evolutionary algorithms k. ganesh research scholar, ph.d., industrial management division,...

Evolutionary Algorithms

K. GaneshResearch Scholar, Ph.D.,

Industrial Management Division,

Humanities and Social Sciences Department,

Indian Institute of Technology Madras,

Chennai, TN, India.

• Optimization

• Darwin’s Theory of evolution.

• Simulated annealing, Tabu Search

• Genetic Algorithms

• Search Space

• NP Problems

Optimization

S = { S1,S2, ….. }

f : S R

Search Space

Fitness Landscape

Obj. fn.

Variables Constraints

Where to start ? Where to look for the solution?

Suitable Solutions ……….Not the best solution but good

Not possible to prove which is Real optimum

NP Problems Story ????????

Continuous and Discrete variablesObj. Fn. Is Continuous

Directional infr. To guide searchStarting point?

Graph colouring problemScale is pregiven

Optimization Techniques

Traditional Methods• Linear Programming • Dynamic Programming• Non Linear Programming

Heuristic Methods• Genetic Algorithm • Simulated Annealing• Tabu search• Ant Colony Optimization

Classes of Search Techniques

F inonacc i N ew ton

D irect m ethods Indirec t m ethods

C alcu lus-based techn iques

E volu tionary s trategies

C entra l ized D is tr ibuted

Para l le l

S teady-s ta te G enera tiona l

S equentia l

G ene tic a lgori thm s

E volutionary a lgori thm s S im u lated annealing

G uided random search techniques

D ynam ic program m ing

E num erative techn iques

S earch techniques

Advantages of Heuristic Method

• Better Solution ( Near optimal )

• Reasonable Computation time

• No requirement of Complex derivatives & careful choice of initial values

Many optimization problems have an enormous search space.We shall examine the algorithms according to:

Time

Space

Soundness

Completeness

Robustness

For example…

1. Start at a random point and climb where you can2. Save the best result, return to step 1

(we shell stop after a while…)

Going in circlesLocal minimum

Run the algorithm:

Hill Climbing

Very fastNo memory neededSimple to program



Exploits the best solution & Ignores exploration of search space

Random Search

Explores the Search space and ignoring the exploitation of search space.

Genetic Search

Can make a remarkable balance between exploration and exploitation of search space

History

• 1960 – Introduced by I. Rechenberg

• 1975 – Popularized by John Holland

• 1975 - book "Adaptation in Natural and Artificial Systems" published

• 1992 – John Koza’s work

“Genetic Algorithms are good at taking large, potentially huge search spaces and navigating

them, looking for optimal combinations of things, solutions you might not

otherwise find in a lifetime.”

- Salvatore Mangano

Computer Design, May 1995

Genetic Algorithms: A TutorialGenetic Algorithms: A Tutorial

Introduction to GA

• Evolution in a changing world!• Defining GA! (Goldberg, 1989)

Search algorithms based on the principle of natural selection and natural genetics

• Survival of the fittest

• A natural Perspective

• Biological Metaphorsis of GAs

In Nature…

The strongest survives…

We shall look for several alternatives simultaneously.

The searchers exchange information during the search, thisInformation is the basis for the decision regarding their next location

I’m redundant here

Good location

I’ll stick around here

GA -Evaluation of solutions

• Directed search algorithms based on the mechanics of biological evolution• To understand the adaptive processes of natural

systems• To design artificial systems software that retains the

robustness of natural systems

• Provide efficient, effective techniques for optimization and machine learning applications

• Widely-used today in business, scientific and engineering circles

Components of a GA

A problem to solve, and ...

• Encoding technique (gene, chromosome)

• Initialization procedure (creation)

• Evaluation function (environment)

• Selection of parents (reproduction)

• Genetic operators (mutation, recombination)

• Parameter settings (practice and art)

Simple Genetic Algorithm

{

initialize population;

evaluate population;

while Termination Criteria Not Satisfied{

select parents for reproduction;

perform recombination and mutation;

evaluate population;}

}

Generate randomly the popsize times of initial solution Get the input data for No of iterations, cross over probability, mutation probability

Solution from Initialization process

Start

Select the chromosome by roulette wheel selection approach

Evaluation process – calculate the objective function and find out the fitness value and Selection process

Apply Arithmetic cross overApply mutation

Find out the off spring

stopTake this off

spring as initial chromosome

Is no of iterations are over

B

B

Flow chart for Genetic Algorithm

Y N

Outline of the Basic GA• [Start] Generate random population of n chromosomes (suitable solutions

for the problem) • [Fitness] Evaluate the fitness f(x) of each chromosome x in the population • [New population] Create a new population by repeating following steps

until the new population is complete • [Selection] Select two parent chromosomes from a population

according to their fitness (the better fitness, the bigger chance to be selected)

• [Crossover] With a crossover probability cross over the parents to form a new offspring (children). If no crossover was performed, offspring is an exact copy of parents.

• [Mutation] With a mutation probability mutate new offspring at each locus (position in chromosome).

• [Accepting] Place new offspring in a new population • [Replace] Use new generated population for a further run of algorithm • [Test] If the end condition is satisfied, stop, and return the best solution in

current population

Vertical lines represent solutions (points in search space). The red line is the best solution, Yellow lines are the other ones.

Example Problem

• Unconstrained Optimization problem

max f(x1,x2) = 45.5 + X1 sin (4X1) + X2sin(20X2)

-3.0 <=X1<= 12.1

4.1 <=X2<=5.8

RepresentationEncode decision variables into binary strings

Length of the strings depends on required precision

Domain of Variable Xj is [aj,bj]

Required precision is five places after the decimal point.

Range of domain should be divided into atleast (bj-aj)*105 size ranges

• The required bits denoted with mj for a variable is calculated :

2mj-1< (bj-aj)*105< 2mj-1 - 1

• Mapping from a binary string to real number for variable xj is:

Xj = aj + decimal (substringj)* bj-aj

2mj-1

• The required bits for variables X1 and X2 are:

((12.1) – (-3.0)) * 10000) = 151000

217 < 151000 <= 218 so, m1 = 18

((5.8) – (4.1)) * 10000) = 170000

214 < 100000 <= 215 so, m2 = 15

So, m = m1 + m2 = 18 + 15 = 33

• The corresponding values for X1 and X2 :

Binary Number Decimal No.

X1 000001010100101001 5417

X2 101111011111110 24318

X1 = -3.0 + 5417 * ((12.1 – (-3.0)) / 218 -1 )

= -2.687969

X2 = 4.1 + 24318 * ((5.8 – (4.1) / 215 -1 )

= -2.687969

Initial Population – Randomly Generated

V1 = [000001010100101001 101111011111110]

V2 = [001110101110011000 000001010100100]

V3 = [111000111000111000 001111011100011]

V4 = [111111010100101001 100000011111111]

V5 = [111101010100101001 101111010000111]

V6 = [001101010111101001 101111011111110]

V7 = [111001010100101001 101001011111110]

V8 = [111111111111101001 100000011111110]

V9 = [111001010100101001 101111010001111]

V10 = [000011111100101001 101100000011110]

Back

• The corresponding decimal values are

V1 = [X1,X2] = [-2.687969, 5.361653]

V2 = [X1,X2] = [0.474116, 4.155566]

V3 = [X1,X2] = [10.419662, 4.772626]

V4 = [X1,X2] = [6.212122, 4.123212]

V5 = [X1,X2] = [-2.345678, 4.361773]

V6 = [X1,X2] = [11.687969, 4.111113]

V7 = [X1,X2] = [9.682229, 5.533333]

V8 = [X1,X2] = [-0.123469, 4.891653]

V9 = [X1,X2] = [11.118889, 4.812453]

V10 = [X1,X2] = [11.447969, 4.171653]

Evaluation• Step 1. Convert the chromosomes’s genotype to

its phenotype.• Binary strings into relative real values• Xk= (Xk

1, Xk2) , k = 1,2,….,pop_size.

• Step 2. Evaluate the Objective fn. f(Xk)

• Step 3.Convert the value of Obj. fn. Into fitness.• For Maximization problem, the fitness equal to

objective function• eval(Vk) = f(Xk), k = 1,2,….., pop_size

• The fitness function values of above chromosomes are

eval (V1) = f (-2.687969, 5.361653) = 19.233331

eval ( V2 ) = f (0.474116, 4.155566) = 17.237891

eval ( V3 ) = f (10.419662, 4.772626) = 9.781262 - W

eval ( V4 ) = f (6.212122, 4.123212) = 29.881177 - Seval ( V5 ) = f (-2.345678, 4.361773) = 15.615118

eval ( V6 ) = f (11.687969, 4.111113) = 11.900118

eval ( V7 ) = f (9.682229, 5.533333) = 17.012030

eval ( V8 ) = f (0.123469, 4.891653) = 19.912734

eval ( V9 ) = f (11.118889, 4.812453) = 26.197641

eval ( V10 ) = f (11.447969, 4.171653) = 10.276541

Selection• Roulette wheel approach• Select a new population w.r.to prob. Distr. based on

fitness values

1.Calculate fitness value eval (Vk) for each chromosome Vk

eval ( Vk ) = f(X), k = 1,2,….pop_size

2.Calculate the total fitness for population

pop_size F = ∑ eval ( Vk ) k = j

3.Calculate selection probability Pk for each chromosome Vk :

eval ( Vk )

Pk = k = 1, 2, ….., pop_size

F

4.Calculate cumulative probability qk for each chromosome Vk :

k

qk = ∑ Pj, k = 1, 2, ….., pop_size

j = 1

The selection process begins by spinning the roulette wheel pop_size times.

Selection1. Generate a random number r from the range [0,1]

2. If r <= q1, then select the first chromosome V1; otherwise select the k th chromosome Vk (2<=k<=pop_size) such that qk-1< r < qk

The total fitness F of the population is

10

F = ∑ eval ( Vk ) = 178.135372

k = 1

The probability of a selection pk for each chromosome Vk

(k=1,…..,10) is as follows:

P1 = 0.111180 P2 = 0.097515 P3 = 0.053839

P4 = 0.165077 P5 = 0.088057 P6 = 0.066806

P7 = 0.100815 P8= 0.110945 P9 = 0.148211

P10= 0.057554

The cumulative probabilities qk for each chromosome Vk

(k=1,…..,10) is as follows:

q1 = 0.111180 q2 = 0.208695 q3 = 0.262534

q4 = 0.427611 q5 = 0.515668 q6 = 0.582475

q7 = 0.683290 q8= 0.794234 q9 = 0.942446

q10= 1.00000

Back

Chromosomes

Back

Chromosomes

Spin the roulette wheel 10 times and each time select a single chromosome for a new population.

Let us assume that a random sequence of 10 numbers from the range [0,1] is as follows

0.301431 0.322062 0.766503 0.881893

0.350871 0.583392 0.177618 0.343242

0.032685 0.197577

The first number r1 = 0.301431 is greater than q3 and smaller than q4 meaning that the chromosome V4 is selected for the new population

The new population after selection process

V’1= [111111010100101001100000011111111] (V4)

V’2= [111111010100101001100000011111111] (V4)

V’3= [111111111111101001100000011111110] (V8)

V’4= [111001010100101001101111010001111] (V9)

V’5= [111111010100101001 100000011111111] (V4)

V’6= [111001010100101001101001011111110] (V7)

V’7= [001110101110011000 000001010100100] (V2)

V’8= [111111010100101001 100000011111111] (V4)

V’9= [000001010100101001 101111011111110] (V1)

V’10= [000001010100101001 101111011111110] (V1)

Back

CrossoverOne cut point crossoverProbability of Crossover Pc = 0.25Generate Random Numbers:0.651234 0.266666 0.288888 0.2999990.166666 0.566666 0.088888 0.3999990.701111 0.544444Choose random numbers less than Pc=0.25

The chromosomes V’5 and V’7 were selected for crossover.

Generate random number between [0,32] for choosing the position in chromosome

Chromosomes selected for crossover

V’5=[11111101010010100 1100000011111111] (V4)

V’7= [00111010111001100 0000001010100100] (V2)

The random position from [0,32] is 17.So, cut from the 17 th gene

Chromosomes after crossover

V’5= [11111101010010100 0000001010100100] (V4)

V’7= [00111010111001100 1100000011111111] (V2)

• V’’1= [11111101010010100 0000001010100100] (V’5)• V’’2= [00111010111001100 1100000011111111] (V’7)• V’’3= [111111111111101001100000011111110] (V’8)• V’’4= [111001010100101001101111010001111] (V’9)• V’’5= [111111010100101001 100000011111111] (V’4)• V’’6= [111001010100101001101001011111110] (V’7)• V’’7= [001110101110011000 000001010100100] (V’2)• V’’8= [111111010100101001 100000011111111] (V’4)• V’’9= [000001010100101001 101111011111110] (V’1)• V’’10= [000001010100101001 101111011111110] (V’2)

The new population after crossover process

Mutation • Flip bit mutation

• m=33, pop_size=10

• Probability of mutation Pm=0.01

• Generate sequence of random numbers rk

(k= 1,….330) from

the range [0 1]

Bit_pos

Ch_num

Bit_no

Rand_no

105 4 6 .009857

164 5 32 .003113

199 7 1 .000946

329 10 32 .001282

The Random position is 105,

4th chromosome,

Bit number = 6

Before Mutation

V’’4= [111001010100101001101111010001111] (V’9)

After Mutation

V’’’4= [111000010100101001101111010001111] (V’9)

• V’’’1= [11111101010010100 1000001010100100] (V’1)• V’’’2= [00111010111001100 0100000011111111] (V2)• V’’’3= [111111111111101001100000011111110] (V3)

• V’’’4= [111000010100101001101111010001111] (V4)

• V’’’5= [111111010100101001100000011111101] (V’5)

• V’’’6= [111001010100101001101001011111110] (V6)

• V’’’7= [101110101110011000100001010100100] (V’7)

• V’’’8= [111111010100101001100000011111111] (V8)• V’’’9= [000001010100101001101111011111110] (V9)

• V’’’10= [000001010100101000001111011111100] (V’10)

The new population after Mutation process

The corresponding decimal values (X1,X2) and fitness

f(6.167919, 4.107171) = 29.173737f(6.424242, 4.104451) = 29.771923f(-3.134919, 4.607171) = 19.171234f(11.617919, 4.807166) = 5.473734f(8.167779, 4.176171) = 19.234737f(9.222219, 5.101271) = 17.155537f(6.654419, 4.189971) = 29.666737f(6.88919, 4.234171) = 29.173777f(-2.117919, 5.87651) = 19.888837f(0.163229, 4.12371) = 17.188837

Termination

• Run the test for 1000 generations• Best chromosome in 419th generation

V*=(11111000000001110001111010001010110)Eval(V*)= f(11.631407, 5.724824) = 38.818208

X1*=11.631407X2*= 5.724824

f(X1*,X2*) = 38.818208

evolutionary algorithms k. ganesh research scholar, ph.d., industrial management division,...

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