genetic algorithms a step by step tutorial

General remarks and comparison with other methodGenetic algorithms step-by-step

Advanced topicsSummary

Genetic AlgorithmsA step by step tutorial

Max Moorkamp

Dublin Institute for Advanced Studies

Barcelona, 29th November 2005

Max Moorkamp Genetic Algorithms



Outline

1 General remarks and comparison with other method

2 Genetic algorithms step-by-stepA hand-calculated exampleSome experiments on simple problems

3 Advanced topicsMulti-objective optimizationA geophysical example




Comparison of different optimization methods

Method Func. Eval. Class of Problems

Analytical Gradient 1 One special problemNumerical Gradient 10− 102 differentiable, lim. no. of min.Genetic Algorithm 103 − 104 limited parameter rangeMonte Carlo 105 noneGrid Search 108 none




Advantages and Disadvantages of Genetic Algorithm

Advantages:

No derivatives needed

Easy to parallelize

Can escape local minima

Works on a wide range of problems

Disadvantages:

Need much more function evaluations than linearized methods

No guaranteed convergence even to local minimum

Have to discretize parameter space




General program flow of a genetic algorithm

Generate random population

Select models into new population

Calculate probabilities from Objective function

Determine parameter encoding

Crossover

Mutation

N It

erat

ions

The general structure is the same for allgenetic algorithms

Individual steps can be done in a varietyof ways

We will only discuss the most basicoperators




A hand-calculated exampleSome experiments on simple problems

Outline








Encoding the parameters

Model parameters are encoded into a binary string

x = (1.2, 3.4, . . . , 100)← (0101110011 . . . 100011)

For each paramter a minimum value, stepsize and encoding lengthhas to be specified. For example two parameters x1, x2:

x1 ∈ [−1 . . . 1] and x2 ∈ [0 . . . 3.1]

5-bit x1 x2

Basevalue -1 0Stepsize 0.0625 0.1

10-bit x1 x2

Basevalue -1 0Stepsize 0.0002 0.003





Generating a population

At the start a population of N random models is generated.

p1 = (0101110011) → (01011) = −1 + 26 ∗ 0.0625 = 0.625

(10011) = 0 + 25 ∗ 0.1 = 2.5

p2 = (1111010110) → (11110) = −1 + 15 ∗ 0.0625 = −0.0625

(10110) = 0 + 13 ∗ 0.1 = 1.3

p3 = (1001010001) → (10010) = −1 + 9 ∗ 0.0625 = −0.4375

(10001) = 0 + 17 ∗ 0.1 = 1.7...

...

pN = (0110100001) → (01101) = −1 + 22 ∗ 0.0625 = 0.375

(00001) = 0 + 16 ∗ 0.1 = 1.6





Calculating the objective function values

For each member of the population calculate the value of theobjective function:

O(x1, x2) = x21 + x2

2

p1 → O(0.625, 2.5) = 6.64

p2 → O(−0.0625, 1.3) = 1.69

p3 → O(−0.4375, 1.7) = 3.08...

...

pN → O(0.375, 1.6) = 2.70





Assigning probabilities

Based on the objective function value, a probability is assigned tomake it into the next generation P(pj) = f (O(xi )). In our example

P(pj) =1∑

i max O(xi ) − O(xi )(max O(xi ) − O(xi ))

In our case:∑

i max O(xi ) − O(xi ) = 12.45 this gives:

P(p1) =6.64− 6.64

12.45= 0.

P(p2) =6.64− 1.69

12.45= 0.40

P(p3) =6.64− 3.08

12.45= 0.28

P(p4) =6.64− 2.70

12.45= 0.32

Now we can generate a new population of the same size.Max Moorkamp Genetic Algorithms




Selection

Each model’s chance to mate is determined by their probability.The offsprings undergo two steps: Crossover and Mutation

Mating Cand. Old Pos. Prob.

m1 p2 0.4m2 p3 0.28m3 p4 0.32m4 p2 0.4





Crossover

With probability Pc two members of the population exchange theirgenes after a randomly chosen point.

(000111) (000001)3−→

(110001) (110111)

Typically 0.2 ≤ Pc ≤ 0.7.





Mutation

With a probability Pm one bit changes its value.

(000111)→ (010111)

Typically Pm is chosen so that 80-90 percent of the population donot mutate.





Combined effect on the reproduction operators

Result of selection, crossover and mutation:

New Father Mother Cross. Mut. Result O(x)

p1 m1 m2 (1111010110) 1.69p2 m1 m2 (1001010001) 2.7p3 m3 m4 4 (0110|100001)→ (0110010110) 2.08p4 m3 m4 4 3 (1111|010110)→ (1101100001) 3.03

The resulting genetic strings make up the new population.





Outline








A slightly more difficult problem

O(x) = (x − 6)(x − 2)(x + 4)(x + 6)

has two minima O(−5.11) = −78.03 and O(4.41) = −335.475.We search x ∈ [−10 . . . 20] with 15-bit and a stepsize of 0.001.N = 20 and we calculate 6 generations i.e 120 forward calculations.The objective function ↔ probability scaling is

P(pi ) =1∑ exp(−O(x)/T ) where T =

1000 g < 3100 g ≥ 3





Statistics of GA performance

0 1 2 3 4Generation

-400

-200

0

200

400

Func

tion

Val

ue

Best memberMean performance

Fitness vs. Generation

25799.3 starting value

0 500 1000 1500 2000

4.1

4.2

4.3

4.4

4.5

Index

GA

Res

ult





Performance of the GA

−10 −5 0 5 10 15 20

050

000

1000

0015

0000

Generation 0

x

y

+

+

+

+

+

+++++ +++ +

+

+

+

−10 −5 0 5 10

020

0040

0060

00

Generation 1

x

y

+

+

+

+

+++++ ++ +++ +

+

+++−10 −5 0 5 10

020

0040

0060

00

Generation 2

x

y

+++++ ++++++ +++ ++++++

−10 −5 0 5 10

020

0040

0060

00

Generation 3

x

y

++ ++

+

+++++ ++++ ++++++−10 −5 0 5 10

020

0040

0060

00

Generation 4

x

y

+++++++

+++++++++

+++−10 −5 0 5 10

020

0040

0060

00

Generation 5

x

y++++++++++++ ++++++++





Looking at some statistics

0 500 1000 1500 2000-6

-4

-2

0

2

4

6

2000 runs, N=20, g=6

Histogram of x

x

Fre

quen

cy

−6 −4 −2 0 2 4 6

010

020

030

040

050

060

070

0

Min. 1st Qu. Median Mean 3rd Qu. Max.

-5.671 4.255 4.383 4.215 4.465 6.389





A hard problem

O(x) = |floor(x)|+ 0.5 (1.− (|x | − |floor(x)|))

Objective function

−10 −5 0 5 10 15 20

05

1015

20

x

y

Histogram of result of 2000 runs

N = 20 6 generations T = 1000Histogram of ga

ga

Fre

quen

cy

−2 0 2 4

050

100

150





Statistics for the hard problem

Statistical results for different GA-parameters

Parms Min. 1st Qu. Median Mean 3rd Qu. Max.

20/6/1000 -2.930 0.221 0.580 0.636 0.920 4.40530/10/1000 -0.9550 0.2467 0.5070 0.5340 0.7930 2.9590

20/6/10 -3.4410 0.2357 0.5550 0.6158 0.8350 4.099030/10/10 -1.0000 0.2350 0.4990 0.5103 0.7552 2.9590

The algorithm fails to find the exact minimum, but identifies thegeneral regionMore iterations do not improve the mean/median performance, butdecrease diversity in the populationIn this case the parameter T has no big influence on the result




Multi-objective optimizationA geophysical example

Outline








Multi-objective optimization

Multi-objective optimization means that we try to minimize severalfunctions simultaneously.

O(xi ) =

O1(xi )O2(xi )

...OM(xi )

Usually the weighted norm of this vector is then minimized e.g

‖W · O(xi )‖2 =M∑

j=1

(WjOj(xi ))2

Genetic algorithms can also directly work on the vector withoutweighting.





Pareto optimality and domination

Domination:

(x <p y)⇔ (∀ i : xi ≤ yi ∧ ∃ i : xi < yi )

If a member of the population is not dominated by any other it iscalled nondominated or pareto-optimal.

10 20 30 40MT RMS

0

0,02

0,04

0,06

Mod

el R

ough

ness

Rank 1Rank 2Rank 3Rank 4

Population fitnessGeneration 1





Transforming rank into probability

Pi =

(1− r

rmax

)αr < rmax

0 r ≥ rmax

0 10 20 30 40Rank

0

0,2

0,4

0,6

0,8

1

α = 1 α = 2 α = 3 α = 4 α = 0.5

Fitness as a function of Rankr max = 40





Designing an objective function

In linearized methods the objective function has to be scalar and isa linear combination of the individual objective functions

Olin(di , si ) = WMTOMT (di , si ) + WRecORec(di , si ) + WRegOReg (xj)

WMT , WRec and Wreg are usually constant, but could be variableduring inversion.In contrast genetic algorithms can handle vector valued objectivefunctions without a need to weight the individual objectivefunctions.

Ogen(di , xj) =

OMT (di , f (xj))ORec(di , f (xj))

OReg (xj)

OMT (di , f (xj)) =

1

n

n∑i=1

(di − si∆di

)2

with si ∈ [ρa, φ, Z ]





Outline








Inversion of synthetic data

A simple two-layer model is inverted to test the inversion.

0 10 20 30 40 50 60 702

3

4

5

6

7

8

S-Vel.DensityP-Vel.

Input model

0 10 20 30 40 50 60 70Depth [km]

20

40

60

80

100

120

140

Res

istiv

ity

0 20 40 602

3

4

5

6

7

8

S-Vel.DensityP-Vel.

Inversion result

0 20 40 60Depth [km]

0

20

40

60

80

100

120

140

Res

istiv

ity

0,01 1 100 10000 Period [s]

0

15

30

45

60

75

90

Pha

se

Input Data

Inversion

0,01 1 100 1000010

100

1000

ρ a [

Ωm

]

Data fitSynthetic Input model





Evolution of misfit and roughness

0,01 0,1 1 10 100MT Misfit

0,0001

Rec

. Fun

c. M

isfi

t

Iteration 0Iteration 20Iteration 40Iteration 60Iteration 80Iteration 100

Best modelsNo Regularization

0 50 100 150 2000,0001

0,001

0,01

0,1

1

10

100

1000

Max MisfitMin MisfitMedian MisfitMin RoughnessMedian Roughness





Inversion of a more realistic model

0,01 1 100 10000 Period [s]

0

15

30

45

60

75

90

Pha

se

Original data

Inversion result

0 1 100 1000010

100

1000

ρ a [

Ωm

]

Inversion fitMT Data

1 10 100Depth [km]

10

100

1000

ρ a [

Ωm

]

Original ModelInversion Result

MT Models

0 100 200 300 400 500Time [s]

0

0,02

0,04

Am

plitu

de

Inversion FitReceiver function data

1 10 100Depth [km]

1,5

2

2,5

3

3,5

4

4,5

5

Vs

[km

/s]

Inversion resultOriginal model

Receiver function model comparison




Summary

Genetic algorithms can be applied to a wide range of problems

Their flexibility comes at the cost of more function evaluations

There are many variants of genetic algorithms available, eachwith its own strengths and weaknesses


Appendix For Further Reading

For Further Reading

D. E. GoldbergGenetic algorithms in search, optimization & machine learningAddison Wesley, 1990.

K. DebA Fast and Elitist Multiobjective Genetic Algorithm: NSGA-IIIEEE Transactions on Evolutionary Computation, Vol. 6, No.2, April 2002

Kanpur Genetic Algorithm LaboratoryNSGA-II and general informationhttp://www.iitk.ac.in/kangal/index.shtml

M. WallGALib programming library in C++http://lancet.mit.edu/ga


genetic algorithms a step by step tutorial

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