GENETIC ALGORITHM FOR CELL DESIGN UNDER SINGLE AND MULTIPLE PERIODS A genetic algorithm is a random search technique for global optimisation in a complex search space. It was originally inspired by an analogy with the process of natural evolution. In evolution, the problem that each species faces is one of searching for beneficial adaptations to a complicated and changing environment. The knowledge that each species has gained is embodied in the make-up of the chromosomes of its members. Genetic operations are used in the search for beneficial adaptation when the parents reproduce and evolution takes place. Likewise, a genetic algorithm combines the survival-of-the-fittest among solution structures with a structured, yet randomised, information exchange and creates offsprings. The offsprings are new sets of solutions using bits and pieces of the fittest of the old solution structures and it displaces weak solutions during each generation [Singh and Rajamani (1996)]. Robust performance, and balance between efficiency and efficacy necessary for survival in many different environments are expected from genetic algorithms. Genetic algorithms work by maintaining a population of possible solutions to the problem. A candidate solution (a point in the search space) is represented by a sequence of genes and is known as a chromosome. A judiciously selected set of chromosomes is called a population and the population at a given time is a generation. The algorithm begins with an initial population of solutions whose potential as solution is determined by its fitness function, which evaluates a chromosome with respect to the objective function of the optimisation problem at hand. The population size remains from generation to generation and has a significant impact on the performance of the genetic algorithm [Gupta et al. (1996)]. The next generation of population is obtained by first selecting parent structures (chromosomes) in proportion to their fitness from the current generation. This operation is called reproduction. By selecting fit parents it is hoped that desirable solution characteristics will be repeated in future generations while undesirable characteristics die out. The fit chromosomes are subjected to operations such as cross over and mutation. The chromosomes resulting from these three operations are often known as offsprings or children. The fitness of the children is measured and the fittest GA Cell/part family (Single and multi periods) 85 March 2013
replaces the old population, and thus new generation is created. This procedure repeats until a stopping criterion is reached. The genetic algorithm design issues such as representation, initialisation, fitness function, reproduction, cross over, and mutation are discussed in the following sections. Representation Unlike more traditional optimisation techniques, genetic algorithms work with a coding of the decision variables themselves. Representation by coding a solution in the form of strings (chromosomes) plays a key role in the development of genetic algorithm. The bits (genes) in the strings could be binary, integers or a combination of characters. The chromosomal structure needs to code the key features of the problem in such way that good solution characteristics are repeated in subsequent generation while undesirable characteristics die out. In the machine cell-part family formation problem of single period, each gene represents a cell number and the positioning of the gene in the chromosome represents the machine number. The length of the chromosome represents the number of machines considered. For example, when the string 32123 represents a three-cell solution with the following machines in each cell: Cell = 1: machine 3 Cell = 2: machines 2, 4 Cell = 3: machines 1, 5. The chromosome used in the multi-period PF/MC formation genetic algorithm is divided into periods, since the composition of machine cells may be different in each period due to machine relocation [Wicks and Reasor (1999), and Pillai and Subbarao (2008)]. Within each period, the chromosome is further divided into cells, representing the composition of each cell in terms of machine type. {Period 1 assignment} {Period 2 assignment} {Period P assignment} {[Cell 1][Cell 2] [Cell C]} {[Cell1][Cell 2] [Cell C]} {[Cell1][Cell 2] [Cell C]} GA Cell/part family (Single and multi periods) 86 March 2013
In each cell sub-string, the string position corresponds to a machine type, and the value in the position (either 1 or 0) indicates whether or not the machine type has been assigned to that cell [Cell k] [ y y y ] 1 kl 2 kl Mkl The length of the chromosome is M C P where M is the number of machine types, C is the number of cells, and P is the number of periods in the planning horizon. Initialisation The initialisation step in genetic algorithm is to create an initial population. The initialisation process can be executed with either a randomly created population or a welladapted population. The appropriate population size, the length of a chromosome, and the number of cells are to be chosen according to the design requirements. The population size depends on the length of a chromosome. In general, the population size should be at least equal to the length of a chromosome. The chromosomes in the population are decoded and it is evaluated according to the objective function. Evaluation or Fitness Function The purpose of evaluation is to measure the fitness of candidate solutions in the population with respect to the design objectives. The fitness values are used to select parent solutions to create the next generation of solutions. The specific form of evaluation function depends on the design objective being considered. The fitness of an individual solution dictates the number of copies of that solution in the mating pool. The more copies an individual receives, the greater is the probability that the characteristics will be repeated in subsequent generations. In a genetic algorithm a fitness function value is computed for each string in the population with the objective of finding a string with maximum value. The objective of the cell formation problem is the minimisation of the inter cell traffic. However, genetic GA Cell/part family (Single and multi periods) 87 March 2013
algorithm works with maximisation functions. Thus, it is necessary to map this objective function to a fitness function. A transformation is: F T T i < T i = max T i, where max = 0, otherwise where, F is the fitness function of string i, i T is the objective function value of a string i i and T is the largest objective function value in the current generation. max Another transformation is: F i T = T min i where, T min is the smallest objective function value in the current generation. If the minimum objective function value is equal to zero, its value is set to 0.1. Now, the maximisation of the fitness function indirectly means the minimisation of the objective function. Some of the chromosomes produced by the genetic algorithm may be illegal (infeasible) as the objective function has constraints. The illegal chromosomes are repaired to satisfy the constraints. Selection and Reproduction A generation of the genetic algorithm begins with reproduction. The reproduction operator is used to select individuals from the current population to become parents of the next generation. Parents are selected according to their fitness value. Strings with higher fitness values are selected for crossover and mutation. A selection process is remainder stochastic sampling without replacement [Goldberg (1989)]. According to this method the probability of selection is calculated as follows: p s = Fi F i where, F is the fitness function of string i. Then the expected number of individuals for i each string is ei which is calculated as follows: GA Cell/part family (Single and multi periods) 88 March 2013
where, ppsz is the population size. ei = ps ppsz In the process of creating a next generation, the genetic algorithm creates an intermediate population of size equal to ppsz. In the intermediate population, each string from the old population has copies equal to the integer part of the e i values, and the fractional parts of the e i are treated as probabilities. If the intermediate population is not filled up, one by one, weighted coin tosses (Bernoulli trials) are performed using the fractional parts as success probabilities. For example, if the expected value ( e i ) of a string is 2.3, the intermediate population will receive two copies of the string. This is done for all the strings of the old population till the intermediate population gets filled up. If the intermediate population is not filled up completely, the fractional part of the expected value of a string is treated as probabilities and by performing Bernoulli trials the string is copied to the intermediate population. Crossover The crossover operator creates new potential solutions by exchanging a portion of the parent solutions available in the intermediate population. The crossover is done with a probability called crossover probability. In this way, child retains some of the features of the parent solution. A selection process for chromosomes to be crossed is stochastic sampling with replacement. The number of chromosomes selected is equal to population size and the crossing points are identified randomly. Several crossover operators are available and two such operators are single-point and two-point operators. The single-point crossover operator randomly generates a single crossover point along the length of the chromosome. This cross over point divides each of the parent chromosomes into two segments. The children are created by swapping the second segments of the parents. The first child consists of segment 1 of parent 1 and segment 2 of parent 2. The second child consists of the remaining segments: segment 1 of parent 2 and segment 2 of parent 1. The two-point crossover operator randomly generates GA Cell/part family (Single and multi periods) 89 March 2013
two crossover points along the length of the chromosome, dividing each parent into three segments. The two children are created by exchanging segment 2 between parents. The probability of selection in the stochastic sampling with replacement is calculated as follows: p s = Fi F i where, F is the fitness function for a string i. Calculate the cumulative value of the i probability. Generate a random number and identify the chromosome corresponding to the random number. Similarly identify the next chromosome and based on the probability of crossover these two chromosomes are crossed. The crossing is explained with an example below. Let the parent identified be Parent 1: 1 3 2 1 2 1 3 2 2 3 Parent 2: 2 1 3 2 3 2 1 2 1 3 and the randomly identified crossing points be 2 and 7 for two point crossing. The crossing points are shown in the parents given above. The crossover operator generates two children given below. Child 1: 1 3 3 2 3 2 1 2 2 3 Child 2: 2 1 2 1 2 1 3 2 1 3 For the multi period problem, in addition to the above cross over operators the problem specific two crossover operators possible are cell-swap operator and period-swap operator. Cell-swap crossover operator randomly chooses a cell number and a period number. The children are created by exchanging the machine assignments in the selected cell within the selected period between the parents. The period-swap crossover operator randomly selects a period in the planning horizon and creates the children by exchanging the machine assignments for the entire selected period between the parents. These operators are very similar to the two-point crossover operator. GA Cell/part family (Single and multi periods) 90 March 2013
Mutation The purpose of mutation operator is to rejuvenate the search and extend it into previously unexplored areas of the solution space. Mutation prevents the value of any parameter from remaining unchanged forever. It is done with low probability called probability of mutation. In a single period problem for the mutation purpose two random integers r 1 and r 2 are generated such that 1 r 1 m and 1 r w where w is the maximum number of cells specified and m is the number of machines. In the mutation operation the cell number corresponding to the machine specified by r 1 is replaced with the cell number r 2. In the multi-period problem, when mutation occurs, parent values of 1 are changed to 0 in the child chromosome and parent values of 0 are changed to 1. To identify the mutation point, choose randomly a cell and a period, and locate a point randomly in the cell identified. Replacement Strategy 2 After genetic operations new strings are created. Poor performing offsprings are replaced in the new generation with a replacement strategy. The offsprings are evaluated with respect to the evaluation function. The goal of the replacement strategy is to create generations of solutions that, on an average, outperform the previous generation. This is accomplished by restricting admittance to the new population to only those children that are better than members of the current population. This can be accomplished by comparing the fitness values of the chromosomes of the old population with the chromosomes of the new population. The fittest among the two will become the old population for the next generation. Terminating the Genetic Algorithm The genetic algorithm iterates, and as the process proceeds, the generation includes chromosomes with higher fitness function values. Termination criterion is used to stop GA Cell/part family (Single and multi periods) 91 March 2013
the iteration. A single criterion or a set of criteria can be used to halt the genetic algorithm. Three termination criteria are test of population convergence, monitoring of improvement from generation to generation, and maximum number of generations. Algorithm The abbreviations used in the algorithm given below are as follows: ppsz Population size xgen Maximum number of generation p cs - Probability of crossover P mut - Probability of mutation POP(g) - Population in generation g Step 0. Set the values for ppsz, xgen, p, cs P and constraints related parameters. mut Initialise the generation counter, g = 0 Step 1. Randomly generate an initial population, POP(g). Step 2. For each potential machine assignment solution, apply the part assignment heuristic to form the part family for each cell. Step 3. Evaluate the fitness of each solution with respect to the design evaluation function. Step 4. Select individuals from the current population to become parents of the next generation according to their fitness values. Step 5. Create children by applying the genetic operators of crossover and mutation. Step 6. Evaluate the fitness of each child with respect to the design evaluation function. Step 7. Increment the generation counter, g = g + 1. Step 8. Form the new generation, POP(g), by replacing the weak solutions in population POP( g 1) with more fit children. Step 9. If the termination criterion has not been met, go to step 4. Reference 1. Goldberg, D., E., 1989, Genetic algorithm in search, optimisation and machine learning. (Addison-Wesley Longman, Inc.). GA Cell/part family (Single and multi periods) 92 March 2013
2. Gupta, Y., Gupta, M., Kumar, A., and Sundaram, C., 1996, A genetic algorithmbased approach to cell composition and layout design problems. International Journal of Production Research, 34, 447-482. 3. Pillai, V. M., and Subbarao, K., (2008), A robust cellular manufacturing system design for dynamic part population using a genetic algorithm. International Journal of Production Research. Vol.46, Issue 18, pp 5191-5210 4. Singh, N., and Rajamani, D., 1996, Cellular manufacturing system: design, planning and control. (Chapman & Hall). 5. Wicks, E. M., and Reasor, R. J., 1999, Designing cellular manufacturing systems with dynamic part populations. IIE Transactions, 31, 11-20. QUESTIONS: 1. What is replacement strategy in a genetic algorithm? After which operation this strategy is carried out? 2. Two chromosomes for the cell design problem are given below. Parent 1: 1 3 2 1 2 1 3 2 2 1 Parent 2: 12 3 2 3 2 1 2 1 3 Demonstrate single point crossover and two point crossover. Take necessary random numbers from the table given below. Random No. Table: Random numbers 1 2 3 4 5 6 7 8 9 10 0.6154 0.9218 0.7919 0.9218 0.7382 0.1763 0.4057 0.9355 0.9169 0.4103 3. Prepare a binary string, which represents three-cell single period cell formation problem. The manufacturing system to be grouped contains 6 machines. Describe twopoint crossover for the above type of string. 4. Prepare a binary string, which represents three-cell single period cell formation problem. The manufacturing system to be grouped contains 8 machines. Take two strings of the above type and show the result of two-point crossover. Two uniform random numbers between 0 and 25 are 7 and 18. Using this random numbers while doing cross over operation. 5. Develop a feasible machine/part grouping for the data given below: Number of cells = 2 Minimum number of machine types per cell = 3 Minimum number of parts per family = 4 GA Cell/part family (Single and multi periods) 93 March 2013
Assume that a machine type is allowed in only one cell. The matrix below shows the operation sequence. For example, the operation sequence of part 1 is 1-2-3 and the corresponding machine sequence is 6-1-3. That is, the machine type used for the first operation of part 1 is 6. Similarly, machine types required for operations 2 and 3 can be established. Part Machine type 1 2 3 4 5 6 7 8 1 2 1 0 0 1 0 0 3 2 0 0 1 2 0 2 1 0 3 3 3 0 0 2 1 0 0 4 0 0 3 1 0 0 2 2 5 0 0 2 3 0 0 3 1 6 1 2 0 0 0 3 0 0 Also determine the inter-cell moves for the grouping you have identified when unit quantity of the parts is produced. If genetic algorithm is used as a solution procedure to this problem, show the chromosome representation for the feasible solution you have identified. 6. A cellular manufacturing system consists of 6 machines and 10 parts. The machines have been designated with numbers 1, 2,..., 6. The system has to be partitioned into two cells. It is required that a machine type should be present in a single cell alone. More than 4 machine types are not allowed in a cell. When genetic algorithm is used as a solution procedure to this problem, two chromosomes generated are given below. Decode the chromosomes. Check the feasibility of the chromosome. If the chromosome is not feasible under the given data, convert it into a feasible chromosome and describe the logic of repair. If any random numbers are required for this purpose use the numbers given in Table of random numbers. Indicate the genes modified. Using the first chromosome given below, do bitwise mutation operation. The mutation probability is 0.06. 1 2 2 2 1 2 2 2 2 1 2 2 If random numbers are required for the mutation operation, use the numbers (exclude the numbers previously used) from Table of random numbers. 0.9501 0.2311 0.6068 0.0185 0.8214 0.4447 0.6154 0.7919 0.9218 Table of random Numbers 7. A multi-period cellular manufacturing system consists of 6 machines and 10 parts. The machines have been designated with numbers 1, 2... 6. The number of period is three. The system has to be partitioned into two cells. It is required that a machine type should be present in a single cell alone. More than 4 machine types are not allowed in GA Cell/part family (Single and multi periods) 94 March 2013
a cell. When genetic algorithm is used as a solution procedure to this problem, two chromosomes generated are given below. Decode the chromosomes. Check the feasibility of the chromosome. If the chromosome is not feasible under the given data, convert it into a feasible chromosome and describe the logic of repair. Indicate the genes modified. Using the first chromosome given below, do bitwise mutation operation. The mutation probability is 0.06. Position of bit 1 2 3 4 5 6 7 8 9 101 12131415161718 Chromosome 1 1 2 2 2 1 2 2 1 2 2 1 1 2 1 2 1 1 2 Chromosome 2 2 2 2 1 2 2 1 1 2 2 1 2 2 2 2 1 1 2 If random numbers are required for the above problems, use the following numbers given below. 0.9501 0.2311 0.6068 0.4860 0.8913 0.7621 0.4565 0.0185 0.8214 0.4447 0.6154 0.7919 0.9218 0.7382 0.1763 0.4057 0.9355 0.9169 0.4103 0.8936 8. A multi-period cellular manufacturing system consists of 6 machines and 10 parts. The number of period is three. The system has to be partitioned into two cells. When genetic algorithm is used as a solution procedure to this problem, two chromosomes generated are given below. These chromosomes have to be crossed over using twopoint crossover operation. Carry out the crossover operation. Use the given random numbers to identify the crossover site. Given random numbers are 0.9501 and 0.2311. Check the feasibility of the off springs considering that all the machines are assigned in every period. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 1 1 1 1 0 1 0 1 0 1 0 1 0 0 1 0 1 1 1 0 1 1 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 0 1 1 0 0 0 1 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0 1 0 0 0 1 1 Also, carry out mutation on one of the off springs. You are provided with three random numbers such as: 0.6068, 0.4860 and 0.8913. (Note: Use appropriate method such that only one of the genes in a chromosome should be mutated.) 9. A production system consists of 6 types of machines, which is to be converted into a cellular manufacturing system. Develop a (feasible and efficient) chromosome for the following cases: (i) Part population changes from period to period and a machine type are allowed be present in a single cell alone. The planning horizon contains 3 periods. (ii) The planning horizon contains a single deterministic product mix. A machine type is allowed to be present in more than one cell. (iii) The planning horizon contains a single deterministic product mix. A machine type is allowed to be present in a single cell alone. to GA Cell/part family (Single and multi periods) 95 March 2013
Following conditions are valid for all the above cases: Number of cells is 2. The maximum number of machine types in cell is 4 and minimum is 2. For all these cases show the machine assignment to cells. 10. A production system consists of 8 types of machines, which is to be converted into a cellular manufacturing system. Develop a chromosome for the following cases (a) Part population changes from period to period. The planning horizon contains 4 periods. (b) Deterministic product mix is considered. Number of cells in both cases is three. 11. A multi-period cellular manufacturing system consists of 6 machines and 10 parts. The machines have been designated with numbers 1, 2... 6. The number of period is three. The system has to be partitioned into two cells. When genetic algorithm is used as a solution procedure to this problem, two chromosomes generated are given below. Decode the first chromosome. Check the feasibility of the chromosome. If the chromosome is not feasible under the given data, convert it into a feasible chromosome. Indicate the genes modified. Using the two chromosomes given below, do the two-point crossover operation. Random No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 1 1 1 1 0 1 0 1 0 1 0 1 0 0 1 0 1 1 1 0 1 1 1 1 1 1 0 0 0 0 1 1 1 1 0 0 First Chromosome 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 0 0 1 1 1 0 1 1 0 0 0 1 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0 1 0 0 0 1 1 Second Chromosome If random numbers are required for the above problems to workout, use the following numbers given in the table below. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0.9501 0.2311 0.6068 0.4860 0.8913 0.7621 0.4565 0.0185 0.8214 0.4447 0.6154 0.7919 0.9218 0.7382 12. Ten chromosomes (population size for a genetic algorithm) and its objective function value (total system cost) are given in the Tables 1 and 2. Generate an intermediate mating pool using remainder stochastic sampling without replacement method. You have to identify the number of copies of a chromosome that has been copied to mating pool. Use the random number given in Table 3. (A random number has to be used only one time.) GA Cell/part family (Single and multi periods) 96 March 2013
Chromosome No. Chromosomes and its cost Table 1. Randomly generated chromosomes 10 numbers Bit position 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 1. 1 0 0 1 0 0 1 1 1 1 1 1 0 0 1 1 1 1 1 0 0 1 1 0 0 0 1 0 1 1 0 1 1 0 0 0 2. 1 1 1 1 0 0 0 0 1 0 1 0 1 0 0 0 0 1 1 1 0 1 1 1 0 1 1 0 1 0 0 0 1 1 1 0 3. 1 1 1 1 0 1 0 1 0 1 0 1 0 0 1 0 1 1 1 0 1 1 1 1 1 1 0 0 0 0 1 1 1 1 0 0 4. 0 0 1 1 1 0 1 1 0 0 0 1 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0 1 0 0 0 1 1 5. 0 0 0 0 1 1 1 1 1 1 1 0 1 0 0 0 0 1 1 1 1 1 0 1 1 0 1 0 0 0 1 0 1 0 1 0 6. 0 1 1 0 1 0 1 0 1 0 1 0 0 0 1 1 1 1 1 0 0 1 1 1 0 0 1 0 0 1 0 1 1 1 1 1 7. 1 1 0 0 1 1 1 0 1 1 0 1 0 1 0 1 0 0 0 1 0 1 0 0 0 1 1 1 0 1 1 0 0 1 1 1 8. 1 0 1 1 1 1 0 1 0 1 1 0 1 0 0 1 0 0 0 1 1 0 1 1 1 1 0 1 1 0 1 0 0 0 1 0 9. 0 0 0 1 0 1 1 0 0 1 1 1 0 1 1 1 0 0 0 0 0 1 1 0 1 1 1 0 0 1 0 0 0 0 1 1 10. 1 0 1 0 1 1 0 0 1 1 0 0 0 1 0 1 0 1 0 1 1 1 1 1 1 0 1 1 0 1 0 0 0 1 1 1 Table 2. Total System Cost Chromosome No. 1 2 3 4 5 6 7 8 9 10 Costs 101800 104350 102150 110500 87350 104500 116250 90400 95300 107250 Table 3. Random Numbers Sl. No. 1 2 3 4 5 6 7 8 9 10 Random No..9501.2311.6068 0.4860.8913 0.7621 0.4565 0.0185 0.8214 0.4447 Sl. No. 11 12 13 14 15 16 17 18 19 20 Random No. 0.6154 0.7919 0.9218 0.7382 0.1763 0.4057 0.9355 0.9169 0.4103 0.8936 13. Ten chromosomes generated at random and their costs are given in Tables 1 5. For the first chromosome given in Table 1, identify machine types belonging to various cells in different periods. Write down the selection process from the chromosomes given in Table 1 using remainder stochastic sampling without replacements. Take first two chromosomes of Table 1 and demonstrate the two-point crossover. Using one of the chromosomes obtained in crossover operation, demonstrate the mutation operation (Clearly write down the steps indicating the random numbers used). GA Cell/part family (Single and multi periods) 97 March 2013
Table 1. Randomly generated chromosomes (10 numbers) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 1 1 0 0 1 0 0 1 1 1 1 1 1 0 0 1 1 1 1 1 0 0 1 1 0 0 0 1 0 1 1 0 1 1 0 0 0 2 1 1 1 1 0 0 0 0 1 0 1 0 1 0 0 0 0 1 1 1 0 1 1 1 0 1 1 0 1 0 0 0 1 1 1 0 3 1 1 1 1 0 1 0 1 0 1 0 1 0 0 1 0 1 1 1 0 1 1 1 1 1 1 0 0 0 0 1 1 1 1 0 0 4 0 0 1 1 1 0 1 1 0 0 0 1 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0 1 0 0 0 1 1 5 0 0 0 0 1 1 1 1 1 1 1 0 1 0 0 0 0 1 1 1 1 1 0 1 1 0 1 0 0 0 1 0 1 0 1 0 6 0 1 1 0 1 0 1 0 1 0 1 0 0 0 1 1 1 1 1 0 0 1 1 1 0 0 1 0 0 1 0 1 1 1 1 1 7 1 1 0 0 1 1 1 0 1 1 0 1 0 1 0 1 0 0 0 1 0 1 0 0 0 1 1 1 0 1 1 0 0 1 1 1 8 1 0 1 1 1 1 0 1 0 1 1 0 1 0 0 1 0 0 0 1 1 0 1 1 1 1 0 1 1 0 1 0 0 0 1 0 9 0 0 0 1 0 1 1 0 0 1 1 1 0 1 1 1 0 0 0 0 0 1 1 0 1 1 1 0 0 1 0 0 0 0 1 1 10 1 0 1 0 1 1 0 0 1 1 0 0 0 1 0 1 0 1 0 1 1 1 1 1 1 0 1 1 0 1 0 0 0 1 1 1 Table 2. Total Intercell Move Costs Chromosome No. 1 2 3 4 5 6 7 8 9 10 Costs 53100 48900 36750 63000 39150 50550 52050 47400 53850 50250 Table 3. Total Acquisition Costs Chromosome No. 1 2 3 4 5 6 7 8 9 10 Costs 41500 50500 52000 31000 36000 41000 52500 38000 30500 47000 Table 4. Total Relocation Cost Chromosome No. 1 2 3 4 5 6 7 8 9 10 Costs 7200 4950 13400 16500 12200 12950 11700 5000 10950 10000 Table 5. Total System Cost Chromosome No. 1 2 3 4 5 6 7 8 9 10 Costs 101800 104350 102150 110500 87350 104500 116250 90400 95300 107250 GA Cell/part family (Single and multi periods) 98 March 2013
14. Consider a production system with details given below: Number of cells (N c ) = 2 Minimum number of machine types per cell = 2 Minimum number of parts per family = 3 Assume that a machine type is allowed in only one cell. The matrix below shows the operation sequence. For example, the operation sequence of part 1 is 1-2-3 and the corresponding machine sequence is 6-3-1. That is, the machine type used for the first operation of part 1 is 6. Similarly, machine types required for operations 2 and 3 can be established. Machine type Part 1 2 3 4 5 6 7 8 1 3 1 0 0 1 0 0 3 2 0 0 1 2 0 2 1 0 3 2 3 0 0 2 1 0 0 4 0 0 3 1 0 0 2 2 5 0 0 2 3 0 0 3 1 6 1 2 0 0 0 3 0 0 A mathematical programming model is developed and solved. One feasible solution is as follows: σ x 11 11 Where, x ij σ Mj = 1, σ = 1, x 22 21 = 1, σ 31 = 1, σ 42 = 1, σ 52 = 1, σ 61 = 1 = 1, x = 1, x = 1, x = 1, x = 1, x 32 42 1 if part i is assigned to cell j = 0 Otherwise 1 if machine type M is assigned to cell j = 0 Otherwise 51 = 1, x a) Determine the cells and part families from the given solution. Develop the block diagonal structure of this solution. Identify an alternative solution that reduces the intercell movements. b) Write the constraints that will ensure the assignment of required machine to the right part. Use the following notations for developing the above constraints. M - Machine type required for the operation k of the part i ik i index of parts, i = 1,2,, j Index of cells, k - Index of operation, j = 1,2,, N p N c k = 1,2,, Ni GA Cell/part family (Single and multi periods) 99 March 2013 61 72 82 = 1
M index of machines, M = 1,2,, N m Random No. Ni - Number of operations required to complete the processing requirement of part i N p number of parts N m number of machines c) If genetic algorithm is used as a solution procedure to this problem, show the chromosome representation for the above feasible solution. d) If a machine type is allowed in more than one cell then, develop a chromosome for the above problem using random numbers given below. Table : Random numbers 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0.9501 0.2311 0.6068 0.4860 0.8913 0.7621 0.4565 0.0185 0.8214 0.4447 0.6154 0.7919 0.9218 0.7382 Analyze the feasibility of the chromosome and identify infeasibility, if any. 15. A production system has the following details: Number of machines = 6 and Number of parts = 8. Table 1: Capacity of machines Machine type 1 2 3 4 5 6 Capacity (time units) 8000 12000 13000 9000 10000 7000 Part number 1 2 3 4 5 6 7 8 Table 2: Part data for the problem Operation sequence machine(processing time) Demand 6 (5)-4(4)-5 (5)-1(4)-3(4)-2(4) 600 4(5)- 2 (4)- 3(5)-1(4) 800 5 (5)-3(4)-6(3)-1(6)-2(5)-4(4) 600 2(6)-3 (6)-1(4)-4(5)-5(4) 100 3 (5)-1 (6)-6(3)-4(4)- 5 (6)-2(5) 100 1(5)-4(4)-3(4)-2(3) 100 3 (4)-5(5)-4(4)-2(5)-1(4) 1000 2(4)-3(5)-4(4)-1(6) 200 The management decided to have cellular manufacturing system with the following details. Number of cells = 3, Minimum number of machine types in each cell = 2, Maximum number of machine type in a cell = 3, and Minimum number of parts in each cell = 2. A typical solution for the problem is as follows: C 1 = {1, 4, 5}, PF 1 = {4, 6, 7, 8} C 2 = {2, 3, 4}, PF 2 = {2, 3} GA Cell/part family (Single and multi periods) 100 March 2013
C 3 = {4, 5, 6}, PF 3 = {5, 1} C stands for cell and PF stands for part family. (a) If genetic algorithm is used as a solution procedure for this problem, code the given solution to get a chromosome. Develop another feasible chromosome for this problem and carryout two point crossover. Find out the crossover site randomly and use random numbers 0.950 and 0.231 for the site identification. (b) If this problem has varying demand and genetic algorithm is used as a solution procedure for the cell formation, generate a chromosome for the case given below. It has been found that if we divide the planning horizon into two periods, the demand in each period is approximately constant but demand in each period is different. A machine type is allowed to be present in more than one cell, if necessary. All other details are same as given above. For the generated chromosome, show the machines assigned to each cell. (c) Calculate the Jacards similarity between machines for the above problem. Calculate the similarity between the three cells given below using average linkage. C 1 = {1, 4}, C 2 = {2, 3} and C 3 = {5, 6}. 16. A production system has the following details: Number of machines = 6, Number of parts = 9, Intercell material handling cost = Rs 1 per unit per transfer. Table 1: Capacity of machines Machine type 1 2 3 4 5 6 Capacity (time units) 8000 12000 13000 9000 10000 7000 Table 2: Part data for the problem Part number 1 2 3 4 5 6 7 8 9 Operation sequence machine (processing time) 6 (5)-4(4)-5 (5)-3(4)-2(4) 4(5)- 2 (4)- 3(5)-1(4) 5 (5)-3(4)-6(3)-1(6)-2(5)-4(4) 2(6)-3 (6)-1(4)-4(5)-5(4) 3 (5)-1 (6)-6(3)-4(4)- 5 (6)-2(5) 1(4)-4(3) -2(6)-6(5)- 3(4) 1(5)-4(4)-3(4)-2(3) 3 (4)-5(5)-4(4)-2(5)-1(4) 2(4)-3(5)-4(4)-1(6) Demand 300 600 1000 300 800 1000 200 100 500 The management decided to have cellular manufacturing system with the following details. Number of cells = 3, Minimum number of machine types in each cell = 2, Maximum number of machine type in a cell = 3, and Minimum number of parts in each cell = 2. GA Cell/part family (Single and multi periods) 101 March 2013
A typical solution for the problem is as follows: C 1 = {1, 4, 5}, PF 1 = {4, 6, 7, 8, 9} C 2 = {2, 3, 4}, PF 2 = {2, 3} C 3 = {4, 5, 6}, PF 3 = {5, 1} C stands for cell and PF stands for part family. (d) What is the intercell material handling cost for above solution? How many units of machine 1 should be present in cell 1? (e) You may notice from the above solution that a type of machine is allowed to present in more than one cell. If a type of machine is allowed to present in one cell alone and genetic algorithm is used as a solution procedure for this problem, generate randomly two chromosomes. Carryout two point crossover. Find out the crossover site randomly. Use the first 18 random numbers (from Table 3) for chromosome generation and next two for crossover site identification. (f) If a mathematical programme is used to define cell formation and part assignment and if a feasible solution as given above is applicable, generate constraints that should ensure sufficient capacity for each machine type in a cell. Use the following notations for generating the constraints. i index of parts, i = 1,2,, N j index of machines, j = 1,2,, M k index of cells, k = 1,2,, C Di - Demand (production volume) for part i Tij C j x ik y jkl n jkl - Processing time of part i on machine type j - Capacity of machine type j 1 if part i is assigned to cell k = 0 Otherwise 1 if machine j is assigned to cell k during period l = 0 Otherwise - Number of type j machines assigned to cell k Table 3: Random numbers Sl. No. 1 2 3 4 5 6 7 8 9 10 Random No. 0.9501 0.2311 0.6068 0.4860 0.8913 0.7621 0.4565 0.0185 0.8214 0.4447 Sl. No. 11 12 13 14 15 16 17 18 19 20 Random No. 0.6154 0.7919 0.9218 0.7382 0.1763 0.4057 0.9355 0.9169 0.4103 0.8936 GA Cell/part family (Single and multi periods) 102 March 2013