Artificial Inteligence System Technique

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Examinations MODEL SOLUTIONS and MARKS Scheme Module Code: EE5614 Module Title: Intelligent Systems Examiner(s): Stationery: Special Requirements: Please start each question on a new sheet. Solutions may be hand written. Question No. 1 Marks a) fuzzy membership i). An interpretation of tallness is usually given by a nondecreasing monotonic function which is piece-wise linear: ii) An interpretation of aboutness can be approximated by using the symmetrical triangular membership function: where D>0 and m ϵ R iii). The in-betweeness condition is approximated by using the so-called trapezoidal function: b) The fuzzy membership functions are defined as follows: Rule base and inference Calculation need to be done plus defuzzification c). Cartesian product The results of the Cartesian product is directly as follows: 0.2 0.2 0.2 0.2 0.2 0.25 0.45 0.5 0.5 0.5 R = 0.25 0.45 0.6 0.6 0.6 0.25 0.45 0.75 0.8 0.8 0.25 0.45 0.75 0.8 0.9 d). Normalisation Normalisation of variables will allow: Use the intervals of the universe of discourse efficiently To generalise the rule base for a wide range of fuzzification Provide more flexibility for tuning. 1 1 1 2 1 4 1 6 3 T: 20 Question No. 2 Marks a) Airliner calculation B = {0.15, 0.15, 0.3, 0.4} (0.15 Ʌ 0.0) V (0.15 Ʌ 0.0) V (0.3 Ʌ 0.1) V (0.4 Ʌ 0.7) (0.15 Ʌ 0.7) V (0.15 Ʌ 0.8) V (0.3 Ʌ 0.5) V (0.4 Ʌ 0.2) (0.15 Ʌ 0.3) V (0.15 Ʌ 0.1) V (0.3 Ʌ 0.4) V (0.4 Ʌ 0.1) This results in (0.0 V 0.0 V 0.1 V 0.4) (0.15 V 0.15 V 0.3 V 0.2) (0.15 V 0.1 V 0.3 V 0.1) Hence for 737: 0.4 Sup 0.2 Eq 0.3 Def b). Calculations t=1 x1 = 7.5, x1 will activate N@0.7, and Z@0.3 x2=-150, x2 will activate N@0.5 and Z@0.5 The rule-base, only four rules will be activated R1: X1 is N@0.7 AND X2 is N@0.5, firing strength = 0.5 THEN V is PB@0.5 R2: X1 is N@0.7 AND X2 is Z@0.5, firing strength = 0.5 THEN V is P@0.5 R3: X1 is Z@0.3 AND X2 is N@0.5, firing strength = 0.3 THEN V is P@0.3 R4: X1 is Z@0.3 AND X2 is Z@0.5, firing strength = 0.3 THEN V is Z@0.3 Max = PB@0.5, P@0.5, Z@0.3 V = (20*0.5 + 10*0.5 + 0*0.3)/(0.5+0.5+0.3) = 15/1.3 = 11.6 t=2, x1(2) = -150*0.01 + 7.5 = 5 x2(2) = 17.5*11.3 + 0.36*-150 = 148 x1 = 5, x1 will activate P@0.5, and Z@0.5 x2=-150, x2 will activate P@0.5 and Z@0.5 The rule-base, only four rules will be activated R1: x1 is P@0.5 AND x2 is P@0.5, firing strength = 0.5 THEN V is NB@0.5 R2: x1 is P@0.5 AND x2 is Z@0.5, firing strength = 0.5 THEN V is Z@0.5 R3: x1 is Z@0.5 AND x2 is P@0.5, firing strength = 0.5 THEN V is N@0.5 R4: x1 is Z@0.5 AND x2 is Z@0.5, firing strength = 0.5 THEN V is Z@0.5 Max = NB@0.5, N@0.5, Z@0.5 V = (-20*0.5 + -10*0.5 + 0*0.5)/(0.5+0.5+0.5) = -15/1.5 = -10 t=3, x1(2) = 148*0.01 + 5 = 7.5 x2(2) = 17.5*-10 + 0.36*148 = -121 x1 = 7.5, x1 will activate P@0.75, and Z@0.25 x2=-121, x2 will activate N@0.25 and Z@0.75 The rule-base, only four rules will be activated R1: x1 is P@0.75 AND x2 is N@0.25, firing strength = 0.25 THEN V is Z@0.25 R2: x1 is P@0.75 AND x2 is Z@0.75, firing strength = 0.75 THEN V is Z@0.75 R3: x1 is Z@0.25 AND x2 is N@0.25, firing strength = 0.25 THEN V is P@0.25 R4: x1 is Z@0.25 AND x2 is Z@0.75, firing strength = 0.25 THEN V is Z@0.25 Max = P@0.25, Z@0.75 V = (10*0.25 + 0*0.75)/(0.25+0.75) = 2.5/1 = 2.5 c). A, B, C fuzzy sets Alpha cuts λ = 0.2 A = 4 ~ 5, B = 0 ~ 2.3, C = 0. ~ 6 5 λ = 0.6 A = 4.5 ~ 5, B = 0 ~ 0.7, C = 2. ~ 6 5 λ = 1.0 A = 0 ~ 0, B = 0 ~ 0, C = 0. ~ 0 3 3 1 1 1 1 1 1 1 1 3 3 T: 20 Please start each question on a new sheet. Solutions may be hand written. Question No. 3 Marks a) (i) Define the chromosome and fitness function for the function minimization problem. Each candidate solution has a set of properties (its chromosomes or genotype) which can be mutated and altered; traditionally. Chromosome: (x1,x2) Fitness function: the calculated value of . (ii) What is population? Explain the process of population construction for Evolutionary Strategy. A population of candidate solutions (called individuals, creatures, or phenotypes) to an optimization problem is evolved toward better solutions. The best chromosome from the population is saved separately. In the new population, the best chromosome is duplicated times. The size of population is . (iii) What are the different steps involve into generation? Define each step for the function minimization problem. Generation includes: 1. Evaluation of the population (Definition of the fitness function, calculated value of .) 2. Keep the best chromosome: compare the best chromosome of the population with saved separately the best chromosome. If chromosome with better fitness value is saved as best chromosome 3. Generate new population: copy the best chromosome 5 times into the new population. 4. Mutation – random change of the genes in the new population with frequency of mutation rate. The mutation is defined as generation randomly of new value for gene xi. 5. Evaluation of the population (Definition of the fitness function, calculated value of .) 6. Keep the best chromosome: compare the best chromosome of the population with saved separately the best chromosome. If chromosome with better fitness value is saved as best chromosome. 7. The algorithm is terminated? If not, get back to step 3. iv) Explain the difference between phenotype and genotype. Identify the phenotype and genotype for the function minimization problem. Every organism has a set of rules, a blueprint so to speak, describing how that organism is built up from the tiny building blocks of life. These rules are encoded in the genes of an organism, which in turn are connected together into long strings called chromosomes. Each gene represents a specific trait of the organism, like eye colour or hair colour, and has several different settings. For example, the settings for a hair colour gene may be blonde, black or auburn. These genes and their settings are usually referred to as an organism's genotype. The physical expression of the genotype - the organism itself - is called the phenotype. b) One-point crossover: A single crossover point on both parents' organism strings is selected. All data beyond that point in either organism string is swapped between the two parent organisms. The resulting organisms are the children: Bit string mutation: The mutation of bit strings ensue through bit flips at random positions. 3 4 6 3 2 2 T:20 Please start each question on a new sheet. Solutions may be hand written. Question No. 4 Marks a). Neural network Assuming that UP=1 and DOWN=0, and HOLD=0 and SELL=1, the student should show the following decision table: Index Price of Tonne of Copper Brown Recommends (OR) Angela Suggests (AND) Brian Argues (XOR) 0 0 0 0 0 0 1 1 0 1 1 0 1 0 1 1 1 1 1 0 Brown’s behaviour can be simulated using an AND network and Angela’s by an OR network. Brian’s behaviour follows that of an XOR network. Brown and Angel’s behaviour can be simulated using a single layer network but Brian’s behaviour requires a multi-layer perceptron: Brown’s Behaviour: w1= w2=1 Angela’s Behaviour: w1= w2=1 b). Neural network propagation In order to find the output of the network it is necessary to calculate weighted sums of hidden nodes 3 and 4: v3 = w13x1 + w23x2 , v4 = w14x1 + w24x2 Then find the outputs from hidden nodes using activation function ': y3 = (v3) , y4 = (v4) . Use the outputs of the hidden nodes y3 and y4 as the input values to the output layer (nodes 5 and 6), and find weighted sums of output nodes 5 and 6: v5 = w35y3 + w45y4 , v6 = w36y3 + w46y4 . Finally, find the outputs from nodes 5 and 6 (also using ): y5 = (v5) , y6 = (v6) . The output pattern will be (y5, y6). Perform these calculation for each input pattern: P1: Input pattern (0, 0) v3 = −2 · 0 + 3 · 0 = 0, y3 = (0) = 1 v4 = 4 · 0 − 1 · 0 = 0, y4 = (0) = 1 v5 = 1 · 1 − 1 · 1 = 0, y5 = (0) = 1 v6 = −1 · 1 + 1 · 1 = 0, y6 = (0) = 1 The output of the network is (1, 1). P2: Input pattern (1, 0) v3 = −2 · 1 + 3 · 0 = −2, y3 = (−2) = 0 v4 = 4 · 1 − 1 · 0 = 4, y4 = (4) = 1 v5 = 1 · 0 − 1 · 1 = −1, y5 = (−1) = 0 v6 = −1 · 0 + 1 · 1 = 1, y6 = (1) = 1 The output of the network is (0, 1). P3: Input pattern (0, 1) v3 = −2 · 0 + 3 · 1 = 3, y3 = (3) = 1 v4 = 4 · 0 − 1 · 1 = −1, y4 = (−1) = 0 v5 = 1 · 1 − 1 · 0 = 1, y5 = (1) = 1 v6 = −1 · 1 + 1 · 0 = −1, y6 = (−1) = 0 The output of the network is (1, 0). P4: Input pattern (1, 1) v3 = −2 · 1 + 3 · 1 = 1, y3 = (1) = 1 v4 = 4 · 1 − 1 · 1 = 3, y4 = (3) = 1 v5 = 1 · 1 − 1 · 1 = 0, y5 = (0) = 1 v6 = −1 · 1 + 1 · 1 = 0, y6 = (0) = 1 The output of the network is (1, 1). c) What is deep learning Deep learning is a class of machine learning algorithms that: · use a cascade of multiple layers of nonlinear processing units for feature extraction and transformation. Each successive layer uses the output from the previous layer as input. · learn in supervised (e.g., classification) and/or unsupervised (e.g., pattern analysis) manners. · learn multiple levels of representations that correspond to different levels of abstraction; the levels form a hierarchy of concepts. · use some form of gradient descent for training via backpropagation. Convolutional neural network is one example which is used for image classifications. 4 2 2 2 1.5 1.5 1.5 1.5 3 1 T: 20 Please start each question on a new sheet. Solutions may be hand written. Question No. 5 Marks a) 1.a different transition rule, 2.Local/global pheromone trail updates, 3.use of local updates of pheromone trail to favor exploration 4.a candidate list to restrict the choice of the next city to visit. b) • with probability exploitation (Edge AD = 8) • with probability (1- ) exploration AB with probability 5/17 AC with probability 4/17 AD with probability 8/17 c. d) As stated above evaporation is simply a mechanism to stop pheromone levels building unbounded. The amount of evaporation, p, is a value between 0 and 1. It is used in the pheromone updating formula. That is (where p is the evaporation term): Tij (t + n) = p . Tij(t) + ΔTij ii) Visibility is, essentially a heuristic. In the TSP it is defined as the distance between one city and the next. That is, an ant should prefer a shorter distance but if this is done all the time, we simply get a greedy algorithm. iii) The transition probability formula is shown above. It is used for an ant to decide which city to visit next (assuming we are modelling the TSP). It is a combination of visibility and the current pheromone levels (each having a weight). The values are normalised over all the allowed values (in the case of the TSP this is the cities that have not yet been visited). 4 4 3 3 2 2 2 T: 20 00.511.522.533.544.55 0 0.2 0.4 0.6 0.8 1 A B C

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