Genetic algorithm: Difference between revisions

Revision as of 23:00, 12 December 2024

Author: Yunchen Huo (yh2244), Ran Yi (ry357), Yanni Xie (yx682), Changlin Huang (ch2269), Jingyao Tong (jt887) (ChemE 6800 Fall 2024)

Stewards: Nathan Preuss, Wei-Han Chen, Tianqi Xiao, Guoqing Hu

Introduction

Algorithm Discussion

The Genetic Algorithm was first introduced by John H. Holland ^[1] in 1973. It is an optimization technique based on Charles Darwin’s theory of evolution by natural selection.

Before diving into the algorithm, here are definitions of the basic terminologies.

Gene: The smallest unit that makes up the chromosome (decision variable)
Chromosome: A group of genes, where each chromosome represents a solution (potential solution)
Population: A group of chromosomes (a group of potential solutions)

GA involves the following seven steps:

Initialization
- Randomly generate the initial population for a predetermined population size
Evaluation
- Evaluate the fitness of every chromosome in the population to see how good it is. Higher fitness implies better solution, making the chromosome more likely to be selected as a parent of next generation
Selection
- Natural selection serves as the main inspiration of GA, where chromosomes are randomly selected from the entire population for mating, and chromosomes with higher fitness values are more likely to be selected ^[2].
Crossover
- The purpose of crossover is to create superior offspring (better solutions) by combining parts from each selected parent chromosome. There are different types of crossover, such as single-point and double-point crossover ^[2]. In single-point crossover, the parent chromosomes are swapped before and after a single point. In double-point crossover, the parent chromosomes are swapped between two points ^[2].
Mutation
- A mutation operator is applied to make random changes to the genes of children's chromosomes, maintaining the diversity of the individual chromosomes in the population and enabling GA to find better solutions^[2].
Insertion
- Insert the mutated children chromosomes back into the population
Repeat 2-6 until a stopping criteria is met
- Maximum number of generations reached
- No significant improvement from newer generations
- Expected fitness value is achieved

Numerical Example

1. Simple Example

We aim to maximize $f(x)=x^{2}$ , where $x\in [0,31]$ . Chromosomes are encoded as 5-bit binary strings since the binary format of the maximum value 31 is 11111.

1.1 Initialization (Generation 0)

The initial population is randomly generated:

Chromosome (Binary)	x (Decimal)
10010	18
00111	7
11001	25
01001	9

1.2 Generation 1

1.2.1 Evaluation

Calculate the fitness values:

Chromosome	$x$	$f(x)=x^{2}$
10010	18	324
00111	7	49
11001	25	625
01001	9	81

1.2.2 Selection

Use roulette wheel selection to choose parents for crossover. Selection probabilities are calculated as:

$P({\text{Chromosome}})={\frac {\text{Fitness}}{\text{Total Fitness}}}.$

Thus,

Total Fitness $=324+49+625+81=1079$

Compute the selection probabilities:

Chromosome	$f(x)$	Selection Probability
10010	$324$	${\frac {324}{1,079}}\approx 0.3003$
00111	$49$	${\frac {49}{1,079}}\approx 0.0454$
11001	$625$	${\frac {625}{1,079}}\approx 0.5792$
01001	$81$	${\frac {81}{1,079}}\approx 0.0751$

Cumulative probabilities:

Chromosome	Cumulative Probability
10010	$0.3003$
00111	$0.3003+0.0454=0.3457$
11001	$0.3457+0.5792=0.9249$
01001	$0.9249+0.0751=1.0000$

Random numbers for selection:

$r_{1}=0.32,\quad r_{2}=0.60,\quad r_{3}=0.85,\quad r_{4}=0.10$

Selected parents:

Pair 1: 00111 and 11001
Pair 2: 11001 and 10010

1.2.3 Crossover

Crossover probability: $P_{c}=0.7$

Pair 1: Crossover occurs at position 2.

Parent 1: $00|111$
Parent 2: $11|001$
Children: $00001\ (x=1)$ , $11111\ (x=31)$

Pair 2: No crossover.

Children: Child 3: $11001\ (x=25)$ , Child 4: $10010\ (x=18)$

1.2.4 Mutation

Mutation probability: $P_{m}=0.1$ Child 1: $00001$ (bits: $0\ 0\ 0\ 0\ 1$ ).

Mutations: Bit 1 and Bit 4 flip.
Resulting chromosome: $10011\ (x=19)$ .

Child 2: $11111$ (bits: $1\ 1\ 1\ 1\ 1$ ).

Mutation: Bit 3 flips.
Resulting chromosome: $11011\ (x=27)$ .

Child 3: $11001$ (bits: $1\ 1\ 0\ 0\ 1$ ).

Mutations: Bit 1 and Bit 5 flip.
Resulting chromosome: $01000\ (x=8)$ .

Child 4: $10010$ (bits: $1\ 0\ 0\ 1\ 0$ ).

Mutations: Bit 1 and Bit 3 flip.
Resulting chromosome: $00110\ (x=6)$ .

1.2.5 Insertion

Chromosome	$x$	$f(x)$
10011	$19$	$361$
11011	$27$	$729$
01000	$8$	$64$
00110	$6$	$36$

1.3 Generation 2

1.3.1 Evaluation

Calculate the fitness values:

Chromosome	$x$	$f(x)$
10011	$19$	$361$
11011	$27$	$729$
01000	$8$	$64$
00110	$6$	$36$

1.3.2 Selection

Total fitness: ${\text{Total Fitness}}=361+729+64+36=1,190$

Compute selection probabilities:

Chromosome	$f(x)$	Selection Probability
10011	$361$	${\frac {361}{1,190}}\approx 0.3034$
11011	$729$	${\frac {729}{1,190}}\approx 0.6126$
01000	$64$	${\frac {64}{1,190}}\approx 0.0538$
00110	$36$	${\frac {36}{1,190}}\approx 0.0303$

Cumulative probabilities:

Chromosome	Cumulative Probability
10011	$0.3034$
11011	$0.3034+0.6126=0.9160$
01000	$0.9160+0.0538=0.9698$
00110	$0.9698+0.0303=1.0000$

Random numbers for selection:

$r_{1}=0.20,\quad r_{2}=0.50,\quad r_{3}=0.80,\quad r_{4}=0.95$

Selected parents:

Pair 1: 10011 and 11011
Pair 2: 11011 and 01000

1.3.3 Crossover

Crossover probability: $P_{c}=0.7$

Pair 1: Crossover occurs at position 2.

Parent 1: $10|011$
Parent 2: $11|011$
Children: $10011\ (x=19)$ , $11011\ (x=27)$

Pair 2: Crossover occurs at position 4.

Parent 1: $11|011$
Parent 2: $01|000$
Children: $11000\ (x=24)$ , $01011\ (x=11)$

1.3.4 Mutation

Mutation probability: $P_{m}=0.1$

Child 1: No mutations. ${\text{Resulting Chromosome: }}10011$ .
Child 2: Bit 1 flips. ${\text{Resulting Chromosome: }}01011\ (x=11)$ .
Child 3: Bit 5 flips. ${\text{Resulting Chromosome: }}11010\ (x=26)$ .
Child 4: No mutations. ${\text{Resulting Chromosome: }}11011$ .

1.3.5 Insertion

Chromosome	$x$	$f(x)$
10011	$19$	$361$
01011	$11$	$121$
11010	$26$	$676$
11011	$27$	$729$

1.4 Conclusion

After 2 iterations, the best optimal solution we find is $f=729,\ x=27$ .

Due to the limitation of the page, we will not perform additional loops. In the following examples, we will show how to use code to perform multiple calculations on complex problems and specify stopping conditions.

2. Complex Example

We aim to maximize the function: $f(x,y)=21.5+x\sin(4\pi x)+y\sin(20\pi y)$ subject to the constraints: $-3.0\leq x\leq 12.1,\quad 4.1\leq y\leq 5.8.$

2.1 Encoding the Variables

Each chromosome represents a pair of variables $x$ and $y$ . We encode these variables into binary strings:

$x$ : $-3.0\leq x\leq 12.1$ , precision $0.1$ , requiring $8$ bits.
$y$ : $4.1\leq y\leq 5.8$ , precision $0.01$ , requiring $8$ bits.

Each chromosome is a 16-bit binary string, where the first 8 bits represent $x$ and the next 8 bits represent $y$ .

References

↑ Holland, J. H. (1973). Genetic algorithms and the optimal allocation of trials. SIAM Journal on Computing, 2(2), 88–105
↑ ^{Jump up to: 2.0} ^2.1 ^2.2 ^2.3 Mirjalili, S. (2018). Genetic Algorithm. Evolutionary Algorithms and Neural Networks, Springer, pp. 43–56

[1] Holland, J. H. (1973). Genetic algorithms and the optimal allocation of trials. SIAM Journal on Computing, 2(2), 88–105

[Mirjalili-2] {Jump up to: 2.0} ^2.1 ^2.2 ^2.3 Mirjalili, S. (2018). Genetic Algorithm. Evolutionary Algorithms and Neural Networks, Springer, pp. 43–56

[1]

[2]

@@ Line 129: / Line 129: @@
 * Pair 2: 11001 and 10010
-===== 3. Crossover =====
+===== 1.2.3 Crossover =====
 Crossover probability: <math>P_c = 0.7</math>
@@ Line 140: / Line 140: @@
 * Children: Child 3: <math>11001 \ (x = 25)</math>, Child 4: <math>10010 \ (x = 18)</math>
 ===== 1.2.4 Mutation =====
 Mutation probability: <math>P_m = 0.1</math>
 '''Child 1:''' <math>00001</math> (bits: <math>0 \ 0 \ 0 \ 0 \ 1</math>).
 * Mutations: Bit 1 and Bit 4 flip.
 * Resulting chromosome: <math>10011 \ (x = 19)</math>.
 '''Child 2:''' <math>11111</math> (bits: <math>1 \ 1 \ 1 \ 1 \ 1</math>).
 * Mutation: Bit 3 flips.
 * Resulting chromosome: <math>11011 \ (x = 27)</math>.
 '''Child 3:''' <math>11001</math> (bits: <math>1 \ 1 \ 0 \ 0 \ 1</math>).
 * Mutations: Bit 1 and Bit 5 flip.
 * Resulting chromosome: <math>01000 \ (x = 8)</math>.
 '''Child 4:''' <math>10010</math> (bits: <math>1 \ 0 \ 0 \ 1 \ 0</math>).
 * Mutations: Bit 1 and Bit 3 flip.
 * Resulting chromosome: <math>00110 \ (x = 6)</math>.
 ===== 1.2.5 Insertion =====
@@ Line 184: / Line 183: @@
 |}
 ===== 1.3.2 Selection =====
-'''Total fitness:'''
+'''Total fitness:'''  <math>\text{Total Fitness} = 361 + 729 + 64 + 36 = 1,190</math>
-<math>\text{Total Fitness} = 361 + 729 + 64 + 36 = 1,190</math>
 Compute selection probabilities:
 {| class="wikitable"
@@ Line 212: / Line 211: @@
 | 00110 || <math>0.9698 + 0.0303 = 1.0000</math>
 |}
 Random numbers for selection:
-<math>r_1 = 0.20, \quad r_2 = 0.50, \quad r_3 = 0.80, \quad r_4 = 0.95</math>
+<math>r_1 = 0.20, \quad r_2 = 0.50, \quad r_3 = 0.80, \quad r_4 = 0.95</math>
 Selected parents:
 * Pair 1: 10011 and 11011
 * Pair 2: 11011 and 01000
 ===== 1.3.3 Crossover =====
 Crossover probability: <math>P_c = 0.7</math>
 '''Pair 1:''' Crossover occurs at position 2.
 * Parent 1: <math>10|011</math>
@@ Line 248: / Line 250: @@
 ==== 1.4 Conclusion ====
 After 2 iterations, the best optimal solution we find is <math>f = 729, \ x = 27</math>.
 Due to the limitation of the page, we will not perform additional loops. In the following examples, we will show how to use code to perform multiple calculations on complex problems and specify stopping conditions.
 === 2. Complex Example ===