Skip to content
Leantensify Learn

Where is the best setting — and is the point I found actually the best?

Fit a curved surface to two factors and locate where the gradient is zero — then find out whether that point is a maximum, a minimum, or a saddle that is not an optimum at all.

Response Surface (Optimisation) · improve · Black Belt

Use this when

  • A factorial has identified the factors and you need the settings
  • Centre points show the surface bending, so a two-level design cannot describe it
  • Somebody has reported an optimum from a quadratic fit and you want to know its kind

Response surface (central composite design)

sandbox

The surface bends — a quadratic is required

The centre runs average 92.0120 where the factorial corners average 86.4925 — a difference of -5.5195 that pure error does not explain (p < 0.001). The surface BENDS over this region, so a first-order model is the wrong shape and steepest ascent computed from one would set off in a direction the surface does not go. Add axial points and fit a second-order model.

Corners
86.4925
Centres
92.0120
F
996.02
p
0.001

The stationary point is a MAXIMUM

The gradient is zero at Temp = 0.4383, Time = -0.4112, where the model predicts 92.7478. Every eigenvalue is negative (-3.5123, -1.8072), so the surface curves downward in every direction and this is a MAXIMUM.

Contour plot of the fitted response surface over Temp and Time, coded from -1.7 to 1.7. Predicted response ranges from 70.96 to 92.75. The stationary point is at Temp 0.438, Time -0.411, marked with a cross, and it is a maximum: the contours form closed rings around it, tightening towards the highest value.edge of the designmaximum−1.7+1.7Temp (coded)Time (coded)
Darker is a higher predicted response. Closed rings mean a maximum or minimum; contours that open in opposite directions mean a saddle. The dashed circle is how far the design actually reached — anything read outside it is extrapolation from a curve.

R²

0.9917

Adjusted R²

0.9858

Residual SE

0.4234

df

7

Fitted second-order model coefficients.
TermCoefficientStd errorp
(intercept)92.0120——
Temp2.26450.14970.001
Time-1.16480.14970.001
Temp^2-3.20100.16050.001
Time^2-2.11850.16050.001
TempTime-1.31750.21170.001
Factor names
Central composite: the cube, the axes, and five centre replicates. Run in random order; enter in this order.
#KindTempTime
1factorial-1.000-1.000
2factorial1.000-1.000
3factorial-1.0001.000
4factorial1.0001.000
5axial-1.4140.000
6axial1.4140.000
7axial0.000-1.414
8axial0.0001.414
9centre0.0000.000
10centre0.0000.000
11centre0.0000.000
12centre0.0000.000
13centre0.0000.000

Load an example

How this is calculated
  • The curvature test compares the centre runs with the average of the factorial corners: SS = nfnc(ȳf − ȳc)² / (nf + nc) on one degree of freedom, against pure error from the replicated centre points. It decides which model belongs, so it runs before the model.
  • The model is y = b₀ + Σbᵢxᵢ + Σbᵢᵢxᵢ² + Σbᵢⱼxᵢxⱼ, fitted by least squares through the normal equations.
  • The stationary point is x = −½B⁻¹b, where B holds the quadratic coefficients with the cross terms halved off the diagonal. Forgetting that halving moves the point without making anything look wrong.
  • The classification comes from the eigenvalues of B, by the Jacobi method. All negative is a maximum, all positive a minimum, mixed signs a SADDLE — which is not an optimum, however tempting the coordinates look.
  • The design is a rotatable central composite: α = (2ᵏ)¼, so prediction variance depends on distance from the centre and not on direction. That matters when you are searching in a direction you do not know.

Source: Montgomery, D.C., Design and Analysis of Experiments 8e, Ch. 11; Box, G.E.P. and Draper, N.R., Empirical Model-Building and Response Surfaces.

How this is calculated

Rotatable central composite design: the 2^k factorial cube, 2k axial points at alpha = (2^k)^(1/4), and replicated centre points. The centre-point curvature test runs FIRST and decides which model belongs: SS = nf*nc*(ybar_f - ybar_c)^2/(nf+nc) on 1 degree of freedom against pure error from the centre replicates. Where curvature is present, a full second-order model y = b0 + sum(bi xi) + sum(bii xi^2) + sum(bij xi xj) is fitted by least squares through the normal equations, solved by Gaussian elimination with partial pivoting. The stationary point is x = -0.5 B^-1 b, where B carries the quadratic coefficients with the cross terms HALVED off the diagonal. Its nature comes from the eigenvalues of B by the cyclic Jacobi method: all negative is a maximum, all positive a minimum, mixed signs a saddle point which is not an optimum. A stationary point beyond the design's own radius is flagged as extrapolation. Where curvature is absent the tool declines to fit a quadratic and gives the path of steepest ascent instead.

Source: Montgomery, D.C., Design and Analysis of Experiments 8e, Ch. 11 — response surface methodology, steepest ascent, central composite designs and canonical analysis; Example 11.1's data verifies the curvature test. Box, G.E.P. and Draper, N.R., Empirical Model-Building and Response Surfaces. The fit itself is verified by exact recovery of constructed quadratics with known optima, eigenvalues and stationary points.

Learn the method