ST 516
Experimental Statistics for Engineers II
The Fitted Response Surface
Graph the fitted surface and its standard error: response.R
1 / 17
Response Surface Methods
Analyzing a Second-Order Response Surface
ST 516
Experimental Statistics for Engineers II
75
.5
78
75
80
79
176
174
75
74
168
78.5
76
170
78
172
Temperature
.5
79.5
178
180
182
Predicted Yield
78
75
.5
76
.5
80
82
77.5
77
84
86
77
88
5
76.
90
76
92
Time
2 / 17
Response Surface Methods
Analyzing a Second-Order Response Surface
ST 516
Experimental Statistics for Engineers II
80
Yield
78
76
74
180
Te
90
m 175
pe
ra
tu
r
85 e
m
Ti
e
170
3 / 17
80
Response Surface Methods
Analyzing a Second-Order Response Surface
ST 516
Experimental Statistics for Engineers II
182
Standard Error
8
0.2
0.
0.2
4
26
0.14
2
0.
0.2
8
0.1
6
0.1
0.
0.
26
32
8
174
176
2
0.1
170
172
Temperature
0.22
0.22
4
0.2
178
180
32
0.
0.
168
32
78
0.2
6
0.3
80
0.24
82
0.22
0.22
84
86
0.24
88
6
0.2
0.3
90
2
3
0.
92
Time
4 / 17
Response Surface Methods
Analyzing a Second-Order Response Surface
ST 516
Experimental Statistics for Engineers II
rd
Standa
0.35
0.30
0.25
0.20
0.15
Error
180
Te
90
m 175
pe
ra
tu
r
85 e
m
Ti
e
170
5 / 17
80
Response Surface Methods
Analyzing a Second-Order Response Surface
ST 516
Experimental Statistics for Engineers II
Finding the Optimum
Write the full quadratic model as
ŷ = β̂0 + x0 b + x0 Bx
where



x=

x1
x2
..
.
xk






,b = 


β̂1
β̂2
..
.
β̂k






 , and B = 


β̂1,1
1
β̂
2 1,2
..
.
1
β̂
2 1,2
β̂2,2
..
.
...
...
..
.
1
β̂
2 1,k
1
β̂
2 2,k
1
β̂
2 1,k
1
β̂
2 2,k
...
β̂k,k
..
.



.

The the location of the stationary point (maximum, minimum, or
saddle-point) is
1
xs = − B−1 b.
2
6 / 17
Response Surface Methods
Analyzing a Second-Order Response Surface
ST 516
Experimental Statistics for Engineers II
In the example
l <- lm(y ~ x1 + x2 + I(x1^2) + I(x2^2) + x1:x2, t11d6)
lc <- coefficients(l)
b <- lc[2:3]
B <- matrix(c(lc[4], lc[6]/2, lc[6]/2, lc[5]), 2, 2)
cat("Stationary point:", -solve(B, b) / 2, "\n")
Output
Stationary point: 0.3892304 0.3058466
In engineering units:
ξ1 = 85 + 5 ∗ 0.3892304 = 86.94615,
ξ2 = 175 + 5 ∗ 0.3058466 = 176.5292.
7 / 17
Response Surface Methods
Analyzing a Second-Order Response Surface
ST 516
Experimental Statistics for Engineers II
We want to round these, but how much?
87 and 177?
85 and 175?
8 / 17
Response Surface Methods
Analyzing a Second-Order Response Surface
ST 516
Experimental Statistics for Engineers II
We can test the hypothesis that the stationary point is x(0) :
the general quadratic with a stationary point at x(0) is
2
2
(0)
(0)
∗
∗
∗
y = β0 + β1,1 x1 − x1
+ β2,2 x2 − x2
(0)
(0)
∗
+ β1,2
x1 − x1
x2 − x2
+ ;
fit this as a reduced model, and compare with the full model
using the extra sum of squares;
reduced model has 4 parameters versus 6 in the full model, so
test statistic is F2,n−p .
Test specific values, or graph the statistic for a grid of values.
9 / 17
Response Surface Methods
Analyzing a Second-Order Response Surface
ST 516
Experimental Statistics for Engineers II
Test whether 87, 177 is optimal
x10 <- (87 - 85) / 5
x20 <- (177 - 175) / 5
summary(aov(y ~ I((x1 - x10)^2) + I((x2 - x20)^2) +
I((x1 - x10)*(x2 - x20)) + x1 + x2, t11d6))
Output
Df Sum Sq Mean Sq
I((x1 - x10)^2)
1 18.901 18.901
I((x2 - x20)^2)
1 8.681
8.681
I((x1 - x10) * (x2 - x20)) 1 0.526
0.526
x1
1 0.004
0.004
x2
1 0.134
0.134
Residuals
7 0.496
0.071
--Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 .
F value
Pr(>F)
266.553 7.88e-07 ***
122.423 1.09e-05 ***
7.415
0.0296 *
0.057
0.8186
1.895
0.2110
0.1
1
F = ((0.004 + 0.134)/2)/0.071 = 0.97, P ≈ .5, do not reject H0 .
10 / 17
Response Surface Methods
Analyzing a Second-Order Response Surface
ST 516
Experimental Statistics for Engineers II
We can test the same hypothesis without fitting the reduced model:
the model is
ŷ (x1 , x2 ) = β0 + β1 x1 + β2 x2 + β1,1 x12 + β1,2 x1 x2 + β2,2 x22 ;
so
∂ ŷ
= β1 + 2β1,1 x1 + β1,2 x2 ,
∂x1
∂ ŷ
= β2 + 2β2,2 x2 + β1,2 x1 .
∂x2
So the hypothesis is equivalent to
(0)
(0)
(0)
(0)
β1 + 2β1,1 x1 + β1,2 x2 = 0,
β2 + 2β2,2 x2 + β1,2 x1 = 0.
11 / 17
Response Surface Methods
Analyzing a Second-Order Response Surface
ST 516
Experimental Statistics for Engineers II
That is, L x(0) β = 0, where
#
"
(0)
(0)
x
0
0
1
0
2x
1
2
L x(0) =
(0)
(0)
2x2
0 0 1
0
x1
and
β = (β0 , β1 , β2 , β1,1 , β1,2 , β2,2 )0 .
Any hypothesis of this form can be tested as follows:
−1
SSL = (Lβ̂)0 L(X0 X)−1 L0
(Lβ̂)
is the sum of squares associated with the hypothesis, MSL = SSL /2 is
the mean square, and F = MSL /MSE is the test statistic.
12 / 17
Response Surface Methods
Analyzing a Second-Order Response Surface
ST 516
Experimental Statistics for Engineers II
R commands
betaHat <- coefficients(l)
V <- summary(l)$cov.unscaled
msE <- summary(l)$sigma^2
L <- rbind(c(0, 1, 0, 2 * x10,
0, x20),
c(0, 0, 1,
0, 2 * x20, x10))
LbetaHat <- L %*% betaHat
ssL <- sum(LbetaHat * solve(L %*% V %*% t(L), LbetaHat))
msL <- ssL / 2
F <- msL / msE
We can calculate F for a grid of x(0) values and graph it.
The region where F < F2,n−p (α) is a 100(1 − α)% confidence region
for the optimal settings.
13 / 17
Response Surface Methods
Analyzing a Second-Order Response Surface
ST 516
Experimental Statistics for Engineers II
60
120
100
168
170
172
174
40
4.7
37
41
4
176
140
20
Temperature
178
180
182
F−Statistic
80
78
80
82
84
86
88
90
92
Time
14 / 17
Response Surface Methods
Analyzing a Second-Order Response Surface
ST 516
Experimental Statistics for Engineers II
F−Stati
150
100
stic
50
180
Te
90
m 175
pe
ra
tu
r
85 e
m
Ti
e
170
15 / 17
80
Response Surface Methods
Analyzing a Second-Order Response Surface
ST 516
Experimental Statistics for Engineers II
180
F−Statistic Detail
73
10
4.
15
177
74
14
178
40
30
175
176
Temperature
179
20
35
5
30
86
25
25
87
35
88
89
90
Time
16 / 17
Response Surface Methods
Analyzing a Second-Order Response Surface
ST 516
Experimental Statistics for Engineers II
We can round to 87 and either 176 or 177.
We cannot round to 85 and 175.
17 / 17
Response Surface Methods
Analyzing a Second-Order Response Surface