Dpatel6903 Assignment 4

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Conestoga College *

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8115

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Industrial Engineering

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Feb 20, 2024

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Assignment 4: RSM and Optimization Dhruv Patel Process Quality Engineering – Conestoga College QUAL 8115: Advance Design of Experiments Iain Hastings July 28, 2023

Part 1- Conversion percent Analysis of Variance Source DF Adj SS Adj MS F-Value P-Value Model 9 2555.73 283.97 12.76 0.000 Linear 3 763.05 254.35 11.43 0.001 Temperature 1 14.44 14.44 0.65 0.439 Time 1 222.96 222.96 10.02 0.010 Catalyst 1 525.64 525.64 23.63 0.001 Square 3 601.30 200.43 9.01 0.003 Temperature*Temperature 1 48.47 48.47 2.18 0.171 Time*Time 1 124.48 124.48 5.60 0.040 Catalyst*Catalyst 1 388.59 388.59 17.47 0.002 2-Way Interaction 3 1191.37 397.12 17.85 0.000 Temperature*Time 1 36.12 36.12 1.62 0.231 Temperature*Catalyst 1 1035.12 1035.12 46.53 0.000 Time*Catalyst 1 120.13 120.13 5.40 0.043 Error 10 222.47 22.25 Lack-of-Fit 5 56.47 11.29 0.34 0.869 Pure Error 5 166.00 33.20 Total 19 2778.20 The overall model is highly significant. Factors Time and Catalyst have p-values less than 0.05, indicating they are significant in the model. For the square case, Time and Catalyst have p- values less than 0.05. Furthermore, in the two-way, the interaction between Temperature and Catalyst and between Time and Catalyst, based on their p-values, is less than 0.05, showing they are significant. However, the Temperature alone and its interactions with Time do not significantly affect the response. Model Summary S R-sq R-sq(adj) R-sq(pred) 4.71669 91.99% 84.79% 75.66% The model has an R-squared value of 91.99%, and a high R-squared value indicates that the model is a good fit for the data. The adjusted R-squared is 84.79%. Since it is slightly lower than the R-squared value, it suggests that the model's fit is still strong even after considering the number of predictors in the model. The predicted R-squared value is 75.66%. It indicated that it could predict responses 75.66 % accurately. Overall, the model appears to have a good fit to the data

We can concentrate on Observation 20, which has a Conversion% of 91.00 and a predicted value (Fit) of 81.09. This observation has a residual of 9.91 and a standardized residual of 2.30, denoted with an "R" to indicate an unusual observation. A standardized residual value of 2.30 is relatively large, indicating that the Conversion% of Observation 20 differs significantly from what the model predicted. This suggests that the model is not accurately capturing the behaviour of this particular data point. Recommendation: The unusual residual for Observation 20 requires careful consideration and investigation. By taking the necessary steps to address the issue, such as data validation and outlier analysis, you can improve the model's performance and ensure that the analysis is based on reliable and meaningful information.

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The partial output for the reduced model: Coded Coefficients Term Coef SE Coef T-Value P-Value VIF Constant 79.59 1.75 45.45 0.000 Temperature 1.03 1.37 0.75 0.467 1.00 Time 4.04 1.37 2.95 0.012 1.00 Catalyst 6.20 1.37 4.53 0.001 1.00 Time*Time 3.12 1.33 2.35 0.036 1.01 Catalyst*Catalyst -5.01 1.33 -3.78 0.003 1.01 Temperature*Catalyst 11.37 1.79 6.36 0.000 1.00 Time*Catalyst -3.88 1.79 -2.17 0.051 1.00 Based on the provided Mode, the proposed model equation for Conversion percent can be expressed as follows: Conversion % = 79.59 + 1.03 * Temperature + 4.04 * Time + 6.20 * Catalyst + 3.12 * Time^2 - 5.01 * Catalyst^2 + 11.37 * Temperature * Catalyst - 3.88 * Time * Catalyst Part 2- A reduced model for viscosity Analysis of Variance Source DF Adj SS Adj MS F-Value P-Value Model 3 253.204 84.401 39.63 0.000 Linear 2 243.264 121.632 57.11 0.000 Temperature 1 175.352 175.352 82.34 0.000 Catalyst 1 67.912 67.912 31.89 0.000 Square 1 9.939 9.939 4.67 0.046 Temperature*Temperature 1 9.939 9.939 4.67 0.046 Error 16 34.074 2.130 Lack-of-Fit 11 30.421 2.766 3.78 0.077 Pure Error 5 3.653 0.731 Total 19 287.278 The table presents the results of an Analysis of Variance (ANOVA) for a reduced model. The total variability in the response variable is 287.278, representing the sum of squares for the total data. The reduced model includes two linear predictors (Temperature and Catalyst) and

one square term (Temperature*Temperature). The model is significant, and both linear predictors are highly significant, indicating that they significantly impact the response variable. The square term is marginally significant. Model Summary S R-sq R-sq(adj) R-sq(pred) 1.45933 88.14% 85.91% 63.02% The model's R-squared adjacent value is 85.91 percent, indicating that its factors explain 85.91 percent of its interpretation. The R-sq value is 88.14 percent, nearly identical to the total percentage of variance in the model. As a result, the model corresponds to the experiment. Coded Coefficients Term Coef SE Coef T-Value P-Value VIF Constant 59.948 0.417 143.65 0.000 Temperature 3.583 0.395 9.07 0.000 1.00 Catalyst 2.230 0.395 5.65 0.000 1.00 Temperature*Temperature 0.823 0.381 2.16 0.046 1.00 Viscosity = 59.948 + 3.583 * Temperature + 2.230 * Catalyst + 0.823 * Temperature^2 Fits and Diagnostics for Unusual Observations Obs Viscosity (y2) Fit Resid Std Resid 9 59.100 56.250 2.850 2.98 R 10 65.900 68.302 -2.402 -2.51 R Both observations 9 and 10 have unusual residuals. Observation 9 has a positive standardized residual (2.98), suggesting its viscosity value is higher than expected based on the model. On the other hand, Observation 10 has a negative standardized residual (-2.51), indicating that its viscosity value is lower than expected.

Based on the information provided, the reduced model appears to be a good fit; however, further analysis of the residuals and investigation of the unusual observations will help ensure the model's validity and improve its predictive capabilities. Part 3 - Optimize the two responses

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This contour plot shows the relationship and interactions between the variables Time, Temperature, and Catalyst Percentage. The first plot depicts the time-temperature relationship. According to this plot, Time and Temperature have a minor effect on the Response Conversion % (y1). The second plot shows the interaction of Time and Catalyst%. This contour plot shows the significant impact that increases in time and catalyst% have on the response conversion percent (y1). The third contour plot presents the interaction of temperature and catalyst percent. This plot shows the interaction between them and the Response.

The Contour Plot for Viscosity illustrates that the maximum value for Y2 can only be achieved when the catalyst percent and temperature are maximized. For the viscosity to be between 55 and 60, we must fix the values of Temperature = (-0.5 to 1.0) and Catalyst = (-0.5 to 1.5). For the viscosity requirement to be a target of 65: Based on this response optimizer result, we can attain the required target of 65 for viscosity if we set the factors at 1.09 for Time, 1.69 for Temperature, and 0.06 for Catalyst percent.

Dpatel6903 Assignment 4

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