Black Belt · 15 min

The gradient is zero here. That is not the same as best.

After this you can

use centre points to decide whether a region needs a curved model at all.
classify a stationary point as a maximum, a minimum or a saddle, and say what each means for the setting to recommend.
say why a predicted optimum outside the experimental region is a direction rather than a setting.
compute a path of steepest ascent from a first-order model and say when it is the right move.

Assumes you have done Half the runs, and knowing exactly what you gave up.

A machining team fits a response surface to speed and feed, solves for the stationary point, and finds it at speed 0.03, feed −0.27 with a predicted 70.2. The software calls it the optimum. The settings go into the standard.

Operators start reporting that the machine does better when they push speed and feed both up, and better again when they pull both down — and worse in between. Nobody can explain it, so it gets written off as operator variation.

The eigenvalues of that fit were −3.12 and +1.39. Mixed signs. The point was a saddle: the response genuinely falls along one direction from it and rises along another. The operators were right, and the analysis had the answer on its second line.

The idea

A factorial finds which factors matter. It cannot find the best setting, because two levels per factor is two points and two points cannot describe a curve — a straight line through them fits perfectly and tells you nothing about what happens in between.

Response surface methods add the third level and fit a quadratic, which can bend.

First, does it bend at all?

Adding centre points to a two-level design costs almost nothing and answers this directly. A plane predicts the centre of the design to be the average of the corners. If the centre runs come out systematically above or below that average, the surface bends.

That test decides which model belongs, so it comes before the model:

No curvature — the region is a plane. There is no optimum inside it, and the finding is a direction. Move.
Curvature — a plane is the wrong shape, and steepest ascent computed from one would set off in a direction the surface does not go.

When it is flat: steepest ascent

Move proportionally to the first-order coefficients: the factor with the largest coefficient moves most. Run the settings along that path, watch the response, and stop when it stops improving. That is where the next experiment goes.

The path is worth nothing unless it is run. A path computed and not run is a prediction from a model that has already left the region it was fitted on.

When it bends: the central composite design

The factorial cube, plus axial points out along each axis, plus replicated centre runs. The axial distance α = (2ᵏ)^¼ makes the design rotatable — prediction is equally precise in every direction from the centre, which matters when you are searching in a direction you do not yet know.

Thirteen runs in two factors buys a full quadratic: linear terms, squared terms, and the cross term.

The stationary point, and the question nobody asks

Set the gradient to zero and solve. You get a point where the surface is flat. That is three different things:

Eigenvalues of B	The point is	What to do
all negative	a maximum	adopt it, if it is inside the region
all positive	a minimum	adopt it only if you want the response low
mixed signs	a saddle	do not adopt it — follow the rising direction

The eigenvalues of the second-order coefficient matrix are what separate them, and a package that prints "optimum" beside the coordinates has not checked. On a contour plot the difference is unmistakable: closed rings around a maximum, hyperbolas opening in opposite directions around a saddle.

A saddle is not a failure. It usually means the best settings lie along a ridge, and the next move is to follow the rising direction to the edge of the region. What it must not do is go into a standard as "the optimum".

And if the point is outside the region

Then it is a direction, not a setting. A fitted quadratic extrapolates worse than a line: a curve continued past its data does not merely drift, it accelerates. A predicted optimum beyond the runs is somewhere to go and look, and the looking is more runs.

Worked example

An adhesive cure process: temperature and time, response is bond strength. A rotatable central composite — 4 factorial corners, 4 axial points at ±1.414, 5 centre runs.

Step 1 — does it bend?

The four corners average 86.4925. The five centre runs average 92.0120 — five and a half units higher. Against the pure error the centre replicates provide, that is F = 996.02, p < 0.001.

Emphatically yes. A plane fitted here would have been badly wrong, and steepest ascent from it would have set off across a dome as if it were a slope.

Step 2 — the fitted surface.

Term	Coefficient
intercept	92.0120
Temp	2.2645
Time	−1.1648
Temp²	−3.2010
Time²	−2.1185
Temp × Time	−1.3175

R² 0.9917, adjusted R² 0.9858, residual standard error 0.4234 on 7 degrees of freedom.

Step 3 — the stationary point, and its kind.

The gradient is zero at Temp = 0.4383, Time = −0.4112, where the model predicts 92.7478.

Eigenvalues: −3.5123 and −1.8072. Both negative, so the surface curves downward in every direction from that point. It is a genuine maximum.

Step 4 — is it somewhere you can actually go?

It sits 0.601 coded units from the centre, and the design reaches 1.414. Comfortably inside. The prediction rests on runs that surround the point rather than on a curve continued past them.

So: a real optimum, at a real setting, with the model interpolating rather than extrapolating. That is the case where you may adopt the answer — and it is worth noticing how many conditions had to hold.

Step 5 — the other preset, which is the lesson.

The machining data from the Hook fits just as well (R² 0.9948) and gives a stationary point at Speed 0.0254, Feed −0.2698. Everything looks the same except two numbers: the eigenvalues are −3.1203 and +1.3941.

Mixed signs. Saddle. Same quality of fit, same confident coordinates, completely different recommendation — and the only thing that distinguishes them is a check most reports never run.

Your turn

The tool opens on the adhesive cure data, with the curvature test above the model.

Read the curvature panel first. Corners at 86.4925, centres at 92.0120, F = 996.02. Notice the tool puts this above the fitted model, because it decides whether that model belongs.
Confirm the stationary point at 0.4383, −0.4112 predicting 92.7478, and the eigenvalues −3.5123 and −1.8072.
Look at the contour plot. Closed rings tightening around the cross — that shape is what a maximum looks like, and it is why the eigenvalues say what they say.
Now load Machining — a saddle. The R² is higher (0.9948). Read the verdict line before the coordinates.
Look at the contour plot again. The rings are gone; the contours open in opposite directions. Trace a path away from the cross along one diagonal and the shading gets darker; along the other it gets lighter. That is the operators' experience, drawn.
Find the dashed circle marking the edge of the design. Anything read outside it is extrapolation from a curve — which accelerates away from its data rather than drifting.

Response surface (central composite design)

practice

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.

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

Fitted second-order model coefficients.
Term	Coefficient	Std error	p
(intercept)	92.0120	—	—
Temp	2.2645	0.1497	0.001
Time	-1.1648	0.1497	0.001
Temp^2	-3.2010	0.1605	0.001
Time^2	-2.1185	0.1605	0.001
TempTime	-1.3175	0.2117	0.001

Factor names

Central composite: the cube, the axes, and five centre replicates. Run in random order; enter in this order.
#	Kind	Temp	Time
1	factorial	-1.000	-1.000
2	factorial	1.000	-1.000
3	factorial	-1.000	1.000
4	factorial	1.000	1.000
5	axial	-1.414	0.000
6	axial	1.414	0.000
7	axial	0.000	-1.414
8	axial	0.000	1.414
9	centre	0.000	0.000
10	centre	0.000	0.000
11	centre	0.000	0.000
12	centre	0.000	0.000
13	centre	0.000	0.000

Responses — one per run

Load an example

How this is calculated

The curvature test compares the centre runs with the average of the factorial corners: SS = n_fn_c(ȳ_f − ȳ_c)² / (n_f + n_c) 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.

Check yourself

No hints. Wrong answers are explained, not softened.

A response surface fit reports a stationary point with eigenvalues −4.1 and +2.3. What should the recommendation be?

A 2² factorial with five centre runs shows the centre runs averaging almost exactly the corner average. What does that license?

A quadratic fitted over coded levels reaching ±1.414 predicts its maximum at (3.8, −2.9). What is the honest report?

A first-order fit over a flat region gives coefficients b₁ = 0.8, b₂ = −0.2. What does the path of steepest ascent look like, and what makes it useful?

Worth remembering

What decides whether a stationary point is a maximum?

The signs of the eigenvalues of the second-order coefficient matrix. All negative is a maximum, all positive a minimum, mixed signs a SADDLE — which is not an optimum however confident the coordinates look.

What are centre points for, and when do you run them?

To test whether a plane is adequate over the region, before fitting anything curved. They also give the only pure-error estimate an unreplicated factorial has. Run them with the factorial, not after it.

Why is a predicted optimum outside the design a direction rather than a setting?

A fitted quadratic extrapolates worse than a line — a curve continued past its data accelerates away from it. The point is where to look next, and the looking is more runs.

When is steepest ascent the right move, and what makes it work?

When the centre points show no curvature: the region is a plane, so there is no optimum in it. Move proportionally to the first-order coefficients, following their signs, RUN the settings, and stop where the response stops improving.

Can you do this now?

Rate yourself honestly. We compare your rating with how you actually answered — the gap is more useful than either number alone.

I can use centre points to decide whether a region needs a curved model at all.
I can classify a stationary point as a maximum, a minimum or a saddle, and say what each means for the setting to recommend.
I can say why a predicted optimum outside the experimental region is a direction rather than a setting.
I can compute a path of steepest ascent from a first-order model and say when it is the right move.

Your rating is recorded alongside your drill results. Neither alone marks the competency as met.

The problem