temporarily integrating the system using a drop of araldite (glue) in 50 problem machines. A test run was pen on all 50 machine and no skip defect was observe This led to the conclusion that drum shaft DOP specifications are not fail-safe against skips. It was now felt necessary to arrive at new specifications for DOP to ensure no excessive play between the drum
• 45 shaft and drive gear. The question arose as to how much play can be permitted. To find an answer, a similar drive system of the very successful two-wheeler Lambretta Scooter was studied, and it was found that the Play varied between 0.04 and 0.07 mm. To be on the safe side, it was decided to allow maximum play of only 0.04 mm between the drum shaft and drive gear. These drum shafts are manufactured by subcontractors, so new specifications were reached by taking into consideration the supplier’ s capability of machining these dimensions and a maximum permissible play of 0.04mm. The old and new specifications for DOP are shown in figure. Confirmatory Trial The implications of the new specifications on other systems of the machine were examined and it was found that the change in specification would not create any problems. The 36 worst-affected machines were selected from the field. Drum shafts with the new specifications were made and then fitted on these machines. Test results of these machines showed a total elimination of skip defects. Ultimately, to give the customer the benefits of the study, 5000 drum shafts with 30 new specifications were made and incorporated in 5000 existing machines with the old design in the field. A sample performance audit of 800 machines (out of those 5000) in the field was carried out and none of these 800 machines indicated skip problems. This provided confidence that the new design had worked successfully. After that, the new design was implemented fully by releasing the new specification. The rate of occurrence of the skip problem in the assembly line dropped from the previous 13% to less than 0.5%. Beating the Benchmark
• 46 Machine specifications released from Rank Xerox (UK) permit the occurrence of skip up to 10mm from the lead edge. Earlier specifications of Modi Xerox permitted the occurrence of skip up to 60 mm from the lead edge, but, to most of the customers, loss of information near the lead edge is not acceptable, as the company logos are located near the lead edge of the letterheads. The exercise was initially taken up to reach the standard of Rank Xerox (skip up to 10mm). Now, the modified design of the drum shaft, evolved through scientific and systematic investigation, has completely eliminated the skip and hence has surpassed even the Rank Xerox benchmark of permitting skip up to 10 mm from the lead edge. This is a great accomplishment towards skip-free copy. A problem is completely solved for which no solution was previously available worldwide. CHAPTER 4 4.1 WHAT IS ARRAY An Array’s name indicates the number of rows and column it has , and also the number of levels in each of the column. Thus the array L4(2³) has four rows and three 2-level column. 4.2 HISTORY OF ORTHOGONAL ARRAY Historically, related methods was developed for agriculture, largely in UK, around the second world war ,Sir R.A.Fisher was particularly associated with this work . F1 F2 F3 F4 Here the fields area has been divided I1 up into rows and column and four fertilizers (F1-F4) and four irrigation levels are I2 Represented. Since all combination are I3 Taken ,sixteen ‘cells’ or ‘plots’ result. I4
• 47 The Fisher field experiment is a full factional experiment since all 4*4 =16 combinations of the experiments factors ,fertilizer and irrigation level ,are included. The number of combinations required may not be feasible or economic. To cut down on the number of experimental combinations included, a Latin Square design of experiment may be used. Here there are three fertilisers, three irrigation levels and three alternative additives (A1-A2) but only nine instead of the 3*3*3 =27 combination, of the full factorial are included. There are ‘pivotal’ combinations, however, that still allow the F F2 F3 identification of the best fertiliser, Irrigation level I1 A1 A2 A3 and additive provided that these no serious non-additives or interactions in the relationship I2 A2 A3 A1 between yield and these control factors. The property of Latin Squares A3 A1 A2 that corresponds to this separability is that each I3 of the labels A1,A2,A3 appears exactly once in each row and column. Difference from agricultural applications is that in agriculture the ‘noise’ Or uncontrollable factors that disturb production also tend to disturb experimentation, such as the weather. In industry, factors that disturb production, or are uneconomic to control in production, can and should be directly manipulated in test. Our desire is to identify a design or line calibration which can best survive the transient effects in the manufacturing process caused by the uncontrolled factors. We wish to have small piece to piece and time to time variations associated with this noise variation. To do this we can force diversity on noise conditions by crossing our orthogonal array of controllable factors by full factorial or orthogonal array of noise factors. Thus in the example, we evaluate our product for each of the nine trials against the background of four different combinations of noise conditions. We are looking one of the nine rows of control factors combinations, or for one of the ‘missing’ 72 rows ({3*3*3*3}=81; 81-9=72), which not only gives the correct mean result on average, but also minimises variation away from the mean. To do this Taguchi introduces the signal-to-noise ratio. 4.3 Introduction to Orthogonal Arrays Engineers and Scientists are often faced with two product or process improvement situations. One development situation is to find a parameter that will improve some performance characteristic to an acceptable and optimum value. This is the most typical situation in most organizations. A 2nd situation is to find a less expensive, alternate design, material, or method which will provide equivalent performance
• 48 When searching for improved or equivalent deigns, the person typically runs some tests, observe some performance of the product and makes a decision to use or reject the new design. In order to improve the quality of this decision, proper test strategies are utilized. Before describing the OA’s let us look at some other test strategies: Most common test plan is to evaluate the effect of one parameter on product performance. This is what is typically called as one factor experiment. This experiment evaluates the effect of one parameter while holding everything else constant The simplest case of testing the effect of one parameter on performance would be to run a test at two different conditions of that parameter. For example: the effect of cutting speed on the finish of a machined part. Two different cutting speeds could be used and the resultant finish measured to determine which cutting speed gave better results. If the first level, the first cutting speed, is symbolized by 1 and the second level by 2, the experimental conditions will look like this: Trial No. Factor Level Test Results 1 1 *,* 2 2 *,* The * symbolizes the value of finish that would be obtained. This sample of two (in this case) could be averaged and compared to the second test. If there happens to be an interaction of this factor with some other factor then this interaction cannot be studied. Several Factors One at a Time If this doesn’t work, the next progression is to evaluate the effect of several parameters on product performance one at a time. Let us assume the experimenter has looked at four different factors A, B, C and D each evaluated one at a time. The resultant test program may appear like the table below: Factor and Factor Level Test Results Trial No. A B C D 1 1 1 1 1 *,* 2 2 1 1 1 *,* 3 1 2 1 1 *,* 4 1 1 2 1 *,* 5 1 1 1 2 *,*
• 49 One can see that the first trial is the base line condition and results of trial 2 can be compared to trial 1 to estimate the effect of factor A on product performance. Similarly results of trial 3 can be compared to trial 1 to estimate the effect of factor B and so on. The main limitation of several factors one at a time is that no interaction among the factors studied can be observed. Also, the strategy makes limited use of data when evaluating factor effects. Of the ten data points we had in the above example, only two were used to compare against two others; and the remaining six data points were temporarily ignored. If we try to use all the data points, then the experiment will not remain orthogonal. One main requirement of orthogonality is a balanced experiment which means equal number of samples under various test conditions. (Equal no. of tests under A1 and A2) For instance, in the above experiment if all the data under A1 and A2 is averaged and compared, then this is not a fair comparison of A1 to A2. Of the four trials under level A1, three were level B1 and one at level B2. The one trial under A2 was at level B1. Therefore one cannot see that if factor B has an effect on the performance it will be part of the observed effect of factor A, and vice versa. Only when trial 1 is compared to other trials one at a time are the factor effects orthogonal. Several factors all at the same time The most desperate and urgent situations finds the experimenter evaluating effect of several parameters on performance all at the same time. Here the experimenter hopes that at least one of the changes will improve the situation sufficiently. Factor and Factor Level Test Results Trial No. A B C D 1 1 1 1 1 *,* 2 2 2 2 2 *,* This situation makes separation of main factor effects impossible, let alone any interaction effects.
• 50 Some factors may be making positive effect and some negative, but we will not get any hint of this information. 4.3.1 Investigating many factors – a case study In most problems, preliminary brainstorming would reveal a large number of factors which may influence the output of the process under study. How are the effects of these factors prioritized? The traditional approach is to - Isolate what is believed to be the most important factor - Investigate this factor by itself, ignoring all others - Make recommendations on changes to this crucial factor - Move on to the next factor and repeat This OFAT (One factor at a time) approach has several critical weaknesses. The factorial approach in which several factors are studied simultaneously in a balanced manner is much better. We will try to understand this through an example. 4.3.1.1 Example A process producing steel springs is generating considerable scrap due to cracking after heat treatment. A study is planned to determine better operating conditions to reduce the cracking problem. There are several ways to measure cracking - Size of the crack - Presence or absence of cracks The response selected was Y: the percentage without cracks in a batch of 100 springs Three major factors were believed to affect the response - T: Steel temperature before quenching - C: carbon content (percent) - O: Oil quenching temperature Levels chosen for the study are: Factor Low (Level 1) High (Level 2) T 1450 °F 1600 °F
• 51 C 0.5% 0.7% O 70 °F 120 °F Classical approach : OFAT Experiment: Four runs at each level of T with C and O at their low levels Steel Tempt. % springs without cracks Average 1450 61 67 68 66 65.5 1600 79 75 71 77 75.5 Conclusion: Increased T reduces cracks by 10% Problem: How general is this conclusion? Does it depend upon - Quench Temperature? - Carbon Content? - Steel chemistry? - Spring type? - Analyst - Etc.?? Carrying out similar OFAT experiments for C and O would require a total of 24 observations and provide limited knowledge. Factorial Approach: - Include all factors in a balanced design: - To increase the generality of the conclusions, use a design that involves all eight combinations of the three factors. The treatments for the eight runs are given as under: Run C T O 1 0.5 1450 70 2 0.7 1450 70 3 0.5 1600 70 4 0.7 1600 70 5 0.5 1450 120 6 0.7 1450 120 7 0.5 1600 120 8 0.7 1600 120 The above eight runs constitute a FULL FACTORIAL DESIGN. The design is balanced for every factor. This means 4 runs have T at 1450 and 4 have T at 1600. Same is true for C and O. IMMEDIATE ADVANTAGES - The effect of each factor can be assessed by comparing the responses from the appropriate sets of four runs.
• 52 - More general conclusions - 8 runs rather than 24 runs. The data for the complete factorial experiment are: Run C T O Y 1 0.5 1450 70 67 2 0.7 1450 70 61 3 0.5 1600 70 79 4 0.7 1600 70 75 5 0.5 1450 120 59 6 0.7 1450 120 52 7 0.5 1600 120 90 8 0.7 1600 120 87 The main effects of each factor can be estimated by the difference between the average of the responses at the high level and the average of the responses at the low level. For example to calculate the O main effect: 67 + 61 + 79 + 75 Avg. of responses with O as 70 = = 70.5 4 59 + 52 + 90 + 87 Avg. of responses with O as 120 = = 72 4 So the main effect of O is = 72.0 − 70.5 = 1.5
• 53 Avg. Y 74 O Main Effect 73 72 71 70 70 O 120 The apparent conclusion is that changing the oil temperature from 70 to 120 has little effect. The use of factorial approach allows examination of two factor interactions. For example we can estimate the effect of factor O at each level of T. At T = 1450 67 + 61 Avg. of responses with O as 70 = = 64.0 2 59 + 52 Avg. of responses with O as 120 = = 55.5 2 So the effect of O is 55.5 – 64 = -8.5 At T = 1600 79 + 75 Avg. of responses with O as 70 = = 77.0 2 90 + 87 Avg. of responses with O as 120 = = 88.5 2 So the effect of O is 88.5 – 77 = 11.5 The conclusion is that at T = 1450, increasing O decreases the average response by 8.5 whereas at T = 1600, increasing O increases the average response by 11.5.
• 54 That is, O has a strong effect but the nature of the effect depends on the value of T. This is called interaction between T and O in their effect on the response. It is convenient to summarize the four averages corresponding to the four combinations of T and O in a table: O 70 120 Average 1450 64 55.5 59.75 T 1600 77 88.5 82.75 Average 70.5 72 71.25 Respo nse Y 90 O = 120 80 O = 70 70 60 50 1450 T 1600 When an interaction is present the lines on the plot will not be parallel. When an interaction is present the effect of the two factors must be considered simultaneously. The lines are added to the plot only to help with the interpretation. We cannot know that the response will increase linearly. Two way tables of averages and plots for the other factor pairs are: C 0.5 0.7 Average 1450 63.0 56.5 59.75 T 1600 84.5 81.0 82.75 Average 73.75 68.75 71.25
• 55 Res Y 90 C = 0.5 80 C = 0.7 70 60 50 1450 T 1600 O 70 120 Average 0.5 73 74.5 73.75 C 0.7 68 69.5 68.75 Average 70.6 72 71.25 Conclusions: - C has little effect - There is an interaction between T and O. Recommendations: - Run the process with T and O at their high levels to produce about 90% crack free product (further investigation at other levels might produce more improvement). - Choose the level of C so that the lowest cost is realized. Comparison with OFAT On the basis of the observed data we can see that OFAT approach leads to different conclusions if the factors are considered in the following order: Fix T = 1450 and C = 0.5 and vary O, conclude O=70 is best Run C T O Y 1 0.5 1450 70 67 2 0.7 1450 70 61 3 0.5 1600 70 79 4 0.7 1600 70 75
• 56 5 0.5 1450 120 59 6 0.7 1450 120 52 7 0.5 1600 120 90 8 0.7 1600 120 87 Fix O = 70 and C = 0.5 and vary T, conclude T = 1600 is best Run C T O Y 1 0.5 1450 70 67 2 0.7 1450 70 61 3 0.5 1600 70 79 4 0.7 1600 70 75 5 0.5 1450 120 59 6 0.7 1450 120 52 7 0.5 1600 120 90 8 0.7 1600 120 87 T = 1600 and vary C, conclude C = 0.5 is Fix O = 70 and best Run C T O Y 1 0.5 1450 70 67 2 0.7 1450 70 61 3 0.5 1600 70 79 4 0.7 1600 70 75 5 0.5 1450 120 59 6 0.7 1450 120 52 7 0.5 1600 120 90 8 0.7 1600 120 87 This approach will incorrectly conclude that T = 1600 C = 0.5 O = 70 Is the best Whereas the factorial approach concluded that T and O should be at their high levels and C has no effect. Looking at the above experimental situations, it will be pertinent to answer a few questions here: • How can poor utilization of data and non orthogonal situation be avoided? • How can interactions be estimated and still have an orthogonal experiment? The use of full factorial experiments is one possibility!!!
• 57 And, the use of some orthogonal arrays is another. Better Test Strategies Let us recall the example we had discussed for the two-way ANOVA discussion. A full factorial experiments is as shown: Factor and Factor Level Trial No. Hardness data (RB) A B 1 1 1 76 78 2 1 2 77 78 3 2 1 73 74 4 2 2 79 80 The above is a full factorial experiment and is orthogonal. Note that under level A1, factor B has two data points under B1 condition and two under B2 condition. The same is true under level A2. The same balanced situation is true when looking at the experiment w.r.t two conditions B1 and B2. Because of the balanced arrangement, factor A does not influence the estimate of effect of factor B and vice versa. All possible combinations of factor A and B at both there levels are represented in the test matrix. Using this information both factor and interaction effects can be estimated. The perfect experimental design is a full factorial, with replications, that is conducted in a random manner. Unfortunately, this type of experimental design may make the number of experimental runs prohibitive, especially if the experiment is conducted on production equipment with lengthy setup times. The number of treatment conditions (TC) for a full factorial experiment is determined by TC = l f Where TC = number of treatment conditions, l is the number of levels and f is the number of factors. Thus for a two level design 2 2 = 4,2 6 = 32.......... and for a three level design 3 2 = 6,3 3 = 27,3 4 = 81........ If each treatment condition is replicated only once, the number of experimental runs is doubled. Thus, for a three-level design with five factors and one replicate, the number of runs is 486. Table below shows a three factor full factorial design. The design space is composed of seven columns with 1 or 2, and the design matrix is composed of the three individual factor columns A, B, and C.
• 58 The design matrix tells us how to run the Treatment Conditions. Treatment Factors Response Condition A B C AB AC BC ABC 1 1 1 1 2 2 2 1 * 2 2 1 1 1 1 2 2 * 3 1 2 1 1 2 1 2 * 4 2 2 1 2 1 1 1 * 5 1 1 2 2 1 1 2 * 6 2 1 2 1 2 1 1 * 7 1 2 2 1 1 2 1 * 8 2 2 2 2 2 2 2 * Three factor interactions with a significant effect on the process are rare, and some two- factor interactions will not occur or can be eliminated by using engineering experience and sound judgment. If our engineering judgment shows that there was no three-factor interaction (AxBxC), we could place a factor D in that column and make it part of the design matrix. Of course, we would need to have a high degree of confidence that factor D does not interact with other columns. Similarly, we could place a factor E in column headed BxC if we thought there was no BxC interaction. This approach keeps the number of runs the same and adds factors. Please note that a full factorial experiments is possible only if there are few factors to be investigated otherwise the matrix may become too large for many factors. Typically most engineering problems will be five or more factors affecting performance (at least initially). For a seven factor experiments with each factor at two levels 2 7 = 128 experiments need to be conducted. However, usual time and financial limitations preclude the use of full factorial experiments. How can an engineer efficiently (economically) investigate these design factors? 4.3.2 A Full Factorial Experiment An actual example of an experiments used in an engine plant to investigate the problem of water pump leaks involved seven factors Factor Level 1 Level 2 Front cover design Production New Gasket Design Production New
• 59 Front bolt torque Low High Gasket Coating Yes No Pump Housing Finish Rough Smooth Rear bolt torque Low High Torque pattern Front rear Rear front If a full factorial is to be used in this situation 2 7 = 128 will have to be conducted. (As shown in figure below) A1 A2 B1 B2 B1 B2 C1 C2 C1 C2 C1 C2 C1 C2 D1 D2 D1 D2 D1 D2 D1 D2 D1 D2 D1 D2 D1 D2 D1 D2 G1 F1 G2 E1 G1 F2 G2 G1 F1 G2 E2 G1 F2 G2 Efficient Test Strategies Statisticians have developed more efficient test plans which are called fractional factorial experiments (FFE’s) FFEs use only a portion of the total combinations to estimate the main factor effects and some, not all, of the interactions. Certain treatment conditions are chosen to maintain the orthogonality among the various factors and interactions. It is obvious that 1/8th FFE with only 16 test combinations or 1/16th with only 8 combinations is much more appealing to the experimenter from a time and cost standpoint. Taguchi has developed a family of FFE matrices which can be utilized in various situations. In this situation a possible matrix is an eight trial OA which is labeled as L8 matrix.
• 60 L8 OA matrix Column No. Trial No. 1 2 3 4 5 6 7 1 1 1 1 1 1 1 1 2 1 1 1 2 2 2 2 3 1 2 2 1 1 2 2 4 1 2 2 2 2 1 1 5 2 1 2 1 2 1 2 6 2 1 2 2 1 2 1 7 2 2 1 1 2 2 1 8 2 2 1 2 1 1 2 th This is 1/16 FFE which has only 8 of the possible 128 combinations represented. One can observe that there are 7 columns in this array which may have a factor assigned to each column. The eight trials provide 7 dof s for the entire experiment allocated to 7 columns of 2 levels, each column with one dof The array allows all the error dofs to be traded for factor dofs and provide the particular test combinations that accommodate that approach. When all columns are assigned a factor, this is known as a saturated design. The levels of factors are designated by 1’s and 2’s. It can be seen that each column provide 4 tests under level 1 and 4 tests under level 2. This is the feature that provides orthogonality to the experiments. This is the real power of OA i.e. the ability to evaluate several factors in a minimum of tests. This is called an efficient experiment since much information is obtained from few trials. The assignment of factors to a saturated design FFE is not difficult; all columns are assigned a factor. However, the experiments which are not fully saturated (when all columns cannot be assigned factors) may be more complicated to design. 4.4 Steps in Designing, Conducting and Analyzing an Experiment The major initial steps are: 1. Selection of factors and/or interactions to be evaluated 2. Selection of number of levels for the factors 3. Selection of the appropriate OA 4. Assignment of factors and/or interactions to columns
• 61 5. Conduct tests 6. Analyze results 7. Confirmation experiment Steps 1 to 4 concern the actual design of experiment Let us try to understand each step 4.4.1 Selection of factors and/or interactions to evaluate Several methods are useful for determining which factors to include in initial experiments. These are 1. Brainstorming 2. Flowcharting 3. Cause and Effect diagrams 4.4.2 Selection of Number of Levels Initial round of experiments should involve many factors at few levels; two are recommended to minimize the size of the initial experiment This is because dof for a factor is the no. of levels minus one; increasing the number of levels increases the total dof in the experiments which is a direct function of total number of tests to be conducted The initial round of experimentation will eliminate many trivial factors from contention and a few remaining can then be investigated with multiple levels without causing an undue inflation in the size of the experiments.
• 62 4.4.3 Selection of OA Degrees of Freedom: The selection of which OA to use depends upon: 1. The number of factors and interactions of interest 2. The number of levels for the factors of interest Recall of ANOVA analysis: Dof for each factor is (Say factor A) v A = k A −1 Dof for interaction is (Say A and B) v A×B = (v A )(vB ) The minimum required dof in the experiment is the sum of all the factor and interaction dofs. Orthogonal Arrays Two basic kind of arrays are available. - Two level arrays: L4, L8, L12, L16, L32 - Three level arrays: L9, L18, L27 The number in the array designates the number of trials in the array. The total dof available in an array vLN = N − 1 where N is the number of trials When an array is selected, the following inequality is to be satisfied: vLN ≥ vrequired for factor and int erations Depending upon number of levels in a factor, a 2 or a 3 level OA can be selected. If some factors are two-level and some three-level, then whichever is predominant should indicate which kind of OA is selected. Once the decision is made about the right OA, then the number of trials for that array must provide an adequate total dof. When required dof fall between the two dof provided by two OAs, the next larger OA must be chosen.
• 63 4.4.4 Assignment of Factors and Interactions Before getting into the details of using some method of assignment of factors and interactions, a demonstration of a mathematical property of OAs is in order. Demonstration of Interaction Columns The simplest OA is an L4 which has an arrangement as shown: Column No. Trial No. First Second Third 1 1 1 1 2 1 2 2 3 2 1 2 4 2 2 1 Recall the two-way ANOVA problem: Let us consider an experimental situation. A student worked at an aluminum casting foundry which manufactured pistons for reciprocating engines. The problem with the process was how to attain the proper hardness (Rockwell B) of the casting for a particular product. Engineers were interested in the effect of Cu and Mg content on casting hardness. According to specifications the copper content could be 3.5 to 4.5% and the magnesium content could be 1.2 to 1.8%. The student runs an experiment to evaluate these factors and conditions simultaneously. If A = % Copper Content A1 = 3.5 A2 = 4.5 If B = % Magnesium Content B1 = 1.2 B2 = 1.8 The experimental conditions for a two level factors is given by 2 f = 4 which are A1 B1 A1 B2 A2 B1 A2 B2 Imagine, four different mixes of metal constituents are prepared, casting poured and hardness measured. Two samples are measured from each mix for hardness. The result will look like: A1 A2 B1 76, 78 73, 74 B2 77, 78 79, 80 To simplify discussion 70 points from each value are subtracted in the above observations from each of the four mixes. Transformed results can be shown as:
• 64 A1 A2 B1 6, 8 3, 4 B2 7, 8 9, 10 Let us adapt this problem to a L4 OA Factor A can be assigned to column 1 and Factor B can be assigned to column 2. The entire experiment can be shown in a L4 OA as under: Column No. y data Trial No. Factor A Factor B 3 (Rb – 70) 1 1 1 1 6, 8 2 1 2 2 7, 8 3 2 1 2 3, 4 4 2 2 1 9, 10 ANOVA for L4 OA The ANOVA for an OA is conducted by calculating the sum of squares for each column. The formula for SS A is the same as done for 2-way ANOVA. ( A1 − A2 ) 2 SS A = N A1 = 6 + 8 + 7 + 8 = 29 A2 = 3 + 4 + 9 + 10 = 26 SS A = ( 29 − 26 ) 2 =1.125 8 The sum of squares for factor B, column 2 is SS B = ( B1 − B2 ) 2 = (21 − 34) 2 = 21.125 N 8 Note that the Sum of Squares for factor A and B are identical to two-way ANOVA. The sum of squares of column 3 is SS 3 = (31 −32 ) 2 = (33 −22) 2 =15.125 N 8 Note that the value of SS3 is equal to SS A× B This is not coincidental but is a mathematical property of OA.
• 65 The calculation is a demonstration that the third column represents the interaction of factors assigned to the column 1 and 2. This particular L4 example is similar to the two-way ANOVA example and has similar test results. Thus L4 with two factors assigned to it is equivalent to a full factorial experiment and the ANOVA is equivalent to a two-way ANOVA because certain columns represent the interaction of two other columns. 4.4.5 Conducting the experiment Once the factors are assigned to a particular column of an OA, the test strategy has been set and physical preparation for performing the experiment can begin. Some decisions need to be made regarding the order of testing the various trials. Factors are assigned to columns, trial test conditions are dictated by rows of the OA. For an L8 OA, one can observe that trial 6 requires the test conditions of Trial 6: A2 B1 C 2 D1 Factors A B AxB C AxC BxC D Column No. Trial # 1 2 3 4 5 6 7 1 1 1 1 1 1 1 1 2 1 1 1 2 2 2 2 3 1 2 2 1 1 2 2 4 1 2 2 2 2 1 1 5 2 1 2 1 2 1 2 6 2 1 2 2 1 2 1 7 2 2 1 1 2 2 1 8 2 2 1 2 1 1 2 The interaction conditions cannot be controlled when conducting a test because they are dependent upon main factor levels. Only the analysis is concerned with the interaction columns. Therefore, it is recommended that test sheets be made up which show only the main factor levels required for each trial. This will minimize mistakes in conducting the experiment which may inadvertently destroy the orthogonality.
• 66 4.4.6 Analysis of Experimental Results The simple example of casting discussed for two-way ANOVA is intended to demonstrate another basic property of OAs, which is that the total variation can be accounted for by summing the variation for all columns. Let us try to put the data from that two-way ANOVA example to an L8OA. Factor A is assigned to column 1 and B to column 2. The first two trials of the OA represent the A1 B1 condition which has corresponding results if 6 and 8 in the example. Similarly, Trials 3 and 4 represent the A1 B2 condition which has results corresponding to 7 and 8. The complete OA is as shown below: Factors and Interactions A B AxB Column No. Y data Trial # 1 2 3 4 5 6 7 Rb =70 1 1 1 1 1 1 1 1 6 2 1 1 1 2 2 2 2 8 3 1 2 2 1 1 2 2 7 4 1 2 2 2 2 1 1 8 5 2 1 2 1 2 1 2 3 6 2 1 2 2 1 2 1 4 7 2 2 1 1 2 2 1 9 8 2 2 1 2 1 1 2 10 ANOVA of Taguchi L8 OA A1 = 6 + 8 + 7 + 8 = 29 A2 = 3 + 4 + 9 + 10 = 26 SS A = ( A1 − A2 ) 2 = ( 29 −26) 2 =1.125 N 8 The sum of squares for factor B SS B = ( B1 − B2 ) 2 = (21 −34) 2 = 21.125 N 8 The sum of squares of column A x B is SS A× B = ( 31 − 32 ) 2 = (33 − 22) 2 = 15.125 N 8
• 67 Note that the SS for factor A, B and interaction A x B are same as two-way ANOVA calculated earlier. Continuing with sum of squares calculations for other columns: SS 4 = ( 41 −4 2 ) 2 = ( 25 −30) 2 =3.125 N 8 SS 5 = ( 51 −52 ) 2 = ( 27 − 28) 2 = 0.125 N 8 SS 6 = ( 61 −6 2 ) 2 = ( 27 − 28) 2 = 0.125 N 8 SS 7 = ( 71 −7 2 ) 2 = ( 27 −28) 2 = 0.125 N 8 SS e = SS 4 + SS 5 + SS 6 + SS 7 = 3.500 The total sum of squares for the unassigned columns is equal to SS e calculated in the two way ANOVA example. Thus the unassigned columns in an OA represent an estimate of error variation. Here the difference of the particular array selected for the experiment changes the analysis approach slightly. The L4 has two data points per trial and the L8 one data point per trial. L4 OA for the same experiment: Column No. y data Trial No. Factor A Factor B 3 (Rb – 70) 1 1 1 1 6, 8 2 1 2 2 7, 8 3 2 1 2 3, 4 4 2 2 1 9, 10 The error variance of the L4 must come from the repetitions in each trial, but the error variance in L8 must come from the columns since there are no repetitions within trials. Note that: SS T = ∑ SS columns This is a demonstration of the property of the total sum of squares being contained within the columns of an OA.
• 68 Column estimate of error variance In the previous example, the unassigned columns were shown to estimate error variance when there was only one test result per trial. This approach of using columns to estimate the error variance will be used even if all the columns have factors assigned to them. Some factors assigned to an experiment will not be significant at all, even though thought to be before experimentation. This is equivalent to saying the color of the car can affect fuel economy and assigning two different colors to a column (2 levels). This column will have small sum of squares because it will be estimating error variance rather than any true color effect. When a column effect turns out to be small in an OA, it means any one of the following: • No assigned factors or interactions like in the L8 example discussed above • Not significant or very small factor and /or interaction effect • Canceling factor and/or interaction effects A fully saturated OA will depend upon some column effects turning out small relative to others and using the small ones as estimate of error variance. Pooling estimates of error variance In the above example, there were four unassigned columns having four degrees of freedom, one for each column, which provides estimate for error variance. A better estimate was the combination of all four columns effects for one overall estimate of error variance with 4 dofs. The combining of column effects to better estimate variance is referred to as ‘pooling’. The pooling up strategy entails F-testing the smallest column effect against the next larger one to see of significance exists. If no significance exists, then these two effects are pooled to test the next larger column effect until some significant F-ratio exists. The ANOVA table for the experimental problem discussed above will appear like this: Source SS Dof V F A 1.125 1 1.125 B 21.125 1 21.125 22.83 AxB 15.125 1 15.125 16.35 Col 4 3.125 1 3.125 Col 5 0.125 1 0.125 Col 6 0.125 1 0.125 Col 7 0.125 1 0.125 T 40.875 7 Error pooled 4.625 5 0.925
• 69 In this situation five smaller column effects have been pooled to form an estimate of error variance with 5 dof associated with that estimate. As a rule of thumb, pooling up to half of the dofs is advisable. Here that rule was exceeded slightly because two of the column effects were substantially larger than the others. The ANOVA summary table could be rewritten to recognize the pooling as shown in Table below; Source SS Dof V F B 21.125 1 21.125 22.83 AxB 15.125 1 15.125 16.35 Error pooled 4.625 5 0.925 T 40.875 7 4.4.7 Confirmation experiment The confirmation experiment is the final step in verifying the conclusions from the previous round of experimentation. The optimum conditions are set for the significant factors and levels and several tests are made under constant conditions. The average of the confirmation experiment results is compared to the anticipated average based on the factors and levels tested. 4.5 EXAMPLE EXPERIMENTAL PROCEDURE Popcorn experiment This is an example to make you walk through the process of designing experiments. The scenario is to develop process (cooking) specifications to go on a bag of popcorns. The owner of the company has developed a new hybrid seed which may or may not use the same cooking process recommended for the current seed. One of the processes used by the customer is the hot oil method, which is addressed in this situation. Statement of problem and objective of experiment To find process factors which influence popcorn quality characteristics relative to customer’s requirement characteristics such as • unpopped kernels in a batch, • the fluffiness or volume of the popped corn, • the color, • the taste, and • the crispiness are typically considered.
• 70 The objective of the experiment is to find the process conditions which optimize the various quality characteristics to provide improved popping, fluffiness, color, taste and texture. Measurement Methods Number of un-popped kernels in a batch can be easily measured, but it assumes there are an equal number of uncooked kernels in a measuring cup The fluffiness or volume can be quantified by placing the popped kernels in a measuring cup; which again assumes there are an equal number of uncooked kernels were used in each batch. The performance of color, taste and texture are somewhat more abstract. These may be addressed by assigning a numerical color rating, taste rating, and texture rating to each batch. Flowchart – Popcorn Cooking Add Corn process Heat Oil Agitation (Preheat) Venting Inspect Popped Corn Quality Improvement: Problem Solving Cause & Effect diagram – Popcorn Experiment Oil Method Type Agitation Amount Venting High Quality Popcorn Shape Preheat Amount Material Pan Heat
• 71 Popcorn factors and levels Factor Level 1 Level 2 A – Type of Oil Corn Oil Peanut Oil B – Amount of Oil Low High C – Amount of heat Medium High D – Preheat No Yes E – Agitation No Yes F – Venting No Yes G – Pan Material Aluminum Steel H – Pan Shape Shallow Deep Assignment of factors to columns – L16 OA The factor list is small enough to fit into an L16 OA at a resolution 2 if 2 levels are used for each factor. Using the Taguchi L16 OA the assignment of factors to the columns can be done by referring to table D2 of the appendix. (Hard copies distributed already) Factors A to H are assigned to columns 1, 2, 4, 7, 8, 11, 13 and 14. The trial data sheets can then be generated from the factor column assignment. Popcorn factors A B C D E F G H Column Number Trial 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 1 1 1 2 2 2 2 1 1 1 1 2 2 2 2 4 1 1 1 2 2 2 2 2 2 2 2 1 1 1 1 5 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2 6 1 2 2 1 1 2 2 2 2 1 1 2 2 1 1 7 1 2 2 2 2 1 1 1 1 2 2 2 2 1 1 8 1 2 2 2 2 1 1 2 2 1 1 1 1 2 2 9 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 10 2 1 2 1 2 1 2 2 1 2 1 2 1 2 1 11 2 1 2 2 1 2 1 1 2 1 2 2 1 2 1 12 2 1 2 2 1 2 1 2 1 2 1 1 2 1 2 13 2 2 1 1 2 2 1 1 2 2 1 1 2 2 1 14 2 2 1 1 2 2 1 2 1 1 2 2 1 1 2 15 2 2 1 2 1 1 2 1 2 2 1 2 1 1 2 16 2 2 1 2 1 1 2 2 1 1 2 1 2 2 1 Trial # 5 data sheet will look like:
• 72 • Corn Oil • High amount of oil • Medium Heat • Preheat before adding popcorn • No agitation during popping • Vented during popping • Aluminum Pan • Deep Pan shape Conducting the experiment: Order may be completely randomized with one test per trial. A batch of 200 seeds could be made for each trial with the specified test conditions. For each trial, un-popped kernels, fluffiness, color, taste and texture would be noted. Popcorn experiment interpretation ANOVA will be used to analyze each performance characteristic separately to determine which factors and levels gave the best result. Problems: 1. Assign factors A,B,C,D and E as well as interactions CxD and CxE to an OA if all factors are using two levels. 2. Assign these factors and interactions to an OA  A,B,C,D, and E (Two levels)  AxB BxC CxE  AxC BxD DxE  AxD BxE  BxF Answer to Problem 1: Several possibilities exist L8 (Resolution 1) Column # 1 2 3 4 5 6 7 Option 1 C D CxD E CxE A B Option 2 D E A C CxD CxE B Option 3 A B C D E CxE CxD L16 (Resolution 3) Column # 1 2 4 8 11 12 15 A B C D CxE CxD E
• 73 4.6 Standard Orthogonal Array