Genetic algorithm

lAuthor: 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

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: $f(x,y)=21.5+x\sin \left(4\pi x\right)+y\sin \left(20\pi y\right)$ 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 $y$.