advanced33 min studyLesson 299 in Mathematics

Derivative-free models and direct search

Extract local information from function values while accounting for geometry, sample cost, and evaluation noise.

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What you’ll understand

  • Build a low-dimensional interpolation model from objective values.
  • Use a positive-spanning poll to bound a gradient after a failed search.
  • Distinguish a smooth deterministic stationarity result from a noisy black-box claim.

Sometimes an objective is available only as a simulation, an experimental measurement, or a legacy executable with no derivative interface. Derivative-free optimization uses values strategically rather than pretending that derivative information is free. Two important approaches fit local models and poll structured directions. Both need assumptions: a black box can still be smooth, deterministic, noisy, discontinuous, or expensive in different ways. The label derivative-free describes what the algorithm requests, not an exemption from regularity, sampling geometry, or evaluation cost.

Value samples can build a local model

A linear interpolation model in n variables has n+1 coefficients; a full quadratic has (n+1)(n+2)/2. At least that many appropriately positioned value samples are required for unique interpolation of the chosen class. The geometry matters as much as the count: nearly coincident or poorly arranged points can make fitted coefficients extremely sensitive to small value errors. Model-based trust-region methods maintain useful interpolation geometry while limiting the radius over which the fit is trusted.

Fully linear is a quantitative model-quality concept: within a ball of radius Δ, objective error is bounded by a constant times Δ squared and gradient error by a constant times Δ. Those bounds depend on smoothness and controlled sample geometry; they are not guaranteed by calling a polynomial an interpolant. A model can be optimized cheaply, but a trial still needs a new objective evaluation to compare predicted and actual reduction. The expensive resource is often that sequence of simulator calls, not the arithmetic used to solve the fitted model.

∣f(x+p)−m(p)∣≤κfΔ2,∥∇f(x+p)−∇m(p)∥≤κgΔ(∥p∥≤Δ)|f(x+p)-m(p)|\leq\kappa_f\Delta^2,\qquad \|\nabla f(x+p)-\nabla m(p)\|\leq\kappa_g\Delta\quad(\|p\|\leq\Delta)
These fully-linear bounds describe model quality for a smooth objective on a trust-region ball. Uniform constants need assumptions on the objective and interpolation geometry.↔ Scroll wide equations horizontally.

A failed poll can be useful evidence

Direct search can poll points along a finite positive-spanning set of directions, such as both signs of every coordinate vector. Positive spanning ensures that a nonzero negative gradient has a useful projection onto at least one poll direction. Accept a point only if its decrease exceeds a forcing amount, for example c times the squared poll radius. If every direction fails this sufficient-decrease test, the failed poll contains information about the possible gradient size.

For unit coordinate directions and an L-Lipschitz gradient, select the sign opposite each gradient component. Combining its failed decrease with the smoothness upper bound yields an absolute component bound of (c+L/2) times the radius. Summing coordinate squares gives a Euclidean norm bound with a square-root dimension factor. This is a mathematical consequence under known smoothness and exact values. An implementation without a valid L cannot print that numerical certificate merely because it observed an unsuccessful poll.

f(x±αei)≥f(x)−cα2 for all i⟹∥∇f(x)∥2≤n (c+L/2)αf(x\pm\alpha e_i)\geq f(x)-c\alpha^2\ \text{for all }i\quad\Longrightarrow\quad \|\nabla f(x)\|_2\leq\sqrt n\,(c+L/2)\alpha
The implication assumes an L-Lipschitz gradient along the poll segments and exact deterministic values. It follows componentwise from the descent lemma and a complete signed-coordinate poll.↔ Scroll wide equations horizontally.

Noise changes what the samples mean

If objective evaluations contain absolute errors of size delta, decreasing the sample spacing does not indefinitely improve a derivative estimate. Differences divide those errors by the spacing. Likewise, a direct-search forcing decrease smaller than the evaluation uncertainty cannot be reliably distinguished from noise. Repetition, common random numbers, statistical acceptance rules, or a noise-aware stopping radius may be necessary. Those modifications require their own analysis instead of borrowing an exact deterministic guarantee.

For smooth, lower-bounded deterministic objectives, suitable positive-spanning polls, sufficient-decrease rules, controlled radius updates, and bounded direction geometry support stationarity and evaluation-complexity results. Their constants include dimension and directional coverage. They do not imply that every popular heuristic, including every variant of simplex search, has the same guarantee. Always specify which function calls count, whether parallel polls change elapsed time rather than total work, and whether the available budget can support the chosen local model dimension.

Recover a quadratic from three values

The black box is f(x)=(x−1)², but fit a model m(p)=a+bp+cp² around x=0 using only values at p=0 and p=±1/2. Then minimize the model inside |p|≤1/2.

  1. The three values are f(0)=1, f(1/2)=1/4, and f(−1/2)=9/4. Thus a=1. Subtracting the two nonzero-point equations gives b=(1/4−9/4)/(1)=−2.

  2. Adding those equations gives 2a+c/2=5/22a+c/2=5/2, hence c=1. The fitted model is 1−2p+p², exactly equal to this particular objective. Its unconstrained minimizer p=1 lies outside the specified radius.

  3. Within [−1/2,1/2] the model decreases toward the right boundary, so choose p=1/2. Predicted reduction is 3/4; a fresh true evaluation gives value 1/4 and the same actual reduction. Exact agreement here follows from a quadratic objective and uncorrupted samples, not from interpolation alone.

The result

Three well-placed values identify this scalar quadratic and a defensible radius-limited trial. A general nonlinear objective would leave model error to be tested.

Explore the model

Interactive lab

Interactive lab · function

Check the signed-coordinate poll in one dimension

Pause, Reset, then set a=1 and x=0.5; record the value 0.25 and slope one. Repeat from Reset with both a=1 and x=−0.5 to see the equal value and opposite slope. These are the signed poll points around zero for x². The instrument supplies exact analytic values and derivatives; it does not fit a black-box model or add evaluation noise.

Check the signed-coordinate poll in one dimensionThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA symmetric poll probes both directions, while the mathematical stationarity bound still needs smoothness and exact-value assumptions.

Check your understanding

Why can shrinking finite-difference spacing eventually worsen a gradient estimate when objective values have fixed absolute noise?

Try it yourself

A complete signed-coordinate poll in dimension n=4 fails to find decrease beyond cα², with c=1, α=0.02, and a known Lipschitz-gradient constant L=6 on all poll segments. What gradient bound follows? State its assumptions.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the bound for each component first, then the Euclidean norm factor √n.

Compare with the solution

Each absolute gradient component is at most (1+6/2)0.02=0.08(1+6/2)0.02=0.08. With four components, the Euclidean norm is at most 4 0.08=0.16\sqrt4\,0.08=0.16. This conclusion needs exact deterministic values, both signed unit-coordinate polls, the stated sufficient-decrease failure, and a valid L on the connecting segments. It is a first-order stationarity bound, not a global objective-gap bound for a nonconvex function.

Take it with you

  • Interpolation needs informative geometry as well as enough values.
  • An unsuccessful structured poll can imply stationarity under explicit smoothness assumptions.
  • Fixed evaluation uncertainty imposes a useful resolution limit on deterministic comparisons.

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