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lecture-09-evolutionary-computation-genetic-algorithms.pdf
- 1. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 1 Lecture 9 Lecture 9 Evolutionary Computation: Evolutionary Computation: Genetic algorithms Genetic algorithms I I Introduction, or Introduction, or can can evolution evolution be be intelligent intelligent? ? I I Simulation of Simulation of natural natural evolution evolution I I Genetic Genetic algorithms algorithms I I Case Case study study: : maintenance maintenance scheduling scheduling with with genetic genetic algorithms algorithms I I Summary Summary
- 2. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 2 I I Intelligence can be defined as the capability of a Intelligence can be defined as the capability of a system to adapt its behaviour to ever system to adapt its behaviour to ever- -changing changing environment. According to Alan Turing, the form environment. According to Alan Turing, the form or appearance of a system is irrelevant to its or appearance of a system is irrelevant to its intelligence. intelligence. I I Evolutionary computation simulates evolution on a Evolutionary computation simulates evolution on a computer. The result of such a simulation is a computer. The result of such a simulation is a series of optimisation algorithms series of optimisation algorithms, usually based on , usually based on a simple set of rules. Optimisation a simple set of rules. Optimisation iteratively iteratively improves the quality of solutions until an optimal, improves the quality of solutions until an optimal, or at least feasible, solution is found. or at least feasible, solution is found. Can evolution be intelligent Can evolution be intelligent? ?
- 3. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 3 I I The behaviour of The behaviour of an individual organism an individual organism is is an an inductive inference about some yet unknown inductive inference about some yet unknown aspects of its environment. aspects of its environment. If, If, over successive over successive generations, the organism survives, we can say generations, the organism survives, we can say that this organism is capable of learning to predict that this organism is capable of learning to predict changes in its environment. changes in its environment. I I The evolutionary approach is based on The evolutionary approach is based on computational models of natural selection and computational models of natural selection and genetics. We call them genetics. We call them evolutionary evolutionary computation computation, an umbrella term that combines , an umbrella term that combines genetic algorithms genetic algorithms, , evolution strategies evolution strategies and and genetic programming genetic programming. .
- 4. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 4 I I On 1 July 1858, On 1 July 1858, Charles Darwin Charles Darwin presented his presented his theory of evolution before the theory of evolution before the Linnean Linnean Society of Society of London. This day marks the beginning of a London. This day marks the beginning of a revolution in biology. revolution in biology. I I Darwin Darwin’ ’s s classical classical theory of evolution theory of evolution, together , together with with Weismann Weismann’ ’s s theory of natural selection theory of natural selection and and Mendel Mendel’ ’s s concept of concept of genetics genetics, now represent the , now represent the neo neo- -Darwinian paradigm. Darwinian paradigm. Simulation of Simulation of natural natural evolution evolution
- 5. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 5 I I Neo Neo- -Darwinism Darwinism is based on processes of is based on processes of reproduction, mutation, competition and selection. reproduction, mutation, competition and selection. The power to reproduce appears to be an essential The power to reproduce appears to be an essential property of life. The power to mutate is also property of life. The power to mutate is also guaranteed in any living organism that reproduces guaranteed in any living organism that reproduces itself in a continuously changing environment. itself in a continuously changing environment. Processes of competition and selection normally Processes of competition and selection normally take place in the natural world, where expanding take place in the natural world, where expanding populations of different species are limited by a populations of different species are limited by a finite space. finite space.
- 6. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 6 I I Evolution can be seen as a process leading to the Evolution can be seen as a process leading to the maintenance maintenance of a population of a population’ ’s s ability to survive ability to survive and reproduce in a specific environment. This and reproduce in a specific environment. This ability is called ability is called evolutionary fitness evolutionary fitness. . I I Evolutionary fitness can also be viewed as a Evolutionary fitness can also be viewed as a measure of the organism measure of the organism’ ’s ability to anticipate s ability to anticipate changes in its environment. changes in its environment. I I The The fitness, or the quantitative measure of the fitness, or the quantitative measure of the ability to predict environmental changes and ability to predict environmental changes and respond adequately, can be considered as the respond adequately, can be considered as the quality that is quality that is optimised optimised in natural life. in natural life.
- 7. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 7 How is a population with increasing How is a population with increasing fitness generated? fitness generated? I I Let us consider Let us consider a population of rabbits. Some a population of rabbits. Some rabbits are faster than others, and we may say that rabbits are faster than others, and we may say that these rabbits possess superior fitness, because they these rabbits possess superior fitness, because they have a greater chance of avoiding foxes, surviving have a greater chance of avoiding foxes, surviving and then breeding and then breeding. . I I If two parents have superior fitness, there is a good If two parents have superior fitness, there is a good chance that a combination of their genes will chance that a combination of their genes will produce an offspring with even higher fitness. produce an offspring with even higher fitness. Over time the entire population of rabbits becomes Over time the entire population of rabbits becomes faster to meet their environmental challenges in the faster to meet their environmental challenges in the face of foxes. face of foxes.
- 8. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 8 I I All methods of evolutionary computation simulate All methods of evolutionary computation simulate natural natural evolution by creating evolution by creating a population of a population of individuals, evaluating their fitness, generating a individuals, evaluating their fitness, generating a new population through genetic operations, and new population through genetic operations, and repeating this process a number of times repeating this process a number of times. . I I We will start with We will start with Genetic Algorithms Genetic Algorithms (GAs) as (GAs) as most of the other evolutionary most of the other evolutionary algorithms can be algorithms can be viewed as variations of viewed as variations of genetic algorithms genetic algorithms. . Simulation of Simulation of natural natural evolution evolution
- 9. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 9 I I In the early 1970s, John In the early 1970s, John Holland introduced the Holland introduced the concept of genetic algorithms. concept of genetic algorithms. I I His His aim was to make computers do what nature aim was to make computers do what nature does. does. Holland was concerned with Holland was concerned with algorithms algorithms that manipulate strings of binary digits that manipulate strings of binary digits. . I I Each artificial Each artificial “ “chromosomes chromosomes” ” consists of a consists of a number of number of “ “genes genes” ”, and each gene is represented , and each gene is represented by 0 or 1: by 0 or 1: Genetic Algorithms Genetic Algorithms 1 1 0 1 0 1 0 0 0 0 0 1 0 1 1 0
- 10. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 10 I I Nature has an ability to adapt and learn without Nature has an ability to adapt and learn without being told what to do. In other words, nature being told what to do. In other words, nature finds good chromosomes blindly. finds good chromosomes blindly. GAs GAs do the do the same. Two mechanisms link a GA to the problem same. Two mechanisms link a GA to the problem it is solving: it is solving: encoding encoding and and evaluation evaluation. . I I The GA uses a measure of fitness of individual The GA uses a measure of fitness of individual chromosomes to carry out reproduction. As chromosomes to carry out reproduction. As reproduction takes place, the crossover operator reproduction takes place, the crossover operator exchanges parts of two single chromosomes, and exchanges parts of two single chromosomes, and the mutation operator changes the gene value in the mutation operator changes the gene value in some randomly chosen location of the some randomly chosen location of the chromosome. chromosome.
- 11. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 11 Step 1 Step 1: : Represent the problem variable domain as Represent the problem variable domain as a chromosome of a fixed length, choose the size a chromosome of a fixed length, choose the size of a chromosome population of a chromosome population N N, the crossover , the crossover probability probability p pc c and the mutation probability and the mutation probability p pm m. . Step Step 2 2: : Define Define a fitness function to measure the a fitness function to measure the performance, or fitness, of an individual performance, or fitness, of an individual chromosome in the problem domain. The fitness chromosome in the problem domain. The fitness function establishes the basis for selecting function establishes the basis for selecting chromosomes that will be mated during chromosomes that will be mated during reproduction. reproduction. Basic Basic genetic genetic algorithms algorithms
- 12. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 12 Step 3 Step 3: : Randomly generate an initial population of Randomly generate an initial population of chromosomes of size chromosomes of size N N: : x x1 1, , x x2 2, . . . , , . . . , x xN N Step 4 Step 4: : Calculate the fitness of each individual Calculate the fitness of each individual chromosome: chromosome: f f ( (x x1 1), ), f f ( (x x2 2), . . . , ), . . . , f f ( (x xN N) ) Step 5 Step 5: : Select a pair of chromosomes for mating Select a pair of chromosomes for mating from the current population. Parent from the current population. Parent chromosomes are selected with a probability chromosomes are selected with a probability related to their fitness. related to their fitness.
- 13. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 13 Step 6 Step 6: : Create a pair of offspring chromosomes by Create a pair of offspring chromosomes by applying the genetic operators applying the genetic operators − − crossover crossover and and mutation mutation. . Step 7 Step 7: : Place the created offspring chromosomes Place the created offspring chromosomes in the new population. in the new population. Step 8 Step 8: : Repeat Repeat Step 5 Step 5 until the size of the new until the size of the new chromosome population becomes chromosome population becomes equal to the equal to the size of the size of the initial population, initial population, N N. . Step 9 Step 9: : Replace Replace the initial (parent) chromosome the initial (parent) chromosome population with the new (offspring) population. population with the new (offspring) population. Step Step 10 10: : Go to Go to Step 4 Step 4, and repeat the process until , and repeat the process until the termination criterion is satisfied. the termination criterion is satisfied.
- 14. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 14 I I GA represents an iterative process. Each iteration is GA represents an iterative process. Each iteration is called a called a generation generation. A typical number of generations . A typical number of generations for a simple GA can range from 50 to over for a simple GA can range from 50 to over 500. The 500. The entire set entire set of generations is called a of generations is called a run run. . I I Because GAs use a stochastic search method, the Because GAs use a stochastic search method, the fitness of a population may fitness of a population may remain stable for a remain stable for a number of generations before a superior chromosome number of generations before a superior chromosome appears. appears. I I A common practice is to terminate a GA after a A common practice is to terminate a GA after a specified number of generations and then examine specified number of generations and then examine the best chromosomes in the population. If no the best chromosomes in the population. If no satisfactory solution is found, the GA is restarted. satisfactory solution is found, the GA is restarted. Genetic Genetic algorithms algorithms
- 15. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 15 A simple example will help us to understand how A simple example will help us to understand how a GA works. Let us find the maximum value of a GA works. Let us find the maximum value of the function (15 the function (15x x − − x x2 2 ) where parameter ) where parameter x x varies varies between 0 and 15. For simplicity, we may between 0 and 15. For simplicity, we may assume that assume that x x takes only integer values. Thus, takes only integer values. Thus, chromosomes can be built with only four chromosomes can be built with only four genes: genes: Genetic Genetic algorithms algorithms: : case case study study Integer Binary code Integer Binary code Integer Binary code 1 0 0 0 1 6 0 1 1 0 11 1 0 1 1 2 0 0 1 0 7 0 1 1 1 12 1 1 0 0 3 0 0 1 1 8 1 0 0 0 13 1 1 0 1 4 0 1 0 0 9 1 0 0 1 14 1 1 1 0 5 0 1 0 1 10 1 0 1 0 15 1 1 1 1
- 16. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 16 Suppose that the size of the chromosome population Suppose that the size of the chromosome population N N is 6, the crossover probability is 6, the crossover probability p pc c equals 0.7, and equals 0.7, and the mutation probability the mutation probability p pm m equals 0.001. The equals 0.001. The fitness function in our example is defined fitness function in our example is defined by by f f( (x x) = ) = 15 15 x x − − − − − − − − x x2 2
- 17. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 17 The fitness function and chromosome locations The fitness function and chromosome locations Chromosome label Chromosome string Decoded integer Chromosome fitness Fitness ratio, % X1 1 1 0 0 12 36 16.5 X2 0 1 0 0 4 44 20.2 X3 0 0 0 1 1 14 6.4 X4 1 1 1 0 14 14 6.4 X5 0 1 1 1 7 56 25.7 X6 1 0 0 1 9 54 24.8 x 50 40 30 20 60 10 0 0 5 10 15 f(x) (a) Chromosome initial locations. x 50 40 30 20 60 10 0 0 5 10 15 (b) Chromosome final locations.
- 18. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 18 I I In natural selection, only the fittest species can In natural selection, only the fittest species can survive, breed, and thereby pass their genes on to survive, breed, and thereby pass their genes on to the next generation. the next generation. GAs GAs use a similar approach, use a similar approach, but unlike nature, the size of the chromosome but unlike nature, the size of the chromosome population remains unchanged from one population remains unchanged from one generation to the next. generation to the next. I I The last column in Table shows the ratio of the The last column in Table shows the ratio of the individual chromosome individual chromosome’ ’s fitness to the s fitness to the population population’ ’s total fitness. This ratio determines s total fitness. This ratio determines the chromosome the chromosome’ ’s chance of being selected for s chance of being selected for mating. mating. The chromosome The chromosome’ ’s average fitness s average fitness improves from one generation to the next. improves from one generation to the next.
- 19. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 19 Roulette Roulette wheel wheel selection selection The most commonly used chromosome selection The most commonly used chromosome selection techniques is the techniques is the roulette wheel selection roulette wheel selection. . 100 0 16.5 36.7 43.1 49.5 75.2 X1: 16.5% X2: 20.2% X3: 6.4% X4: 6.4% X5: 25.3% X6: 24.8%
- 20. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 20 Crossover operator Crossover operator I I In our example, we have an initial population of 6 In our example, we have an initial population of 6 chromosomes. Thus, to establish the same chromosomes. Thus, to establish the same population in the next generation, the roulette population in the next generation, the roulette wheel would be spun six times wheel would be spun six times. . I I Once a pair of parent chromosomes is selected, Once a pair of parent chromosomes is selected, the the crossover crossover operator is applied. operator is applied.
- 21. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 21 I I First, the crossover operator randomly chooses a First, the crossover operator randomly chooses a crossover point where two parent chromosomes crossover point where two parent chromosomes “ “break break” ”, and then exchanges the chromosome , and then exchanges the chromosome parts after that point. As a result, two new parts after that point. As a result, two new offspring are created. offspring are created. I I If a pair of chromosomes does not cross over, If a pair of chromosomes does not cross over, then then the chromosome cloning takes place, and the the chromosome cloning takes place, and the offspring are created as exact copies of each offspring are created as exact copies of each parent. parent.
- 22. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 22 X6i 1 0 0 0 0 1 0 X2i 0 0 1 0 X2i 0 1 1 1 X5i 0 X1i 0 1 1 1 X5i 1 0 1 0 0 1 0 0 1 1 1 0 1 0 Crossover Crossover
- 23. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 23 I I Mutation represents a change in Mutation represents a change in the gene. the gene. I I Mutation is a background operator. Its role is to Mutation is a background operator. Its role is to provide a guarantee that provide a guarantee that the search algorithm is the search algorithm is not trapped on a local optimum. not trapped on a local optimum. I I The mutation operator flips a randomly selected The mutation operator flips a randomly selected gene in a chromosome. gene in a chromosome. I I The mutation probability is quite small in nature, The mutation probability is quite small in nature, and is kept low for and is kept low for GAs GAs, typically in the range , typically in the range between 0.001 and 0.01. between 0.001 and 0.01. Mutation operator Mutation operator
- 24. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 24 Mutation Mutation 0 1 1 1 X5'i 0 1 0 X6'i 1 0 0 0 0 1 0 X2'i 0 1 0 0 0 1 1 1 1 X5i 1 1 1 X1"i 1 1 X2"i 0 1 0 0 X1'i 1 1 1 0 1 0 X2i
- 25. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 25 The The genetic algorithm cycle genetic algorithm cycle 1 0 1 0 X1i Generation i 0 0 1 0 X2i 0 0 0 1 X3i 1 1 1 0 X4i 0 1 1 1 X5i f = 56 1 0 0 1 X6i f = 54 f = 36 f = 44 f = 14 f = 14 1 0 0 0 X1i+1 Generation (i + 1) 0 0 1 1 X2i+1 1 1 0 1 X3i+1 0 0 1 0 X4i+1 0 1 1 0 X5i+1 f = 54 0 1 1 1 X6i+1 f = 56 f = 56 f = 50 f = 44 f = 44 Crossover X6i 1 0 0 0 0 1 0 X2i 0 0 1 0 X2i 0 1 1 1 X5i 0 X1i 0 1 1 1 X5i 1 0 1 0 0 1 0 0 1 1 1 0 1 0 Mutation 0 1 1 1 X5'i 0 1 0 X6'i 1 0 0 0 0 1 0 X2'i 0 1 0 0 0 1 1 1 1 X5i 1 1 1 X1"i 1 1 X2"i 0 1 0 0 X1'i 1 1 1 0 1 0 X2i
- 26. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 26 Genetic Genetic algorithms algorithms: : case case study study I I Suppose it is desired to find the maximum of the Suppose it is desired to find the maximum of the “ “peak peak” ” function of two variables: function of two variables: where parameters where parameters x x and and y y vary between vary between − −3 and 3. 3 and 3. I I The first step is to represent the problem variables The first step is to represent the problem variables as a as a chromosome chromosome − − parameters parameters x x and and y y as a as a concatenated binary string: concatenated binary string: 2 2 2 2 ) ( ) 1 ( ) , ( 3 3 ) 1 ( 2 y x y x e y x x e x y x f − − + − − − − − − = 1 0 0 0 1 1 0 0 0 1 0 1 1 1 0 1 y x
- 27. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 27 I I We also choose We also choose the size of the chromosome the size of the chromosome population, for instance 6, and randomly generate population, for instance 6, and randomly generate an initial population. an initial population. I I The next step is to calculate the fitness of each The next step is to calculate the fitness of each chromosome. This is done in two stages. chromosome. This is done in two stages. I I First, a chromosome, that is a string of 16 bits, is First, a chromosome, that is a string of 16 bits, is partitioned into two 8 partitioned into two 8- -bit strings: bit strings: 10 0 1 2 3 4 5 6 7 2 ) 138 ( 2 0 2 1 2 0 2 1 2 0 2 0 2 0 2 1 ) 10001010 ( = × + × + × + × + × + × + × + × = and 10 0 1 2 3 4 5 6 7 2 ) 59 ( 2 1 2 1 2 0 2 1 2 1 2 1 2 0 2 0 ) 00111011 ( = × + × + × + × + × + × + × + × = I I Then these strings are converted from binary Then these strings are converted from binary (base 2) to decimal (base 10): (base 2) to decimal (base 10): 1 0 0 0 1 1 0 0 0 1 0 1 1 1 0 1 and
- 28. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 28 I I Now the range of integers that can be handled by Now the range of integers that can be handled by 8 8- -bits, that is the range from 0 to (2 bits, that is the range from 0 to (28 8 − − 1), is 1), is mapped to the actual range of parameters mapped to the actual range of parameters x x and and y y, , that is the range from that is the range from − −3 to 3: 3 to 3: I I To To obtain the actual values of obtain the actual values of x x and and y y, we multiply , we multiply their decimal values by 0.0235294 and subtract 3 their decimal values by 0.0235294 and subtract 3 from the results: from the results: 0235294 . 0 1 256 6 = − 2470588 . 0 3 0235294 . 0 ) 138 ( 10 = − × = x and 6117647 . 1 3 0235294 . 0 ) 59 ( 10 − = − × = y
- 29. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 29 I I Using decoded values of Using decoded values of x x and and y y as inputs in the as inputs in the mathematical function, the GA calculates the mathematical function, the GA calculates the fitness of each chromosome. fitness of each chromosome. I I To To find the maximum of the find the maximum of the “ “peak peak” ” function, we function, we will use crossover with the probability equal to 0.7 will use crossover with the probability equal to 0.7 and mutation with the probability equal to 0.001. and mutation with the probability equal to 0.001. As we mentioned earlier, a common practice in As we mentioned earlier, a common practice in GAs GAs is to specify the number of generations. is to specify the number of generations. Suppose the desired number of generations is 100. Suppose the desired number of generations is 100. That is, the GA will create 100 generations of 6 That is, the GA will create 100 generations of 6 chromosomes before stopping. chromosomes before stopping.
- 30. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 30 Chromosome locations on the surface of the Chromosome locations on the surface of the “ “peak peak” ” function: initial population function: initial population
- 31. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 31 Chromosome locations on the surface of the Chromosome locations on the surface of the “ “peak peak” ” function: first generation function: first generation
- 32. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 32 Chromosome locations on the surface of the Chromosome locations on the surface of the “ “peak peak” ” function: local maximum function: local maximum
- 33. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 33 Chromosome locations on the surface of the Chromosome locations on the surface of the “ “peak peak” ” function: global maximum function: global maximum
- 34. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 34 Performance graphs for 100 generations of 6 Performance graphs for 100 generations of 6 chromosomes chromosomes: : local maximum local maximum pc = 0.7, pm = 0.001 G e n e r a t i o n s Best Average 80 90 100 60 70 40 50 20 30 10 0 -0.1 0.5 0.6 0.7 F i t n e s s 0 0.1 0.2 0.3 0.4
- 35. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 35 Performance graphs for 100 generations of 6 Performance graphs for 100 generations of 6 chromosomes chromosomes: : global maximum global maximum Best Average 100 G e n e r a t i o n s 80 90 60 70 40 50 20 30 10 pc = 0.7, pm = 0.01 1.8 F i t n e s s 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6
- 36. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 36 pc = 0.7, pm = 0.001 Best Average 20 G e n e r a t i o n s 16 18 12 14 8 10 4 6 2 0 F i t n e s s 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 Performance graphs for Performance graphs for 20 20 generations of generations of 60 60 chromosomes chromosomes
- 37. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 37 Case Case study study: : maintenance maintenance scheduling scheduling I I Maintenance scheduling problems are usually Maintenance scheduling problems are usually solved using a combination of search techniques solved using a combination of search techniques and and heuristics. heuristics. I I These problems are complex and difficult to These problems are complex and difficult to solve. solve. I I They are NP They are NP- -complete and cannot be solved by complete and cannot be solved by combinatorial search combinatorial search techniques. techniques. I I Scheduling involves competition for limited Scheduling involves competition for limited resources, and is complicated by a great number resources, and is complicated by a great number of badly formalised constraints. of badly formalised constraints.
- 38. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 38 Steps in the GA development Steps in the GA development 1. Specify 1. Specify the problem, define constraints the problem, define constraints and and optimum optimum criteria; criteria; 2 2. Represent . Represent the problem domain as the problem domain as a a chromosome chromosome; ; 3. 3. Define Define a fitness function to evaluate the a fitness function to evaluate the chromosome chromosome performance; performance; 4. Construct 4. Construct the genetic the genetic operators; operators; 5. Run 5. Run the GA and tune its the GA and tune its parameters. parameters.
- 39. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 39 Case Case study study Scheduling of 7 units in 4 equal intervals Scheduling of 7 units in 4 equal intervals The problem constraints: The problem constraints: I I The maximum loads expected during four intervals are The maximum loads expected during four intervals are 80, 90, 65 and 70 MW; 80, 90, 65 and 70 MW; I I Maintenance of any unit starts at the beginning of an Maintenance of any unit starts at the beginning of an interval and finishes at the end of the same or adjacent interval and finishes at the end of the same or adjacent interval. The maintenance cannot be aborted or finished interval. The maintenance cannot be aborted or finished earlier than scheduled; earlier than scheduled; I I The net reserve of the power system must be greater or The net reserve of the power system must be greater or equal to zero at any interval. equal to zero at any interval. The optimum criterion is the maximum of the net The optimum criterion is the maximum of the net reserve at any maintenance period. reserve at any maintenance period.
- 40. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 40 Case Case study study Unit data and maintenance requirements Unit data and maintenance requirements Unit number Unit capacity, MW Number of intervals required for unit maintenance 1 20 2 2 15 2 3 35 1 4 40 1 5 15 1 6 15 1 7 10 1
- 41. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 41 Case Case study study Unit gene pools Unit gene pools Unit 1: 1 0 1 0 0 1 1 0 0 1 0 1 Unit 2: 1 0 1 0 0 1 1 0 0 1 0 1 Unit 3: 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 Unit 4: 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 Unit 5: 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 Unit 6: 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 Unit 7: 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 Chromosome for the scheduling problem Chromosome for the scheduling problem 0 1 1 0 0 1 0 1 0 0 0 1 1 0 0 0 0 0 1 0 0 1 0 0 1 0 0 0 Unit 1 Unit 3 Unit 2 Unit 4 Unit 6 Unit 5 Unit 7
- 42. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 42 Case Case study study The crossover operator The crossover operator 0 1 1 0 0 1 0 1 0 0 0 1 1 0 0 0 0 0 1 0 0 1 0 0 1 0 0 0 Parent 1 1 0 1 0 0 1 1 0 0 0 1 0 0 0 0 1 0 1 0 0 1 0 0 0 0 0 1 0 Parent 2 0 1 1 0 0 1 0 1 0 0 0 1 1 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 Child 1 1 0 1 0 0 1 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 Child 2
- 43. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 43 Case Case study study The mutation operator The mutation operator 1 0 1 0 0 1 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 0 0 1 0 1 0 1 0 0 1 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 1
- 44. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 44 Performance Performance graphs and the best maintenance graphs and the best maintenance schedules created in a population of 20 chromosomes schedules created in a population of 20 chromosomes G e n e r a t i o n s 0 30 60 90 120 150 Unit 2 Unit 2 Unit 7 Unit 1 Unit 6 Unit 1 Unit 3 Unit 5 Unit 4 1 2 3 4 T i m e i n t e r v a l M W N e t r e s e r v e s: 15 35 35 25 N = 20, pc = 0.7, pm = 0.001 Best Average 5 15 25 35 10 30 20 40 45 50 G e n e r a t i o n s 0 F i t n e s s -10 -5 0 5 10 15 ( (a a) 50 ) 50 generations generations
- 45. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 45 Performance Performance graphs and the best maintenance graphs and the best maintenance schedules created in a population of 20 chromosomes schedules created in a population of 20 chromosomes G e n e r a t i o n s 0 30 60 1 2 3 4 T i m e i n t e r v a l M W N e t r e s e r v e s : 40 25 20 25 Unit 2 Unit 2 Unit 7 Unit 1 Unit 6 Unit 1 Unit 3 Unit 5 Unit 4 90 120 150 N = 20, pc = 0.7, pm = 0.001 Best Average 10 30 50 70 20 0 60 40 80 90 100 F i t n e s s -10 0 10 20 G e n e r a t i o n s ( (b b) ) 100 100 generations generations
- 46. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 46 Performance Performance graphs and the best maintenance graphs and the best maintenance schedules created in a population of schedules created in a population of 100 100 chromosomes chromosomes ( (a a) ) Mutation rate is 0.001 Mutation rate is 0.001 1 2 3 4 T i m e i n t e r v a l 35 25 25 25 Unit 2 Unit 2 Unit 7 Unit 1 Unit 6 Unit 1 Unit 3 Unit 5 Unit 4 0 30 60 M W 90 120 150 N = 100, pc = 0.7, pm = 0.001 G e n e r a t i o n s N e t r e s e r v e s : Best Average 10 30 50 70 20 0 60 40 80 90 100 -10 F i t n e s s 0 10 20 30
- 47. Negnevitsky, PearsonEducation, 2011 Negnevitsky, Pearson Education, 2011 47 Performance Performance graphs and the best maintenance graphs and the best maintenance schedules created in a population of schedules created in a population of 100 100 chromosomes chromosomes ( (b b) ) Mutation rate is 0.01 Mutation rate is 0.01 1 2 3 4 T i m e i n t e r v a l N e t r e s e r v e s : 25 25 30 30 Unit 2 Unit 2 Unit 7 Unit 6 Unit 1 Unit 3 Unit 5 Unit 4 0 30 60 M W 90 120 150 Unit 1 N = 100, pc = 0.7, pm = 0.01 G e n e r a t i o n s 10 30 50 70 20 0 60 40 80 90 100 Best Average F i t n e s s -20 -10 0 10 30 20