Derivative Free Optimization class Part I

Exercice 1 - partial correctionAnne Auger (Inria) anne.auger_AT_inria.fr

https://www.lri.fr/~auger/teaching.html

Adaptive step-size algorithms Questions 2 - 3

• you should obtain something similar to the plot already in the class slides:

Adaptive step-size algorithms Question 4

• Three phases:

1.step-size too small compared to distance to optimum

2.step-size adapted to distance to optimum

3.step-size too large compared to distance to optimum

Adaptive step-size algorithms Question 5

c < 0

Adaptive step-size algorithms Question 5

• adaptation of the step-size takes place

1.initial step-size is too small, the algorithm increases it

2.once step-size reaches the order of distance to the optimum (after 50 f-evals), both the step-size and distance to optimum decrease linearly

Adaptive step-size algorithms Question 5

• We see linear convergence (after the initial stage), formally: There exists

limt!1

1

tln

kXtkkX0k

= c

limt!1

1

tln

�t�0

= c

c < 0

Adaptive step-size algorithms Question 5

Adaptive step-size algorithms Question 5

kXtk�t

is a stable homogeneous Markov chains

difficult to prove but feasible

Implies linear convergence of algorithm

PhD thesis possible related to those aspects

Anne Auger, Nikolaus Hansen (2013), Linear Convergence of Comparison-based Step-size Adaptive Randomized Search via Stability of Markov Chains

Anne Auger, Nikolaus Hansen (2013), Linear Convergence on Positively Homogeneous Functions of a Comparison Based Step-Size Adaptive Randomized Search: the (1+1) ES with Generalized One-fifth Success Rule

References:

Adaptive step-size algorithms Question 6

•(1+1)-ES slower than on the sphere because the ellipsoid is a ill-conditioned function

•need for covariance matrix adaptation

sphere

ellipsoid

Adaptive step-size algorithms Question 7

Rosenbrock function (banana shape) “ill-conditioned” level sets

(1+1)-ES slower than on the sphere because of the shape of the function

Remark: Optimum of the function in (1,…,1), so change initial point

Adaptive step-size algorithms Question 8

The function g being strictly increasing and the algorithm being rank-based, you should theoretically observe the same convergence plot on f and g o f

But: numerical precision issue observed due to floating point representation

Typically: with matlab significand with 15 numbers such that 1+10^(-14) = 1 (for the computer)

Adaptive step-size algorithms Question 8

g1(y) = ln(1 + y2)

g2(y) = ln(1 + y1/2)

g3(y) = y1/4

Complementary question: Observe that for the three following increasing functions the numerical problems appears at a different moment (explain why)

y � 0