More on Heuristics

Genetic Algorithms (GA)

Terminology

• Chromosome – candidate solution - {x1, x2, .... , xn }

• Gene– variable - xj

• Allele– numerical value of the gene

• Fitness measure– objective function + penalty function

• Generation– new population created from current population

• Reproduction– producing next generation

Reproduction

• Cloning– copying the best chromosomes into the new

population

• Mating – creating a new chromosome (child) from two

current chromosomes (parents)

• Mutation– randomly altering one or more genes to form a

new chromosome

Characteristic of GAs

• Global search procedure (search from a population rather than a single point)

• Technique “evolves” towards better solutions• Population may stay constant (by eliminating

the less fit) or be allowed to grow• Objective function does not need to be

continuous, differentiable, or unimodal• Initial population may be randomly generated

or selected using some criteria• Terminate procedure when several

generations produce no further improvement

Cloning

f(xi1 , xi2, xi3, . . . xin )

deemed fit

Xi = { xi1 , xi2, xi3, . . . xin }

generation k+1:

Xi = { xi1 , xi2, xi3, . . . xin }

generation k:

i th chromosome

Mutation

Xi = { xi1 , xi2, xi3, . . . xin }

generation k:

i th chromosome

1. randomly select r1 < r2 2. For j = r1 to r2

x’i,j = randomly selected value

Xi = { xi1 , x’i,r1, x’i, r1+1, . . x’i, r2, ... xin }

generation k+1:

Mating - one point crossover

X = { x1 , x2, x3, . . . xn }

generation k:

Y = { y1 , y2, y3, . . . yn }

Select random r such that 1 <= r <= n

C1 = { x1 , x2, . . . yr , yr+1 , . . . yn }

C2= { y1 , y2, . . . xr , xr+1, . . . xn }

Mating - two point crossover

X = { x1 , x2, x3, . . . xn }

generation k:

Y = { y1 , y2, y3, . . . yn }

Select random r such that 1 <= r < s <= n

C1 = { x1 , x2, ... yr , yr+1 , ... ys, xs+1, ... xn }

C2= { y1 , y2, ... xr , xr+1, …xs, ys+1, ... yn }

Mating - random crossover

X = { x1 , x2, x3, . . . xn }

generation k:

Y = { y1 , y2, y3, . . . yn }

Create:C1 = { u1, u2, … un } and C2 = { v1, v2, … vn }

where

ux with prob

y with probj

i

j

RST

1v

y with prob

x with prob

with

j

i

j

RST

1

0 1

Mating - general strategies

• May include both children in next generation• May select better child and discard (abort) the

other• Usually 60 to 90 percent of the next

generation is determined by mating pairs of parents

• Parents to mate may be selected randomly from the best of the current generation

• Initial population sizes vary from 40 to 250+

The Fitness Function

f(x1, x2, … xn ) + < g(x1, x2, … xn) >

where

RSTg xg x if g x

if g x( )

( ) ( )

( )

2 0

0 0

Min f (x1, x2, … xn )

Subj to: g(x1, x2, … xn) <=0

To Use Genetic Algorithms

• Solutions must be represented by a numeric string of constant length (vector)

• There must be a suitable fitness function

• A suitable crossover (mating) operator must exist

Example - knapsack problem

Let xj = 1 if the jth item is included; 0 otherwise

Max

subj. to:

z c x

w x W

j jj

n

j jj

n

1

1

fitness c x w x Wj jj

n

j jj

n

LNM

OQP

1 1

2

100

If w x Wj jj

n

1

Else fitness c xj jj

n

1

Implementing the Algorithm

Select initialpopulation atrandom

Evaluatefitnessfunction

Clone thebest solutions

Mutate selectedsolutions

Mate thebettersolutions

Continuedimprovement?

Stop

no

Evaluatefitnessfunction

yes

create next generation

Genetic Algorithm - KNAPSACK problem Generation Nbr= 1 CHROMOSOME VALUE WEIGHT

1 FITNESS= 84 0 0 1 0 0 0 1 1 0 0 1 0 0 84 822 FITNESS= 84 0 1 0 0 0 0 0 1 0 1 1 0 1 84 873 FITNESS= -14286 0 1 1 0 1 1 0 1 1 0 0 0 0 114 1124 FITNESS= -19476 1 0 1 0 1 0 1 1 1 1 0 0 0 124 1145 FITNESS= -35971 1 1 1 0 0 1 0 0 0 0 1 0 1 129 1196 FITNESS= -62371 0 0 1 1 0 1 1 0 0 1 0 0 1 129 1257 FITNESS= -102248 1 1 1 1 1 1 0 1 0 1 1 0 1 152 1328 FITNESS= -249847 1 0 1 1 0 0 0 1 1 1 1 1 1 153 1509 FITNESS= -336215 0 1 0 1 0 0 0 1 0 1 1 1 1 185 15810 FITNESS= -623935 0 1 1 1 1 1 0 1 1 1 1 0 1 165 179

Genetic Algorithm - KNAPSACK problemGeneration Nbr= 2 CHROMOSOME VALUE WEIGHT

1 FITNESS= 103 1 1 1 0 0 1 0 0 0 0 1 0 0 103 902 FITNESS= 96 1 0 1 0 0 0 1 1 0 0 1 0 0 96 853 FITNESS= 84 0 0 1 0 0 0 1 1 0 0 1 0 0 84 824 FITNESS= 84 0 0 1 0 0 0 1 1 0 0 1 0 0 84 825 FITNESS= 82 1 1 1 0 0 1 0 0 0 0 1 0 0 82 756 FITNESS= -11988 0 0 1 0 1 0 1 1 1 1 0 0 0 112 1117 FITNESS= -11990 0 0 1 0 0 0 1 1 0 0 1 0 1 110 1118 FITNESS= -14286 0 1 1 0 1 1 0 1 1 0 0 0 0 114 1129 FITNESS= -67469 0 0 1 0 0 0 1 1 0 0 1 0 1 131 12610 FITNESS= -623935 0 1 1 1 1 1 0 1 1 1 1 0 1 165 179

Genetic Algorithm - KNAPSACK problemgeneration Nbr= 4 CHROMOSOME VALUE WEIGHT

1 FITNESS= 106 1 1 1 0 0 0 1 1 0 0 1 0 0 106 942 FITNESS= 103 1 1 1 0 0 1 0 0 0 0 1 0 0 103 903 FITNESS= 89 1 0 1 0 0 0 1 1 0 0 1 0 0 89 734 FITNESS= 89 1 0 1 0 0 0 1 1 0 0 1 0 0 89 735 FITNESS= 89 1 0 1 0 0 0 1 1 0 0 1 0 0 89 736 FITNESS= 89 1 0 1 0 0 0 1 1 0 0 1 0 0 89 737 FITNESS= 86 1 1 1 0 0 0 1 0 0 0 1 0 0 86 728 FITNESS= 86 1 1 1 0 0 0 1 0 0 0 1 0 0 86 729 FITNESS= 69 1 0 1 0 0 0 1 0 0 0 1 0 0 69 5110 FITNESS= -623935 0 1 1 1 1 1 0 1 1 1 1 0 1 165 179

Traveling Salesman Problem

6 - city tour 6! = 720 solutionsX1 = {1,2,3,4,5,6}X2 = {4,3,6,5,2,1}a permutation is a feasible solution

fitness1 = d12 + d23 + d34 + d45 + d56 + d61

fitness2 = d43+ d36 + d65 + d52 + d21 + d14

Crossover with Permutations

Partially matched crossover operator - PMXFor t = r to s; if xt = yt , then swap xt, yt in permutation x

X = { 4, 5, 6, 2, 1, 3 }Y = { 1, 2, 6, 4, 3, 5 } r = 2 and s = 5

C1 = { 4, 2, 6, 5, 1, 3 } = { 2, 4, 6, 5, 1, 3 } = { 2, 4, 6, 5, 3, 1 }

C2 = { 1, 5, 6, 4, 3, 2 } = { 1, 5, 6, 2, 3, 4 } = { 3, 5, 6, 2, 1, 4 }

More on Heuristics

Genetic Algorithms (GA)

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