1. **Problem Statement:** Perform the following genetic algorithm operations on the given chromosomes: (a) Uniform Crossover (b) Rank Selection (c) Flipping Mutation Given: Chromosome 1: 100101011110 Chromosome 2: 010101101011 2. **Uniform Crossover:** Uniform crossover exchanges genes between two chromosomes based on a crossover mask, usually generated randomly. Assuming a random mask (1 means take gene from Chromosome 1, 0 means from Chromosome 2): Example mask: 101010101010 Apply mask: Child 1 gene i = Chromosome 1 gene i if mask i = 1 else Chromosome 2 gene i Child 2 gene i = Chromosome 2 gene i if mask i = 1 else Chromosome 1 gene i Child 1: Positions with mask 1: take from Chromosome 1 Positions with mask 0: take from Chromosome 2 Child 1 = 1 1 0 1 0 1 1 0 1 0 1 0 Child 2 = 0 0 0 0 1 0 0 1 1 1 1 1 3. **Rank Selection:** Rank selection assigns selection probabilities based on the rank of fitness rather than raw fitness. Assuming fitness values (for example): Chromosome 1 fitness = 8 Chromosome 2 fitness = 7 Rank 1: Chromosome 1 Rank 2: Chromosome 2 Selection probability for rank $r$ in population size $N$ is: $$ P(r) = \frac{2 (N + 1 - r)}{N (N + 1)} $$ For $N=2$: $$ P(1) = \frac{2 (2 + 1 - 1)}{2 \times 3} = \frac{4}{6} = \frac{2}{3} $$ $$ P(2) = \frac{2 (2 + 1 - 2)}{2 \times 3} = \frac{2}{6} = \frac{1}{3} $$ So Chromosome 1 has a 2/3 chance to be selected, Chromosome 2 has 1/3. 4. **Flipping Mutation:** Flipping mutation inverts bits at certain positions. Assuming mutation at position 3 (0-based index): Chromosome 1 before mutation: 1 0 0 1 0 1 0 1 1 1 1 0 Flip bit at position 3 (4th bit): 1 -> 0 Chromosome 1 after mutation: 1 0 0 0 0 1 0 1 1 1 1 0 **Final answers:** - Uniform Crossover children: Child 1: 110101101010 Child 2: 000010011111 - Rank Selection probabilities: Chromosome 1: 2/3 Chromosome 2: 1/3 - Flipping Mutation on Chromosome 1 at position 3: Result: 100001011110