DOE / optimisation

Response surface optimisation

Once a screening design has found the important factors, the next question is where the response peaks. A central composite design puts five levels on two factors, a quadratic is fitted by least squares, and the canonical analysis says whether the stationary point is a peak, a bowl or a saddle.

NatureMaximumDF5Ascent(0.86, 0.51)

Stationary point

(1.00, 0.88)

Inside the design region

Predicted response

65.2

Maximum of the fitted surface

Lack of fit

p 0.225

F 3.60 on 3 and 2 df

R²

0.996

Adjusted 0.992 · s 0.48

Central composite design

Four factorial points, four axial points and the centre, with the fitted value and residual.

11 runs

Central composite design
Run	Type	x₁	x₂	Response	Fitted	Residual
1	factorial	-1.00	-1.00	51.7	51.1	0.53
2	factorial	-1.00	1.00	53.9	53.6	0.32
3	factorial	1.00	-1.00	57.0	57.1	-0.09
4	factorial	1.00	1.00	64.8	65.1	-0.30
5	axial	-1.41	0.00	49.5	50.0	-0.56
6	axial	1.41	0.00	62.7	62.4	0.32
7	axial	0.00	-1.41	53.3	53.5	-0.27
8	axial	0.00	1.41	61.0	61.0	0.03
9	centre	0.00	0.00	62.1	61.8	0.29
10	centre	0.00	0.00	61.5	61.8	-0.31
11	centre	0.00	0.00	61.8	61.8	0.01

Coded units: ±1 is the factorial level, ±α the axial level, 0 the centre.

Fitted response surface

Rows run from +1.6 to −1.6 on x₂; columns from −1.6 to +1.6 on x₁.

Coefficient t statistics

Terms beyond about ±2 are doing real work; the rest are inside the noise.

Canonical equations

The exact relationships implemented by this engine.

y = b₀ + b₁x₁ + b₂x₂ + b₁₂x₁x₂ + b₁₁x₁² + b₂₂x₂² · Six parameters, fitted by least squares.

x_s = −½ B⁻¹ b · Where the gradient vanishes.

λ₁, λ₂ from |B − λI| = 0 · Both negative: a peak. Mixed signs: a saddle.

F = (SS_lof/df) / (SS_pe/df) · Pure error comes from the centre replicates.

direction ∝ (b₁, b₂) · Where to move before curvature dominates.

Show model notation and computationOpen the symbolic form, the variable definitions, and the step-by-step derivation when you need the detail.

A clear path from inputs to results.

Once screening has found the important factors, the question becomes where the response turns. A central composite design puts five levels on each factor without running the full grid, a quadratic is fitted by least squares, and the canonical analysis says whether the stationary point is a peak, a bowl or a saddle — and whether it sits inside the region you actually tested.

Step-by-step methodWorked exampleAssumptions explained

How to use

Set the quadratic surface the process follows: linear terms, the interaction and both curvatures.

Choose the number of centre points to set the error degrees of freedom.

Add measurement noise to see how the fit degrades.

Read the stationary point, the predicted response and the nature of the surface.

Check R², the residuals and the t statistics before trusting the optimum.

The formula

y = b_0 + b_1x_1 + b_2x_2 + b_{12}x_1x_2 + b_{11}x_1^2 + b_{22}x_2^2 \;;\; x_s = -\tfrac{1}{2}B^{-1}b

Reference: Box & Wilson (1951), 'On the Experimental Attainment of Optimum Conditions'; Myers, Montgomery & Anderson-Cook, Response Surface Methodology.

Worked example

Two factors, eleven runs, σ = 0.4

Inputs

Surface: 62 + 4.2x₁ + 2.6x₂ + 1.4x₁x₂ − 3.1x₁² − 2.2x₂².
Central composite with three centre points, rotatable α = 1.414.
Measurement noise σ = 0.4.

Calculation path

1
Eleven runs: four factorial, four axial and three centre.
2
The least-squares fit returns b₁ = 4.38, b₂ = 2.62, b₁₂ = 1.41, b₁₁ = −2.82, b₂₂ = −2.29.
3
R² is 0.996 and the residual standard deviation is 0.48, close to the noise added.
4
Both eigenvalues are negative, so the surface is a peak.
5
The stationary point is (1.00, 0.88) in coded units, predicting 65.2 — just inside the design region.

Result

11 runs · R² 0.996 · peak at (1.00, 0.88) · predicted 65.2.

Interpreting the result

A stationary point inside the region is an optimum; outside it, the fitted surface is still climbing and the next experiment should move in the direction of steepest ascent rather than sit on the design. The t statistics matter more than R²: a high R² with a coefficient t of 0.4 means the surface is being fitted to noise.

Assumptions & limitations

Two factors, one response, a full quadratic model with six parameters.

Rotatable central composite design with α = 1.414 and a fixed number of centre points.

Responses generated from the surface you set plus seeded noise; the fit is a real least-squares fit to those runs.

Ordinary least squares with independent errors of constant variance.

Frequently asked questions

Why a central composite design?

It adds curvature information to a two-level factorial with only four extra axial runs, and at α = 1.414 the prediction variance is constant at a fixed distance from the centre.

How many centre points?

Three to five is usual. They give degrees of freedom for pure error and a check on lack of fit.

What does a saddle mean?

The surface rises in one direction and falls in another, so there is no single optimum. The practical response is to move along the ridge or re-scope the region.

What if the stationary point is outside the region?

Treat it as a direction, not a location. Run a steepest-ascent experiment until the response stops improving, then re-fit.