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  • 1. M6141 QUALITY ENGINEERING CONTENTS 1. Introduction DESIGN OF EXPERIMENTS 2. Fundamentals of Probability and Statistics (cancelled) A/P Wu Zhang 3. Single Factor Experiments MAE, NTU 4. Randomized Blocks (cancelled) Office: N3.2-02-14 Tel: 6790 4445 5. Factorial Experiments Email: mzwu@ntu.edu.sg 6. 2k Experiments 7. Fractional ExperimentsTEXT 8. A Case Study Montgomery, D. C., (2009), Design and Analysis of Experiments, John Wiley & Sons. 1-1 1-21 INTRODUCTION Example 1.1: Two factors: Machine, Feed rate Response: Process capabilityA designed experiment is a test or series of tests, in which, theexperimenter chooses certain factors for study. He purposely varies Table 1.1those factors in a controlled fashion, and then observes the effect of Process capability Cpsuch factors on the response function. Feed rate (mm/sec.) Machine 0.2 0.4 0.6 USA 2.1, 2.2 1.7, 1.6 1.2, 1.3 Japan 0.9, 1.0 0.7, 0.6 0.4, 0.3 Conclusion: using USA machine at lowest feed rate will result in max. process capability. 1-3 1-4
  • 2. Factor (input variable) (3) Run: A run is one set of levels for all factors in an experiment. In above experiment, there are three factors: (A) temperature,(1) Types (B) machine and (C) switch. The total number of runs is Quantitative: Temperature, time: They can be set at any value. R = 4 × 3 × 2 = 24 Qualitative: Machines, operators: They have different categories Switch: It has different status For example, [1] A = 10, B = Japan-made, C = on ..........(2) Levels [24] A = 40, B = UK-made, C = off Temperature (oC): 10, 20, 30, 40 [4 levels] Machine: made by Japan, by USA, by UK [3 levels] For single factor experiments, run is equivalent to level of the factor. The number R of runs is equal to the number a of levels. Switch: on, or off [2 levels] (4) Replicates (n): number of observations at each run. A larger value of n increases the accuracy of the experimental results, but makes the experiment more expensive. 1-5 1-6Run A B C Run A B C Response (or output, observation)No Temp. Machine Switch No Temp. Machine Switch 1 10 Japan On 13 30 Japan On Response is dependent on the values of the factors. 2 10 Japan Off 14 30 Japan Off 3 10 USA On 15 30 USA On Types of response: 4 10 USA Off 16 30 USA Off 5 10 UK On 17 30 UK On Larger the better: product yield, process capability, student’s 6 10 UK Off 18 30 UK Off mark 7 20 Japan On 19 40 Japan On Smaller the better: product cost, number of errors 8 20 Japan Off 20 40 Japan Off 9 20 USA On 21 40 USA On Nominal the best: diameter of a shaft 10 20 USA Off 22 40 USA Off 11 20 UK On 23 40 UK On By DOE, we can find out which factors have effects on the response, 12 20 UK Off 24 40 UK Off and the direction of the influence (i.e., whether the increase of a factor will increase or decrease the response ?) 1-7 1-8
  • 3. Example 1.2: Single factor: A text book (a = 2, n = 3, R = 2) Example 1.3: Two factors: A text book, B studying hours Response: Students’ marks Response: Students’ marks Table 1.2 (a = 2, b = 3, n = 2, R = 6) Students’ marks Table 1.3 Text 1 76, 74, 78 Ave = 76 Students’ marks Text 2 48, 50, 52 Ave = 50 1 hr 2 hr 3 hr Text 1 45, 47 59, 58 76, 75 Text 2 32, 33 51, 54 70, 71 Text book (a qualitative factor) Text 1: book by Montgomery Text 2: book by Taguchi 1-9 1-10The objectives of an experiment may include What Makes Designed Experiment (DOE) Special?(1) Determining which factors xs are most influential on the response, (1) It conducts the experiments in a systematic and efficient way. y. Reduce design and development time, as well as the cost. (2) It presents experiment results in the simplest and clearest way.(2) Determining where to set the influential xs, so that y is near the desired requirement. (3) It extracts the maximum amount of information from a given set of experiments.(3) Determining where to set the influential xs, so that variability in y is small. (4) It draws the right conclusions, despite the variability presented in the data. Examples: Increase process capability Improve students’ study Reduce output variability. 1-11 1-12
  • 4. Example 1.4 Reducing Defect Level in a Process In this application, DOEA manufacturing engineer is going to apply DOE to a process for (1) Determines which factors affect the occurrence of defectssoldering electronic components to printed circuit boards, in order toreduce defect levels even further. (2) Determines the direction of adjustment for the effective factors toFactors to be investigated: further reduce the number of defects per unit. Solder temperature Preheat temperature (2) Decides whether changing the factors together produces different Conveyor speed results than just adjusting individual factor separately? Flux type Flux specific gravity Solder wave depth Conveyor angle Thickness of the printed circuit board Types of components used on the board 1-13 1-14
  • 5. 3 SINGLE FACTOR EXPERIMENTS Table 3.1 Tensile strength of paper ___________________________________________________________________________________________________________ Hardwood Observations3.1 Introduction _____________________________________________________________________________________________ Concen. (%) 1 2 3 4 5 6 Total Aver. _________________________________________________________________________________________________________________________________Single factor experiment can be used in the situation where the effect(influence) of a single factor on the response is dominant. The effects 5 7 8 15 11 9 10 60 10.00of all other factors are negligible. 10 12 17 13 18 19 15 94 15.67The procedures and formulae developed for the single factor 15 14 18 19 17 16 18 102 17.00experiments can be easily modified, and then used for the several 20 19 25 22 23 18 20 127 21.17factors experiments. __________________________________________________ 383 15.96 Study the effect (influence) of hardwood concentration on the tensile strength of paper. 3-1 3-2(1) One factor: hardwood concentration. Table 3.3 General data table for a single factor experiment(2) Four levels: 5%, 10%, 15%, 20%: a = 4. Level Observation Total Average 1 y11 y12 . . y1n y1. y1.(3) Response is the tensile strength of paper. 2 y21 y22 . . y2n y2. y2.(4) Each level has six observations or replicates: n = 6. . . . . . .(5) Total number of observations: N = an = 24. . . . . . . a ya1 ya2 . . yan ya. ya. y.. y.. yij the jth observation taken under the ith level of the factor. 3-3 3-4
  • 6. yi. the total of the observations at the ith level 3.2 The Analysis of Variance (ANOVA) n ANOVA is the most important statistical tool used in DOE. Its task yi. = ∑ yij (3.1) j =1 is to decide whether and which factors or interactions have significant effect (influence) on the response.yi. the average of the observations at the ith level yi. = yi. / n (3.2) Model of ANOVA Yij = μ + τ i + ε ij = μi + ε ij (3.5)y.. the grand total of all observations a n a y.. = ∑∑ yij = ∑ yi. (3.3) μ the overall mean. μ ≈ y.. (mean ≈ average) i =1 j =1 i =1 μi the ith level mean. μ i ≈ yi.y.. the grand average of all observations. τi the ith level effect. τ i = μi − μ ≈ yi. − y.. y.. = y.. / N (3.4) ε ij the random error. ε ij = yij − μ i ≈ yij − yi.N = an is the total number of observations 3-5 3-6Machining Example (1) y1. = 2.1 + 2.0 + 1.9 = 6.0, y1. = y1. / 3 = 2.0Study the effect of machine on the process capability y 2. = 0.9 + 1.0 + 1.1 = 3.0, y 2. = y 2. / 3 = 1.0 Table 3.4 y.. = (2.1 + 2.0 + 1.9) + (0.9 + 1.0 + 1.1) = 9.0 Machine Process capability Cp y.. = 9.0 / 6 = 1.5 USA 2.1, 2.0, 1.9 Japan 0.9, 1.0, 1.1 μ1 ≈ y1. = 2.0, μ 2 ≈ y 2. = 1.0 μ ≈ y.. = 1.5(1) One factor: machine (qualitative). τ1 = μ1 − μ ≈ y1. − y.. = 2.0 − 1.5 = 0.5(2) Two levels: USA-made, Japan-made, a = 2. τ 2 = μ 2 − μ ≈ y2. − y.. = 1.0 − 1.5 = −0.5(3) Response is the process capability.(4) Each level has 3 observations or replicates: n = 3.(5) Total number of observations: N = an = 2 × 3 = 6. 3-7 3-8
  • 7. Evaluation of Errors y ε11 = 0.1 For instance, if i = 1, j = 1, yij = y11 = 2.1 ε ij = yij − μi μ1 = 2 τ1 = 0.5 ε11 = y11 − μ1 ≈ y11 − y1. = 2.1 − 2.0 = 0.1 μ = 1.5 y11 = 2.1 τ2 = -0.5 Errors for the machining example (1) μ2 = 1 Table 3.5 Machine Errors USA 0.1, 0.0, -0.1 y11 = μ + τ1 + ε11 Japan -0.1, 0.0, 0.1 = 1.5 + 0.5 + 0.1 = 2.1 Figure 3.2 3-9 3-10 μ1 μ2 μ1 τ1 τ2 τ1μ1 μ2 μ1 μ2 μ3 μ μ τ1 τ2 τ3 μμ3 μ4 μ τ2 τ3 τ4 μ2 μ3 μ4 τ3 τ4 τ4 μ3 μ4 μ4 Figure 3.3 Figure 3.4 3-11 3-12
  • 8. Hypothesis Test: to test if a factor has effect on the response. Null: Ho : the factor has no effect on the responseIf a factor has no effect on the response, the response is always the Alternative: H1: the factor has effect on the responsesame, regardless the change of the factor, Ho : τ1 = τ 2 =L = τ a = 0 (|τi| is small, no effect) μ1 = μ 2 = L = μ a = μ H1 : τ i = o for at lease one i / (|τi| is larger, has effect) (3.6) ( μ1 − μ ) = ( μ 2 − μ ) = L = ( μ a − μ ) = 0 τ 1 = τ 2 = Lτ a = 0 The larger the value of any |τi|, the more effective the factor will be. Here, the magnitude of τi, rather than its sign, makes sense.If a factor has effect, the value of the response will be different, alongwith the change of the factor. The more the response differs, the Type I error: the null hypothesis Ho is rejected when it is actuallygreater the effect of the factor is. true (The factor is concluded effective, when it is not) τ i = o for at least one i / Type II error: the null hypothesis is accepted when it is actually false (The factor is concluded ineffective, when it is effective). 3-13 3-14Machining Example (2) y Table 3.6 Machine Process capability Cp τ1 = τ2 = 0 USA 1.6, 1.5, 1.4 μ = μ1 = μ2 =1.5 Japan 1.4, 1.5, 1.6 Minor variation is caused by errors, rather than the machines μ1 ≈ 1.5, μ 2 ≈1.5 μ ≈ 1.5 τ 1 = μ1 − μ = 15 − 15 = 0 . . τ 2 = μ2 − μ = 15 − 15 = 0 . .Conclusion: in this example, the factor (machine) has no effect on the process capability. Because, when using different Figure 3.5 machines, the response is always the same, on average. 3-15 3-16
  • 9. Analysis of Variance (ANOVA) (1) Calculate sum of squares (SS)Major Steps Three types of differences: [1] Difference between an individual observation and the grand(1) Calculate sum of squares (SS). average: yij - y..(2) Determine degrees of freedom (D.O.F.). [2] Difference between a level average and the grand average:(3) Calculate mean square (MS). yi. - y.. = τ i(4) Calculate the test ratio F0. [3] Difference between an individual observation and the corresponding level average:(5) Make conclusion. yij - yi. = ε ij Note: yij - y.. = ( yi. - y.. ) + ( yij - yi. ) 3-17 3-18 a n a n a n Sum of squares is the sum of squares of the above three differences.∑∑ ( yi =1 j =1 ij - y.. ) ∑∑ ( yi. - y.. ) i =1 j =1 ∑∑ ( y i =1 j =1 ij - yi. ) SST The total sum of squares, which is a measure of total a n a n a n variability in the data.∑∑ ( y - y.. ) ∑∑ ( yi. - y.. ) ∑∑ ( y - yi . ) 2 2 2 ij iji =1 j =1 i =1 j =1 i =1 j =1 a n SS T = ∑ ∑ ( yij - y.. ) 2 (3.7) i =1 j =1 SST = SSF + SSE SSF the sum of squares due to the factor, that is the sum of squares of differences between factor level averages and the grand average (difference between levels). a n a a SS F = ∑∑ ( yi. − y.. ) 2 = n∑ ( yi. − y.. ) 2 ≈ n∑ τ i2 (3.8) i =1 j =1 i =1 i =1 3-19 3-20
  • 10. SSE the sum of squares due to error, that is the sum of squares Alternative Formulae of Calculating the Sum of Squares of differences between observations and their level averages (difference within levels). a n y . .2 a n a n SS T = ∑ ∑ yij 2 - i =1 j =1 an (3.11) SS E = ∑∑ ( yij − yi. ) 2 ≈∑∑ ε ij 2 (3.9) i =1 j =1 i =1 j =1 a yi. 2 y . .2 SS F = ∑ n an i =1 - (3.12) SS T = SS F + SS E (3.10) SS E = SST - SS F (3.13) Formulae (3.7) to (3.9) are used to describe the underlying ideas. Formulae (3.11) to (3.13) are used for actually computation. 3-21 3-22Machining Example (1) If SSF is large, it is due to differences among the means at the different factor levels. See equation (3.8), a large SSF means that τ2i Table 3.7 (or |τi|) is great. It is an indication that the factor has significant effect Machine Process capability Cp on the response. USA 2.1, 2.0, 1.9 Usually, SSF is standardized by taking SSF / SSE . By comparing SSF Japan 0.9, 1.0, 1.1 to SSE, we can see how much variability is due to changing factor levels and how much is due to error. a = 2, n = 3 y1⋅ = 6 y2⋅ = 3 y⋅⋅ = 9 Machining Examples y1⋅ = 2 y2⋅ = 1 y⋅⋅ = 15 . Table 3.8 Effective ? SSF/SSE SST = (2.12 + 2.0 2 + 1.9 2 + 0.9 2 + 1.0 2 + 1.12 ) − 9 2 / 6 = 1.54 Example 1 Yes 37.5 SS F = (6 2 + 32 ) / 3 − 9 2 / 6 = 1.50 Example 2 No 0 SS E = 1.54 − 1.50 = 0.04 3-23 3-24
  • 11. (2) Determine degrees of freedom (D.O.F.). Machining Example (1) Table 3.9 total number of observations N: an Machine Process capability Cp degrees of freedom of SST: N - 1 = an - 1 (3.14) USA 2.1, 2.0, 1.9 degrees of freedom of SSF: a-1 (3.15) Japan 0.9, 1.0, 1.1 degrees of freedom of SSE: R (n - 1) = a (n – 1) (3.16) a = 2, n = 3 where, (n - 1) is the degrees of freedom of errors in a run. total number of observations N: 2×3=6 degrees of freedom of SST: N–1 =6–1=5Equation of DOF degrees of freedom of SSF: a–1 =2–1=1 an - 1 = (a - 1) + a (n - 1) (3.17) degrees of freedom of SSE: a (n-1) = 2 × (3 - 1) = 4 DOF of SST = DOF of SSF + DOF of SSE Check equation (3.17): 5 = 1 + 4 3-25 3-26(3) Calculate mean square (MS). (4) Calculate the test ratio F0. The ratio between a sum of squares and its DOF. MS F Fo = (3.20) MS E SS F MS F = (3.18) a − 1 If H0 is true, F0 follows a theoretical F distribution, which is completely determined by the two parameters ν1 and ν2. SS E MS E = (3.19) ν1 = a - 1 the D.O.F. of MSF (numerator) a ( n - 1) ν2 = a(n - 1) the D.O.F. of MSE (denominator) There is no need to calculate MS for the SST 3-27 3-28
  • 12. f(x) (5) Make conclusion. ν1 = 1 ν2 = 4 (1) Decide the curve from SS F ν1 and ν2. MS F a (n − 1) SS F SS (2) Decide Fα,ν1,ν2 from α Fo = = a −1 = ⋅ = Q⋅ F MS E SS E a − 1 SS E SS E a (n − 1) a (n − 1) α Where, Q = is a constant. a −1 SS F Fα ,ν1 ,ν 2 F0 is equivalent to (the standardized SSF) except for a constant Q. SS E Figure 3.6 SS The density function curve of an example of the F distribution However, while F0 follows the theoretical F distribution, F does SS E not. 3-29 3-30 SS F Table 3.10 Control Limit Fα ,ν 1 ,ν 2 found from the F TableIf F0 is large ----> is large ---> SSF is large ----> τ2i is large ---> SS E |τi| is large ---> the factor has significant effect on the response. α = 0.05Control limit: Fα ,ν 1 ,ν 2 ν1 1 2 3 4 5 . ∞ ν2 ν1 = a - 1 and ν2 = a(n - 1) are the two parameters of an F 1 161.4 199.5 distribution 2 18.51 19.00 α is the probability of Type I error, which means concluding that the factor has significant effect on the response when it in fact has 3 10.13 9.55 Examples: no effect. Usually, set α = 1% or 5% 4 7.71 6.94 F0.05,2,1 = 199.5If Fo < Fα ,ν 1 ,ν 2 , conclude that the factor has no effect. . F0.05,1,4 = 7.71 ∞If Fo > Fα ,ν 1 ,ν 2 , conclude that the factor has significant effect. 3-31 3-32
  • 13. Table 3.11 Table 3.12 3-33 3-34 Table 3.13 Machining Example (1) (a = 2, n = 3, N = 6), from Table 3.7 Analysis of Variance for a Single-Factor Experiment___________________________________________________ Table 3.14 ___________________________________________________Source of Sum of Degrees of Mean Fo Source of Sum of Degrees of Mean FoVariation Squares Freedom Square Variation Squares Freedom Square ______________________________________________________________________________________________________ Machine 1.50 1 (= ν1) 1.50 150Factor SSF a-1 MSF MS F Error 0.04 4 (= ν2) 0.01 MS E Total 1.54 5 ___________________________________________________Error SSE a(n-1) MSETotal SST an-1___________________________________________________ 3-35 3-36
  • 14. Specify α = 0.05, Control limit Fα ,ν 1 ,ν 2 = F0.05,1, 4 = 7.71 Machining Example (2) (a = 2, n = 3, N = 6), Table 3.15Since F0 > Fα ,ν 1 ,ν 2 , machines have significant effect on the process Source of Sum of Degrees of Mean Fo capability. Variation Squares Freedom SquareSum of squares: 1.50 + 0.04 = 1.54 Machines 0 1 (= ν1) 0 0Degrees of freedom 1 + 4 = 5 Error 0.04 4 (= ν2) 0.01 Total 0.04 5 ___________________________________________________ Specify α = 0.05, Control limit Fα ,ν 1 ,ν 2 = F0.05,1, 4 = 7.71 Since F0 < Fα ,ν1 ,ν2 , machines don’t have significant effect on the process capability. 3-37 3-38 Table 3.16 Example: Tensile strength of paper 4 6 y.. 2___________________________________________________________________________________________________________ SS T = ∑ ∑ yij 2 - Hardwood Observations i =1 j =1 an _____________________________________________________________________________________________Concen. (%) 1 2 3 4 5 6 Total Aver. (383) 2_________________________________________________________________________________________________________________________________ = (7) 2 + (8) 2 + L + (20) 2 - = 512.96 24 5 7 8 15 11 9 10 60 10.00 4 yi.2 y..2 10 12 17 13 18 19 15 94 15.67 SS F = ∑ - i =1 n an 15 14 18 19 17 16 18 102 17.00 (60) 2 + (94) 2 + (102) 2 + (127) 2 (383) 2 = - = 382.79 20 19 25 22 23 18 20 127 21.17 6 24__________________________________________________ SS E = SS T - SS F 383 15.96 = 512.96 - 382.79 = 130.17a = 4, n = 6, N = an = 24. 3-39 3-40
  • 15. Table 3.17 ANOVA for tensile strength of paper SUMMARY: ANOVA for Single Factor Experiment___________________________________________________________________________________________________________________________________Source of Sum of Degrees of Mean FoVariation Squares Freedom Square 1. Decide the factor to be investigated and the following two__________________________________________________________________________________________________________________________________ parameters:Hardwood 382.79 3 (= ν1) 127.60 19.61 1) number of levels a;concentration 2) replicate n.Error 130.17 20 (= ν2) 6.51 2. Carry out the experiments, obtain N (= an) observations yij by aTotal 512.96 23 random manner.______________________________________________________________________________________________________________________________________Specify α = 0.01, Control limit Fα ,ν 1 ,ν 2 = F0.01,3, 20 = 4.94 3. Check if the residuals (estimates of errors) satisfy the requirements.Since F0 > Fα ,ν 1 ,ν 2 , hardwood concentration has significant effect on the tensile strength of paper. 4. Calculate sum of squares SST, SSF and SSE (3.11, 3.12, 3.13). 3-41 3-425. Calculate the degrees of freedom for each of SST, SSF, SSE (3.14, 3.3 Test on Individual Level Means 3.15, 3.16). ANOVA results in a single parameter F0, which can tell whether a6. Calculate the mean squares MSF and MSE (3.18, 3.19). factor has effect on the response, or whether the mean response values will be different at different levels of the factor from an7. Calculate the ratio F0 between MSF and MSE (3.20). overall viewpoint.8. Specify a type I error α, and find the control value Fα ,ν 1 ,ν 2 from However, ANOVA cannot decide the direction of influence of the the F distribution Tables (ν1 = a - 1, ν2 = a (n – 1) ). factors, nor identify which factor level mean is different from another level mean.9. Conclude if the factor has significant effect on the response. 3-43 3-44
  • 16. The plots indicate that changing the hardwood concentration has a strong effect on the paper strength; specifically, higher hardwood concentrations produce higher observed paper strength. Box plots show the variability of the observations within a factor level and the variability between factor levels. However, a less subjective approach is required to test the individual level means. Figure 3.7 Box plots of paper strength example 3-45 3-46Multiple Comparison Procedures One of the major problems with making comparisons among level means is that unrestricted use of these comparisons can lead to anIt is the techniques for making comparisons between two or more excessively high probability of a Type I error.level means subsequent to an analysis of variance.When we run an analysis of variance and obtain a significant F0 For example, if we have 10 levels in which the complete nullvalue, we have shown simply that the overall null hypothesis is false. hypothesis is true (with α = .05)We do not know which of a number of possible alternativehypotheses is true. Ho: µ1 = µ2 = µ3 = … = µ10 H1 : µ1 ≠ µ2 ≠ µ3 ≠ µ4 ≠ µ5 A series t tests between all pairs of level means will lead to making atOr H1: µ1 ≠ µ2 = µ3 = µ4 = µ5 least one Type I error 57.8% of the time. In other words, the experimenter who thinks he is working at the (α = .05) level ofMultiple comparison techniques allow us to investigate hypotheses significance is actually working at (α = 0.578).that involve means of individual levels. For example, we might beinterested in whether µ1 is different from µ2, or whether µ2 isdifferent from µ3. 3-47 3-48
  • 17. The probability of making at least one Type I error increases as we Least Significant Difference (LSD) Test is a useful tool to test theincrease the number of independent t tests we make between pairs of differences between the individual level means.level means. While it is nice to find significant differences, it is notnice to find ones that are not really there. We need to find some way The basic requirement for a LSD test is that the F0 for the overallto make the comparisons we need but keep the probability of analysis of variance (ANOVA) must be significant. If the F0 was notincorrect rejections of Ho under control. significant, no comparisons between level means are allowed. You simply declare that there are no differences among the level means and stop right there. On the other hand, if the overall F0 is significant, you can proceed to make any or all pairwise comparisons between individual level means by the use of a t test. 3-49 3-50The t distributions To test between the ith level mean (µi) and jth level mean (µj), Ho: µi = µj H1: µi ≠ µj (3.21) we calculate yi⋅ − y j⋅ t= (3.22) 2 MS E / n The value of MSE has already been obtained during the overall ANOVA. The DOF of t is equal to the DOF of MSE, i.e., a(n-1) ort(30) is a t distribution with a DOF of 30. (N-a).t(∞) means a standard normal distribution with (µ = 0 and σ = 1). Then, using a two-tailed t test for a specified level of type I error α (e.g., α = 0.05), we can decide a pair of two-sided control limits ±tα(dof) from the t table. 3-51 3-52
  • 18. If the t value determined by Eq (3.22) falls within the limits ±tα(dof), we cannot reject the null hypothesis in (3.21). We will therefore conclude that there is no difference between µi and µj. On the other hand, if the t value falls beyond the limits ±tα(dof), we will reject the null hypothesis and conclude that there is a difference between µi and µj. 3-53 3-54 The Example of Tensile Strength of Paper Problem: if hardwood concentrations 5% and 20% produce the Table 3.18 same paper strength ?___________________________________________________________________________________________________________ Hardwood Observations Ho : μ 1 = μ 4 _____________________________________________________________________________________________ Hypothesis:Concen. (%) 1 2 3 4 5 6 Total Aver. H1 : μ1 = μ 4 /_________________________________________________________________________________________________________________________________ 5 7 8 15 11 9 10 60 10.00 Five Steps: 10 12 17 13 18 19 15 94 15.67 (1) Use Eq (3.22) 15 14 18 19 17 16 18 102 17.00 20 19 25 22 23 18 20 127 21.17 yi ⋅ − y j ⋅ 10.00 − 21.17 t= = = −7.58__________________________________________________ 2 MS E / n 2 × 6.51 / 6a = 4, n = 6 (Note: MSE = 6.51 was calculated during ANOVA, which isy1⋅ = 10.00, y2⋅ = 15.67, y3⋅ = 17.00, y4⋅ = 21.17 always carried out before the test on individual level means) 3-55 3-56
  • 19. (2) Determine the DOF, DOF = a (n – 1) = 4 × (6 – 1) = 20 General Steps to Test the Individual Level Means(3) Decide a type I error level. We use α = 0.01 as in the ANOVA for (1) Calculate t using Eq (3.22) and find the MSE value obtained this example. during ANOVA.(4) Find the limits from the t table: ±tα(dof) = ±t0.01(20) ≈ 3.00 (2) Determine the DOF, DOF = a (n – 1).(5) Make conclusion: since the calculated t value (= -7.58) falls (3) Specify a type I error α. beyond the limits ±3.00, we will reject the null hypothesis. (4) Find the control limits from the t table: ±tα(dof) based on α and Conclusion: There is a significant difference in paper strength DOF. between using hardwood concentrations 5% and 20%. Moreover, since t < 0, µ1 < µ4. (5) Make conclusion by comparing the calculated t value against the control limits. 3-57 3-58The Bonferroni Procedure To put this in a way that is slightly more useful to us, if you want the overall family-wise error rate to be no more than (α = 0.05), and youIn a Bonferroni procedure the family-wise error rate (Type I error) is want to run c tests, then each run of them should be at (α = 0.05/c).divided by the number of comparisons. To run these tests, you do exactly what you did in Fishers LSD test, though you omit any requirement about the significance of overall F.The basic idea behind this procedure is that if you run several tests(say c tests), each at a significance level (type I error) represented byα, the probability of at least one Type I error can never exceed cα.Thus, for example, if you ran 5 tests, each at α = 0.05, the family-wise error rate would be at most 5(0.05) = 0.25. But that is a too highType I error rate to make anyone happy.But suppose that you ran each of those 5 tests at the α = 0.01. Thenthe maximum family-wise error rate would be 5(0.01) = 0.05, whichis certainly acceptable. 3-59 3-60
  • 20. a ni 2 y.. SST = ∑∑ yij −3.5 Unbalanced Experiments 2 DOF = N -1 i =1 j =1 NIn an unbalanced experiment, the number of observations taken undereach run is different. a yi2. y.. 2 SS F = ∑ − DOF = a -1ni is the number of observations taken under the ith run (for i =1 ni N single factor experiment, it is simply the ith level). a a SS E = SST − SS F DOF = ∑ (ni − 1) N = ∑ ni = n1 + n2 + L + na i =1 i =1 is the total number of observations.Note, in unbalanced experiments, N ≠ an. 3-61 3-62Machining Example SST = (2.12 + 2.0 2 + 1.9 2 ) + (0.9 2 + 1.0 2 ) - 7.9 2 / 5 = 1.348 DOET = N – 1 = 5 – 1 = 4 Machine Process capability Cp USA 2.1, 2.0, 1.9 SSF = (6.02/3 + 1.92/2) – 7.92/5 = 1.323 Japan 0.9, 1.0, DOEF = a – 1 = 2 – 1 = 1n1 = 3, n2 = 2, N = 3 + 2 = 5 SSE = 1.348 – 1.323 = 0.025y1. = 6.0, y2. = 1.9, y.. = 7.9 DOEE = (n1 – 1) + (n2 – 1) = (3 – 1) + (2 – 1) = 3 3-63 3-64
  • 21. ANOVA for Machining Example (a = 2, n1 = 3, n2 = 2, N = 5) 3.7 Guidelines for Designing Experiments (1) Recognition and statement of the problem.Source of Sum of Degrees of Mean Fo Fully develop all ideas about the problem and the specificVariation Squares Freedom Square objectives of the experiment; Solicit input from all concerned parties -- engineering, quality,Machine 1.323 1 (= ν1) 1.323 158.8 marketing, customer, management and operators.Error 0.025 3 (= ν2) 0.0083 (2) Choice of factors and levels.Total 1.348 4 Choose the factors, their ranges and levels. Process knowledge is___________________________________________________ required.Specify α = 0.05, Control limit: Fα ,ν 1 ,ν 2 = F0.05,1,3 = 10.13 Investigate all factors that may be of importance and avoid being overly influenced by past experience.Since F0 > Fα ,ν 1 ,ν 2 , machines have significant effect on the process Keep the number of factor levels low (Most often two levels are capability. used.) in the early stage 3-65 3-66(3) Selection of the response (6) Data analysis Most often, the average or standard deviation (or both) of the Analyze the data so that results and conclusions are objective rather measured characteristic will be the response variable. than judgmental. Use software packages and simple graphical methods(4) Choice of experimental design. Carry out residual analysis. Decide the factorial fraction experiment, select the replicate (sample size) and run order for the experimental trials. (7) Conclusions and recommendations. Decide whether or not blocking or other randomization methods are involved. (8) Experimentation is an important part of the learning process. New hypotheses are formulated based on the investigation of the(5) Performing the experiment. tentative hypotheses. Don’t design a single, large comprehensive experiment at the start of a study. As a rule of thumb, the first Ensure that everything is being done accordingly to plan. experiment should spend no more than 25% of the total budget of the study. 3-67 3-68
  • 22. 4 RANDOMIZED BLOCKS Response: coded roughness of a machined surface.4.1 Randomized Block Design Main factor (factor to be studied in the experiment)Nuisance factor probably has an effect on the response, but we are Cutter of a machine toolnot interested in that effect. Known nuisance factor (factor not to be studied in the experiment,For unknown and uncontrollable nuisance factors, use Complete but we are well aware of the existence of its effect on the response)Randomization technique to get rid of them. Different machine tools of the same typeFor known and controllable nuisance factors, use RandomizedBlocking technique instead. Unknown nuisance factor (unknown and uncontrollable factors, which may have some effect on the response)The variability of the nuisance factor will contribute to the variabilityobserved in the experimental data. As a result, the experimental error Temperature, humiditywill reflect both the random error of the unknown nuisance factorsand variability of the known nuisance factor. 4-1 4-2 Table 4.1 MS F MS F DOEE F0 = = = MS F (4.4) Machine tool (Block) MS E SS E SS E Cutter 1 2 3 4 DOEE 1 -2 -1 1 5 2 -1 -2 3 4 So, whether F0 will be increased by the addition of the known 3 -3 -1 0 2 nuisance factors depends on which of SSE or DOEE will increase 4 2 1 5 7 more. If SSE increases more by including the known nuisance factors, F0SST = SS F + SS known + SSunknown (4.1) becomes less sensitive. Therefore, Randomized Blocking technique is preferred.If the effects of the known nuisance factors are included in the error If DOEE increases more, F0 becomes more sensitive. Therefore, Complete Randomization technique is preferred.SS E = SS known + SSunknown (4.2) Note, the increase of DOEE also reduces the control limit Fα ,ν 1 ,ν 2 (ν2DOEE = DOEknown + DOEunknown (4.3) = DOEE), and therefore, makes F0 relatively more sensitive. 4-3 4-4
  • 23. Randomized Block Design Block 1 Block 2 Block b It means blocking all the runs according to the different levels of a known nuisance factor. Within a block, the order in which the y11 y12 y1b runs of the main factors are tested is randomly determined. y21 y22 y2b The blocks form a more homogeneous experimental unit in terms of the known nuisance factor. ............. y31 y32 y3b Effectively, this design strategy improves the accuracy of the . . . comparisons of the responses by eliminating the variability of the . . . known nuisance factor. . . . ya1 ya2 yabExamples of blocks Equipment, machinery, batches of raw material, people, time slot Figure 4.1 4-5 4-6The statistical model Hypotheses of interest are ⎧ i = 1,2,L, a H0: μ1 = μ2 = = … μa yij = μ + τ i + β j + ε ij ⎨ (4.5) ⎩ j = 1,2,L, b H1: at least one μi ≠ μj (4.6)where, μ is an overall mean, τi is the effect of the ith run, βj is the Oreffect of the jth block, and εij is the usual NID(0, σ2) random error H0: τ1 = τ2 = … = τa = 0term. H1: at least one τi ≠ 0 (4.7) βj will not be studied, because it is the effect of the known nuisance factor. 4-7 4-8
  • 24. yi. the total of all observations taken under run i (total of row) yi. the average of the observations taken under the ith runy.j the total of all observations in block j (total of column) y. j the average of the observations in block jy.. the grand total of all observations y.. the grand average of all observations.N = ab the total number of observations. b yi. = yi . / b y. j = y . j / a y.. = y.. / N (4.9) yi. = ∑ yij i = 1,2,L, a j =1 a y. j = ∑ yij j = 1,2,L, b (4.8) i =1 a b y.. = ∑∑ yij i =1 j =1 4-9 4-10Sum of square Degrees of Freedom SST = SS F + SS Block + SS E (4.10) SST N - 1 = ab - 1 a b y..2 SSF a-1 SST = ∑∑ yij 2 - (4.11) i =1 j =1 N SSBlocks b-1 a 2 1 y.. SS F = ∑ yi.2 - N b i=1 (4.12) SSE (ab − 1) − (a − 1) − (b − 1) = ab − a − b + 1 (4.15) 1 b 2 y..2 SS Block = ∑ y. j - (4.13) a j =1 N SS E = SST - SS F − SS Block (4.14) 4-11 4-12
  • 25. Test Statistic Table 4.2 ANOVA Table _______________________________________________________ MS F Source of Sum of Degrees of Mean F0 F0 = MS E Variation Squares Freedom Square SS F _______________________________________________________ MS F = (4.16) a −1 SS F MS F Factor SS F a -1 Fo = SS E a −1 MS E MS E = ab − a − b + 1 SS Blocks Blocks SSBlocks b–1We would reject Ho in (4.7), if Fo > Fα,(a-1)(ab-a-b+1). b −1 SS E Error SS E ab – a – b + 1 ab − a − b + 1 Total SST N -1 _______________________________________________________ 4-13 4-14We may also examine the ratio of MSBlocks (= SSBlocks/(b-1)) to MSE. If Example 4.1this ratio is large, it implies that the blocking factor has a large effect Table 4.3and that the noise reduction obtained by blocking was probably Machine tool (Block) Cutter 1 2 3 4 yi.helpful in improving the effectiveness of the experiment. Otherwise, 1 -2 -1 1 5 3blocking may actually be unnecessary, i.e, the known nuisance factor 2 -1 -2 3 4 4should be handled as random error. 3 -3 -1 0 2 -2 4 2 1 5 7 16 y.j -4 -3 9 18 y.. = 20 4-15 4-16
  • 26. The sums of squares are obtained as follows: Table 4.4 ANOVA Table _______________________________________________________ a b y..2 20 2 Source of Sum of Degrees of Mean F0 SST = ∑∑ yij 2 - = 154 − = 129 i =1 j =1 N 16 Variation Squares Freedom Square _______________________________________________________ = (3 + 4 + (−2) + 15 ) − 1 a 2 y..2 1 2 20 2 SS F = ∑ yi. - 2 2 2 = 38.5 Cutter 38.50 3 12.83 14.44 b i=1 N 4 16 Machine 82.50 3 27.50 ∑ y.2j - N = 4 ((−4) 2 + (−3) 2 + 92 + 182 ) − 16 = 82.5 b 2 2 (Block) 1 y.. 1 20 SS Block = a j =1 Error 8.00 9 0.89 SS E = SST - SS F − SS Block = 129 − 38.5 − 82.5 = 8 Total 129.00 15 _______________________________________________________ Specify α = 0.05, F0.05,3,9 = 3.86. Since 14.44 > 3.86, we conclude that the type of cutter affects the mean hardness reading. 4-17 4-18If the randomized block design is not used, F0.05,3,12 = 3.49, thehypothesis of equal mean values from the four cutters cannot berejected. Table 4.5 ANOVA TableSource of Sum of Degrees of Mean F0Variation Squares Freedom SquareCutter 38.50 3 12.83 1.70Error 90.50 12 7.54Total 129.00 15Here, the error is actually the sum of error and block in Table 4.4. 4-19
  • 27. 5 FACTORIAL EXPERIMENTS Case one Two factors: A: Machine, B: Feed rate5.1 Introduction Response: Process capabilityFactorial experiments study the effects of several factors on theresponse. Table 5.1 Process capability CpIn factorial experiments, all possible runs (combinations of the levels Feed rate (mm/sec.)of the factors) are investigated. Machine 0.2 0.4 0.6For example, for two factors A: temperature (with a levels) and B: USA 2.1, 2.2 1.7, 1.6 1.2, 1.3humidity (with b levels), the factorial experiment has ab possible Japan 0.9, 1.0 0.7, 0.6 0.4, 0.3runs. a = 2, b = 3, n = 2, R =ab = 6If for each run, n observations are taken, then, the total number N of N = Rn = abn = 2 × 3 × 2 = 12the observations is equal to abn. 5-1 5-2Case two 5.2 ANOVA for Two-Factor Factorial Experiments Three factors: A, B and C. Mathematical model a = 2, b = 3, c = 4, n = 2, R = abc = 24 Yijk = μ + τ i + β j + (τβ )ij + ε ijk (5.1) N = Rn = abcn = 2 × 3 × 4 × 2 = 48 μ the overall mean τi the effect incurred at the ith level of factor ANumber of runs is equal to the product of the levels of all factors. βj the effect incurred at the jth level of factor B (τβ)ij the effect incurred at the interaction between the ith level of A and the jth level of B εijk the random error (normally and independently distributed, zero mean and constant variance) 5-3 5-4
  • 28. Interaction indicates the influence of one factor on the effect of Table 5.2 Data Arrangement for a Two-Factor Factorial Design another factor, and vice versa. Factor B 1 2 … b Total AverageFactors A: temperature 1 y111, y112, y121, y122, … y1b1, y1b2, y1.. y1.. B: pressure Factor A …, y11n …, y12n …, y1bn 2 y211, y212, y221, y222, … Y2b1, y2b2, y2.. y 2..Response Process yield …, y21n …, y22n …, y2bn : … … … … … …If: (1) Increasing temperature alone cannot increase the yield; and a ya11, ya12, ya21, ya22, … yab1, yab2, ya.. y a.. (2) Increasing pressure alone cannot increase the yield; and …, ya1n …, ya2n …, yabn (3) Increasing temperature and pressure at the same time does Total y.1. y.2. … y.b. y… increase the yield Average y .1. y .2. … y .b. y ...Then: There is interaction between temperature (factor A) and yijk is the kth observation taken at the ith level of factor A and the jth pressure (factor B) level of factor B. The total number of observations: N = abn 5-5 5-6Total and average at the ith level of factor A: Hypotheses b n y yi .. = Σ Σ yijk y i .. = i .. i = 1,2,...,a Factor A: j =1 k =1 bn (5.2) H o : τ 1 = τ 2 = ... = τ a = 0 Factor A has no effect on yTotal and average at the jth level of factor B: a n y. j . H 1 : at least one τ i ≠ 0 Factor A has effect on y y. j . = Σ Σ yijk y. j. = j = 1,2,..., b i =1 k =1 an (5.3) Factor B:Total and average when A at the ith level and B at the jth level: H 0 : β1 = β2 =L = βb = 0 Factor B has no effect on y n y H 1 : at least one β j ≠ 0 yij . = Σ yijk y ij . = ij . Factor B has effect on y k =1 n (5.4) Interaction AB:Grand total and average a b n y... H o : (τβ ) ij = 0 for all i, j Interaction AB has no effect y y...= ΣΣΣ yijk y...= (5.5) i =1 j =1 k =1 abn H1 : at least one (τβ ) ij ≠ 0 Interaction AB has effect on y 5-7 5-8
  • 29. Equation of the Sum of Squares Sum of squares for interaction AB: a b SST = SSA + SSB + SSAB + SSE (5.6) SSAB = n ΣΣ ( y i =1 j =1 ij . − yi.. − y. j . + y... )2Total sum of squares: (5.10) a b 2 2 yij . y... SS Τ = Σ Σ Σ ( yijk − y... ) = Σ Σ Σ yijk − a b n 2 2 a b n 2 y... (5.7) = ΣΣ i =1 j =1 − n abn − SSA − SSB i =1 j =1 k =1 i =1 j =1 k =1 abnSums of squares for main effects: Sum of squares for error: yi2.. y... SSE = SSΤ − SSAB − SSA − SSB a a 2 SS A = bn∑ ( yi .. − y... ) (5.11) =Σ − 2 (5.8) i =1 i =1 bn abn y.2j . 2 SS B = an∑ ( y. j . − y... ) = Σ b 2 b y... − (5.9) j =1 j =1 an abn 5-9 5-10Degrees of freedom Mean Square SST abn - 1 (or N - 1) MS = SS / D.O.F. (5.13) SSA a-1 F0 = MS / MSE (5.14) SSB b-1 SSA MSA Mean square of A: MSA = F0 = SSAB (a - 1)(b - 1) a −1 MSE SSE ab(n - 1) SSB MSB Mean square of B: MSB = F0 = b −1 MSEEquation of D.O.F. DOFSST = DOFSS A + DOFSSB + DOFSS AB + DOFSSE SSAB MSAB Mean square of AB: MSAB = F0 = ( a − 1)(b − 1) MSE abn − 1 = (a − 1) + (b − 1) + (a − 1)(b − 1) + ab(n − 1) (5.12) 5-11 5-12
  • 30. SSE Table 5.3 ANOVA for a two-factor factorial designMean square of error: MSE = ab(n − 1) Source of Sum of Degrees of Mean square F0 variation squares freedomThe denominator of the F0 ratio is always equal to MSE A SSA a-1 SS A MS A MS A = a −1 MS EControl limit: Fα ,ν 1 ,ν 2 B SSB b-1 SS B MS B MS B = b −1 α is the probability of Type I error. MS E AB SSAB (a - 1)( b- 1) SS AB MS AB ν1 = D.O.F. of the numerator of the F0 ratio. MS AB = ( a − 1)( b − 1) MS E ν2 = ab(n - 1), D.O.F. of MSE Error SSE ab(n - 1) SS E MS E = ab( n − 1) Fα ,ν 1 ,ν 2 may differ for different effects or interactions as ν1 may Total SST abn - 1 not be the same. 5-13 5-14How to find whether a factor (or an interaction) has effect on the Example 5.1response? A: machine B: temperature Response: process capability Cp(1) Calculate the sum of squares SS and then the mean square MS a = 3, b = 2, n = 3, N = 3 × 2 × 3 = 18 of the factor (or interaction) to be studied. Table 5.4 o 1 10 C 2 20o C yi..(2) Calculate the ratio F0 between this mean square MS and the mean square error MSE. 1 USA 4.0, 4.5, 4.3 12.8 5.4, 4.9, 5.6 15.9 28.7 2 Japan 5.6, 4.9, 5.4 15.9 5.8, 6.1, 6.3 18.2 34.1(3) Decide the control limit Fα ,ν 1 ,ν 2 , where, α is the type I error probability, ν1 is the D.O.F. of the factor (or the interaction), 3 UK 3.8, 3.7, 4.0 11.5 5.5, 5.0, 5.0 15.5 27.0 ν2 is the D.O.F. of the error. y.j. 40.2 49.6 89.8 = y…(4) If F0 is larger than Fα ,ν 1 ,ν 2 , the factor (or interaction) has Number inside the rectangle is the total observations in a run. significant effect on the response; otherwise it has no effect. 5-15 5-16
  • 31. a b n 2 2 ΣΣΣ y... a b yij . 2 y...SSΤ = 2 yijk − SS AB = ∑ ∑ − − SS A − SS B i =1 j =1 k =1 abn i =1 j =1 n abn 89.82 12.82 15.92 15.52 89.82 2 = 4.0 + 4.5 +L+5.0 − 2 = 10.722 = + +L+ − − 4.58 − 4.91 = 0.24 3×2 ×3 3 3 3 3× 2 × 3 a yi2.. y... 2 SS E = SS Τ − SS AB − SS A − SS BSS A = Σ − i =1 bn abn = 10.72 − 0.24 − 4.58 − 4.91 = 0.99 28.72 3412 27.02 . 89.82 = + + − = 4.58 2 × 3 2 × 3 2 × 3 3× 2 × 3 b y.2j .2 y...SS B = Σ − j =1 an abn 40.22 49.62 89.82 = + − = 4.91 3× 3 3× 3 3× 2 × 3 5-17 5-18 Table 5.5 ANOVA Table Specify α = 0.05 (1) Factor A (machine) Source of Sum of Degrees of Mean F0 variation squares freedom square Fα ,ν 1 ,ν 2 = F0.05,2,12 = 3.89 machine 4.58 2 2.29 27.758 F0 > Fα ,ν 1 ,ν 2 , so, machine has effect on the Cp. temp. 4.91 1 4.91 59.515 (2) Factor B (temperature) Fα ,ν 1 ,ν 2 = F0.05,1,12 = 4.75 AB 0.24 2 0.12 1.455 F0 > Fα ,ν 1 ,ν 2 , so, temperature has effect on the Cp. Error 0.99 12 0.0825 (3) Interaction AB (between machine and temperature) Total 10.72 17 Fα ,ν 1 ,ν 2 = F0.05,2,12 = 3.89 F0 < Fα ,ν 1 ,ν 2 , so, interaction almost has no effect on the Cp. 5-19 5-20
  • 32. 6 2K EXPERIMENTS Advantages:6.1 Introduction (1) 2k experiment can substantially reduce the total number of runs in2k experiment (or called 2k design) is a special and simplest, but the an experiment for given number of factors, therefore, more factorsmost widely used, factorial experiment with k factors, each at two can be considered at the same time. 2k design is especially usefullevels. to the screening experiments in the early stage when we are not certain which factor (or factors) may have significant effect on theThe value of 2k by itself is equal to the total number of runs in the response.experiment. For example, due to the time limit, we can only conduct 65 runs inFor example, if there are three factors A (temperature), B (humidity) an experiment.and C (feed rate) in an experiment, i.e., k = 3. If each factor is to be investigated in 3 levels, only 3 factors can beNumber of levels for each factor: a = b = c = 2 studied, because 33 < 65 < 34Number of runs: abc = 2 × 2 × 2 = 23 = 8 If each factor is to be investigated in 2 levels, 6 factors can beThe replicate n is the number of observations at each run. studied, because 26 < 65 < 27 6-1 6-2(2) In 2k experiments, the formulae for calculating the sum of squares In 2k experiments, each factor has only two levels. They are called are much simpler than for other types of factorial experiments. low (-) level and high (+) level, respectively.(3) In 2k experiments, we can easily use a simple statistics effect to For examples (response is Cp) estimate the effect (or influence) of a factor on the response. The complicated ANOVA procedure may be saved under certain Table 6.1 circumstances. Factor Low (-) High (+) A temperature 10 oC 30 oC B machine USA-made Japan-made The definition of low and high levels for the qualitative factor machine (B) is arbitrary. If they are defined in another way (Japan- made is low level and USA-made is high level), the conclusions resulting from a DOE will not be changed 6-3 6-4
  • 33. Special Notations Used in the 2k Designs: High b ab B(1) A run is represented by a series of lower case letters. (Japan) (-) (+)(2) If a letter is present, the corresponding factor is set at the high (-) b (1) level in that run. first diagonal B A(3) If a letter is absent, the corresponding factor is set at the low level. a ab (+) Low A run is indicated (USA) (1) a by a point Low A High (10oC) (30oC) Figure 6.1 6-5 6-6The Four Runs in a 22 Design: In addition, each of the notations (1), a, b, and ab has a value, which is equal to the sum of all the n observations taken at the(1) factor A at low level; factor B at low level. corresponding run.a factor A at high level; factor B at low level. Table 6.3 Run A B Responseb factor A at low level; factor B at high level. temp. machine Cp (1) 10 USA 1.2 1.3 (1) = 2.5ab factor A at high level; factor B at high level. a 30 USA 2.1 2.2 a = 4.3 Table 6.2 b 10 Japan 1.7 1.8 b = 3.5 ab 30 Japan 2.7 2.8 ab = 5.5 Run A B temp. machine Two usages of the lower-case-letter notations: (1) - - a + - (1) represent a run; b - + (2)indicate the total of the observed values of the response in that run. ab + + 6-7 6-8
  • 34. The special notation is used throughout 2k experiments (designs). 6.2 22 DesignExample: A 23 design with three factors A, B, C y+ is the response value when a factor is at its high level Table 6.4 y- is the response value when a factor is at its low level No Run A B C 1 (1) - - - The difference (y+ - y-) is called the effect of that factor. It is defined 2 a + - - as the change in response produced by a change in the level of the 3 b - + - factor. The larger the effect (the difference (y+ - y-)), the more 4 ab + + - influential the factor is on the response. 5 c - - + 6 ac + - + 7 bc - + + 8 abc + + + 6-9 6-10For example, Y(Cp) Y(Cp)(1) If temperature is the factor, and Cp is the response, Y+ y+ = 2.5 when temperature is at the high level of 30oC; 2.5 1.7 Y+ y- = 0.8 when temperature is at the low level of 10oC; 1.25 Y- Then, the effect of the temperature is: (y+ - y-) = 1.7. This means Y- 0.8 temperature has significant effect on the Cp. Increasing the T machine temperature will increase Cp. low high low high 10 30 USA Japan Figure 6.2 6-11 6-12
  • 35. (2) If machine is the factor, and Cp is the response, Usually, n (n ≥ 1) observations are taken at the low level or high level of a factor, so effect should be generalized as y+ = 1.25 when machine is made by Japan (high level); Effect = y+ − y− (6.1) y- = 1.24 when machine is made by USA (low level); y+ : Average of all the observations taken when the factor is at the Then, the effect of the machine: (y+ - y-) = 0.01. This means high level. machine has very minor effect on the Cp. The value of Cp is almost constant no matter which machine is used. y− : Average of all the observations taken when the factor is at the low level. 6-13 6-14Example 6.1 The main effect (or simply the effect) of A is denoted by A.Response is the process capability Cp A = y A+ − y A− a + ab b + (1) 1 Table 6.5 = − = [a + ab − b − (1)] (6.2) 2n 2n 2n Factor A Factor B (temperature) 1 (machine) = [8 + 6 − 2 − 4] = 2 B- (100 C) B+ (200 C) 2× 2 A- (USA) (1) 1.8, 2.2 b 0.8, 1.2 The main effect of B is denoted by B. A+ (Japan) a 3.8, 4.2 ab 2.8, 3.2 B = yB + − yB − (1) = 1.8 + 2.2 = 4 b + ab a + (1) 1 = − = [b + ab − a − (1)] (6.3) a = 3.8 + 4.2 = 8 2n 2n 2n 1 b = 0.8 + 1.2 = 2 = [2 + 6 − 8 − 4] = −1 2×2 ab = 2.8 + 3.2 = 6 6-15 6-16
  • 36. A positive effect means that the response will be greater when thefactor is at the high level. Cp Cp A = 2 > 0, means Cp is greater if Japanese machine (A+) is y+ used. y− -1A negative effect means that the response will be greater when the 2 y+factor is at the low level. y− B = -1 < 0, means Cp is greater if temperature is low. A B USA machine temp Japan 10 20When the magnitude of an effect is large, its influence on response issignificant, no matter the effect is positive or negative. Figure 6.3 6-17 6-18For Example 6.1, if we set A+ as USA machine and A- as Japanese From (6.2)machine, the observed values are as follows: 1 A= [a + ab − b − (1)] Table 6.6 2n Factor A Factor B (temperature) 1 (machine) = [4 + 2 − 6 − 8] = −2 2× 2 B- (100 C) B+ (200 C) A- (Japan) (1) 3.8, 4.2 b 2.8, 3.2 A+ (USA) a 1.8, 2.2 ab 0.8, 1.2 Since A = -2 < 0, (1) = 3.8 + 4.2 = 8 Cp will be greater if Japanese machine (A-) is used. a = 1.8 + 2.2 = 4 This conclusion is the same as obtained before. b = 2.8 + 3.2 = 6 ab = 0.8 + 1.2 = 2 6-19 6-20
  • 37. Interaction Example 6.2 (without interaction)If the effects of one factor are different at different levels of another Response is the process capability Cpfactor, there is interaction between these two factors. For example, if the effects of A are different at different levels offactor B, there is interaction between A and B. This interaction is Table 6.7denoted as AB. Factor A Factor B (temperature) (machine)For example, a particular medicine has effect on children, but no B- (100 C) B+ (200 C)effect on adults. This means the age influence the effect of the A- (USA) (1) 2 b 1medicine. Or in other words, there is interaction between the factor A+ (Japan) a 4 ab 3medicine and the factor age. 6-21 6-22The A effect (with the condition that B is low), Example 6.3 (with interaction) Output: the process capability Cp A(B-) = 4 – 2 = 2The A effect (with the condition that B is high), Table 6.8 A(B+) = 3 – 1 = 2 Factor A Factor B (temperature) (machine)The effect of A is identical, no matter B is at the low or high lever. B- (100 C) B+ (200 C) A- (USA) (1) 3 b 2Conclusion: there is no interaction between factors A and B. No A+ (Japan) a 1 ab 4matter temperature is low or high, Japanese machine is always better. 6-23 6-24
  • 38. The A effect (with the condition that B is low), The greater the difference between A(B+) and A(B-), the more significant the interaction AB is . So, A(B+) - A(B-) is a measure of the A(B-) = 1 – 3 = -2 interaction AB (refer to Figure 6.1).The A effect (with the condition that B is high), AB = 1 [A( B + ) − A( B −) ] 2 [ ] A(B+) = 4 – 2 = 2 1 = ( y A+ − y A− )( B+ ) − ( y A+ − y A− )( B− ) 2The effects of A are different at different levels of factor B. 1 ⎡⎛ ab b ⎞ ⎛ a (1) ⎞⎤ = ⎢⎜ − ⎟ − ⎜ − ⎟⎥Conclusion: there is interaction between factors A and B. When 2 ⎣⎝ n n ⎠ ⎝ n n ⎠⎦temperature is low, USA machine is better; when temperature is high, 1Japanese machine is better. = [ab − b − a + (1)] 2n (6.4) 6-25 6-26When the interaction AB is positive, the response will be augmented Alternative Approach (equivalent)if both A and B are increased.When the interaction AB is negative, the response will be augmented AB = 1 [B( A+ ) − B( A−) ] 2if A or B is increased and another is decreased. 1 = [ab − b − a + (1)] 2nOnly the magnitude (not the sign) of AB is important. Interaction AB is mutual and equivalent to factors A and B. Diagonal method (see Figure 6.1): The expression inside the bracket of (6.4) is equal to the difference between the total of observations at the first diagonal and the total of observations at the second diagonal. The first diagonal includes the run of (1) and ab (for interaction AB) 6-27 6-28
  • 39. For example 6.2 (without interaction) When interaction is large, the main effects are less meaningful. In Example 6.3 (with interaction) 1 AB = [ab − b − a + (1)] 2n Table 6.9 1 = [3 − 1 − 4 + 2] = 0 Factor A Factor B (temperature) 2 ×1 (machine) B- (100 C) B+ (200 C)For example 6.3 (with interaction) A- (USA) (1) 3 b 2 A+ (Japan) a 1 ab 4 1 AB = [ab − b − a + (1)] 2n (1) = 3, a = 1, b = 2, ab = 4 1 = [4 − 2 − 1 + 3] = 2 2 ×1 6-29 6-30From (6.2), the main effect of factor A is Guidelines for Handling the Interaction 1 A= [a + ab − b − (1)] (1) First calculate the interaction. 2n 1 = [1 + 4 − 2 − 3] = 0 (2) If the interaction is equal or close to zero, the effect of each 2 ×1 factor can be studied by simply calculating its main effect.Can we conclude that factor A (machine) has no effect ??? NO! (3) Otherwise, the effect of a factor should be studied separately at oIn fact, at the low temperature (10 C), USA machine is better different levels of another factor. A (B-) = 1 - 3 = -2At the high temperature (30oC), Japanese machine is better A (B+) = 4 - 2 = 2That is, factor A (machine) does have significant effect on Cp. But,this effect differs when temperature is different. 6-31 6-32
  • 40. Contrast 1 C A = [− (1) + a − b + ab] = A (6.5) Contrast is a linear combination of run totals: 2n 2n R 2k 1 C = ∑ ci y i = ∑ ci y i ⋅ B= [− (1) − a + b + ab] = CB (6.6) i =1 i =1 2n 2n With the restriction on the coefficients ci: 1 AB = [(1) − a − b + ab] = C AB a 2n 2n (6.7) ∑c i =1 i = 0. where, yi. is the total of the ith run.Evaluating effects is a simple way to estimate the influence of thefactors and interactions on the response. But, it sometime fails to give For example, if k = 2, R = 22 = 4,a clear-cut conclusion and cannot provide a confidence level for the Contrast: C A = −(1) + a − b + abconclusion. Therefore, more complicated ANOVA may have to becarried out. Restriction: c1 = -1, c2 = 1, c3 = -1, c4 = 1, c1 + c2 + c3 + c4 = 0 6-33 6-34In (6.5) to (6.7) The contrast of a main effect or an interaction can be easily obtained from the plus and minus sign Table. C A = −(1) + a − b + ab Table 6.10 Plus and Minus Signs Table for the 22 Design C B = −(1) − a + b + ab Factor C AB = (1) − a − b + ab Run I A B AB (1) + - - +CA, CB and CAB are contrasts, because each of them is a linear a + + - -combination of run totals and the sum of the coefficients is equal tozero. b + - + - ab + + + + 6-35 6-36
  • 41. (1) The signs under column A or B are determined by the run (3) The contrast of a factor or an interaction can be obtained by notation. summing up the products of each run and the sign under the corresponding column. For example, for run (1), both factors A and B are at the low level, so, both columns A and B have a minus sign. CA = (-)(1) + (+)(a) + (-)(b) + (+)(ab) For run (b), factor A is at the low level and column A has a minus = -(1) + a - b + ab (6.8) sign; factor B is at the high level and column B has a plus sign. CB = (-)(1) + (-)(a) + (+)(b) + (+)(ab)(2) The signs under column AB are decided by the products of the = -(1) - a + b + ab (6.9) signs under columns A and B. CAB = (+)(1) + (-)(a) + (-)(b) + (+)(ab) = (1) - a - b + ab (6.10) I is the grand total I = + (1) + a + b + ab (6.11) 6-37 6-38The Sum of Squares for the Contrast (6.12) can be simplified as: ⎛ R ⎞ 2 C2 C2 ⎜ ∑ ci yi. ⎟ 2 SS = = N 4n (6.14) SS = ⎝ i=1 R ⎠ =C , R = 2k (6.12) n ∑ ci Q C A [−(1) + a − b + ab]2 2 2 i =1 SS A = = 4n 4nFor two-factor case, k = 2. The coefficients ci of the contrasts in a 2 k C 2 [−(1) − a + b + ab]2 SS B = B = (6.15)design is equal to either 1 or (-1), therefore, (ci)2 ≡ 1. So, 4n 4n C 2 [(1) − a − b + ab]2 R R SS AB = AB = Q = n∑ ci2 = n∑1 = nR = N (6.13) 4n 4n i =1 i =1 (6.15) is based on the concept of contrast, and can be applied to the 2k(D.O.F. of the contrast ) ≡ 1 designs only. If any factor has more than two levels, general formulae should be used. 6-39 6-40
  • 42. Example 6.4 A 22 Design (n = 4) Use contrast formulae:A: Spindle speed (A- = 1000 rpm, A+ = 3500 rpm) C A [−(1) + a − b + ab]2 2B: Feed rate (B- = 0.1 μm/s, B+ = 0.15 μm/s) SS A = = 4n 4nResponse: Thickness of a component Table 6.11 SS A = ( − 56.081 + 59.299 − 55.686 + 59156)2 = 2.7956 . Factor 4×4 ( − 56.081 − 59.299 + 55.686 + 59156)2 = 0.0181 Run A B AB Thickness (μm) Total Average . SS B = (1) - - + 14.037 14.165 13.972 13.907 56.081 14.020 4×4 a + - - 14.821 14.757 14.843 14.878 59.299 14.825 SS AB = ( + 56.081 − 59.299 − 55.686 + 59156)2 = 0.0040 . b - + - 13.880 13.860 14.032 13.914 55.686 13.922 4×4 ab + + + 14.888 14.921 14.415 14.932 59.156 14.789 y… = 230.222(1) = 56.081 a = 59.299 b = 55.686 ab = 59.156 6-41 6-42Use conventional formulae Table 6.12 Table of ANOVA Source of Sum of Degrees of Mean square F0 a b n y2SS Τ = Σ Σ Σ y − ... 2 variation squares freedom ijk i =1 j =1 k =1 abn (6.16) A 2.7956 1 2.7956 134.40 230.2222 = 14.0372 +...+14.9322 − = 3.0672 2×2×4 B 0.0181 1 0.0181 0.87SS E = SST − SS A − SS B − SS AB AB 0.0040 1 0.0040 0.19 (6.17) = 30672 − 2.7956 − 0.0181 − 0.0040 = 0.2495 . Error 0.2495 12 0.0208 Total 3.0672 15 D.O.F.total = N – 1 = 15 D.O.F.error = 2k(n – 1) = 4 × (4 – 1) = 12 6-43 6-44
  • 43. Steps to Calculate the Sum of Squares for 2k Design 6.3 2k Design for k≥3 Factors(1) Establish the plus and minus signs table. For example, 23 design(2) Decide the contrast (6.8), (6.9), (6.10). k = 3, 3 factors, each at two levels(3) Calculate the sum of squares of the main effects and interactions from the contrasts (6.15). 8 runs: (1) a b c ab ac bc abc(4) Calculate the total sum of squares by the conventional formula (6.16). 3 main effects: A, B, C(5) Calculate the sum of squares of the error by subtraction (6.17). 3 two-factor interactions: AB, AC, BC 1 three-factor interaction: ABC(6) Complete the ANOVA. 6-45 6-46 23 Design bc abc Main effect of A c ac + a + ab + ac + abc (1) + b + c + bc A = y A+ − y A− = − 4n 4n (6.18) 1 C = [− (1) + a − b + ab − c + ac − bc + abc ] 4n b ab - (1) Main effect of B a + b + ab + bc + abc (1) + a + c + ac B = y B+ − y B− = − - B 4n 4n - + (6.19) A 1 = [− (1) − a + b + ab − c − ac + bc + abc ] 4n Figure 6.4 The 23 factorial design 6-47 6-48
  • 44. Main effect of C When calculating the effects, the denominator is the total number of observations at the low (or high) level of a factor. It is always equal c + ac + bc + abc (1) + a + b + ab to half of the total observations N in a 2k design. Because, for eachC = y C + − yC − = − 4n 4n factor, one half of the observations are taken at its high level and (6.20) 1 = [− (1) − a − b − ab + c + ac + bc + abc ] another half at its low level. 4n N 2k n = = 2 k −1 n (6.21) 2 2 k R=2k N denominator 2k-1n 2 4 4n 2n 3 8 8n 4n 4 16 16n 8n 5 32 32n 16n 6-49 6-50Interaction AB Interaction AC 1AB = [ A (B + ) - A (B-) ] 1 4n AC = [ A (C + ) - A (C-) ] 1 4n (6.23) = [(ab + abc − bc − b) − (a + ac − (1) − c)] (6.22) 1 4n = [(ac + abc − bc − c) − (a + ab − b − (1))] 1 4n = [ab + abc + (1) + c − bc − b − a − ac] 4n bc c ac bc abc C high abc B low B high c ac b ab Figure 6.5 C low Figure 6.6 (1) (1) a a b ab 6-51 6-52
  • 45. Interaction BC Interaction ABC (refer to Figure 6.6) 1 Let T = ABBC = [ B (C + ) - B (C-) ] 4n 1 (6.24) ABC = TC = [(bc + abc − c − ac) − (b + ab − (1) − a )] = { ( C + ) − T( C − ) } 4n 1 T 4n = {AB( C + ) − AB( C − ) } 1 4n 1 = {[(abc + c) − (ac + bc)] − [(ab + (1) ) − (a + b )]} 4n 1 = [abc - bc - ac + c - ab + b + a - (1)] (6.25) 4n 6-53 6-54 Table 6.13 Plus and Minus Signs Table for the 23 design Systematic way to write the standard run order: Factor effect Run I A B C AB AC BC ABC (1) Convert the run number (from 0 to 7) from a decimal number into a binary number. (1) + - - - + + + - a + + - - - - + + (2) If the first digit (from right) is 1, a is present; if it is 0, a is absent. b + - + - - + - + ab + + + - + - - - (3) If the second digit is 1, b is present; if it is 0, b is absent. c + - - + + - - + ac + + - + - + - - (4) If the third digit is 1, c is present; if it is 0, c is absent. bc + - + + - - + - abc + + + + + + + +A (1) a [ Standard Order ]B b abC c ac bc abc 6-55 6-56
  • 46. Table 6.14 (1) The signs under columns A, B and C are determined by the runDecimal run number Binary run number Run notation notation. (2) The signs under column AB are decided by the products of the 0 000 (1) signs under columns A and B. The signs under columns AC and 1 001 a BC can be decided similarly. 2 010 b (3) The signs under column ABC can be decided by several different 3 011 ab ways, e.g., the products of the signs under columns A, B and C; or 4 100 c the products of the signs under column A and column BC. 5 101 ac 6 110 bc (4) The effect of a factor or an interaction can be obtained by 7 111 abc summing up the products of each run and the sign under the corresponding column. For example,Standard run order is required only for a few methods, for example CAB = (+)(1) + (-)(a) + (-)(b) + (+)(ab) + (+)(c) + (-)(ac) + (-)(bc)the Yates’ Method + (+)(abc) = (1) - a - b + ab + c – ac – bc + abc 6-57 6-58General Formulae to Calculate effect and sum of squares from Example 6.5 Response: Surface roughness (n = 2)the contrast A: Spindle speed (A- = 1000 rpm, A+ = 3500 rpm) B: Feed rate (B- = 0.1 μm/s, B+ = 0.15 μm/s) Contrast C: Machine (C- = USA machine, C+ = UK machine) Effect = (6.26) N /2 Table 6.14 Factors Surface (Contrast) 2 Run A B C roughness Total SS = (6.27) N (1) - - - 9, 7 16 a + - - 10, 12 22The total number of observations in a 2k design: b - + - 9, 11 20 ab + + - 12, 15 27 N = Rn = 2 k n (6.28) c - - + 11, 10 21 ac + - + 10, 13 23 bc - + + 10, 8 18 abc + + + 16, 14 30 6-59 6-60
  • 47. CA = -(1) + a - b + ab - c + ac - bc + abc Values of all main effects and interactions = -16 + 22 - 20 + 27 -21 + 23 - 18 + 30 = 27 A = 3.375N = 2kn = 23 × 2 = 16 B = 1.625 C = 0.875 CA 27A= = = 3.375 AB = 1.375 N / 2 16 / 2 AC = 0.125 C 2 27 2SS A = A = = 45.563 BC = -0.625 N 16 ABC = 1.125 The highlighted values are the significant effects or interactions. A, B and AB are dominant. 6-61 6-62 Table 6.15 Table of ANOVA Specify α = 0.01, control limit Fα ,ν1 ,ν 2 = F0.01,1,8 = 11.26 Source of Sum of Degrees of Mean F0 variation squares freedom square Only factor A (spindle speed) is considered by ANOVA having significant influence on the response roughness A 45.563 1 45.563 18.69 B 10.563 1 10.563 4.33 C 3.063 1 3.063 1.26 Usually, the conclusions from the values of effects are similar to that AB 7.563 1 7.563 3.10 from ANOVA. AC 0.063 1 0.063 0.03 The calculation of effects is easier than ANOVA. However, the BC 1.563 1 1.563 0.64 information from the effects only has relative or comparative ABC 5.063 1 5.063 2.08 meaning, and cannot give the confidence level as in ANOVA. Error 19.500 8 2.438 Total 92.938 15 6-63 6-64
  • 48. 6.4 Project of 2k Design C bc abcAny 2k design can be projected into a smaller 2r design (k ≥ r), if oneor more of the original factors are negligible and can be dropped. c ac 23 22 21 B b abProjecting reduces the number of factors, so that it can simplifyANOVA and provide more insight into the remaining important (1)factors. a A Figure 6.7 Projecting the 23 design (in terms of factors A, B and C) into a 22 design (in terms of A and B), suppose factor C is negligible and can be dropped. 6-65 6-66Use Example 6.5 again Response: Surface roughness (n = 2) Projecting this 23 design (in terms of A, B and C) into a 22 design (inA: Spindle speed (A- = 1000 rpm, A+ = 3500 rpm) terms of A and B). Because, effect C is very minor.B: Feed rate (B- = 0.1 μm/s, B+ = 0.15 μm/s) Table 6.17 Run A B Surface Run afterC: Machine (C- = USA machine, C+ = UK machine) roughness dropping B Table 6.16 (1) - - 9, 7, 11, 10 (1) Factors Surface Run after a + - 10, 12, 10, 13 a Run A B C roughness Total dropping C b - + 9, 11, 10, 8 (1) (1) - - - 9, 7 16 (1) ab + + 12, 15, 16, 14 a a + - - 10, 12 22 a b - + - 9, 11 20 b (1) Start from the table for the 23 design. Erase the column C. ab + + - 12, 15 27 ab (2) In column Run, for every run notation including letter c, drop c - - + 11, 10 21 (1) letter c (i.e., change c to (1), ac to a, bc to b, abc to ab). ac + - + 10, 13 23 a bc - + + 10, 8 18 b (3) Work out a table for the 22 design in terms of A and B. abc + + + 16, 14 30 ab (4) Merge the rows that have identical run notation. 6-67 6-68
  • 49. Further projecting the 22 design (in terms of A and B) into a 21 design (1) Start from the table for the 22 design. Erase the column B.(in terms of A). Since, effect B is also relatively smaller than A. (2) In column Run, for every run notation including letter b, drop Table 6.18 letter b (i.e., change b to (1), ab to a). Run A Surface roughness (3) Work out a table for the 21 design in terms of A. (1) - 9, 7, 11, 10, 9, 11, 10, 8 a + 10, 12, 10, 13, 12, 15, 16, 14 (4) Merge the rows that have identical run notation. After projection, the number of factors and number of runs will be reduced, but the total number of observations is always the same. In projecting, no extra experiments are needed. Only the available observations have to be re-organized. 6-69 6-706.5 Single Replicate of 2k Design For a 24 experiment: Total number of runs = 24 = 16Single replicate: n = 1 Let p indicate the number of factors involved in an interaction.In single replicate design, the degrees of freedom for error is equal tozero. Table 6.19 p p-factor interaction list of interactions DOFE = R (n - 1) = R (1 -1) = 0 1 1-factor interaction (main effect) A, B, C, D 2 2-factor interaction AB, AC, AD, BC, BD, CD SS E SS 3 3-factor interaction ABC, ABD, ACD, BCD MS E = = E 4 4-factor interaction ABCD DOFE 0 Number of p-factor interactions in a 2k design can be calculated by MS F k! F0 = ANOVA cannot be carried out !! Cp = k = (k − p + 1)(k − p + 2 )L k (6.29) MS E p!(k − p )! 6-71 6-72
  • 50. Interactions having three or more factors (p ≥ 3) are called high Guideline for dealing with the single replicate experimentsorder interactions. The magnitudes of the high order interactionsare usually negligible and their physical meaning is hard to explain. (1) When the number of factors is larger than 3 or 4, the number of runs (=2k) is very large. In order to minimize the number ofThe conclusions of an experiment are usually dominated by the main observations (=2kn), single replicate design is widely used.effects and 2-factor interactions. (2) Under this circumstance, in order to carry out the ANOVA, a common practice is to pool the higher order interactions as an estimate of error. SSE = SSABC + SSABD + SSACD + SSBCD + SSABCD (6.30) DOFE = DOFABC+DOFABD+DOFACD+DOFBCD+DOFABCD (6.31) Then, ANOVA can be carried out as usual. 6-73 6-74(3) A check into the high-order interaction effects is necessary, Example of Single Replicate 24 Experiment because, occasionally one or two high order interactions may have relatively high values. If such non-trivial high order A: Gap dimension (A- = 0.01 μm, A+ = 0.02 μm) interactions have been found, they should be excluded from the sums in (6.30) and (6.31). B: Pressure (B- = 300 mpa, B+ = 700 mpa) C: Flow rate (C- = 1.2 m2/min, C- = 3.2 m2/min) D: Power (D- = 1000 kw, D+ = 3500 kw) Response: Etch rate (Å / min) 6-75 6-76
  • 51. Table 6.20 CA = -(1) + a - b + … + abcd Run A B C D Observation = -550 + 669 - 604 + … + 729 = -813 (1) - - - - 550 a + - - - 669 b - + - - 604 N = 2 k n = 2 4 × 1 = 16 ab + + - - 650 c - - + - 633 CA − 813 ac + - + - 642 A= = = −101.625 bc - + + - 601 N / 2 16 / 2 abc + + + - 635 C A (− 813) 2 2 d - - - + 1037 ad + - - + 749 SS A = = = 41311 bd - + - + 1052 N 16 abd + + - + 868 cd - - + + 1075 acd + - + + 860 bcd - + + + 1063 abcd + + + + 729 6-77 6-78Main and 2-factor interactions High order interactions (D.O.F. = 1. for all of them): A = -101.625 ABC = -15.625 SSABC = 977 B = -1.625 ABD = 4.125 SSABD = 68 AB = -7.875 ACD = 5.625 SSACD = 127 C = 7.375 BCD = -25.375 SSBCD = 2576 AC = -24.875 ABCD = -40.125 SSABCD = 6440 BC = -43.875 All of the high order interactions are small compared to the most D = 306.125 dominant main effects A and D, and 2-factor interaction AD. AD = -153. 625 SSE = 977 + 68 + 127 + 2576 + 6440 = 10187 BD = -0.625 CD = -2.125 DOFE = 1+ 1 + 1 + 1 + 1 = 5 6-79 6-80
  • 52. Table 6.21 ANOVA Table 6.6 Center Points in 2k Design Source of Sum of Degrees of Mean F0 Assumption in 2k design: The response y is a linear function of the variation squares freedom square A 41311 1 41311 20.28 factor, say A. The curve y versus A is a straight line (no-curvature) B 11 1 11 0.01 which can be completely decided by two points (y-, A-) and (y+, A+). C 218 1 218 0.11 Therefore, the 2k design is adequate. D 374850 1 374850 183.99 AB 248 1 248 0.12 yc y y+ AC 2475 1 2475 1.21 AD 94403 1 94403 46.34 BC 7700 1 7700 3.78 yF BD 2 1 2 0.00 y- CD 18 1 18 0.01 Error 10187 5 2037 Total 531421 15 Figure 6.8 A- AC A+ 6-81 6-82AC is the center point between A- and A+: Example 6.6 Response: yield per hour AC = 0.5 ×[(A-) + (A+)]yC is the observed response value at point AC. 160 +1 40.0 41.5yF is calculated by 40.3 B 40.5 y + y+ temperature 155 0 40.7 yF = − 2 40.2 40.6(1) If the distance yC − y F is close to zero, the response function is 40.9 150 -1 39.3 likely to be a linear function of the factor, and the 2k design is adequate. 0 -1 +1(2) If this distance is great, the response function is likely to be a Figure 6.9 nonlinear function of the factor, and the 2k design is inadequate. 30 35 40 More levels for the factor may have to be investigated A reaction time 6-83 6-84
  • 53. nc number of observations at the center point. Hypothesis:nF = 2kn, number of observations in the ordinary factorial design. H0: y F − yc = 0 response is linear, non-curvature, 2k design is adequate.yc the average of the nc observations in the center point. H1: y F − yc ≠ 0 response is nonlinear, curvature, 2k design isyF the average of the nF observations in the ordinary factorial inadequate design. Sum of square nF nc SScurvature = ( yF − yc )2 = R( yF − yc )2 (6.32) nF + nc R is a constant. D.O.F. = 1 6-85 6-86 SS curvature See Example 6.6 MS curvature 1 SSF0 = = = curvature (6.33) MS E MS E MS E nc = 5, nF = 4 40.3 + 40.5 + 40.7 + 40.2 + 40.6 yc = = 40.460If F0 is large ----> SScurvature is large ----> ( y F − yc ) is large ---> 5 2 y F − yc is large ---> response is nonlinear, 2k design is inadequate 39.3 + 40.9 + 40.0 + 415 . yF = = 40.425 4 nF nc (4)(5) 20 R= = = nF + nc 4 + 5 9 6-87 6-88
  • 54. SS curvature = R( y F − yc ) 2 Table 6.22 Table of ANOVA 20 Source of Sum of Degrees of Mean F0 = (40.425 − 40.460)2 = 0.0027 variation squares freedom square 9 A (time) 2.4025 1 2.4025 55.87 SS 0.0027F0 = curvature = = 0.06 B (temp.) 0.4225 1 0.4225 9.83 MS E 0.043 AB 0.0025 1 0.0025 0.06 Curvature 0.0027 1 0.0027 0.06 Error 0.1720 4 0.0430 Total 3.0022 8 D.O.F.total = N – 1 = (nc + nF) – 1 = (5 + 4) – 1 = 8 D.O.F.error = nc – 1 = 5 – 1 = 4 6-89 6-90Specify α = 0.01, the control limit Fα,1,4 = 21.2. Another Application of the Center Points If single replicate is used (as in example 6.6) in a 2k design, we maySince (for curvature) F0 < Fα,1,4 have to pool the high order interactions to estimate the error. However, in example 6.6, there are only main effects A, B and 2-Conclusion: H0 is true. Response is linear, 2k design is adequate. factor interaction AB. Under such circumstance, we may use the nc center observations to estimate MSE for the error. MSE is actually the sample variance σ2 inIn another example, if F0 > Fα,1,4 (for curvature), H1 is true. probability and statistics.Response is nonlinear. Under such situation, we may have to carry nc ∑(yout the more complicated experiment, in which each factor has more − yc ) 2 ithan 2 levels. MS E = σ 2 = i =1 (6.34) nc − 1 And D.O.F .E = D.O.F .σ 2 = nc − 1 (6.35) 6-91 6-92
  • 55. For example 6.6 6.8 The Advantages of Factorial Designs (1) They are more efficient than one-factor-at-a-time experiments.MS E = (40.3 − 40.46)2 +...+(40.6 − 40.46)2 = 0.043 5 −1 Table 6.23D.O.F .E = 5 − 1 = 4 B- B+ A- A-B- A-B+Two usages of the center points A+ A+B- k(1) Check if the response function is linear, i.e., if the 2 design is adequate?(2) Estimate the errors. 6-93 6-94 For experiments with same degree of accuracy, (2) Factorial design is necessary to avoid misleading conclusions when interactions may be present. N one− factor −at −a−time (k + 1)2k n / 2 k + 1 R= = = (6.36) (3) Factorial design allows the effect of a factor to be estimated at N factorial −2k 2k n 2 several levels of the other factors, yielding conclusions that are valid over a range of experimental conditions. Where, 2kn/2 is the number of pairs for each factor. It is a measure of the accuracy of experiments. The relative efficiency increases along with the increase of the number k of factors. 6-95 6-96
  • 56. 7 FRACTIONAL EXPERIMENTS Fractional factorial design allows us to conduct only a fraction (1/2, 1/4, 1/8, and so on) of the complete set of 2k runs in order to save the7.1 Introduction number of runs.A full 2k design can provide information of all the main effects andinteractions. For a 25 design: However, when using fractional design, information for some main effects or interactions may be lost. 5 main effects A, B, C, D, E Good fractional design: 10 two-factor interactions. (1) Important information for main effects is reserved; 16 high-order interactions, which are usually negligible and little (2) Insignificant information for high order interactions is lost. meaningful. Poor fractional design:But full 2k design requires a large number of runs, which may beimpossible to carry out due to the limits in time and resource. For (1) Important information for main effects is lost;example, a full 25 design requires 52 = 32 runs. Even if n = 1, we (2) Insignificant information for high order interactions ishave to take 32 observations. reserved. 7-1 7-27.2 One Half Fraction A 23 designA one-half fraction of a 2k design is called a 2k-1 fractional factorial A: Spindle speed (A- = 1000 rpm, A+ = 3500 rpm)design. It reduces the number of runs by half, from 2k to 2k-1. B: Feed rate (B- = 0.1 μm/s, B+ = 0.15 μm/s) 2k = 2 k −1 (7.1) 2 C: Machine (C- = USA machine, C+ = UK machine) Table 7.1 k Full design Original One-half Reduced y: Response: Surface roughness (n = 1) number fractional number of runs design of runs 3 23 8 23-1 4 4 24 16 24-1 8 5 25 32 25-1 16In fractional design, single replicate (n = 1) is usually used. 7-3 7-4
  • 57. Table 7.2 Plus and minus signs for the full 23 design Now, a one-half 23-1 fractional design is to be constructed from the full 23 design. The key problem is how to select the 4 runs from the Factor effect y original 8 runs.Run I A B C AB AC BC ABC a + + - - - - + + 16 If we select the top four runs that have a plus sign under column b + - + - - + - + 22 ABC, the information for the highest interaction ABC will be lost. c + - - + + - - + 20abc + + + + + + + + 27 It is what we prefer to. Because, by sacrificing ABC, we mayab + + + - + - - - 21 hopefully preserve the information for the more important mainac + + - + - + - - 23 effects.bc + - + + - - + - 18(1) + - - - + + + - 30Dividing the contrast from column A by N/2 + a − b − c + abc + ab + ac − bc − (1)A= (7.2) N /2 7-5 7-6 3-1 Table 7.3 Plus and minus signs for the fractional 23-1 design In the 2 design, we may be tempted to calculate the main effects and interactions by (7.2) as for the full 2k design, i.e., “Dividing the Factor effect y contrast from a column by N/2”.Run I A B C AB AC BC ABC The main effects. a + + - - - - + + 16 b + - + - - + - + 22 A = [a − b − c + abc ] / 2 = 0.5 c + - - + + - - + 20 B = [− a + b − c + abc ] / 2 = 6.5 (7.3) C = [− a − b + c + abc ] / 2 = 4.5abc + + + + + + + + 27Here, ABC is called the generator. In the selected fraction, the signs The two-factor interactionsare identical under the column of the generator. BC = [a − b − c + abc ] / 2 = 0.5In the selected fraction: I = ABC ˆ (equal by sign) AC = [− a + b − c + abc ] / 2 = 6.5 (7.4)Equation I = ABC is called the defining relation. ˆ AB = [− a − b + c + abc ] / 2 = 4.5 7-7 7-8
  • 58. Unfortunately, neither (7.3) nor (7.4) is correct. The formulation Applying (7.2) (dividing the contrast from a column by N/2) to a full(7.2) used by the full 2k design cannot be applied to fractional design. 2k design (Table 7.2), we obtain the corresponding effect, sayFrom (7.3) and (7.4), it is found: + a − b − c + abc + ab + ac − bc − (1) A= A = BC , ˆ B = AC , ˆ C = AB ˆ (7.5) N /2Aliases: A and BC B and AC C and AB Applying (7.2) to a fractional 2k-1 design (Table 7.3), we obtain theEqual by Sign corresponding joint effect, say lA. Example: A = BCD ˆ a − b − c + abc a − b − c + abc lA = = (7.6) Meaning: The signs in columns A and BDC are the same, in N /2 2 the plus and minus signs table for the fractional design. It does not mean that the value of the main lA is neither the pure effect of A nor the pure effect of BC. Instead, lA effect A is equal to the value of the interaction BCD. is the joint effect of A and BC. More generally speaking, lA is the joint effect of A and all of its aliases 7-9 7-10A: Spindle speed (rpm) Go back to Table 7.3B: Feed rate (μm/s) lA = A + BC = [ a − b − c + abc] / 2 (7.7)y: yield (kg) lB = B + AC = [ − a + b − c + abc] / 2 (7.8) Table 7.4 lC = C + AB = [ − a − b + c + abc] / 2 (7.9)No A B y In fractional design, there is no way to calculate the pure value of the 1 1000 0.1 2000 main effects, such as A, B, C. What we can calculate are just the joint 2 3500 0.15 3200 effects lA, lB, lA.The difference of 1200kg in the yield indicates the joint effect of A In fractional design, we usually use the joint effect to estimate the(spindle speed) and B (feed rate) main effect, say A B A ≈ lA, B ≈ lB, C ≈ lC (7.10) A + B = 3200 – 2000 = 1200 600 600 1000 200 A good fractional design will ensure that the error of the estimation is A ≠ 1200, B ≠ 1200 1200 0 very small. 7-11 7-12
  • 59. Rules for interaction The aliases of a factor can be found from the defining relation.(1) IQ = Q ˆ (7.11) Defining relation I = ABC ˆ(2) QQ = Q 2 = I ˆ ˆ (7.12) A = AI = A ⋅ ABC = A2 BC = BC ˆ ˆ ˆ ˆThey can be easily verified from Table 7.2 B = BI = B ⋅ ABC = AB 2C = AC ˆ ˆ ˆ ˆQ is any effect or interaction C = CI = C ⋅ ABC = ABC 2 = AB ˆ ˆ ˆ ˆExample They are exactly the same as (7.5) IA = A ˆ IAB = AB ˆ A2 = I ˆ AB × ABC = A2 B 2C = IIC = IC = C ˆ ˆ ˆ ˆ 7-13 7-14Example 7.1 A 24 design Table 7.5 − (1) + a − b + L + abcd − 2 + 99 − 80 + L + 178 Run A B C D AB ABCD y A= = = 99.875 8 8 (1) - - - - + + 2 a + - - - - - 99 (7.13) b - + - - - - 80 ab + + - - + + 181 c - - + - + - -2 ac + - + - - + 98 bc - + + - - + 82 abc + + + - + - 179 d - - - + + - -1 ad + - - + - + 102 bd - + - + - + 81 abd + + - + + - 182 cd - - + + + + 1 acd + - + + - - 102 bcd - + + + - - 79 abcd + + + + + + 178 7-15 7-16
  • 60. (1) Good fractional design: ABCD is used as the generator. Defining relation is: I = ABCD ˆ Table 7.6 A = AI = A ⋅ ABCD = A2 BCD = BCD ˆ ˆ ˆ ˆ Run A B C D ABCD y B = BI = B ⋅ ABCD = AB 2CD = ACD ˆ ˆ ˆ ˆ (1) - - - - + 2 ab + + - - + 181 C = CI = C ⋅ ABCD = ABC 2 D = ABD ˆ ˆ ˆ ˆ ac + - + - + 98 bc - + + - + 82 D = DI = D ⋅ ABCD = ABCD 2 = ABC ˆ ˆ ˆ ˆ ad + - - + + 102 bd - + - + + 81 cd - - + + + 1 abcd + + + + + 178 All main effects are aliased with high-order interactions.From Table 7.5, select all the runs with a plus sign under columnABCD 7-17 7-18Since we are quite sure that the high order interactions are negligible Similarly(≈ 0), so A ≈ lA = A + BCD lA = A + BCD ≈ A + 0 = A B ≈ lB = B + ACDFrom Table 7.6 C ≈ lC = C + ABDA≈ lA D ≈ lD = D + ABC (7.15) − (1) + ab + ac − bc + ad − bd − cd + abcd = (7.14) 4 − 2 + 181 + 98 − 82 + 102 − 81 − 1 + 178 = = 98.25 4The estimated A resulting from (7.14) is very close to the accurate Aresulting from (7.13) 7-19 7-20
  • 61. (2) Poor fractional design: AB is used as the generator Defining relation is: I = AB ˆ Table 7.7 A = AI = A ⋅ AB = A2 B = B ˆ ˆ ˆ ˆ Run A B C D AB y B = BI = B ⋅ AB = AB 2 = A ˆ ˆ ˆ ˆ (1) - - - - + 2 ab + + - - + 181 C = CI = C ⋅ AB = ABC ˆ ˆ ˆ c - - + - + -2 abc + + + - + 179 D = DI = D ⋅ AB = ABD ˆ ˆ ˆ d - - - + + -1 abd + + - + + 182 cd - - + + + 1 abcd + + + + + 178 Some main effects (i.e., A and B) are aliased with each other.From Table 7.5, select all the runs with a plus sign under column AB 7-21 7-22Since we are not sure that a particular main effect is always Summarynegligible, we may not be able to use (1) If the main effects are aliased with high order interactions, we can A ≈ lA = A + B useFrom Table 7.7 A ≈ lA, B ≈ lB, C ≈ lC − (1) + ab − c + abc − d + abd − cd + abcdlA = to get very accurate estimates of the main effects. The higher the 4 order of the interactions, the more likely they are close to zero, (7.16) − 2 + 181 − (−2) + 179 − (−1) + 182 − 1 + 178 and the more accurate the estimates are. = = 180 4 (2) The order of the interactions aliased with the main effects in alA resulting from (7.16) is quite different from the accurate A fractional design is completely determined by the generator. So,resulting from (7.13) to develop a good fractional design, the key problem is to select the generator. 7-23 7-24
  • 62. (3) For one-half fractional design, we always select the highest order A 2k-1 design can be constructed by first working out a full factorial interaction as the generator. It guarantees the best 2k-1 design. design (the Basic Design) in terms of (k-1) basic factors, and then adding the last factor. Table 7.8 The number of basic factors is (3 – 1) = 2. The two basic factors are Fractional design Generator A and B. For example, for a 23-1 design, defining relation: I = ABC ˆ 23-1 ABC The basic design is a full 22 design in terms of A and B: 4-1 2 ABCD Table 7.9 2 5-1 ABCDE Run A B AB (1) - - + 26-1 ABCDEF a + - - b - + - ab + + + 7-25 7-26C = CI = C ⋅ ABC = ABC 2 = AB Design of a 2k-1 One-Half Fractional Experiment ˆ ˆ ˆ ˆReplace AB by its alias C, we obtain the required 23-1 design. (1) Decide the (k-1) basic factors. Table 7.10 (2) Work out the basic design in terms of the basic factors. Run A B C y c - - + 3 (3) Decide the alias of the last factor from the generator, which is the a + - - 4 highest order interaction. b - + - 2 abc + + + 1 (4) Replace the last factor for the alias in the basic design. The table for the 2k-1 design is completed.This is the same 23-1 design as given in Table 7.3 (5) The entire aliases structure of the 2k-1 design can be determined from the defining relation. 7-27 7-28
  • 63. (6) Evaluate the joint effects l from the plus and minus signs table of Example: Conduct a 24-1 design with factors A, B, C, D (k = 4) the 2k-1 design. Response is the number of defects per lot(7) Since the high-order interactions are usually negligible, these joint (1) Decide the 3 basic factors: A, B, C. effects can be used as the estimates of the main effects and the (2) Work out the basic design in terms of the basic factors. two-factor interactions. Table 7.11 3 Runs for 2 A B C ABC (1) - - - - a + - - + b - + - + ab + + - - c - - + + ac + - + - bc - + + - abc + + + + 7-29 7-30(3) The generator is the highest order interaction ABCD, the defining (4) Replace D for ABC in the basic design, the table for the 24-1 relation is I = ABCD ˆ design is completed. D = D ⋅ I = D ⋅ ABCD = ABCD 2 = ABC ˆ ˆ ˆ ˆ Table 7.12 Runs No of for 24-1 A B C D = ABC ˆ AB defects (1) - - - - + 550 ad + - - + - 749 bd - + - + - 1052 ab + + - - + 650 cd - - + + + 1075 ac + - + - - 642 bc - + + - - 601 abcd + + + + + 729 7-31 7-32
  • 64. (5) The entire aliase structure of the 24-1 design can be determined (6) Evaluate the joint effects. For examplesfrom the defining relation I = ABCD ˆ l A = A + BCD Examples 1 = ( −(1) + ad − bd + ab − cd + ac − bc + abcd ) A = A ⋅ I = A ⋅ ABCD = A2 BCD = BCD ˆ ˆ ˆ ˆ 4 1 AB = AB ⋅ I = AB ⋅ ABCD = A2 B 2CD = CD ˆ ˆ ˆ ˆ = ( −550 + 749 − 1052 + 650 − 1075 + 642 − 601 + 729) 4 ABC = ABC ⋅ I = ABC ⋅ ABCD = A2 B 2C 2 D = D ˆ ˆ ˆ ˆ = −127.00 Entire aliases structure l B = B + ACD = 4.00 A = BCD , B = ACD , C = ABD , D = ABC , ˆ ˆ ˆ ˆ l C = C + ABD = 11.50 l D = D + ABC = 290.51 AB = CD , AC = BD , AD = BC , ˆ ˆ ˆ 7-33 7-34 l AB = AB + CD (7) Use the joint effects to estimate the main effects and the two- 1 factor interactions. = ((1) − ad − bd + ab + cd − ac − bc + abcd ) 4 Since all of the main effects are aliased with the high order 1 = (550 − 749 − 1052 + 650 + 1075 − 642 − 601 + 729) interactions which are usually negligible, we have 4 = −10.00 A ≈ lA, B ≈ lB, C ≈ lC, D ≈ lD l AC = AC + BD = −25.50 However, 2-factor interactions may not be negligible, usually, l AD = AD + BC = −197.50 AB ≠ l AB and CD ≠ l AB We can only assert that the joint effect of AB and CD is (-10) 7-35 7-36
  • 65. Project of the 2k-1 Design ab B bAfter a one-half fractional 2k-1 experiment has been completed, if onefactor and all of its interactions are found negligible, this factor canbe dropped. That is, the design will be projected into a full factorial b (1) a bdesign in terms of the remaining (k - 1) factors. bc abc A (1) a c c (1) a Figure 7.1 C c ac 7-37 7-38A 23-1 design A ≈ l A = [a − b − c + abc ] / 2 = 0.5A: Spindle speed (A- = 1000 rpm, A+ = 3500 rpm) B ≈ l B = [− a + b − c + abc ] / 2 = 6.5 C ≈ l C = [− a − b + c + abc ] / 2 = 4.5B: Feed rate (B- = 0.1 μm/s, B+ = 0.15 μm/s)C: Machine (C- = USA machine, C+ = UK machine) Since A is much smaller than B and C, we can drop it and project they: Response: Surface roughness (n = 1) 23-1 fractional design into a full 22 design in terms of B and C. Table 7.13 (1) Erase the column A.Run A B C y (2) In column Run, for every run notation including letter a, drop letter a (i.e., change a to (1), abc to bc). a + - - 16 b - + - 22 c - - + 20abc + + + 27 7-39 7-40
  • 66. Table 7.14 Design Resolution A design is resolution R if q-factor effect is only aliased with theRun B C y effects containing (R – q) or more factors (Note: 1-factor effect is the (1) - - 16 main effect). b + - 22 In a one-half design, R is equal to the number k of the factors. c - + 20 bc + + 27 Designs of resolution III: 2III3-1 with I = ABC ˆ Designs of resolution V: 2V5-1 with I = ABCDE ˆ Resolution R is an indicator of the effectiveness (goodness) of the fractional design. The greater the R, the better the fractional design. If R is greater, the main effects will be aliased with higher order interactions. Since the higher order interactions tend to be more negligible, the estimate of the main effects from the joint effects (e.g., A ≈ lA) will be more accurate. 7-41 7-427.3 Smaller Fraction (e.g., 1/4 fraction, 1/8 fraction) An experiment with six factors:A 2k design may run in a 1/2p fraction called a 2k-p fractional factorial Full design: 26 64 runsdesign. k−p 2k 1/4 fraction: 26-2 16 runs 2 = p (7.17) 2 1/8 fraction: 26-3 8 runs k k 2 2 p = 1, 1/2 Fraction 2 k −1 = 1 = 2 2 The larger the p value, the smaller the fraction design, the more 2k 2k number of runs can be saved, but, more information will be lost. p = 2, 1/4 Fraction 2 k −2 = 2= 2 4 2k 2k p = 3, 1/8 Fraction 2 k −3 = = 23 8 7-43 7-44
  • 67. 26-2 Design (p = 2, 1/4 Fraction) In the 26-2 fractional design, the total number of generators is three, so each effect has 3 aliases, each of them can be obtained from a Number of runs can be reduced from 64 to 16. particular generator. For example, for main effect A(1) Two basic generators: ABCE, BCDF I = ABCE ˆ I = BCDF ˆ (7.18) From the first generator ABCE A = AI = A( ABCE ) = BCE ˆ ˆ ˆ(2) One induced generator: ADEF From the second generator BCDF A = AI = A( BCDF ) = ABCDF ˆ ˆ ˆ ABCE ( BCDF ) = AB 2C 2 DEF = ADEF ˆ ˆ From the third generator ADEF A = AI = A( ADEF ) = DEF ˆ ˆ ˆ ABCE ( BCDF ) = I ⋅ I = I ˆ ˆ ∴ I = ADEF ˆ (7.19) Number of aliases for an effect = number of generators(3) Total number of generators is 3 (for one-quarter fraction) 7-45 7-46 Table 7.15 The Complete Alias Structure for the 26-2 Design: (1) The main effects are aliased with three-factor or higher order interactions.A = BCE = DEF = ABCDF ˆ ˆ ˆ AB = CE = ACDF = BDEF ˆ ˆ ˆ l A = A + BCE + DEF + ABCDF ≈ A + 0 + 0 + 0 = AB = ACE = CDF = ABDEF ˆ ˆ ˆ AC = BE = ABDF = CDEF ˆ ˆ ˆC = ABE = BDF = ACDEF ˆ ˆ ˆ AD = EF = BCDE = ABCF ˆ ˆ ˆ (2) The two-factor interactions are aliased with each other.D = BCF = AEF = ABCDE ˆ ˆ ˆ AE = BC = DF = ABCDEF ˆ ˆ ˆ Even though only one quarter of the original runs have been carriedE = ABC = ADF = BCDEF ˆ ˆ ˆ AF = DE = BCEF = ABCD ˆ ˆ ˆ out, this design would provide very good estimates on the main effects and give some idea about the two-factor interactions.F = BCD = ADE = ABCEF ˆ ˆ ˆ BD = CF = ACDF = ABEF ˆ ˆ ˆABD = CDE = ACF = BEF ˆ ˆ ˆ BF = CD = ACEF = ABDE ˆ ˆ ˆACD = BDE = ABF = CEF ˆ ˆ ˆ 7-47 7-48
  • 68. CREATION OF A 26-2 DESIGN Table 7.16 Basic design. No Run 24 A B C D E = ABC ˆ F = BCD ˆSix factors: A, B, C, D, E, F 1 (1) - - - - - - 2 a + - - - + -(1) Decide the basic generators ABCE and BCDF, and work out 3 b - + - - + + the induced generator ADEF. 4 ab + + - - - + 5 c - - + - + + 6 ac + - + - - +(2) From the three generators, work out the alias structure table – 7 bc - + + - - - Table 7.15. 8 abc + + + - + - 9 d - - - + - +(3) Decide (6-2) = 4 basic factors A, B, C, D, and work out the 10 ad + - - + + + 11 bd - + - + + - basic design, i.e., the full 24 design in terms of A, B, C, D (for 12 abd + + - + - - general 2k-p design, the number of basic factors is equal to 13 cd - - + + + - (k - p)). 14 acd + - + + - - 15 bcd - + + + - + 16 abcd + + + + + + 7-49 7-50(4) From Table 7.15, Table 7.17 26-2 design 6-2 No Run 2 A B C D E F No of defects E = ABC = ADF = BCDEF ˆ ˆ ˆ 1 (1) - - - - - - 6 The interaction ABC is composed of basic factors only. So, the 2 ae + - - - + - 10 signs of this interaction can be determined from the signs of A, 3 bef - + - - + + 32 4 abf + + - - - + 60 B and C in Table 7.16. Then, replace ABC by E, the signs for 5 cef - - + - + + 4 E is determined. 6 acf + - + - - + 15 7 bc - + + - - - 26(5) Similarly, from Table 7.15, 8 abce + + + - + - 60 F = BCD = ADE = ABCEF ˆ ˆ ˆ 9 df - - - + - + 8 10 adef + - - + + + 12 The interaction BCD is composed of basic factors only. So, 11 bde - + - + + - 34 the signs of this interaction can be determined from the signs 12 abd + + - + - - 60 of B, C and D in Table 7.16. Then, replace BCD by F, the 13 cde - - + + + - 16 14 acd + - + + - - 5 signs for F is determined. 15 bcdf - + + + - + 37 16 abcdef + + + + + + 52(6) Carry out the experiment and record all of the observations. 7-51 7-52
  • 69. (7) The joint effects l can be evaluated from the plus and minus 7.4 The General 2k-p Fractional Factorial Desing signs table. For example, In general 2k-p fractional design, l A = A + BCE + DEF + ABCDF 1 (1) Number of basic generators = p = ( -(1) + ae – bef + abf – cef + … + abcdef ) 8 1 (2) Number of induced generators = 2p - 1 - p = (-6 + 10 - 32 + 60 - 4 + … +52) 8 (3) Total number of generators = p + (2p - 1 - p) = 2p - 1 = 13.88 (4) Each generator corresponds to an alias, so the number of the(8) Since the high-order interactions are thought negligible, these aliases for each effect is equal to the total number of joint effects can be used as the estimates of the main effects. generators, i.e., 2p - 1 For example, A ≈ l A = 13.88 7-53 7-54A reasonable criterion is to select the generators such that resuling Table 7.18 Selected 2k-p fractional factorial designs2k-p design has the highest possible design resolution. We usually Input Outputassume that higher-order interactions (say, third- or fourth-order and Number of Number of Fraction Basichigher) to be negligible. factors k Runs design generators 3 4 2III3-1 ABC 4 8 2IV4-1 ABCDFractional design in Table 7.18 is always the best one. 5 16 2V5-1 ABCDE 8 2III5-2 ABD, ACEIn a 2k-p design, the main effects can be estimated by the joint effects. 6 32 2VI6-1 ABCDEFFor example, 16 2IV6-2 ABCE, BCDF 8 2III6-3 ABD, ACE, BCF Contrastunder column 7 64 2VII7-1 ABCDEFGA ≈ lA = A (7.20) 32 2IV7-2 ABCDF, ABCE N /2 2IV7-3 16 ABCE, BCDF, ACDG 8 2III7-4 ABD, ACE, BCF, ABCGProject technique can be used to eliminate the insignificant factorsbased on the estimated effects. 7-55 7-56
  • 70. Example In total, there are 7 generatorsSix factors A, B, C, D, E, F have to be tested, k = 6. I = ABD = ACE = BCF = BCDE = ACDF = ABEF = DEFBut only 8 runs with (n = 1) is allowed. Alias structureSelect 2III6-3 from Table 7.18. p = 3, the resolution is 3. A = BD = CE = ABCF = ABCDE = CDF = BEF = ADEFThe three basic generators are ABD, ACE, BCF. B = AD = ABCE = CF = CDE = ABCDF = AEF = BDEF C = ABCD = AE = BF = BDE = ADF = ABCEF = CDEFThe number of induced generators = 2p - 1 – p = 23 – 1 – 3 = 4 D = AB = ACED = BCDF = BCE = ACF = ABDEF = EF(ABD)(ACE) = BCDE(ABD)(BCF) = ACDF E = ABDE = AC = BCEF = BCD = ACDEF = ABF = DF(ACE)(BCF) = ABEF F = ABDF = ACEF = BC = BCDEF = ACD = ABE = DE(ABD)(ACE)(BCF) = DEF Since main effects are aliased with two-factor interactions, this fractional design is not very desired. The problem is due to the small number of allowable runs. 7-57 7-58The number of basic factors are (k – p) = 6 – 3 = 3. Simply use A, B, Use D = AB, E = AC, F = BC, we establish the 26-3 designC as the basic factors. Run A B C D E F yThe basic design: = AB = AC = BC def - - - + + + 7 Run A B C af + - - - - + 4 b - + - - + - 13 (1) - - - abde + + - + - - 8 a + - - cde - - + + - - 3 b - + - ac + - + - + - 6 ab + + - bcf - + + - - + 10 c - - + abcdef + + + + + + 13 ac + - + bc - + + abc + + + 7-59 7-60
  • 71. − def + af − b + abde − cde + ac − bcf + abcdef Drop factors A, C, D, E, the following full 22 design is obtainedA ≈ lA = 8/2 − 7 + 4 − 13 + 8 − 3 + 6 − 10 + 13 Run B F y = = −0.50 (1) - - 3, 6 8/ 2 b + - 13, 8B ≈ l B = 6.00 f - + 7, 4 bf + + 10, 13C ≈ l C = 0.00D ≈ l D = −0.50E ≈ l E = −0.25F ≈ l F = 1.00 7-61 7-62Advantages of fractional designs1. It can significantly reduce the number of runs2. It can eliminate the negligible factors by projecting a fractional factorial design into a full factorial design. It results in a stronger experiment in the active factors that remain.3. By combining a sequence of small fractional factorial designs, it can isolate both the main effects and interactions, and also takes the advantage of learning about the process as it goes along. 7-63
  • 72. 8 A CASE STUDY (1) Decide the factors to be investigated and the ranges of their valuesProblem: At the early stage, it is desired to include as many factors as possible. So, 2k experiment is used.For the diameter (a key dimension) of a shaft, the current processcapability Cp is only 0.8 Table 8.1 No. Factors Low HighThe QA engineers decide to conduct a designed experiment in hope 1 A feed rate (liters/min) 10 15to increase Cp substantially (an application of Larger the better). 2 B carbon element (%) 1 2 3 C cooling rate (m3/min) 100 120 4 D machine Japan USA 5 E speed (m/min) 10 20 6 F temperature (o C) 15 25 The central point values over the ranges are the current values for the factors. 8-1 8-2(2) Decide a fractional factorial design (3) Decide the generators Due to the limitations of time and resources, only 16 observations, Basic generators: ABCE and BCDF at most, can be taken. Table 8.2 Selected 2k-p fractional factorial designs Induced generator: ( ABCE )( BCDF ) = ADEF ˆ Input Output Number of Number of Fraction Basic factors k Runs design generators 3 4 2III3-1 ABC 4 8 2IV4-1 ABCD 5 16 2V5-1 ABCDE 8 2III5-2 ABD, ACE 6 32 2VI6-1 ABCDEF 16 2IV6-2 ABCE, BCDF 8 2III6-3 ABD, ACE, BCFThe 26-2 fractional design is selected, basic generators: ABCE andBCDF 8-3 8-4
  • 73. (4) determine the alias structure (5) Decide the basic factors and basic design (Table 8.3) A = AI = A( ABCE ) = BCE ˆ ˆ ˆ (6-2) = 4 basic factors: A, B, C, D. Basic design is a 24 design A = AI = A( BCDF ) = ABCDF ˆ ˆ ˆ 24 Run A B C D ABC BCD (1) - - - - - - A = AI = A( ADEF ) = DEF ˆ ˆ ˆ a + - - - + - b - + - - + + A = BCE = DEF = ABCDF ˆ ˆ ˆ ab + + - - - + c - - + - + + B = ACE = CDF = ABDEF ˆ ˆ ˆ ac + - + - - + bc - + + - - - C = ABE = BDF = ACDEF ˆ ˆ ˆ abc + + + - + - D = BCF = AEF = ABCDE ˆ ˆ ˆ d - - - + - + ad + - - + + + E = ABC = ADF = BCDEF ˆ ˆ ˆ bd - + - + + - abd + + - + - - F = BCD = ADE = ABCEF ˆ ˆ ˆ cd - - + + + - acd + - + + - - All main effects are aliased with the high order interactions. bcd - + + + - + abcd + + + + + + 8-5 8-6(6) Decide the signs for columns E and F Table 8.4 Run order 26-2 Run A B C D E = ABC F = BCD ˆ ˆ Cp From the alias structure, select 4 (1) - - - - - - 0.6 10 ae + - - - + - 1.0 E = ABC , ˆ F = BCD ˆ 14 bef - + - - + + 3.2 1 abf + + - - - + 6.0 The signs of ABC and BCD can be determined from that of the 13 cef - - + - + + 0.4 16 acf + - + - - + 1.5 basic factors. And then, columns ABC is replaced by E, and 7 bc - + + - - - 2.6 column BCD by F. 6 abce + + + - + - 6.0 2 df - - - + - + 0.8 9 adef + - - + + + 1.2 3 bde - + - + + - 3.4 15 abd + + - + - - 6.0 11 cde - - + + + - 1.6 5 acd + - + + - - 0.5 12 bcdf - + + + - + 3.7 8 abcdef + + + + + + 5.2 8-7 8-8
  • 74. (7) Carry out the experiment and take the observations (response (8) Calculate the joint effects Cp ) 1 Note, run order (the order of taking the observations) must be lA = (− (1) + ae − bef + abf − L + abcdef ) 8 random (see the first column). 1 = (− 0.6 + 1.0 − 3.2 + 6.0 − L + 5.2 ) = 1.39 8 l B = 3.56 l C = −0.09 l D = 0.14 l E = 0.04 l F = 0.04 8-9 8-10(9) Estimate the main effects (10) Decide the effective factors and project the 26-2 design From the alias structure, it is found that all of the main effects are Compare the estimated main effects, it is found that only factors aliased only with high order interactions which are negligible. So, A and B are significant. So, the data of the 26-2 experiment can be projected into a 22 design in terms of A and B. A ≈ l A = 139 . [1] In Table 8.4, erase columns C, D, E and F. B ≈ l B = 356 . [2] Drop c, d, e and f from all of the run notations, e.g., C ≈ l C = −0.09 cef --> (1) D ≈ l D = 014 . abcdef --> ab. E ≈ l E = 0.04 [3] In the 22 design in terms of A and B, merge all of the F ≈ l F = 0.04 observations under a same run notation. 8-11 8-12
  • 75. Table 8.5 (11) Calculate the totals of the runs Run A B AB Cp For the remaining factors A and B, ANOVA is to be carried out. (1) - - + 0.6, 0.4, 0.8, 1.6 (1) = 0.6, + 0.4 + 0.8 + 1.6 = 3.4 a + - - 1.0, 1.5, 1.2, 0.5 a = 1.0 + 1.5 + 1.2 + 0.5 = 4.2 b - + - 3.2, 2.6, 3.4, 3.7 b = 3.2 + 2.6 + 3.4 + 3.7 = 12.9 ab + + + 6.0, 6.0, 6.0, 5.2 ab = 6.0, + 6.0 + 6.0 + 5.2 = 23.2 (12) Calculate the contrasts k = 2, a = 2, b = 2, n = 4, N = 16 CA = -(1) + a - b + ab = -3.4 + 4.2 - 12.9 + 23.2 = 11.1 CB = -(1) - a + b + ab = -3.4 - 4.2 + 12.9 + 23.2 = 28.5 CAB = (1) - a - b + ab = 3.4 - 4.2 - 12.9 + 23.2 = 9.5 y… = I = (1) + a + b + ab = 3.4 + 4.2 + 12.9 + 23.2 = 43.7 8-13 8-14(13) Calculate the sum of squares (14) Conduct the analysis of variance C A 11.12 2 Table 8.6 SS A = = = 7.7 N 16 Source of Sum of DOF Mean F0 variation squares square C B 28.52 2 SS B = = = 50.8 N 16 A 7.7 1 7.7 37.0 C AB 9.52 2 SS AB = = = 5.6 B 50.8 1 50.8 243.8 N 16 2 AB 5.6 1 5.6 26.9 2 2 4 43.72 = ( 0.6 + 0.4 +L+5.2 ) − y... SST = ∑ ∑ ∑ y − 2 ijk 2 2 2 = 66.6 i =1 j =1 k =1 N 16 Error 2.5 12 0.208 SS E = SST − SS A − SS B − SS AB = 66.6 − 7.7 − 50.8 − 5.6 = 2.5 Total 66.6 15 8-15 8-16
  • 76. (15) Decide control limit Fα ,ν ,ν (17) Check the residuals 1 2 Specify type I error α = 0.01. For main effects A, B and Residuals eij should satisfy following requirements: interaction AB, ν1 = 1, ν2 = 12. So (1) has normal distribution Fα ,ν1 ,ν2 = 9.33 (2) has independent distribution (3) has zero mean value(16) Decide the significance of the main effects and interaction (4) has constant variance σ 2 for all runs A: F0 = 37.0 > Fα ,ν ,ν , factor A has significant effect. 1 2 B: F0 = 243.8 > Fα ,ν ,ν , factor B has significant effect. 1 2 AB: F0 = 26.9 > Fα ,ν ,ν , interaction AB has significant effect. 1 2 8-17 8-18(18) Checking curvature nF nc 16 × 5 SS curvature = ( y F − yc ) 2 = (2.73 − 2.70)2 = 0.0034 nF + nc 16 + 5 Take five additional observations at the center point, i.e., Feed rate A: 12.5 liter/min SS curvature 0.0034 F0 = = = 0.0165 Carbon element B: 1.5 % MS E 0.208 The observed process capability: Specify α = 0.01. Also, ν1 = 1, ν2 = 12. Fα ,ν ,ν = 9.33 1 2 2.70, 2.74, 2.71, 2.68, 2.69 Since F0 < Fα ,ν1 ,ν2 , there is no obvious curvature, the response is nc = 5, nF = 16 linear function of the factors, 22 design is adequate. yc = ( 2.70 + 2.74 + 2.71 + 2.68 + 2.69) / 5 = 2.70 y F = y... / 16 = 437 / 16 = 2.73 . 8-19 8-20
  • 77. (19) Illustrate the interaction (20) Conclusions A1B1 = (1) / 4 = 0.85, A2B1 = a / 4 = 1.05 [1] Factors C (cooling rate), D (machine), E (speed) and F A1B2 = b / 4 = 3.23, A2B2 = ab / 4 = 5.80 (temperature) have negligible effect on the process capability. [2] Both factor A (feed rate) and B (carbon element) have 6 significant, positive effect on the process capability. B2 5 A 2B 2 [3] At the low level of B, factor A has little influence on the 4 process capability. 3 B2 A 1B 2 [4] If we set B at the high level (2%) and A also at the high level (15 liter/min), the average process capability will be as high as 2 A2B1 5.80 on average (compared to the original value of 0.8). A1B1 B1 1 B1 Figure 8.1 0 A1 A2 8-21 8-22