Assignment Week 14_Amit Apte_Final

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MECH 5750 DoE Final Exam. Name---Amit Apte UML ID: 01693139 QUIZ E DATE: 04/30/2017 UML ID: 0169313 9 QUIZ E TO BE ANSWERED BY STUDENT ID LAST DIGIT 0 OR 9 PROBLEM 1 If you start with the bicycle hill climb problem in week 9 assignment, factor 6 was folded over to remove the confounding of other factors to factor 6, as shown in the notes. Note you do not have to calculate any values just note down the resulting interactions. Problem 1, Part 1. Please only answer one version depending on your last student ID digit (use right chart above showing the isolated interaction[s]). Please only write down the confounding interactions (no calculations values are required). E. If the DoE was folded with two factors at the same time (1 and 6), What is the confounding (aliasing) table? Factor Trial # 1 2 3 4 5 6 7 Cofounding 1 1 1 1 1 1 1 1 2*3, 4*5, 6*7 2 1 1 2 1 2 2 2 1*3, 4*6, 5*7 3 1 2 1 2 1 2 2 1*2, 4*7, 5*6 4 1 2 2 2 2 1 1 1*5, 2*6, 3*7 5 2 1 1 2 2 1 2 1*4, 2*7, 3*6 6 2 1 2 2 1 2 1 1*7, 2*4, 3*5 7 2 2 1 1 2 2 1 1*6, 2*5, 3*4 8 2 2 2 1 1 1 2 - Here 1 & 6 are folded at same time then, Cofounding of Factor 1 No CONFOUNDING. It disappears. Cofounding of Factor 2 1*3, 4*6 is dropped, we have one, two-way interactions Cofounding of Factor 3 1*2, 5*6 is dropped, we have one, two-way interactions Cofounding of Factor 4 1*5, 2*6 is dropped, we have one, two-way interactions Cofounding of Factor 5 1*4, 3*6 is dropped, we have one, two-way interactions Cofounding of Factor 6 No CONFOUNDING. It disappears.

MECH 5750 DoE Final Exam. Name---Amit Apte UML ID: 01693139 QUIZ E DATE: 04/30/2017 Cofounding of Factor 7 1*6 is dropped, we have two, two-way interactions Factor Trial # 1 2 3 4 5 6 7 Cofounding 1 1 1 1 1 1 1 1 -2*3, -4*5, 6*7 2 1 1 2 1 2 2 2 -1*3, -4*6, 5*7 3 1 2 1 2 1 2 2 -1*2, 4*7, -5*6 4 1 2 2 2 2 1 1 -1*5, -2*6, 3*7 5 2 1 1 2 2 1 2 -1*4, 2*7, -3*6 6 2 1 2 2 1 2 1 1*7, -2*4, -3*5 7 2 2 1 1 2 2 1 1*6, 2*5, 3*4 8 2 2 2 1 1 1 2 - But After Adding, Average (1+1’) = 6*7 ___ Average (1-1’) = __2*3, 4*5___ Average (2+2’) = 5*7 Average (2-2’) = 1*3, 4*6___ Average (3+3’) = 4*7 Average (3-3’) = ___1*2, 5*6___ Average (4 +4’) = 3*7 Average (4-4’) = ___1*5, 2*6___ Average (5+5’) = 2*7 Average (5-5’) = ___1*4, 3*6____ Average (6+6’) = 1*7 Average (6-6’) = ____2*4, 3*5___ Average (7+7’) = 1*6, 2*5, 3*4 Average (7-7’) = ____No Factors__ PROBLEM 1 PART 2 All students Please answer this question for all versions (A, B, C, D, E). If the DoE design team folded all factors at the same time (1-7), what is the confounding (aliasing) table? Show all factors and all interactions confounding (aliasing). Average (1+1’) = No Factors Average (1-1’) = 2*3, 4*5, 6*7 Average (2+2’) = No Factors Average (2-2’) = 1*3, 4*6, 5*7 Average (3+3’) = No Factors Average (3-3’) = 1*2, 4*7, 5*6 Average (4 +4’) = No Factors Average (4-4’) = 1*5, 2*6, 3*7 Average (5+5’) = No Factors Average (5-5’) = 1*4, 2*7, 3*6

MECH 5750 DoE Final Exam. Name---Amit Apte UML ID: 01693139 QUIZ E DATE: 04/30/2017 Average (6+6’) = No Factors Average (6-6’) = 1*7, 2*4, 3*5 Average (7+7’) = No Factors Average (7-7’) = 1*6, 2*5, 3*4 PROBLEM 2 (All Students answer All questions a – j) In an industrial waste treatment plant, metals are precipitated by mixing lime in a tank to produce metal hydroxides. Design an experiment to maximize the metal precipitation @Five factors at two levels 2 lime flows [FLOW] (10, 20 L/m) 2 mixing speeds [SPEED] (20 or 25 rpm) 2 tank levels [LVL] (50% or 100%) 2 mixing Times [MIX] (10, 20 rpm) 2 Effluent Temperatures [TEMP] (ambient, 50’F) Select best fit/smallest orthogonal array experiments and show column assignments (in numbers) only: a. Considering all interactions for the five factors at two levels: Array _ L32 _ Factor Assignments: Flow_ 1 _, Speed_ 2 _, LVL_ 4 _, MIX_ 8 _, TEMP_ 16 _ b. Ignore all interactions for the five factors at two levels: Array _ L8 _ Factor Assignments: Flow_ 1 _, Speed_ 2 _, LVL_ 3 _, MIX_ 4 _, TEMP_ 5 _ c. Considering 2 way interactions only among the five factors and ignoring all other interactions: Array _ L16 _ Factor Assignments: Flow_ 1 _, Speed_ 2 _, LVL_ 4 _, MIX_ 8 _, TEMP_ 15 _ d. Considering only these two (two–way interactions) for the five factors, and ignoring all other interactions. Answer your versions only (A, B C D or E – Please circle your letter below) A. [TEMP] x [MIX] and [TEMP] x [FLOW] B. SPEED x TEMP and SPEED x FLOW C. FLOW x MIX and FLOW x LVL D. MIX x LVL and Mix x TEMP E. TEMP x FLOW and TEMP x SPEED

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MECH 5750 DoE Final Exam. Name---Amit Apte UML ID: 01693139 QUIZ E DATE: 04/30/2017 Array _ L8 _ Factor Assignments: Flow_ 2 _, Speed_ 4 _, LVL_ 6 _, MIX_ 7 _, TEMP_ 1 _ e. Considering only four factors at two levels: Flow, Speed, LVL and MIX, and consider all interactions Array _ L16 _ Factor Assignments: Flow_ 1 _, Speed_ 2 _, LVL_ 4 _, MIX_ 8 __ f. Considering only four factors at two levels Flow, Speed, LVL and MIX, and ignore all interactions Array _ L8 _ Factor Assignments: Flow_ 1 _, Speed_ 2 _, LVL_ 3 _, MIX_ 4 _ g. Considering only four factors at two levels: Flow, Speed, LVL and MIX, and consider only two-way interactions while ignoring all others Array _ L8 _ Factor Assignments: Flow_ 1 _, Speed_ 2 _, LVL_ 4 _, MIX_ 7 _ h. Considering only three factors at two levels: Flow, Speed and LVL, and consider all interactions Array _ L8 _ Factor Assignments: Flow_ 1 _, Speed_ 2 _, LVL_ 4 _ i. Considering only three factors at two levels: Flow, Speed and LVL, and ignore all interactions Array _ L4 _ Factor Assignments: Flow_ 1 _, Speed_ 2 _, LVL_ 3 _ j. Considering only three factors at two levels: Flow, Speed and LVL, and consider only two-way interactions while ignoring all others Array _ L8 _ Factor Assignments: Flow_ 1 _, Speed_ 2 _, LVL_ 4 _ PROBLEM 3 (All Students answer All questions a – k). In an industrial waste treatment plant, metals are precipitated by mixing lime in a tank to produce metal hydroxides. Design an experiment to maximize the metal precipitation: Five factors at three levels. 3 lime flows [FLOW] (10, 20 or 30 L/m) 3 mixing speeds [SPEED] (20 or 25 or 35 rpm) 3 tank levels [LVL] (50% 75% or 100%) 3 mixing Times [MIX] (10, 20 or 30rpm)

MECH 5750 DoE Final Exam. Name---Amit Apte UML ID: 01693139 QUIZ E DATE: 04/30/2017 3 Effluent Temperatures [TEMP] (ambient, 25 or 50’F) Select best fit/smallest orthogonal array experiments and show column assignments (in numbers) only: a Considering all 5 factors have three levels and ignore all interactions: Array _ L27 _ Factor Assignments: Flow_ 1 _, Speed_ 2 _, LVL_ 3 _, MIX_ 4 _, TEMP_ 5 _ b. Considering all five factors with three levels and selecting 2 way interactions only among three selected factors (Flow Speed and LVL), while ignoring all other interactions: Array _ L27 _ Factor Assignments: Flow_ 1 _, Speed_ 2 _, LVL_ 5 _, MIX_ 9 _, TEMP_ 10 _ c. Considering only 3 factors with three levels (FLOW, SPEED, and LVL) and ignore all interactions Array _ L9 _ Factor Assignments: Flow_ 1 _, Speed _2 _, LVL _3 _ d. Considering only 4 factors with three levels (FLOW, SPEED, LVL and MIX) and ignore all interactions Array _ L9 _ Factor Assignments: Flow_ 1 _, Speed_ 2 _, LVL_ 3 _, MIX_ 4 _ e. Considering only 4 factors with three levels (FLOW, SPEED, LVL and MIX) and consider all interactions Array _ L81 _ f. Considering all 5 factors and you wanted to select a non-interacting array to reduce confusion Array _ L12 _ Factor Assignments: Flow_ 1 _, Speed_ 2 _, LVL_ 3 _, MIX_ 4 _, TEMP_ 5 _ g. If all five factors had 4 levels, and ignore all interactions (please select a two levels array only) Array _ L16 _ Factor Assignments: Flow 1,2 , Speed_ 3,4 _, LVL_ 5,6 _, MIX_ 7,8 _, TEMP_ 9,10 _ h. If two factors (Flow and Speed) had 4 levels and three factors (LVL, Mix and TEMP) had two levels and ignore all interactions (please select a two levels array only)

MECH 5750 DoE Final Exam. Name---Amit Apte UML ID: 01693139 QUIZ E DATE: 04/30/2017 Array _ L8 _ Factor Assignments: Flow_ 1,2 _, Speed_ 3,4 _, LVL_ 5 _, MIX_ 6 _, TEMP_ 7 _ i. If one factor (Flow) had 4 levels, the rest of the 4 factors at two levels, considering only 2-way interactions (Please select a two levels array only) Array _ L32 _ Factor Assignments: Flow_ 1,2 _, Speed_ 4 _, LVL_ 8 _, MIX_ 16 _, TEMP_ 31 _ j. If one factor (Flow) had 8 levels, the rest of the four factors at two levels, and ignore all interactions (Please select a two levels array only) Array _ L8 _ Factor Assignments: Flow_ 1,2,3 _, Speed_ 4 _, LVL_ 5 _, MIX_ 6 _, TEMP_ 7 _ k. If two factors (Flow and Speed) had 8 levels and three factors (LVL, Mix and TEMP) had two levels and ignore all interactions (please select a two levels array only) Array _ L16 _ Factor Assignments: Flow_ 1,2,3 _, Speed_ 4,5,6 _, LVL_ 7 _, MIX_ 8 _, TEMP_ 9 _ PROBLEM 4 In a waste treatment plant that precipitates metals by adding/mixing lime in a tank to the effluent stream to produce metal hydroxides. Design an experiment to minimize the metal precipitation. 2 lime flows [FLOW] (10, 20 L/m) 2 mixing speeds [SPEED] (20 or 25 rpm) 2 tank levels [LVL] (50% or 100%) 2 mixing Times [MIX] (10, 20 rpm) 2 Effluent Temperatures [TEMP] (ambient, 50’F] For each answer (A, B, C, D or E), We want to repeat the experiment to investigate the effect of different conditions for the repetition noise factors I. Show typical repetition and level of noise factors to test all conditions (example O1S1F1 O1S1F2 etc….) II. For the preceding question I above, show a minimum set of repetitions (if possible) and levels of noise factors. If not possible, indicate no change. E. Use the following conditions (noise factors) as a guide for selecting the number of repetitions, since these noise factors are not under the control experiment design team: 2 Operators (O) - (Mike or Jim); 2 shifts (S) – Shift (day, night)

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MECH 5750 DoE Final Exam. Name---Amit Apte UML ID: 01693139 QUIZ E DATE: 04/30/2017 I. For the above problem, we have 5-factors with two levels. For this case, we have two noise factors as given in part E. Using two noise factors we have 4 interactions which are, 1. O 1 S 1 2. O 1 S 2 3. O 2 S 1 4. O 2 S 2 Now using the above interactions and the five factors we get, FLOW SPEED LVL MIX TEMP NOISE 10 20 50 10 AMB O 1 S 1 10 20 50 10 AMB O 1 S 2 10 20 50 10 AMB O 2 S 1 10 20 50 10 AMB O 2 S 2 20 25 100 20 50 O 1 S 1 20 25 100 20 50 O 1 S 2 20 25 100 20 50 O 2 S 1 20 25 100 20 50 O 2 S 2 II. For this above case if we need minimum set of repetitions and levels of noise factors, then we can say that L4 is the least possible array. So, there is no change. PROBLEM 5 Please answer the questions below depending on your version (A – E). Design an experiment to study the best material for conformal coating. Select best fit/smallest orthogonal array experiments and show column assignments: Use 2 or 3 level arrays and you might assign more than one column to a factor with multiple levels Version E 2 Material Types [M] 3M1, 3M2 3 Material mixing speeds [SP] (10, 20 or 25 rpm) 2 polymer specific gravities [SG] (.9 or 1.1) 1. (Version E) If you used the 3 factors at the levels chosen with two repetitions , fill in the blanks: Total # of Experiments (including two repetitions) _ 24 _ , DoF of all Factors (M, SP, SG) _ 4 _ , DoF of Error_ 14 _ , Total DoF_ 23 _ DoF of M_ 1 _ ; DoF of SP_ 2 _ , DoF of SG_ 1 _ , DoF of ALL Interactions_ 5 _ , 2. (Version E ) Ignoring all interactions: Array _L9_ Factor Assignments: M_1_, SP_3_, SG_2_

MECH 5750 DoE Final Exam. Name---Amit Apte UML ID: 01693139 QUIZ E DATE: 04/30/2017 3. (Version E ) Considering all interactions: Array _L12_ 4. (Version E ) Considering 2 way interactions only Array _L8_ Factor Assignments: M_1_, SP_4_, SG_2_ 5. (Version E) Considering the interaction of Material [M] x [SG] only and ignoring all other interactions between factors: Array _L8_ Factor Assignments: M_1_, SP_4_, SG_2_ PROBLEM 6 Minitab Questions. All Students answer all Questions. Please be brief and to the point, with one concise sentence answer. i. When you use Minitab plots for S/N for selecting the levels, which level would you select to minimize the outcome: Most positive level______________, or Most negative level______________ To minimize the response, we use most positive level. ii. How many formula choices does Minitab have in order to compute the S/N values? Please list all choices by name only (no formulas required). There are four formula choices which are, 1. Larger is better 2. Nominal is best 3. Nominal is best (default) 4. Smaller is better iii. Describe the way Minitab allows for repetitions of the experiment, describe how it creates the experiment matrix based on the repetition types? In Minitab, we have 1 repetition as default value. If we change the value from default to 2 then, all the above values with certain number of experiments will be repeated. Example, if we create full factorial design using 3 factors with 2 level design (Default), we get 8 experiments with one repetition as default value. If we change this repetition value to 2 we get 16 factors as the above 8 experiments are repeated. iv. Describe the way Minitab allows for repetitions of the centerpoint, describe how it creates the experiment matrix based on the repetition types? Minitab adds the center point to the design based on the type of format like text or numbers and depends on the blocks. For example, if we consider only numeric case which have a block then Minitab adds specified number of points to the block. And for the same case if there is no block then Minitab adds specified number of points to design.

MECH 5750 DoE Final Exam. Name---Amit Apte UML ID: 01693139 QUIZ E DATE: 04/30/2017 v. When would you use the Optimizer in Minitab, and what does it provide for an output? (List the single most important output, no formulas please) Optimizer is useful when we need to evaluate the impact of multiple variables on a response. To use it, fitting in a model for each response need to be done. Here the response optimizer uses each response requirements to  Minimize the response (Smaller the better)  Target the response (Target is best)  Maximize the response (Higher the better) It analyses each factor based on the importance given by the user and provides him with the optimal solution for the problem. vi. What is the meaning of the regression equation in Minitab (no formulas required)? The regression equation determines a statistical relationship between the one or more predictor or response variables. vii. What is the meaning of the fit in Minitab (no formulas required)? The fit represents the plot for results in simple regression equation where one predictor variable and response are compared. viii. How does Minitab compute predicted values (in Mean and S/N) in the Taguchi analysis (describe how it does, no formulas or examples), and how do you use the data provided from the predicted values for analysing the experiment? In a static Taguchi design, the mean is the average response for each combination of control factors and control factors that reduce variability in a process by minimizing the effects of uncontrollable factors (noise factors). For analysing the experiment,  Uses the S/N ratio to identify those control factors which reduce variability.  Identify control factors that move the mean to target and have a small or no effect on the S/N ratio. ix. What is the Minitab maximum P-Value for a factor to be significant in the analysis of Coefficients ____ Max P-value is determined at α = 0.05 x. What is the Minitab maximum P-Value for a factor to be significant in the ANOVA analysis _______ Even in ANOVA analysis, Max P-value is determined at α = 0.05 or 95% confidence. xi. What is the recommended Minitab minimum for R-sq (R squared) after the ANOVA analysis _______ For Minitab, R-sq is between 0% and 100%. So, the minimum is 0% xii. What is the purpose for R-sq (pred) = R squared predicted - after the ANOVA analysis

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MECH 5750 DoE Final Exam. Name---Amit Apte UML ID: 01693139 QUIZ E DATE: 04/30/2017 Use predicted R 2 to determine how well the model predicts the response for new observations. Models that have larger predicted R 2 values have better predictive ability. xiii. What is the difference between R-sq (R squared) and R –sq (adj = adjusted) R-squared R-sq (adj = adjusted) R-squared supposes that every independent variable in the model explains the variation in the dependent variable. It gives the percentage of explained variation as if all independent variables in the model affect the dependent variable The adjusted R-squared gives the percentage of variation explained by only those independent variables that affect the dependent variable.