Neural Networks and Fuzzy Systems - Assignment Example

Running head: Neural Networks and Fuzzy Systems Neural Networks and Fuzzy Systems Name Course Tutor Date QUESTION ONE perceptron Dichotomiser Training) The Multilayer Perceptron neural network is possibly the most well-known and well used neural network and was developed to solve more challenging problems, “The Multilayer Perceptron neural networks are constituted by a set of sensor units forming the input layer, one or more hidden layers, and an output layer of computational nodes” (Nedjah and Mourelle, 2005, p. 185). Each neuron within a MLP can be connected to one or all neurons in the following layer. There is a unique scalar weight which is given to each link between the varying neurons. A MLP must have three characteristics which are: 1. The model of each neuron within the network has a smooth activation function which is non-linear thus can be different at any point. 2. There are one or more hidden layers of neurons which function as the input/output of the network. The hidden neurons function in a manner which allows the network to undertake difficult tasks while extracting the most important aspects of the characteristics in the vector. 3. There is a high level of connectivity within the network. If the networks connectivity is modified, there needs to be a change within the synaptic connections (Nedjah and Mourelle, 2005). Let us consider two perceptron dichotomisers are trained to recognise the following classification of six patterns x with known membership d. X1 =, x2 = , x3 = , x4 = , x5 = , x6 = d1 =, d2 =,d3 =, d4 =, d5= , d6 = 1.1). the first dichotomiser is a discrete percepron and assigned-1 to all augmented inputs where the training task of this task of this dichotomiser, the fixed correction rule is used, with an arbitrary selection of learning constant n = 0.05 and the initial weight vector as shown below; Using the Matlab code below Final weight vector is arrived as 0.00136 1.0321 -0.3267 0.9231 -0.72314]; Classification Cycle error curve 1.2). If the second dichotomiser is a continuous perceptron with a bipolar logistic activation function z = f2 = as shown below with all augmented inputs Assigned 1 If the learning constant n = 0.5 with initial weight vector w1 of Using Matlab codes that follow vector w7 after one cycle and the weight vector w301 after 50 cycles is calculated Cycle error when 'Weight Matrix after 7 epoch' When the Cycle error when 'Weight Matrix after 301 epoch' Classification changes from haphazard to well define when it 301 epochs or cycles thus network training improves. QUESTION TWO [50 marks] 2.1 [Engineering Management][15 marks] In engineering management it is often necessary to borrow money to finance projects of significant size. Engineering firms are therefore quite sensitive about their financing capacity, or their credit limits. Credit limits are often related to previous borrowing practices and to current credit account average balances. In turn, average balances are related to the profitability of the engineering venture. Suppose we have a fuzzy set C for credit limits, a fuzzy set B for average account balance, and fuzzy set P for profits, all in units of thousands of dollars. Let the universes for these fuzzy sets be: Credit limits = {500, 1000, 1500, 2000} Average account balance = {20, 50, 100, 500, 1000, 1200} Profits = {-50, 0, 50, 100, 500} Suppose that the following relation has been established between the credit limits of the firm and its average account balance A = And the following relation has been established between the average account balances of the firm and its profit margins. E = Find a relation between C and P by computing R =A0 E 2.11 using max-min composition x = Max {min (0.8, 0.8), min (1.0, 0.7),min (0.6, 0.5),min (0.2, 0.2),min (0, 0.1)min (0, 0)} = max {0.8, 0.7, 0.5, 0.2, 0, 0} = 0.8 Max {min (0.8, 0.9), min (1.0, 1.0),min (0.6, 0.9),min (0.2, 0.5),min (0, 0.4) min (0, 0.3)} = max {0.8, 1.0, 0.6, 0.2, 0, 0} = 1.0 Max {min (0.8, 0.7), min (1.0, 0.8),min (0.6, 0.9),min (0.2, 0.7),min (0, 0.6) min (0, 0.5)} = max {0.7, 0.8, 0.6, 0.2, 0, 0} = 0.8 Max {min (0.8, 0.1), min (1.0, 0.2),min (0.6, 0.5),min (0.2, 0.8),min (0, 0.9) min (0, 0.8)} = max {0.1, 0.2, 0.5, 0.2, 0, 0} = 0.5 Max {min (0.8, 0), min (1.0, 0),min (0.6, 0.1),min (0.2, 0.9),min (0, 0.8) min (0, 0.7)} = max {0, 0, 0.1, 0.2, 0, 0} = 0.2 Max {min (0.2, 0.8), min (0.3, 0.7),min (0.5, 0.5),min (0.8, 0.2),min (0.1, 0.1) min (0, 0)} = max {0.2, 0.3, 0.5, 0.2, 0.1, 0} = 0.5 Max {min (0.2, 0.9), min (0.3, 1.0),min (0.5, 0.9),min (0.8, 0.5),min (0.1, 0.4) min (0, 0.3)} = max {0.2, 0.3, 0.5, 0.5, 0.1, 0} = 0.5 Max {min (0.2, 0.7), min (0.3, 0.8),min (0.5, 0.9),min (0.8, 0.7),min (0.1, 0.6) min (0, 0.5)} = max {0.2, 0.3, 0.5, 0.7, 0.1, 0} = 0.7 Max {min (0.2, 0.1), min (0.3, 0.2),min (0.5, 0.5),min (0.8, 0.8),min (0.1, 0.9) min (0, 0.8)} = max {0.1, 0.2, 0.5, 0.8, 0.1, 0} = 0.8 Max {min (0.2, 0), min (0.3, 0),min (0.5, 0.1),min (0.8, 0.9),min (0.1, 0.8) min (0, 0.7)} = max {0, 0, 0.1, 0.8, 0.1, 0} = 0.8 Max {min (0.1, 0.8), min (0.2, 0.7),min (0.4, 0.5),min (0.9, 0.2),min (0.6, 0.1) min (0.1, 0)} = max {0.1, 0.2, 0.4, 0.2, 0.1, 0} = 0.4 Max {min (0.1, 0.9), min (0.2, 1.0),min (0.4, 0.9),min (0.9, 0.5),min (0.6, 0.4) min (0.1, 0.3)} = max {0.1, 0.2, 0.4, 0.5, 0.4, 0.1} = 0.5 Max {min (0.1, 0.7), min (0.2, 0.8),min (0.4, 0.9),min (0.9, 0.7),min (0.6, 0.6) min (0.1, 0.5)} = max {0.1, 0.2, 0.4, 0.7, 0.6, 0.1} = 0.7 Max {min (0.1, 0.1), min (0.2, 0.2),min (0.4, 0.5),min (0.9, 0.8),min (0.6, 0.9) min (0.1, 0.8)} = max {0.1, 0.2, 0.4, 0.8, 0.6, 0.1} = 0.8 Max {min (0.1, 0), min (0.2, 0),min (0.4, 0.1),min (0.9, 0.9),min (0.6, 0.8) min (0.1, 0.7)} = max {0, 0, 0.1, 0.9, 0.6, 0.1} = 0.9 Max {min (0.1, 0.8), min (0.1, 0.7),min (0.4, 0.5),min (0.8, 0.2),min (0.8, 0.1) min (0.3, 0)} = max {0.1, 0.1, 0.4, 0.2, 0.1, 0} = 0.4 Max {min (0.1, 0.9), min (0.1, 1.0),min (0.4, 0.9),min (0.8, 0.5),min (0.8, 0.4) min (0.3, 0.3)} = max {0.1, 0.1, 0.4, 0.5, 0.4, 0.3} = 0.5 Max {min (0.1, 0.7), min (0.1, 0.8),min (0.4, 0.9),min (0.8, 0.7),min (0.8, 0.6) min (0.3, 0.5)} = max {0.1, 0.1, 0.4, 0.7, 0.6, 0.3} = 0.7 Max {min (0.1, 0.1), min (0.1, 0.2),min (0.4, 0.5),min (0.8, 0.8),min (0.8, 0.9) min (0.3, 0.8)} = max {0.1, 0.1, 0.4, 0.8, 0.8, 0.3} = 0.8 Max {min (0.1, 0), min (0.1, 0),min (0.4, 0.1),min (0.8, 0.9),min (0.8, 0.8) min (0.3, 0.7)} = max {0, 0, 0.1, 0.8, 0.8, 0.3} = 0.8 2.12 using sum-product composition x = sum {product (0.8, 0.8), product (1.0, 0.7),product (0.6, 0.5),product (0.2, 0.2),product (0, 0.1) product (0, 0)} = sum {0.64, 0.07, 0.3, 0.04, 0, 0} = 1.05 sum {product (0.8, 0.9), product (1.0, 1.0),product (0.6, 0.9),product (0.2, 0.5),product (0, 0.4) product (0, 0.3)} = sum {0.72, 0.01, 0.54, 0.1, 0, 0} = 1.37 sum {product (0.8, 0.7), product (1.0, 0.8),product(0.6, 0.9),product (0.2, 0.7),product (0, 0.6) product (0, 0.5)} = sum{0.56, 0.8, 0.54, 0.14, 0, 0} = 2.04 Sum{product (0.8, 0.1), product (1.0, 0.2),product (0.6, 0.5),product (0.2, 0.8),product (0, 0.9) product (0, 0.8)} = Sum{0.08, 0.2, 0.3, 0.16, 0, 0} = 0.74 Sum{product (0.8, 0), product (1.0, 0),product (0.6, 0.1),product (0.2, 0.9),product (0, 0.8) product (0, 0.7)} = Sum{0, 0, 0.06, 0.18, 0, 0} = 0.24 Sum{product (0.2, 0.8), product (0.3, 0.7),product (0.5, 0.5),product (0.8, 0.2),product (0.1, 0.1) product (0, 0)} = Sum {0.16, 0.21, 0.25, 0.16, 0.01, 0} = 0.79 Sum{product (0.2, 0.9), product (0.3, 1.0),product (0.5, 0.9),product (0.8, 0.5),product (0.1, 0.4) product (0, 0.3)} = Sum {0.18, 0.03, 0.45, 0.40, 0.04, 0} = 1.1 Sum{product (0.2, 0.7), product (0.3, 0.8),product (0.5, 0.9),product (0.8, 0.7),product (0.1, 0.6) product (0, 0.5)} = Sum{0.14, 0.24, 0.45, 0.56, 0.06, 0} = 1.45 Max {product (0.2, 0.1), product (0.3, 0.2),product (0.5, 0.5),product (0.8, 0.8),product (0.1, 0.9) product (0, 0.8)} = sum{0.02, 0.06, 0.25, 0.64, 0.09, 0} = 1.06 Sum{product (0.2, 0), product (0.3, 0),product(0.5, 0.1),product (0.8, 0.9),product (0.1, 0.8) product (0, 0.7)} = Sum{0, 0, 0.05, 0.72, 0.08, 0} = 0.85 Sum{product (0.1, 0.8), product (0.2, 0.7),product (0.4, 0.5),product (0.9, 0.2),product (0.6, 0.1) product (0.1, 0)} = Sum {0.08, 0.14, 0.2, 0.18, 0.06, 0} = 0.66 Sum{product (0.1, 0.9), product (0.2, 1.0),product (0.4, 0.9),product (0.9, 0.5),product (0.6, 0.4) product (0.1, 0.3)} = Sum {0.09, 0.2, 0.36, 0.45, 0.24, 0.03} = 1.37 Sum{product (0.1, 0.7), product (0.2, 0.8),product (0.4, 0.9),product (0.9, 0.7),product (0.6, 0.6) product(0.1, 0.5)} = Sum{0.07, 0.16, 0.36, 0.63, 0.36, 0.05} = 1.58 Sum{product (0.1, 0.1), product (0.2, 0.2),product (0.4, 0.5),product (0.9, 0.8),product (0.6, 0.9) product (0.1, 0.8)} = Sum{0.01, 0.04, 0.2, 0.72, 0.54, 0.08} = 1.59 Sum{product(0.1, 0), product (0.2, 0),product (0.4, 0.1),product (0.9, 0.9),product (0.6, 0.8) product(0.1, 0.7)} = Sum{0, 0, 0.04, 0.81, 0.48, 0.07} = 1.4 Sum {product (0.1, 0.8), product(0.1, 0.7),product (0.4, 0.5),product (0.8, 0.2),product (0.8, 0.1) product (0.3, 0)} = sum {0.08, 0.07, 0.2, 0.16, 0.08, 0} = 0.59 Sum{product(0.1, 0.9), product (0.1, 1.0),min (0.4, 0.9),product (0.8, 0.5),product (0.8, 0.4) product (0.3, 0.3)} = Sum {0.09, 0.01, 0.36, 0.4, 0.32, 0.09} = 1.27 Sum{product (0.1, 0.7), product (0.1, 0.8),product (0.4, 0.9),product (0.8, 0.7),product (0.8, 0.6) product (0.3, 0.5)} = sum{0.07, 0.08, 0.36, 0.56, 0.48, 0.15} = 1.7 sun {product (0.1, 0.1), product(0.1, 0.2),product (0.4, 0.5),product (0.8, 0.8),product (0.8, 0.9) product (0.3, 0.8)} = sum{0.01, 0.02, 0.2, 0.64, 0.72, 0.24} = 1.83 sum {product (0.1, 0), product (0.1, 0),product (0.4, 0.1),product (0.8, 0.9),product (0.8, 0.8) product (0.3, 0.7)} = sum{0, 0, 0.04, 0.72, 0.64, 0.21} = 1.61 2.2 [LASER Bean Alignment] [35 marks] Fuzzy logic is used to control a two-axis mirror gimball for aligning a laser beam using a quadrant detector. Electronics sense the error in the position of the beam relative to the centre of the detector and produces two signals representing the x and y direction errors. The controller processes the error information using fuzzy logic and provides appropriate control voltages to run the motors which reposition the beam. The fuzzy logic controller for this system is shown in Figure 2.1 To represent the error input to the controller, a set of linguistic variables is chosen to represents 5 degrees of error, 3 degrees of change of error, and 5 degrees of a mature voltage. Membership functions are constructed to represent the input and output values` grades of membership as shown in Figure 2.2. The rule set in the form of “fuzzy Associative Memories” is shown in Figure 2.3. The controller gains are assumed to be GE = 1, GCE = 1, GU =1. 2.21 If the Mean of Maximum (MOM) defuzzification strategy (sum-product inference) is used with the fire strength of the i-th rule calculated from = (e).(Ce) Calculate the defuzzified output voltages of this fuzzy controller at a particular instant. The error and the change of error at this instant are e = 3.25 and Ce =-0.2. [10 marks] Apply this to the third composite set v 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 (e) -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1.0 1.2 v 0 -0.03 -0.4 -0.3 0 0.5 1.2 2.1 3.2 4.5 6 (ce) 2.5 2.75 3.0 3.25 3.5 3.75 4.0 4.25 4.50 4.75 5.0 0 1.375 3 4.875 7 9.375 12 16.25 18.25 21.32 25 Output -0 0.0413 -1.2 -1.423 0 4.688 14.4 16.32 19.25 23.65 29.2 = (e).(Ce) 2.2.2 If the Centre of Area (COA) defuzzification strategy (max-min inference) is used with the fire strength of the i-th rule calculated from = min ( (Ce)) Calculate the corresponding defuzzified output voltage at a particular instant when the error and the change of error are e = 3.25 and Ce =-0.2. [21 marks] Figure 2.1 Fuzzy Logic control system v 0 0.5 1.0 1.5 2 2.5 3.0 3.5 4 4.5 5 (v)t 0 0.16 0.32 0.48 0.54 0.8 0.64 0.48 0.32 0.24 0.36 (v)t 0 0.16 0.64 1.44 2.56 4 3.34 3.36 2.56 2.16 3.6 v 13 14 15 16 17 18 19 20 21 22 23 (v)t 0.48 0.36 0.24 0.14 0.28 0.42 0.56 0.7 0.56 0.42 0.28 (v)t 6.24 5.04 3.6 2.24 4.76 7.56 10.54 14 11.76 9.24 6.44 = min ( (Ce)) References Bouchon-Meunier, B. (1995). Fuzzy logic and soft computing. World Scientific. Braga, N. (2002). Robotics, mechatronics, and artificial intelligence: experimental circuit blocks for designers. New York: Newnes. Liu, P and Li, H. (2004). Fuzzy Neural Network Theory and Application. World Scientific. Mitra, S et al. (1997). Knowledge-Based Fuzzy MLP for Classification and Rule Generation. Retrieved from Munakata, T. (2008). Fundamentals of the new artificial intelligence: neural, evolutionary, fuzzy and more. Springer. Nedjah, N and Mourelle, L. (2005). Evolvable machines: theory & practice. Springer. Nikravesh, M et al. (2003). Soft computing and intelligent data analysis in oil exploration. Elsevier. Rajasekaran, S and Pai, G. (2004). Neural Networks, Fuzzy Logic and Genetic Algorithms: Synthesis and Applications. PHI Learning PVt. Read More