[R Studio][13.2.2][DS13.2.2] 반응표면모형 (Response Surface) 적합 및 최대값 구하기 예시

 

 < 문제 출처 >

이공학도를 위한 확률과 통계 3판 한글판

 

 

 

ds13.2.2-optimal-fiber-strength.txt

0.00MB

 

 

> model <- lm(Fiber_Strength ~ Concentration_of_Polymer1 + Concentration_of_Polymer2 + I(Concentration_of_Polymer1 * Concentration_of_Polymer1) + I(Concentration_of_Polymer2 * Concentration_of_Polymer2) + Concentration_of_Polymer1 * Concentration_of_Polymer2, data = raw_datas)
> summary(model)

Call:
lm(formula = Fiber_Strength ~ Concentration_of_Polymer1 + Concentration_of_Polymer2 + 
    I(Concentration_of_Polymer1 * Concentration_of_Polymer1) + 
    I(Concentration_of_Polymer2 * Concentration_of_Polymer2) + 
    Concentration_of_Polymer1 * Concentration_of_Polymer2, data = raw_datas)

Residuals:
       1        2        3        4        5        6        7 
-0.41111  0.48889  0.32222 -0.07778  0.08889 -0.30278 -0.18611 
       8        9 
-0.01944  0.09722 

Coefficients:
                                                         Estimate
(Intercept)                                               20.0111
Concentration_of_Polymer1                                 -0.6333
Concentration_of_Polymer2                                  2.0833
I(Concentration_of_Polymer1 * Concentration_of_Polymer1)  -1.4667
I(Concentration_of_Polymer2 * Concentration_of_Polymer2)  -1.7167
Concentration_of_Polymer1:Concentration_of_Polymer2        0.9250
                                                         Std. Error
(Intercept)                                                  0.3501
Concentration_of_Polymer1                                    0.1918
Concentration_of_Polymer2                                    0.1918
I(Concentration_of_Polymer1 * Concentration_of_Polymer1)     0.3322
I(Concentration_of_Polymer2 * Concentration_of_Polymer2)     0.3322
Concentration_of_Polymer1:Concentration_of_Polymer2          0.2349
                                                         t value
(Intercept)                                               57.155
Concentration_of_Polymer1                                 -3.303
Concentration_of_Polymer2                                 10.864
I(Concentration_of_Polymer1 * Concentration_of_Polymer1)  -4.416
I(Concentration_of_Polymer2 * Concentration_of_Polymer2)  -5.168
Concentration_of_Polymer1:Concentration_of_Polymer2        3.938
                                                         Pr(>|t|)
(Intercept)                                              1.18e-05
Concentration_of_Polymer1                                 0.04564
Concentration_of_Polymer2                                 0.00167
I(Concentration_of_Polymer1 * Concentration_of_Polymer1)  0.02156
I(Concentration_of_Polymer2 * Concentration_of_Polymer2)  0.01405
Concentration_of_Polymer1:Concentration_of_Polymer2       0.02917
                                                            
(Intercept)                                              ***
Concentration_of_Polymer1                                *  
Concentration_of_Polymer2                                ** 
I(Concentration_of_Polymer1 * Concentration_of_Polymer1) *  
I(Concentration_of_Polymer2 * Concentration_of_Polymer2) *  
Concentration_of_Polymer1:Concentration_of_Polymer2      *  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.4697 on 3 degrees of freedom
Multiple R-squared:  0.9845,	Adjusted R-squared:  0.9587 
F-statistic: 38.13 on 5 and 3 DF,  p-value: 0.006456

 

문제에서 제시된 모델은 다음과 같기 때문에

 

추정값

βˆ 0 = 20.011

βˆ 1 = −0.633

βˆ 2 = −1.467

βˆ 3 = 2.083

βˆ 4 = −1.717

βˆ 5 = 0.925

임을 알 수 있습니다.

 

 

섬유의 강도가 가장 강해지는 농도는 위 추정식을 x1, x2에 대해 각각 편미분한 식을 0이라 두고 연립방정식을 풀면 구할 수 있습니다.

 

답

x1 = −0.027

x2 = 0.600